Scientiam Dei

Existential Generalization:  “Brian is mortal” leads to “Something is mortal.”  In other words, M(B) leads to ∃x(M(x)).  Let’s now present the abstract rule -

φ_ζ leads to ∃α(φ_α) where α is a variable, ζ is symbolic term, φ_α and φ_ζ are symbolic formulas, and φ_ζ comes from φ_α by proper substitution of ζ for α. 

“Brian is mortal” translates to M(B).  Now our rule is that φ_ζ leads to ∃α(φ_α). In this case, φ_ζ stands for M(B).  α stands for ‘x’, φ_α stands for M(x).  Since ζ is a proper substitution for α in φ_α, then we are allowed to conclude to ∃α(φ_α) , which is ∃x(M(x)).

Now that we have identified what constitutes a proper substitution for variables and name letters we are able to define three new inference rules in the predicate calculus.  This post will deal with the first one: universal instantiation. 

Universal Instantiation:  “Everything is mortal” leads to “Brian is mortal.”  In other words, ∀x(M(x)) leads to M(B).  Let’s now present the abstract rule -

∀α(φ_α) leads to φ_ζ where α is a variable, ζ is symbolic term, φ_α and φ_ζ are symbolic formulas, and φ_ζ comes from φ_α by proper substitution of ζ for α. 

The key point here is that the resulting φ_ζ comes from φ_α by proper substitution of ζ for α.  That means the resulting formula only substitutes ζ for α in those cases where α is free within φ_α.  Notice, it does not say within ∀α(φ_α), for then α is not free.  Do not confuse φ_α with ∀α(φ_α) - they are not the same.  Let’s go back and illustrate this using our previous example.

“Everything is mortal” translates to ∀x(M(x)).  Now our rule is that ∀α(φ_α) leads to φ_ζ. In this case, α stands for ‘x‘, φ_α stands M(x).  We now let ζ stand for ’B', which is itself the name place for ‘Brian’. Since we want to do a proper substitution of  ζ for α in φ_α, we are doing a proper substitution of ‘B’ for ‘x’ in M(x).  This means our φ_ζ becomes M(B).   M(B) says, “Brian is mortal.”  As such, we can go from “Everything is mortal” to “Brian is mortal” by the universal instantiation of ‘Brian’.

Proper substitutions for variables are just like proper substitutions for name letters with one exception.  A name letter can be substituted for all free occurrences of a variable.  A variable can be substituted for all free occurrences of another variable as long as the substituting variable remains free.  This last phrase is the only difference between variable substitution and name substitution.  Let me use an example from our last post.

1. ∀x(F(x) → G(y))

The only free occurrence of a variable is the ‘y‘.  As such, I could take the variable ‘z‘ and substitute it as follows…

2. ∀x(F(x) → G(z))

However, I cannot substitute the variable ‘x‘ for the variable ‘y‘ because the result would be a bound variable…

3. ∀x(F(x) → G(x))

Notice, the ‘x‘ in G(x) is bound by the quantifier ∀x.  Therefore, 3 is not a proper substitution.  Let’s provide one more example…

4. ∃(z)(F(x) ∧ G(y)) → ∃(x)(H(x)) ∨ G(x)

Now, I want to substitute a variable for ‘x‘.  The free occurrences of ‘x‘ are in F(x), and G(x).  Can I use the variable ‘z‘?  No I can’t.  The reason is that when I substitue ’z‘ for ‘x‘ in F(x) the resulting variable is bound by ∃(z).  As such, I need to choose a different variable.  Let’s choose ‘y‘.  

5. ∃(z)(F(y) ∧ G(y)) → ∃(x)(H(x)) ∨ G(y)

Here is our formal definition for the proper substitution of a variable…

Definition: A symbolic formula φ_β comes from a symbolic formula φ_α by proper substitution of a variable β for a variable α if φ_β is like φ_αexcept for having free occurrences of β wherever φ_αhas free occurrences of α.

Notice, in our definition we made sure to say that the substituting variable β is always a free occurrence.

Some Conventions

1.The capital letters ‘P’ through ‘Z’ with or without numerical subscripts will represent symbolic sentences.

2. A variable is a lower-case italicized Latin letter with or without numerical subscripts.

3. The capital letters ‘A’ through ‘E’ with or without numerical subscripts represent name letters.

4.The capital letters ‘F’ through ‘O’ with or without numerical subscripts represent predicate letters.

5. Variables (2) and name letters (3) will be called symbolic terms.

Proper Substitution - Name Letter

Definition: A symbolic formula φ_ζ comes from a symbolic formula φ_α by proper substitution of a letter name ζ for a variable α if φ_ζ is like φ_αexcept for having occurrences of ζ wherever φ_αhas free occurrences of α.

This is very abstract, and as such I will try and explain this.  The first question I will answer is what it means when we speak of φ_αhaving free occurrences of α.  Consider the following symbolic formula:

∀x(F(x) → G(y))   

You will notice that the universal quantifier of x ranges over F(x) but does not range over G(y).  If the ‘y‘ were an ‘x‘, then it would.  But ‘y‘ is not ‘x‘ so ‘∀x‘ cannot range over G(y).  As such, G(y) is said to be a free occurrence of y; whereas, F(x) contains a bound occurrence of x. In short, if a quanitfier does not range over a variable, then that variable is free. Let’s illustrate this with a couple of examples:

Ex. 1 x is an even number.  

If we let ‘F’ stand for the predicate “is an even number,” then the statement above can be symbolically represented as F(x).   Now, F(x) is not a proposition because “x is an even number” is not a proposition.  The reason for this is because we cannot say F(x) is true or false.  F(x) contains a free occurrence of x.   

Ex. 2 For all x, x is an even number. 

This is symbolized by the following formula: ∀x(F(x)).  This statement is clearly false.  However, ‘x‘ has been ranged over by the quantifier ‘∀’.  As such, ‘x‘ is bound. 

Ex. 3 There exists an x such that x is an even number.

This is symbolized by the formula: ∃(x)F(x).  This statement is true. ‘x‘ has been ranged over by the quantifier ‘∃’.  As such, ‘x‘ is bound.

We are now ready to tackle name letter substitution.  A proper name letter substitution occurs in a formula φ_α whenever all free occurrences of α (and only free occurrences of α) are substituted with a name letter. Consider Ex.1 above. If we let A be a name letter, then a proper substitution of F(x) for a name letter is F(A). In our definition above we have,

φ is F
ζ is A
α is x
φ_α is F(x)    
φ_ζ is F(A)

Here is how the definition reads with our example 1 above: A symbolic formula F(A) comes from a symbolic formula F(x) by proper substitution of a letter name A for a variable x if F(A) is like F(x) except for having occurrences of A wherever F(x) has free occurrences of (x).  As one can see, all of the free occurrences of x in F(x) have been replaced by A in F(A).  As such, F(A) comes from a symbolic formula F(x) by proper substitution.

Here is a little more complicated example of name letter substitution.  We let our φ_x stand for the symbolic formula:

 ∃(z)(F(x) ∧ G(y)) → ∃(x)(H(x)) ∨ G(x)  

Now, let’s say that I want to do a proper substitution of the variable ’x‘ for the name letter ‘A’ in the above formula.  The first thing to do is to identify all of the free occurrences of ‘x‘.  They are F(x) and G(x).  H(x) is not a free occurrence of ‘x‘ because H(x) is ranged over by the existential quantifier ‘∃(x)’.  The ‘x‘ in F(x) is free because the existential quantifier ‘∃(z)’ ranges only over the variable ‘z‘.  Now, I just replace the ‘x‘ with ‘A’ in those two places…

∃(z)(F(A) ∧ G(y)) → ∃(x)(H(x)) ∨ G(A)

This formula stands for our φ_A and φ_Ais in fact a proper substitution.

The definition for Libertarian Free Will (LFW) in terms of possible world semantics is as follows:

Defintion (LFW): Person A performs action X given circumstance C with Libertarian freedom if and only if it is possible that person A can perform action ¬X given circumstance C.    

