Solutions to 3.6E starred exercises
3.6 #1b) If P ≡ Q is truth functionally true, then P ≡ Q is true on every truth value assignment (by
definition of t-f truth). If it is true on every truth value assignment then P and Q must receive the same
truth value on every truth value assignment (by ≡ rule). If P and Q receive the same truth value on every
tva, they must be t-f equivalent (by definition of t-f equivalence). So if P ≡ Q is t-f true, P and Q must be t-
f equivalent.
#2b) If Γ t-f entails P ⊃ Q, then there is no truth value assignment on which all the sentences in Γ are true
and P ⊃ Q is false (by definition of t-f entailment). According to the ⊃ rule, this means that there is no
truth value assignment on which all the sentences in Γ are true, P is true, and Q is false. So there is no
truth value assignment on which all the sentences in Γ are true, all the sentences in {P} are true, and Q is
false. So there is no truth value assignment on which all the sentences in Γ ∪ {P} are true and Q is false.
So, by definition of t-f entailment, Γ ∪ {P} t-f entails Q. This is half of what needs to be proven.
Now let’s prove the other half. If Γ ∪ {P} t-f entails Q then there is no tva on which all the
sentences in Γ ∪ {P} are true and Q is false (by def. of t-f entailment). Since P is the only sentence in {P},
this means there there is no tva on which all the sentences in Γ are true, P is true, and Q is false. So (by
⊃ rule), there is no tva on which all the sentences in Γ are true and P ⊃ Q is false. So (by def of t-f
entailment), Γ t-f entails P ⊃ Q.
An alternative version of the argument:
If Γ t-f entails P ⊃ Q, then there is no truth value assignment on which all the sentences in Γ are true and
P ⊃ Q is false (by definition of t-f entailment). According to the ⊃ rule, this means that on any truth value
assignment on which all the sentences in Γ are true, either P is false or Q is true. Suppose P is false.
Then at least one of the sentences in Γ ∪ {P} is false. So it follows trivially that either at least one of the
sentences in Γ ∪ {P} is false or Q is true (because the first of these disjuncts is true). Now suppose Q is
true. Then once again it follows trivially that either at least one of the sentences in Γ ∪ {P} is false or Q is
true (because Q is true). So either way, on any tva, either at least one of the sentences in Γ ∪ {P} is false
or Q is true. But then, by definition of t-f entailment, Γ ∪ {P} t-f entails Q. This is half of what needs to be
proven.
Now let’s prove the other half. If Γ ∪ {P} t

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