> RTQ[ ObjbjYY >^;p\;p\-%&&&8^\D&^," Y[[[[[[$@ n YY +NE.0^ ^ >: Homework 2- Propositional Logic- Semantic Proofs
One of the most interesting issues in the philosophy of language concerns the notion of compositionality. It starts with a puzzle raised by Descartes. Given that the overwhelming majority of sentences you hear and speak have never been spoken before and will never be spoken before, how do you understand what they mean? Some philosophers, like Donald Davidson, pose this issue in terms of human finitude. For any natural language there is no upper bound on the length of sentences. But that means that every natural language in some sense includes an infinite number of sentences. But how do finite beings like us grasp such an infinity?
The standard answer to both Descartes and Davidsons questions is compositionality. Languages contain a finite number of words and a finite set of principles for combining these words. Some of these combinatory principles are merely synctactic, concerning whether sentences are well formed or grammatical. For propositional logic weve already learned all of the syntax. Some of the combinatory principles are semantic, concerning how the meaning of larger units of language, such as sentences, is derived from the meaning of parts (there are many other kinds of principles such as phonetic, morphological, and pragmantic; but in this class we only concern ourselves with syntax and semantics).
Weirdly, in logic there are two approaches to the semantics of the formal languages, the model theoretic approach and the proof theoretic approach. Sometimes by semantics logicians just mean the model theoretic approach, because it is very clear just from looking at the model theory that it is a model of how certain facets of the meaning of sentences are determined via a finite set of principles from certain facets of the meaning of parts. The truth table semantics we are doing today is the model theory for classical propositional logic, and we will learn the proof theory later.
Before shortening all of this via truth-tables, we will first develop a system of semantic proofs parasitic on our system of syntax proofs. For each rule of syntactic proof, there will be two to four semantic rules.
The syntax proof system began with the following rule:
Where G is a propositional variable,
________
n) G rule 1
We replace this with the following two rules:
Where G is a propositional variable,
________
n. T G assignment
and,
________
n. F G assignment
These rules allow you to assign truth-values (T stands for truth and F for falsity) to the propositional variables within a sentence. Then the other rules will tell you how to derive the truth-values of more complicated sentences containing the propositional variables in question.
Now lets look at our negation rules and do a couple of problems.
n. T G
F G n
n. F G
T G n
These rules just say that negation switches the truth-value of sentences.
So here is a sample problem:
Where P is true, what is the truth value of P?
Proof:
1. T P assignment
2. F P 1
3. T P 2
Or sometimes you will need to compute all possible assignments:
Do a semantic proof for all possible assignments of truth values to the propositional variables in P?
Proof: Proof:
1. T P assignment 1. F P assignment
2. F P 1 2. T P 1
3. T P 2 3. F P 2
4. F P 3 5. T P 3
These proofs tell us that if P is true, then P is false, and if P is false, then P is true.
For our other connectives there are four versions of each rule. Let us do first and give a couple of examples.
m. T G m. T G
n. T F n. F F
T (G F) m,n F (G F) m,n
m. F G m. F G
n. T F n. F F
F (G F) m,n F (G F) m,n
The only difference between this and the negation rule is that we are dealing with two lines. Here are three sample problems.
Where P is true and Q is false, what is the truth-value of (P Q)?
Proof:
1. T P assignment
2. F Q assignment
3. F P Q 1,2
Where P is true, what is the truth-value of (P P)?
Proof:
1. T P assignment
2. F (P P) 1,1
Where P is true and Q is false, what is the truth value of ((P Q) P)?
Proof:
1. T P assignment
2. F Q assignment
3. F (P Q) 1,2
4. F P 1
5. F ((P Q) P) 3,4
Proofs involving disjunction look much like the above, except the truth functions are different. Here are our rules for disjunction:
m. T G m. T G
n. T F n. F F
T (G F) m,n T (G F) m,n
m. F G m. F G
n. T F n. F F
T (G F) m,n F (G F) m,n
Many people, when seeing this the first time, think that this is weird in the case where both sentences are true. What we are defining here is called inclusive disjunction and when we get to truth tables we will present the reasons why we use this disjunction. For now its good just to get comfortable with the system. Heres a problem using disjunction:
When P is false and Q also false, what is the truth value of ((P
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Proof:
1. F P assignment
2. F Q assignment
3. T P 1
4. F (P P) 1,3
5. T (P P) 4
6. T ((P P) Q) 5
Finally, here are our four versions of the conditional rule.
m. T G m. T G
n. T F n. F F
T (G ! F) m,n F (G ! F) m,n
m. F G m. F G
n. T F n. F F
T (G ! F) m,n T (G ! F) m,n
As with treating disjunction inclusively, many people find some aspects of this weird, in particular the fact that a conditional is counted as true whenever the antecedent (the first term, G above) is false. Again, let s just get comfortable doing the proofs for now. When we do truth tables we will discuss why this is the case for propositional logic. In both cases, the problem reveals fundamental limitations of the logic, though more severe in the case of the conditional.
At this point it is very, very important to note one difference between the conditional and the other operators. Conjunction and disjunction are commutative, which in this context means that (G F) is true (false) in exactly the same situations as (F G). Similarly with disjunction. But this is not the case with respect to disjunction. Where G is true and F is false then by the above rules (G ! F) will be false and (F! G) will be true.
As a result of this, it is extremely important when doing these proofs with conditionals to make sure what line your G is on and what line your F is on, and to use the correct version of the rule. Otherwise, everything is the same as before.
When P is false, Q is true, and R false what is the truth value of ((P ! (Q R))?
1. F P assignment
2. T Q assignment
3. F R assignment
4. F Q 2
5. F (Q R) 4,3
6. T ((P ! (Q R)) 5 !
----------------------------------------------------------------------------------------------------------------------
Homework 2
Let P be true, Q false, and R true. Construct semantic proofs for the following.
1. (P Q)
2. (P ! Q)
3. (P Q)
4. ((P Q) R)
5. (P ! (P R))
6. P
7. (P P)
8. (Q (P ! (Q R)))
9. (R ! ((P Q) ! R))
10. (P ! (P P))
PAGE
PAGE 4
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