Now, let’s raise a possible objection.  Consider the denial of Christ by Peter.  If Peter’s action is free in the libertarian sense, then there exists a possible world where everything is exactly the same with the exception of Peter’s denial.  The problem with this is that the “everything is exactly the same” includes the prediction by Jesus that Peter would deny Him three times.  In other words, if Peter’s denial is truly free, then there exists a possible world where Jesus falsely predicts that Peter will deny Him three times! 

Someone might argue that possible events are only those events that logically cohere.  Since Jesus, who is infallible, predicted that Peter would deny Him three times, then the event of Peter not denying Jesus does not logically cohere and as such is not a possible world.  This would leave intact Jesus’ infallibility, but then it removes Peter’s libertarian freedom regarding his denial of Christ.  I do not know how the Libertarian can overcome this objection.  Now, one defense concerning this very critique is given as follows…

If God believed at time t1 that Peter would deny Christ at time t2, then if Peter really could have not denied Christ, then at least one of three consequences results:

(1) It was within Peter’s power at time t2 to do something that would bring about God having a false belief.
(2) It was within Peter’s power at time t2 to change God’s belief at t1.
(3) It was within Peter’s power at time t2 to make it such that any person who believed that Peter would deny Christ at time t1 held a false belief.

All three of these possible consequences are very problematic.  Alvin Plantinga (a famous Christian philosopher) argues that we possess a type of “power over the past” that does not entail the ability to change past events by retro-causation.  That is, we have counterfactual power over the past.  In other words, Alvin Plantinga argues for (2) above in some sense.  Consider the proposition: “God believes at time t1 that Peter will deny Christ at time t2.”  If LFW is true, then at time t2 Peter must be able to not deny Christ.  According to Plantinga, Peter can refrain from denying Christ, and if he were to do so, then God would have always foreknown that Peter would not deny Christ.  It is de facto true that Peter will deny Christ, but had he done otherwise, God’s past knowledge would always have been different, and Jeus would not have made the prediction.  Essentially, what is being stated is the following:

P1: If Peter does A at t2, then God’s knowledge at t1 is K.

and

P2: If Peter does ¬A at t2, then God’s knowledge at t1 is ¬K. 

It is like saying whatever happens God knew it would happen.  Plantinga refers to this situation as “counterfactual power over God’s knowledge, ” and he claims this “power over the past” does not entail retro-causation.  Rather, it is the power to assert the truth of various backtracking counterfactuals.  Frankly, this answer seems absurd.   Dr. Laing in a correspondence dated 6/17/07 called it a weak position or claim.  However, Alvin Plantinga is much smarter than I am.  As such, I probably do not fully understand the argument, or am missing a key component.  Therefore, I will leave this as it is, and if I learn more, then I will pick it back up.  

Part I
Part II
Part III
Part IV
PartV

I am going to take a short excursion that I hope will lay a foundation. We are going to create our own system of possible worlds. This system will be very simple and hopefully will allow us to grasp the much more complicated system making up possible world semantics. The construction of our system will parallel the system of possible world semantics.

Our system begins with only two possible initial stages, which we will call IS(1) and ¬IS(1). What is interesting, this is a logically necessary situation. Think of it as: you either start with IS(1) or you don’t. It is just an application of the law of non-contradiction. So, what is ¬IS(1)? Well, for our purposes, we will say that there is only one other possible initial state, which we will call IS(2).  That is to say, in our little system ¬IS(1) is IS(2).   

IS(1) and IS(2) each make up a complete set of events that we are calling the initial stage of a possible world. In real possible world semantics we have the same situation with one major exception. We have IS(1) and ¬IS(1), but ¬IS(1) is made up of a huge number of possible worlds, rather than just one. Think of all the possible starting points there could have been for creation. This would be our initial set. Of course, this is much too large for us to work with, and as such we just pretend that there are only two initial states: IS(1) and IS(2).

Now, there are a number of possible consequences to these initial stages. Think of them as the next stage as time moves forward. In other words, they are what happens next. For IS(1) the possible consequences are MS(1) and ¬MS(1). IS(2) has the possible consequences of MS(2) and ¬MS(2). (‘MS’ stands for middle stage, which is not to be confused with Middle Knowledge. In real possible world semantics the parallel is exact with the exception that there is a huge number of initial stages for which there are two possible consequences, namely MS(n) and ¬MS(n). In our system, we will define ¬MS(1) to be MS(2) and ¬MS(2) to be MS(1) for the sake of simplicity. Of course, this is not the same for real possible world semantics. ¬MS(n) stands for an even larger set of possible consequences than the huge set of possible initial stages. Two stages into real possible world semantics leaves us with a mind boggling number of possibilities.

We will end our possible world system with the final stage FS(1) and FS(2) with all of the corresponding possibilities. In the real possible world semantics, we would not stop here. In fact, there would be numerous stages making the number of possible combinations truly staggering. However, it is not too staggering for God to comprehend. In our little possible world system, the number of possible worlds is only eight. Here are our eight possible worlds in our little bitty system:

P(1) IS(1) → MS(1) → FS(1)
P(2) IS(1) → MS(1) → FS(2)
P(3) IS(1) → MS(2) → FS(2)
P(4) IS(1) → MS(2) → FS(1)
P(5) IS(2) → MS(2) → FS(2)
P(6) IS(2) → MS(2) → FS(1)
P(7) IS(2) → MS(1) → FS(1)
P(8) IS(2) → MS(1) → FS(2)

Each of these possible worlds has an event path that is unique. For instance our P(6) is made up of the following path: <2,2,1>. There is no other possible world that has this path. <1,2,1> is P(4), <2,2,2> is P(5), and <1,1,1> is P(1), etc… The parallel in possible world theory is exactly the same. Each possible world has its own unique event path. Different possible worlds may share many events in common, but each event path will all be unique - just like P(1) and P(2) share the first two stages the same, but end differently.

In Molinism, God considers the humongous set of event paths that make up the humongous set of possible worlds in possible world semantics, and He chooses to actuate one of these possible worlds. It is in this sense that God ordains everything that comes to pass. He actuates one possible world and this in turn actuates the event path of this possible world. All events can then be spoken of as being ordained by God on the basis of Him choosing this particular possible world. So, if God were to actuate P(4), then it would be said that God ordained all events that make up the event path <1,2,1>.

Part I
Part II
Part III
Part IV

It is now time to explain what Middle Knowledge (MK) is relative to Natural Knowledge (NK) and Free Knowledge (FK).  There are two characteristics we are going to use to describe these three types of knowledge.  They are ‘volitional’ and ‘necessary’.

Volitional

Volitional knowledge is that knowledge of God whose foundation is found in the will of God. In other words, God knows ‘X’ volitionally precisely because God wills ‘X’. For example, God knows the proposition, “Brian Bosse exists.” This knowledge is due to God’s will (choice) to create Brian Bosse. As such, this knowledge is volitional. All FK is volitional. On the other hand, all NK is pre-volitional. The laws of logic would be an example of knowledge that is pre-volitional. The law of non-contradiction does not find its foundation in the will of God, but rather in the very nature of God. These laws are a reflection of who God is in His thinking. As such, they exist because God exists, and not because God chose to create them.

Necessary

All NK is necessary knowledge because it is based upon the existence of God who is a necessary being. However, the assumption made by proponents of Middle Knowledge (and most theologians in general) is that God’s choice to create is not necessary. That is to say, God could have chosen not to create or could have chosen to create some other possible world. As such FK is not necessary. Rather, it is contingent - contingent upon God’s will to create.

Middle Knowledge

NK can be characterized as pre-volitional and necessary. FK can be characterized as volitional and contingent. Middle Knowledge (MK) can be characterized as pre-volitional (like NK) and contingent (like FK). It is a knowledge that is between NK and FK. That is why it is called MK. It is in the middle of NK and FK. Because it is pre-volitional it is independent of God’s will.

So, what are the objects of this knowledge? Dr. John Laing describes it this way:

“The doctrine of middle knowledge proposes that God has knowledge of metaphysically necessary states of affairs via natural knowledge, of what He intends to do via free knowledge, and in addition, of what free creatures would do if they were instantiated (via middle knowledge). Thus, the content of middle knowledge is made up of truths which refer to what would be the case if various states of affairs were to obtain. For example, the statement, ‘If John Laing were given the opportunity to write an article on middle knowledge for the Internet Encyclopedia of Philosophy, he would freely do so,’ although true, is certainly not necessarily so. I could easily have refrained from writing, if I were so inclined (or too busy, etc.). Likewise, its truth does not seem to be dependent upon God’s will in the same way that ‘John Laing exists’ is. Even if God chose to not create me, the statement regarding my writing the article could still be true. In fact, its truth does not seem to be dependent upon God’s will at all, but rather upon my will. One of the basic assumptions of the doctrine of middle knowledge outlined above is that God cannot will a creature to freely choose anything. Thus, the content of middle knowledge can be thought of as including a virtually infinite number of propositions of the form, If person, P, were in situation, S, then P would freely perform action, A.”[1]

The idea is that God can look at these truths of counterfactuals of freedom, which are truths about the actions of free creatures, and His actualizing a possible world where these truths are actualized does not violate libertarian freedom. For example, let’s say the following proposition is a counterfactual of freedom…

P: If Brian were in situation X, then he will freely choose to write about Molinism.

Notice, the consequence of this conditional statement that I would choose to write about Molinism is that I would do so in a free manner. That is to say, if God wanted me to write about Molinism all He would need to do was put me in situation X, and I would so freely! So, God knows ‘P’ as part of His MK, and then if He wills that I write about Molinism (FK), He actuates the world that puts me in situation X. God’s will is done and His sovereignty is maintained while my freedom is left intact.

This is the essence of Middle Knowledge and Molinism. However, I think it is incoherent within the framework of possible world theory. I developed my argument for this through a series of email exchanges with Dr. Laing over the summer of 2007. He has gracious allowed me to share this exchange noting that his responses were “off the cuff and not as well thought out as a published article would be.”

[1] Dr. John Laing, Middle Knowledge, The Internet Encyclopedia of Philosophy.

Part I
Part II
Part III

This series is simply to help me clarify my thinking and understanding of Gödel’s Theorems.  To begin, I am going to consider a very simple formal system found in David Hofstadter’s Gödel, Escher, Bach: An Eternal Golden Braid.  What is a formal system?  In the context of Gödel’s Theorem (GT), a formal system can be thought of as a game made up of rules on how to manipulate symbols.  We begin with the pq-system.     

The Language

All formal systems have a language.  A key component about a formal language is that the language does not need to mean anything.  Initially, we are not going to worry at all about meaning, but we will mainly concern ourselves with how the language is developed, i.e., the symbols are manipulated.  There will be only three symbols that make up the pq-system: ‘-’, ‘p’ and ‘q’.  Any combination of these symbols represents a formula of the system.  For example, ‘- - -p’ and ‘qpqpq’ and ‘-’ all represent formulas in our formal system.  However, not all formulas are well-formed.  For instance, ‘lkqw’ is a series of letters in the English alphabet, but it is not considered an actual word according to the rules of the language - there is no vowel.  In other words, ‘lkqw’ is not well-formed.  As such, for our formal language we need to define what is a well-formed formula. 

Definition: A formula is well-formed in the pq-system if and ony if it begins with at least one ‘-’ followed by one and only one ‘p’ followed by at least one ‘-’ followed by one and only one ‘q’ followed by at least one ‘-’.    

Another way to say this is that all well-formed formulas look like xpyqz when ‘x‘ ‘y‘ and ‘z‘ stand for some number of hyphens.  Here is a list of well-formed formulas: ‘-p-q-’ and ‘- - -p-q-’ and ‘-p- - - -q- -’.   On the other hand, ‘- - -p’ and ‘qpqpq’ and ‘-’ are all formulas, but they are not well-formed.  Be very careful here.  ‘x‘ ‘y‘ and ‘z‘ are not part of the pq-system.  They are just place holders that we use for our convenience.  In other words, they assist us in talking about the pq-system, but they are not part of the pq-system itself.  

Axioms 

At this point, we need to further define our language by listing the starting point(s).  That is to say, we are going to decide which well-formed formulas are going to be part of our system.  We call this starting point (or staring points) the axioms of the system.  

Definition: A formula is an axiom of the pq-system if and only if it has the form xp-qx- whenever ‘x‘ is composed of a number of hyphens.   

As such, we have allowed ourselves an infinite number of axioms.  The following are all axioms of the pq-system: ‘-p-q- -’ and ‘- -p-q- - -’ and ‘- - - - -p-q- - - - - -’.  None of the well-formed formulas above are axioms.  Again, note that the ‘x‘ in the above definition is not part of the pq-system.  The only symbols in the pq-system are  ‘-’, ‘p’ and ‘q’. 

Rule of Production

We now need to add to our language a way to take an axiom and create another well-formed formula in our system. These formulas created from the axioms are called theorems of the system. 

Rule: Given that ‘x‘ ‘y‘ and ‘z‘ each stand for some number of hyphens, and given that xpyqz is either an axiom of the system or a theorem of the system, then xpy-qz- is a theorem of the system. 

For example, ‘-p-q- -’ is an axiom of the system.  Let ‘x‘ stand for ‘-’, ‘y‘ stand for ‘-’ and ‘z‘ stand for ‘- -’.  According to our rule of production, then ‘-p- -q- - -’ is a theorem of our system.  This concludes our formal system. 

As an exercise, which of the following are theorems or axioms of the pq-system? 

1. - - - - -p-q- - - - - -

2. -xpy- -qz- - -

3. - - -p- - -q- - - - - -

4. - -p- - - -q- - - - -

For those examples above that are theorems of the system explain how the theorem was derived.

You are on a game show where there are two doors in front of you - door #1 and door #2.  The host of the show tells you that behind one of the doors is one million dollars and behind the other door is nothing.  It is your job to figure out behind which door is the million dollars.  The host says to you, “You may make one statement (proposition) to my assistant to which she will answer either ‘true’ or ‘false’.  My assistant either always tells the truth or always lies.”

Problem: Given the above information, construct a sentence so that after the assistant answers you know behind which door is the million dollars.

Solution: The assistant is a truth teller if and only if the million dollars is behind door #1. (Highlight the prior area to see one possible answer.)

There is a way to construct such a sentence on purpose.  To begin, let’s use the following scheme of abbreviation:

P: The assistant always tells the truth.

Q: The million dollars is behind door #1. 

 To begin, we will construct the following truth-table…

P   |   Q  |   Desired Response   |   Value of Sentence

T       T                   

T       F                   

F       T 

F       F

At this point we have simply listed all of the possible truth-value combinations for P and Q.  The column labled ‘desired response’ is simply what we want the assistant to answer in each particular situation given our sentence.  As such, we choose to create a sentence that if the assistant always tells the truth and the million dollars is behind door #1, then the assistant answers ‘true’.   In other words, if P=T and Q=T, then the desired response is ‘true’. (Note: we could just as easily choose the desired response to be ‘false’.) 

In like manner, if the assistant always tells the truth and the million dollars is not behind door #1, then the assistant answers ‘false’ to our sentence.  In other words, if P=T and Q=F, then the desired response is ‘false’.  Here is what this looks like…

P   |   Q  |   Desired Response   |   Value of Sentence

T       T                True   

T       F                False

F       T                

F       F               

At this point, if we are fortunate to get a truth telling assistant, then we have a way of knowing behind which door contains a million dollars.  All we need to do is construct a sentence using P and Q such that if Q=T, then the sentence is true, and if Q=F, then the sentence is false.  So, if the assistant were to answer ‘true’ to our statment, then we would know that the million dollars is behind door #1.  If the assistant were to answer ‘false’ to our statment, then we would know that the million dollars is behind door #2.  This is what this looks like…

P   |   Q  |   Desired Response   |   Value of Sentence

T       T                True                              True

T       F                False                              False

F       T                

F       F                 

This good as far as things go, but we still have a problem.  We do not know if the assistant is in fact a truth teller.  She may well be a liar.  What do we do now?  Well, we still want the assistant to answer ‘true’ when the million dollars is behind door #1 and false when it is not.  That is to say, if P=F (the assistant is not a truth teller), then she answers ‘true’ if Q=T, and she answers ‘false’ is Q is false.  This looks like this…

P   |   Q  |   Desired Response   |   Value of Sentence

T       T                True                              True

T       F                False                              False

F       T                True

F       F                False

If we can find a sentence that would produce these kinds of results, then we would be good.  In this situation, no matter whether or not the assistant is a truth teller, a response of ‘true’ tells you that the million dollars is behind door #1 and a response of false is behind door #2.  So, what would be the truth value of a given sentence if the assistant was a liar (P=F), and the person answered ‘true’?  The sentence must be false.  By the same token, if the assistant answers ’false’, then the sentence must be true.  This looks like this…

P   |   Q  |   Desired Response   |   Value of Sentence

T       T                True                              True

T       F                False                              False

F       T                True                               False

F       F                False                               True

Consider the column with the red values.  If we can construct a sentence using P and Q such that its truth value is <T, F, F, T>, then we have our answer.  A way to determine a sentence is to consider its Disjunctive Normal Form (DNF).  That just means that we look at the values for P and Q when the sentence is true.  Notice, the sentence is true only when both P and Q are true (P ∧ Q) or when both P and Q are false (¬P ∧ ¬Q) .  This is symbolized as: ((P ∧ Q) ∨ (¬P ∧ ¬Q)), and is one answer.  It says, “either the assistant is a truth teller and the million dollars is behind door #1 or the assistant is not a truth teller and the million dollars is not behind door  #1.” 

Another answer is to use the Conjunctive Normal Form.  The first step is to take the dual of the DNF, which means exchanging every ’∧’ for  ’∨’ and vice versa, along with changing every sentence letter to its negation.  This gives us: ((¬P ∨ ¬Q) ∧ (P ∨ Q)).  The negation of the dual is tautologically equivalent to the DNF.  So, we negate the dual for the answer: ¬((¬P ∨ ¬Q) ∧ (P ∨ Q)).  This sentence says, “It is not the case that (either the assistant is not a truth teller or the million dollars is not behind door#1) and (either the assistant is a truth teller or the million dollars is behind door #1).”  Of course, both of these sentences are a little complicated.  Is there a more simple answer?  Yes, there is…

By definition, here is a list of some truth values for our various logical connectives…

P   |   Q  |   ¬P   |  P → Q  |   P ∧ Q   |   P ∨ Q   |   P ↔ Q 

T       T         F           T             T               T              T

T       F         T           F             F               T               F

F       T         F           T             F               T               F

F       F         T           T             F               F               T

As you can see,  P ↔ Q has the very truth value we are looking for.  As such, this is a very simple answer.  It says, “The assistant is a truth teller if and only if the million dollars is behind door #1.”

What exactly is Middle Knowledge (MK)? Thomas Aquinas presented two categories of knowledge that God was said to have. One was “natural knowledge” (NK) and the other “free knowledge” (FK). MK is a third type of knowledge that God is said to have that fits in-between God’s NK and His FK.

Natural Knowledge

Natural knowledge is that knowledge God has based on His very nature. It includes all necessary truths. For instance, the laws of logic would be part of God’s NK. The set of all possible worlds would be included in this, and along with this all of the possible events that make up these possible worlds. All mathematical truths would be part of this. Here are some examples of this…

God knows that “If P, then Q” is the case, then “If ¬Q, then ¬P” is the case.
God knows that our actual world is possible.
God knows that Germany winning WWII is true in some possible world.
God knows that 2+2=4.
God knows that He is love.

Any propositions like this are considered natural knowledge. What is interesting about this knowledge is that it is knowledge independent of God’s will – that is to say, it is pre-volitional. God did not will the law of non-contradiction. Rather, the law of non-contradiction is part of God’s essence. NK is grounded in the very nature of God.

Free Knowledge

Free knowledge is that knowledge that God has concerning what He wills - desires. For instance, “It is possible that I exist” is true whether or not God creates me. That is to say, I do exist in some possible world even if God does not choose to actuate that world. However, the actuality that I exist depends upon God creating me, or more properly in God actualizing the possible worlds where I exist. All knowledge that is a result of God’s will is FK. Here are several examples of FK…

God knows who will be saved.
God knows that Christ died.
God knows that Germany lost WWII.
God knows that the president of the United States is George Bush.

All propositions of the actual world are part of God’s FK. This knowledge depends upon God’s will. It requires God doing something – actuating a particular possible world. It is “post-volitional” in nature.

Middle Knowledge

In our next post we will explain this. Suffice it to say it comes between FK and NK. One can see that NK is logically prior to FK. NK is not dependent upon God’s will. NK is necessary.  FK is contingent. MK is said to come between God’s NK and His FK. The characteristic of MK is that it is not dependent upon God’s will (it is pre-volitional like NK) yet, it is contingent like FN. We will explain the type of knowledge that makes up this category, and this will explain how the Molinist brings God’s sovereignty and man’s LFW together.

Part I

Part II

Part IV 

In the first post in this series (here), I introduced modal logic and possible world semantics which will be the vehicles for our analysis.  In this post, we will state the three assumptions as given by Dr. Laing in his article “Middle Knowledge“.

Assumption 1

“First, it is assumed that for an action to be free, it must be determined by the agent performing the action. This means that God cannot will a free creature to act in a particular way and the act still be free. Free actions must be self-determinative. This assumption may appear self-evident to some, and quite controversial to others. While it must be admitted that God could certainly desire a creature act in a particular way and the choice remain free, it is difficult to see how He could cause the choice and it still be free in a meaningful way. Proponents of middle knowledge do not deny that God may influence a free choice or persuade an agent to act in a particular way, but such influence and persuasion cannot be determinative if the action performed is to be free. In addition, middle knowledge requires freedom of a libertarian nature. That is, free creatures have the ability to choose between competing alternatives, and really could choose one or the other of the alternatives.”

Regarding this, I put forth the following two definitions:

Definition (free action): Person A performs action X freely if and only if all influences not intrinsic to person A’s will do not constitute a set of influences sufficient to cause person A to perform action X.  (Note: All influences not intrinsic to person A’s will is going to be referred to as C, which stands for the complete set of circumstances apart form the person’s will.) 

Definition (libertarian free will): Person A freely performs action X given circumstance C with libertarian free will if and only if there exists a possible world where Person A freely performs action ¬X given circumstance C. 

Back in the summer of 2007 Dr. Laing and I corresponded several times via email.  In an email dated 6/15/2007, Dr. Laing stated that my definitions “capture the idea.”  He went on to recommend Peter van Inwagen’s An Essay on Free Will as a more thorough treatment of the matter. 

Assumption 2

“Second, it has become customary to speak of a logical priority in divine thoughts. This is not to deny the simplicity or omniscience of God, or to say that He gains knowledge that He did not previously possess. Rather, it is simply to acknowledge that dependency relationships exist between certain kinds of knowledge. It is also to acknowledge that something analogous to deliberation may take place in the divine mind. For example, in order for God to know that one plus one equals two, He must first comprehend the meaning of the concepts represented by the numbers, mathematical symbols, and formulaic expressions; they serve as a basis by which the truthfulness of the formula may be evaluated. But this is not to say that there was a time when God did not know 1+1=2. Thus, a relationship of logical priority, but not necessarily temporal priority exists between some of the content of divine knowledge.”

This is rather straightforward, and certainly is necessary for a God whose existence is independent of time.

Assumption 3 

“Third, proponents of the doctrine of middle knowledge believe that things could have been different than they, in fact, are.”

Essentially, this states that there are events that are contingent (not necessary).  For example, God could have chosen not to create the world.  I could have stayed home today rather than going to a spring training baseball game between the Diamondbacks and the Giants. (Diamondbacks won 6-5.)  This is why possible world semantics is such a good tool when analyzing Middle Knowledge. It accounts for these possibilities.   

Of these three assumptions, pay particular attention to assumption 1.  This is critical in our examination.

Part I

Part III

Part IV 

The following is taken from Logic - Techniques of Formal Reasoning, pg. 104.

Alfred, Kurt and Rudolf are suspected of civil disobedience.  They testify under oath as follows:

Alfred: Kurt is guilty and Rudolf is innocent.

Kurt: If Alfred is guilty, then so is Rudolf.

Rudolf: I am innocent but at least one of the others is guilty. 

Assuming that one is guilty if and only if they are not innocent, answer the following questions:

(a) Assuming everyone’s testimony is true, who is innocent and who is guilty?

(b) Assuming everyone is innocent, who commited perjury?

(c) Assuming that only the innocent told the truth, who is guilty?

__________________________________

Let A: “Alfred is innocent.” 

Let K: “Kurt is innocent.”

Let R: “Randolf is innocent. 

The three statements of Alfred, Kurt and Rudolf can be symbolized as follows:

Alfred: (¬K ∧ R)

Kurt: (¬A → ¬R)

Rudolf: (R ∧ (¬A ∨ ¬K))

(a) Since we are assuming all of these statements are true (see T), we can construct the following abbreviated truth-table:

(¬K ∧ R)  (¬A → ¬R)  (R ∧ (¬A ∨ ¬K))

    T T T      F  T   F     T T    F  T   T

Based on this: Alfred and Rudolf are innocent, and Kurt is guilty.

(b) Since we are assuming everyone is innocent (see T and F), then we can construct the following abbreviated truth-table:

(¬K ∧ R)   (¬A → ¬R)     (R ∧ (¬A ∨ ¬K))

    F F T        F  T   F       T  F    F F   F 

According to this table, Alfred and Rudolf lied, and Kurt told the truth.

(c) At this point, we are told that only the innocent told the truth.  That means we need to find a truth-table where K is true if and only if (¬A → ¬R) is true (T); where R is true if and only if (R ∧ (¬A ∨ ¬K)) is true (T); and A is true if and only if (¬K ∧ R) is true (T).  This can be symbolized as follows:

 Case 1 (assume K is true)

K ↔ (¬A → ¬R)  R ↔ (R ∧ (¬A ∨ ¬K))  A ↔ (¬K ∧ R)

T T    T  T   T    F T  F  F    T  T  F     F  T    F  F F

In this situation Kurt is innocent and told the truth,  and both Alfred and Rudolf are guilty and lied.  This meets our requirements, and as such we can stop.  But let’s look at the other cases. 

Case 2 (assume K is false and R is false)

K ↔ (¬A → ¬R)  R ↔ (R ∧ (¬A ∨ ¬K))  A ↔ (¬K ∧ R)

F T    F   T   T    F T  F  F    F  T  T     T  T    T T F

In this situation we end up with a contradiction.  K is false, and (¬A → ¬R) is true.  This contradicts our assumption that K ↔ (¬A → ¬R)  is true.  As such we discard this case, and move on to case 3.

Case 3 (assume K is false and R is true)

K ↔ (¬A → ¬R)  R ↔ (R ∧ (¬A ∨ ¬K))  A ↔ (¬K ∧ R)

F T    F  T  F     T  T  T T   F   T  T     T  T    T T T

In this situation we end up with a contradiction.  K is false, and (¬A → ¬R) is true.  This contradicts our assumption that K ↔ (¬A → ¬R)  is true.  As such we discard this case, and conclude with only case 1.

There is river on whose south bank is a village called Governance.  The only way to the village is over a bridge that spans the river.  The inhabitants of this village pride themselves as being an orderly society governed by laws.  Those who do not follow the laws are swiftly executed.  (Governorians take their orderliness very seriously.)    

The village appointed a watchman on the north side of the river.  His job was two fold: (1) He was to ask everyone who wanted to cross the bridge where they were going, and the purpose for their visit; and (2) He was to enforce the law of the bridge (LB).  The LB is as follows… 

LB: A person is executed only on the condition that they lied to the watchman and still crossed the bridge.   

Along came a man by the name of Defiance.  The watchman asked him where he was going and the purpose of the visit.  Mr. Defiance answered and then proceeded to cross the bridge.  However, it was not possible for the watchman to enforce the bridge rules.  

Question: What did Mr. Defiance tell the watchman?

_____________________________________

Answer: Mr. Defiance told the watchman that he was going to cross the bridge to be executed. (Highlight the previous section to reveal the answer.)  

Let’s assume the following abbreviation…

T: Mr. Defiance told the truth.

C: Mr. Defiance crossed the bridge.

E: Mr. Defiance is to be executed.

Given this, LB in regards to Mr. Defiance can be symbolized as follows…

(LB) (E ↔ (¬T ∧ C))

In addition to these rules we know that Mr. Defiance did cross the bridge (Q). 

 (1) Q 

He also said where he was going and why.  Since we do not know what he said we will lable these two answers as ‘X’ and ‘Y’.  What he said is true if and only if both X and Y are the case. This can be symbolized as follows… 

(2) (T ↔ (X ∧ Y))

The only other bit of information we are given is that the watchman was unable to follow the bridge rules once he was told (X ∧ Y) and Mr. Defiance crossed the bridge (Q).  In other words, (1) and (2) led to the contradiction of LB.  (The contradiction of LB is what is meant when we say the watchman is unable to enforce LB.)  This can symbolized as follows…

(3) (((T ↔ (X ∧ Y)) ∧ C) → ¬(E ↔ (¬T ∧ C)))

At this point, our question has been boiled down to the following:  What are ‘X’ and ‘Y’ such that (3) is a tautology?  The technique we will use is a reductio technique that assumes (3) is false.  The idea is this, if (3) is a tautology, and we assume it is false, then we should end up with some kind of contradiction.  We will create this senario and see what ‘X’ and ‘Y’ must be to have a contradiction.  Here is the truth table…

Case 1 (T=true)

(((T ↔ (X ∧ Y)) ∧ C) → ¬(E ↔ (¬T ∧ C)))

    T  T      T       T T  F  F F T     F F T

At this point, T is true, C is true and E is false.  In order to have a contradiction, then either X or Y must be false.

Case 2 (T=false)

(((T ↔ (X ∧ Y)) ∧ C) → ¬(E ↔ (¬T ∧ C)))

    F  T      F       T T  F  F T T     T T T

At this point, T is false, C is true and E is true.  In order to have a contradiction, then X and Y must be true.

Now our question has transformed one more time.  We are now asking: What is X and Y such that in case 1 one at least one of them is false and in case two both of them are true?  Mr. Defiance told the watchman (1) where he was going and (2) why.  Of our variables (T, C and E) only C and E could be answers to the watchman’s question.  If we assume that Mr. Defiance answered with some form of C and E, then what form must they be?  In both cases, C is true.  So, let’s let Y stand for C.  We now have the following two cases…

Case 1a (T=true)

(((T ↔ (X ∧ C)) ∧ C) → ¬(E ↔ (¬T ∧ C)))

    T  T      T T   T T  F  F F T     F F T

At this point, T is true, C is true and E is false.  In order to have a contradiction, then X must be false.  Since E is a false statement, letting X be E we satisfy case 1a. However, will this work for case 2a?

Case 2a (T=false)

(((T ↔ (X ∧ C)) ∧ C) → ¬(E ↔ (¬T ∧ C)))

    F  T      F  T   T T  F  F T T     T T T

At this point, T is false, C is true and E is true.  In order to have a contradiction, then X must be true.  We have already set X=E and E is true, therefore X is true in this case.  As such, we have satisfied the contradiction in both cases.  In other words, if we substitute E for X and C for Y, then we have come up with our answer.  In short, Mr. Defiance told the watchman that he was going to cross the bridge to be executed.

Immediately, you can see the dilemma the watchman is in.  If Mr. Defiance told the truth that he was to be executed, then according to LB he should not be executed.  However, if he is not executed, then Mr. Defiance lied, and he should be executed.  And around and around we go.  As you can see, the watchman was unable to do anything.  Here is the formal derivation of  (((T ↔ (E ∧ C)) ∧ C) → ¬(E ↔ (¬T ∧ C)))…

1. Show (((T ↔ (E ∧ C)) ∧ C) → ¬(E ↔ (¬T ∧ C)))

2. Assume ((T ↔ (E ∧ C)) ∧ C)

3. Show ¬(E ↔ (¬T ∧ C)) 

4. Assume (E ↔ (¬T ∧ C))

5. Show (T → E)

6. Assume T

7. (T → (E ∧ C)) [BC/C - 2]

8. (E ∧ C) [MT - 6, 7]

9. E [Simpl. - 8]

                    10. Show (¬T → E)

11. Assume ¬T

12. (¬T ∧ C) → E [BC/C - 4]

13. C [Simpl. - 2]

14. (¬T ∧ C) [Adj. - 11, 13]

15. E [MP - 12, 14]

16.  Show E

17. Assume ¬E

18. (T → E) [R - 5]

19. (¬T → E) [R - 10]

20. ¬T [MT - 17, 18]

21. ¬¬T [MT - 17, 19]

22. Show E → ¬T

23. (E → (¬T ∧ C)) [BC/C - 4]

24. (¬T ∧ C)) [MP - 16, 23]

25. ¬T [Simpl. - 24]

26. Show E → T

27. (T ↔ (E ∧ C)) [Simp. - 2]

28. ((E ∧ C) → T) [BC/C - 27]

29. C [Simpl. - 2]

30. (E ∧ C) [Adj. - 16, 29]

31. T [MP - 28, 30]

32. T [MP - 16, 26]

33. ¬T [MP - 16, 22]

In part I, we saw that on the assumption R is countable, then there exists a function, f, whose domain is the set of natural numbers and whose range is the set of real numbers.  In other words, the elements in N are uniquely mapped to elements in R such that there is no element in R that is not mapped to by an element in N.  If we could find an element in R that is not mapped to by an element in N, then this would contradict our assumption that f is bijective.  Let’s look at our mapping function from part I…

Domain (N)                                        Range (R)

       1                          f(1)                       256      

       2                          f(2)                        

       3                          f(3)                     

       4                          f(4)                        5/9

        .                              .                            .

        .                              .                            . 

        .                              .                            .

        n                          f(n)                         r

        .                              .                            .

        .                              .                            .

        .                              .                            .

By the assumption that R is countable, and that f is bijective, this listing above represents a complete listing of the real numbers as they correspond to the natural numbers: ’1′ corresponds to ‘256′ - ‘2′ corresponds to ’‘ - ‘3′ corresponds to  - and so on and so forth.  Consider the numbers in the range of f, i.e. the real numbers ‘r‘.  I would like to represent these numbers with their full decimal expansions.  The following table shows this…

Domain (N)                                        Range (R)

       1                          f(1)                     256.0000…     

       2                          f(2)                         3.1415…

       3                          f(3)                         1.4142…

       4                          f(4)                          .5555….

        .                              .                            . 

        .                              .                            .

        n                         f(n)                         r

        .                              .                            .

        .                              .                            .

        .                              .                            . 

Consider each real in the list above, but only that part of the real after the decimal point.   In other words, our first listed real is ‘256.000…’ We are only going to consider the ’.0000…’. Regarding our second listed real, we are only going to consider the ‘.1415…’. And so on and so forth.  In other words, we have the following…

1  ↔   .0000…     

2  ↔   .1415…

3  ↔   .4142…

4  ↔   .5555… 

 …

As can be seen, I have highlighted the first digit after the decimal place that corresponds to ‘1′.  I have highlighted the second digit after the decimal place that corresponds to ‘2′. I have highlighted the third digit after the decimal place that corresponds to ‘3′.  And so on and so forth.  At this point, we now perform the action called diagonalization on all of the highlighted numbers in the above array to create a new real number.  We take a number in the diagonal and if it is ‘5′, then we change it to a ‘6′.  If the number in the digonal is any number other than ‘5′, then it is changed to a ‘5′.  Since the first highlighted number in our above diagonal is a ‘0′, then we change it to a ‘5′.  Since the second highlighted number is a ‘4′, then we change it to a ‘5′.  Since the third highlighted number is a ‘4′, then we change it to a ‘5′.  Since the fourth highlighted number is a ‘5′, then we change it to a ‘6′.  And so on and so forth.  This creates the following real number…

 DN = .5556…

Is DN in the list above?  Remember, the list above is suppose to be a complete listing of all real numbers, and as such is supposed to contain DN.   However, DN is different than the first number because the first digit after the decimal in the DN is different than the first digit after the decimal of ’256.0000…’.  DN is different than the second number because the second digit after the decimal in the DN is different than the second digit after the decimal of ’3.1415…’.  In fact, given any ’n‘ and its corresponding ‘r‘ in the list, the DN differs from this ‘r‘ precisely at the nth digit after the decimal place of the ‘r‘.  As such, DN is a real number that is not in our list.  This contradicts our assumption that f is bijective.   Based on this, we conclude that R is uncountable.  Q.E.D.  

Prove:  R is not countable.

Assume: R is countable. 

By the definition of Countability, if R is countable, then |R|≤|N|.  Since N is a proper subset of R, then we know that |R| is not less than |N|.  Therefore, if R is countable, then |R|=|N|. 

We also know by the definition of same Cardinality that if |R|=|N|, then there exists a bijection between R and N.  Specifically, there exists a function whose domain is N and whose range is R.  

Domain (N)                                        Range (R)

       1                          f(1)                       256      

       2                          f(2)                        

       3                          f(3)                     

       4                          f(4)                        5/9

        .                              .                            .

        .                              .                            . 

        .                              .                            .

        n                          f(n)                         r

        .                              .                            .

        .                              .                            .

        .                              .                            .

What Cantor’s proof will do is show that there exists an element of R that is not in the range of f.  That is to say, in the above listing there is an r that is not listed.  This contradicts the claim that f is bijective.  Since we have reached a contradiction from the assumption that R is countable, then the conclusion is that R is not countable. In part II we will look at how Cantor does this. 

The following is taken from Logic - Techniques of Formal Reasoning, pg. 103.

According to ancient legends the Oracle of Delphi was infallible.  The Oracle was once asked whether any of the triumvirate of Pompey, Caesar, and Crassus would be assassinated.  The Oracle gave a two-sentence reply:

More than one of the triumvirate will be assassinated.  Crassus will be assassinated and either Caesar will not be assassinated or Pompey will be assassinated, if and only if, Pompey will not be assassinated and either Crassus or Caesar will be assassintated.

Question:  According to the Oracle, who among the triumvirate will be assassinated?

Answer:   

P = “Pompey will be assassinated.”

Q = “Caesar will be assassinated.”

R = “Crassus will be assassinated.”

The second sentence of the Oracle can be symbolized as follows:

(R ∧ ((¬Q ∨ P) ↔ (¬P ∧ (R ∨ Q))))

Since we are assuming that this statement is true and because the statement is a bi-conditional, then we know each part of the bi-conditional has the same truth value.  Let’s consider the case where each part of the bi-conditional is true.

Case 1:

((R ∧ (¬Q ∨ P)) ↔ (¬P ∧ (R ∨ Q)))

   T T     T T F    T    T T  T T F 

This yields that only Crassus will be assassinated.  However, we know from the Oracle’s first sentence that there will be at least two assassinations.  As such, case 1 fails.  We now turn to case 2 where each part of the bi-conditional is false.

Case 2:

((R ∧ (¬Q ∨ P)) ↔ (¬P ∧ (R ∨ Q)))

       F                  T        F   

At this point we need to consider further cases.  Let’s assume that R is true, and call this case 2a. 

Case 2a:

((R ∧ (¬Q ∨ P)) ↔ (¬P ∧ (R ∨ Q)))

   T F     F F T    T   F  F  T T T

We end up with a contradiction.  (¬Q ∨ P) is false, but ‘P’ is true.  The only time a disjunction is false is when both parts of the disjunct are false.  As such, case 2a fails.  This means ‘R’ must be false.  ’R’ being false simply means that Crassus will not be assassinated.  Since the Oracle told us at least two would be assassignated and we know that Crassus is not one of the two, then we can conclude that Caesar and Pompey will be assassignated. Q.E.D.

Question: In the debate between theists and atheists/agnostics, which side has the burden of proof? Are believers supposed to prove that God must exist, or must atheists demonstrate that God cannot exist?

Richard Heck’s Answer [1]:  What is the purpose of this “debate”? Is there a trophy? a financial reward? Or is the purpose supposed to be to determine the truth? or to determine what we should believe? I think the answer to your question very much depends upon the answer to this question.

Let’s suppose the purpose of the debate is to determine what one should rationally believe. What who should rationally believe? Does the person already have a view on this question? That is: Is she already a believer or a non-believer? Or is he or she utterly agnostic? This question, too, matters, at least according to some epistemologists, since these philosophers would take seriously the idea that the question we thinkers face is always whether to change our existing beliefs.

And there’s another crucial question, at least on some religious epistemologies: If one bases one’s belief upon religious experience, how is that supposed to enter the debate?

Peter Smith’s Answer [2]: “Burden tennis”, batting the burden of proof to and fro over the net, is rarely a very profitable pastime!

But still, maybe this case is an exception. After all, conventional theists when you come down to it are making some pretty exotic claims (claims that make the beliefs, say, of ancient Greek religion look very modest and humdrum). Not just powerful gods, but an omnipotent God. Not just intermittently casting an amused eye over mortal folly, but omniscient. Not just occasionally taking a passing interest in some of us, for good or ill (and occasionally, understandably, running off with a particularly pretty nymph) but incomprehensibly loving us all equally. And so it goes (for example, perhaps add claims about the Trinity here!). By the workaday epistemic standards we use in most of our lives, those extravagant claims look very fanciful indeed. So we can reasonably insist that someone who advances such claims literally, and expects to be taken seriously, had better have some very good arguments. Pending such arguments, the atheist doesn’t have much to do.

Of course, some will say that that is all thumpingly crass, and that religious beliefs aren’t to be construed literally, as involving exotic claims about what or Who exists. Indeed. I have some considerable sympathy with that. But the question as posed is naturally read as one concerning traditional debates about the existence of God: and in this case, the burden of proof surely does lie with the proponent of the extravagant ontological claim.

Scientiam Dei Commentary:  Let’s say I assert some proposition ‘A’ without any evidence or proof.  In Ask Philosophers 1 we considered this situation in light of Richard Dawkin’s maxim: That which can be asserted without evidence, can be dismissed without evidence.  If ‘A’ is simply asserted, then Richard Dawkins says that one is not epistemologically bound to accept ‘A’ simply on the basis of the assertion (there may be other reasons or evidences why one should accept ‘A’).  However, we pointed out that even though one may not be epistemologically bound to accept ‘A’, this says nothing as to whether or not ‘A’ is true. 

Peter Smith concludes that in those cases where claims are ”ontologically extravagant” the burden of proof lies with the proponent.  He says concerning the Christian God, “By the workaday epistemic standards we use in most of our lives, those extravagant claims look very fanciful indeed.”  I find this very interesting.  First off, note the loaded term ‘fanciful’.  Secondly, note that he speaks of the epistemic standards that we all commonly use in everyday life.  Romans 1:19-ff tells us that the evidence for God and His attributes is so overwhelming through the created order that no man is without excuse.  In other words, the Bible says that because our workaday epistemic standards clearly reveal God we are without excuse when we surpress this knowledge.  Now, it may be that Peter Smith would deny that our workaday epistemic standards do reveal such a God.  However, the simple point I am making is that Peter Smith is not neutral in his approach to this question (and neither am I for that matter).  He is bringing all kinds of baggage to the table that he is implicitly asking us just to accept. 

Richard Heck’s answer is a good answer.  Context certainly will weigh in on who in fact has the burden of proof.  However, when we are asking whether or not ‘A’ is true something else comes into play.  If I affirm that ‘A’ is true, I am at the same time denying that ‘¬A’ is true.  By the same token, if I deny ‘A’ is true, then I am affirming ‘¬A’.  Consider this in light of the question of God’s existence.  Is someone really epistemically justified in not believing in God’s existence (A) without appropriate evidence?  In other words, is someone espitemically justified in believing that God does not exist (¬A) without appropriate evidence?  Why is someone justified in believing in one proposition (¬A) without appropriate evidence while they can deny another (A) because of a lack of evidence?  To simply say that ¬A is the default position is to beg the question.      

[1] Richard Heck teaches at Brown University. He attended Duke as an undergraduate, studied for two years at Oxford, and received his Ph.D. from MIT. He taught at Harvard for 14 years before his present position at Brown.  His main interests are logic, language, mathematics, and mind.

[2] Peter Smith is a lecturer in the Faculty of Philosophy at the University of Cambridge. Previously he taught at Aberystwyth and Sheffield, and for twelve years he was editor of the journal Analysis. He has written books in the past on the philosophy of mind and on chaos theory; now he mostly works on logic-related matters, and his last book was on Gödel’s theorems.

Show that (((P → Q) ∧ (Q → P)) → (P ↔ Q)) is a tautology. 

(((P → Q) ∧ (Q → P)) → (P ↔ Q))

        T      T       T       F     F

         3      2       3        1      2

1. We assume the statement is false and attempt to derive a contradiction.

2. Since the statement is assumed to be false and since it is an implication, then the antecedent must be true and the consequent must be false.  This is the only time an implication is false. 

3. Since the antedecent is true and is itself a conjunction, then both parts of the conjunction must be true.

At this point we need to consider cases.  Let us consider the case when ‘P’ is true.

Case 1: 

(((P → Q) ∧ (Q → P)) → (P ↔ Q))

     T T T  T   T T T    F  T F T

      4 3  5  2    5 3 4     1  4  2 5

4. We are considering the case where ‘P’ is true. 

5. If ‘P’ is true and (P → Q) is true, then ‘Q’ must be true.  However, we end up with a contradiction at the red ‘F’.  If both ‘P’ and ‘Q’ are true, then (P ↔ Q) must be true, but it is already labeled false.  As such, we have determined that case 1 gives us contradiction.  So, we turn to case 2 where we assume ‘P’ is false.

Case 2:

(((P → Q) ∧ (Q → P)) → (P ↔ Q))

    F T  F   T   F T F   F   F F F

    6 3  7    2    7 3 6    1   6  2 7

6. We are considering the case where ‘P’ is false. 

7. If ‘P’ is false and (Q → P) is true, then ‘Q’ must be false.  However, we end up with a contradiction at the red ‘F’.  If both ‘P’ and ‘Q’ are false, then (P ↔ Q) must be true, but it is already labeled false.  As such, we have determined that case 2 gives us a contradiction. Since cases 1 and 2 exhaust all possibilities, and each possibility yields a contradiction, then we have shown that (((P → Q) ∧ (Q → P)) → (P ↔ Q)) is a tautology.  Q.E.D.

Using truth tables verify that (((P ∧ Q) → R) → (P → R)) is not a tautology.  A toutology is a symbolic sentence that is true for every possible truth value assignment.  Consider the following example: (((P → Q) ∧ P) → Q).  This is a symbolic representation of Modus Ponens.  As such, we know this statement is a tautology. That is to say, for any truth value assignments of ‘P’ and ‘Q’, the statement will always be true.  Here is the table…

P  Q  (((P → Q) ∧ P) → Q) 

T  T          T      T  T  T  T  

T  F          F      F  T   T  F

F  T          T      F  F   T  T

F  F          T      F  F   T  F

Now, in order to show that a symbolic statement is not a tautology, all one needs to do is to find one case (instance) when the symbolic statement comes out false.  This will be the approach we take with (((P ∧ Q) → R) → (P → R)) - rather than producing the complete table.  The approach we will take is to assume that the statement is false, and then see if we can make each part work consistently.  That is to say, if we do not end up with a contradiction, then we will have identified one truth value assignment were the symbolic statement comes out false.   

(((P ∧ Q) → R) → (P → R))

    T F F   T F   F  T F F

    4 5  6   2  4   1   3 2 3 

1. We assume the statement is false. 

2. Since this false statement is an implication we know that the antecedent is true and the consequent is false.  This is the only case where implications are false. 

3. This is the same reasoning as 2.

4. This assignment follows from 3.

5. Since ((P ∧ Q) → R) is true and the consequent is false, then the antecedent must be false.  If the antecedent were true, then the implication would be false.

6. Since the antecedent must be false, then the conjunction cannot be made of two true statements.  Since ‘P’ is already true, then ‘Q’ must be false. 

We were able to make all the assignments without a contradiction.  That means this particular assigment (P=true, Q=false, R=false) makes (((P ∧ Q) → R) → (P → R)) false, and as such it cannot be a tautology. Q.E.D.

I recently purchased the Cambridge Introduction to Philosphy text: An Introduction to Gödel’s Theorems by Peter Smith.  Dr. Smith is a contributor to a relatively new and intriguing website called Ask Philosophers.  I am going to interact with some of the questions that have been asked along with the various responses given.  The first question is as follows:

Question: Richard Dawkins has written: That which can be asserted without evidence, can be dismissed without evidence.  Is this valid, logically?  If not, what are the consequences?  He is talking about religious belief, i.e., belief in some God or other.  Dawkins’ statement makes sense to me but can any logical argument invalidate it?  Would he then have to retract his statement, or is there a gray area between semantics and logic?

Richard Heck’s [1] Answer: I don’t know the context of this claim, nor why Dawkins thinks—I take it he does think this—that no-one has any “evidence” for religious belief. Most theistically inclined epistemologists of religion, in the analytic tradition, anyway, think we do have certain kinds of evidence for belief in God. Dawkins might not find the evidence impressive, or he might disagree as to the evidential facts themselves, but it would be a parody of religious faith to think people believe on absolutely no basis. Just for example, suppose one is some kind of coherentist. Then you might think belief in God forms part of an overal “theory” of the world, and the evidence one has for it is that this theory is coherent, more successful than alternative theories, etc. You’ve got the same kind of evidence for your belief in God, ultimately, as for anything else you might believe, though belief in God, in such a system, will be deeply embedded, like very high-level theoretical claims, rather than towards the periphery, where experience impinges upon it more directly—to borrow some imagery from Quine.

But anyway, yes, if something is (forget about can be) asserted on absolutely no basis whatsoever, then, well, the person doing the asserting might as well just be making stuff up, and we can safely ignore them.

Scientiam Dei Commentary: The claim, “That which can be asserted without evidence, can be dismissed without evidence,” is not a logical rule.  Rather, the issue speaks more to what acceptable assertions ought to be.  He is saying one is not bound to accept ‘A’ in the case that ‘A’ is asserted without evidence.  Note the ethical norm that is being presupposed by Dawkins when he makes such a claim. 

The next issue is to consider what is counted as evidence for determining acceptable assertions.  As Dr. Heck pointed out, most theists do have reasons or evidence for their belief in God.  Even though Richard Dawkin’s rule does not really speak to this, one needs to make explicit what in fact constitutes as evidence in order to be able to use this rule. 

Assuming that we have an adequate foundation to determine what constitutes as evidence, should we accept the rule itself?  Presumably, if the statement itself is merely asserted (without evidence), then one can simply dismiss it based on the rule.  Therefore, if one wants to hold to this rule, then whatever grounds (evidences) that are used to validate this statement are avaliable to those wanting to validate other statements.  I am wondering what kind of evidence Richard Dawkins would appeal to for us to accept his statement?

Putting this all aside, given the Christian worldview should we accept Dawkins’ universal claim?  Consider God’s word.  What are His expectations of us when He asserts ‘A’ without anything further than His assertion?  He expects us to accept it.            

Lastly, if I simply assert ‘A’ with no evidence, and someone dismisses ‘A’ based on Dawkins’ rule, none of this speaks to whether or not ‘A’ is true.  In other words, even if someone clearly defines what evidences are appropriate in validating acceptable claims, and is able to gound Dawkin’s rule in an appropriate manner, the application of the rule to a claim does not speak to the truth-value of the claim.  Therefore, Dawkins’ dismissal of theistic claims based on his belief that they are not supported with evidence does not mean that such theistic claims are false.  

[1] Richard Heck teaches at Brown University. He attended Duke as an undergraduate, studied for two years at Oxford, and received his Ph.D. from MIT. He taught at Harvard for 14 years before his present position at Brown.  His main interests are logic, language, mathematics, and mind.

Truth tables are one way to verify the validity of arguments.   Consider the following standard derivation:

1. Show ((P → Q) ∨ (Q → R))

2. Show (¬(P → Q) → (Q → R))

3.  Assume ¬(P → Q)

4. Show (Q → R)

5. Assume Q

6. Show R

7. Assume ¬R

8. ¬(P → Q) → ¬Q [See Post - 3]

9. ¬Q [MP - 3, 8]

10. Q [R - 5]

11. ((¬(P → Q) → (Q → R)) ↔ ((P → Q) ∨ (Q → R))) [CD]

12. ((¬(P → Q) → (Q → R)) → ((P → Q) ∨ (Q → R))) [BC/C - 10]

13. ((P → Q) ∨ (Q → R)) [MP - 2, 11] 

Now, consider the following truth-table verification:

(P → Q) ∨ (Q → R)

  T F F  F   F F  ?

  4 2  5  1    6 3

1. We have assumed that the symbolic sentence is false.  This is symbolized by the ‘F’ above the ‘1′ and below the ‘∨’.  The idea is to attempt to derive a contradiction by assuming this sentence is false. 

2 and 3. The only way for a disjuction to be false is when both parts of the disjunct are false.  

4, 5 and 6.  Since (P → Q) is false under this assumption, then P must be true and Q must be false.  However, we have now arrived at our contradiction.  Under this senario, (Q → R) is false and ‘Q’ is false.  But the only way for (Q → R) to be false is for ‘Q’ to be true and ‘R’ be false.  Q.E.D.

Cardinality 

 The cardinality of a set is the size of a set, i.e., the number of elements that make up a set.   For instance, the set A={1, 2, 3} has a cardinality of 3. The cardinality of the natural numbers is:

This is called aleph-0 or aleph-null.  The cardinality of a set A can be symbolized as |A|.  Therefore, given set A above, |A| = 3.  The set of natural numbers, N, |N| = aleph -0.

Cardinality can be determined by using mapping functions.  Here is a key definition that will play a very important role as we consider Cantor’s Theorem.  

Definition: Two sets have the same cardinality if and only if there exists a bijection between them.

Consider the sets A={1, 3, 5} and B={2, 4, 6}.  We can see that both sets have the same number of elements, and therefore they have the same cardinality.  Our definition above implies that there exists a function taking A to B that is both an injection and a surjection (see Mapping Functions II).  This function is as follows:

f: A → B, where f(x)=x+1

The reason this function is an injection is because the only way f(x)=f(y) is when x=y. Proof. Part 1: Assume f(x)=f(y) and x≠y.  If f(x)=f(y), then x+1=y+1. By simple algebra, this leads to x=y, which contradicts x≠y.   Part 11: Assume f(x)≠f(y) and x=y. If f(x)≠f(y), then x+1≠y+1, and x≠y, which contradicts x=y. The reason this function is a surjection is because for any b ∈ B, there exists an a ∈ A such that b=f(a).  I will leave the demonstration of this to the reader.   

Countability 

Definition: A set A is countable if and only if  |A|≤|N|. 

Consider the set of even numbers E and the set of natural numbers N.  The following function is bijective:

 f: N → E, where f(x)=2x

Therefore, the set of even numbers and the set of natural numbers have the same cardinality. Since |E|≤|N|, then E is countable.  Consider the following sets:

The set of natural numbers N={1, 2, 3, …} is countable.

The set of integers Z={…, -3, -2, -1, 0, 1, 2, 3, …} is countable. 

The set of rational numbers Q={a/b: where a and b are elements of Z} is countable.

The set of real numbers (R) is the set of rational numbers plus the set of irrational numbers (ex. π and e).  Cantor’s Theorem is that R is uncountable. 

1. Show ((¬P → Q) ↔ (P ∨ Q))

2. Show ((¬P → Q) → (P ∨ Q))

3. Assume (¬P → Q)

4. Show (P ∨ Q)

5. Assume ¬(P ∨ Q)

6. Show ¬P

7. Assume ¬¬P

8. P [DN - 7]

9. (P ∨ Q) [Add - 8]

10. ¬(P ∨ Q) [R - 5]

11. Q [MP - 3, 6]

12. (P ∨ Q) [Add - 11]

13. Show ((P ∨ Q) → (¬P → Q))

14. Assume (P ∨ Q)

15. Show (¬P → Q)

16. Assume ¬P

17. Q [MTP - 14, 16]

18. ((¬P → Q) ↔ (P ∨ Q)) [C/BC - 2, 13]