Prove that if S tautological consequence of P, S tautological consequence of Q, then S tautological consequence of P | Q

Question

Consider the following argument:

S is a tautological consequence of P.

S is a tautological consequence of Q.

Therefore, S is a tautological consequence of P | Q.

I wish to give an informal proof of this argument.

First I build some joint truth tables:

P S

T T

F ?

Q S

T T

F ?

If P is false, we don't know the truth value of S. Same for Q and S.

Now consider the joint truth table of P | Q and S

P Q P | Q S

T T T ?

T F T ?

F T T ?

F F F ?

To determine whether S is a tautological consequence of P | Q, we need to consider the three cases in which P | Q is true. This suggests a proof by cases.

Case 1: P & Q is true

S follows from P. It also follows from Q.

Case 2: P & ~Q is true

S follows from P. We don't know what Q being false implies. What if the joint truth table for S and Q is such that when Q is false, S is false? Is this possible? If so, what does that imply for this particular case where we would have: P implies S, ~Q implies ~S?

Here's my guess: we have a subcase.

Case 2.1: S is true when Q is false.

Therefore, P implies S, and ~Q implies S. Therefore S.

Case 2.2: S is false when Q is false.

Therefore, P implies S, ~Q implies ~S. Therefore S & ~S, a logical impossibility. Therefore, this case is not possible. S must be true when Q is false.

Case 3: ~P & Q is true

Analogous to Case 2: we have two subcases, one of which is a logical impossibility. S must be true when P is false.

Therefore, we have proved that in each case, S is true. This makes it a tautological consequence of P | Q.

It seems we have also proved that S is true when ~P, and also when ~Q.

Is this proof correct?

Answer 1

So, the conditional theorem below is correct.

   P ⊢ S
   Q ⊢ S
---------
P∨Q ⊢ S

The conditional theorem is only false if there's a case where the premises are true and conclusion is false.

And similarly, in order for the conclusion to be false, P∨Q must be true.

You are examining all the cases you need to examine by looking at P,Q, P,¬Q, and ¬P,Q. To be honest though, I can't follow your argument for why it's sufficient to examine these cases. It would help to make it more explicit.

Your case analysis arguments mostly work, but they can be simplified a bit. You really just need to show that P or Q, whichever one happens to be true in each case, forces S to be true as well.

For case 1 P,Q:

Since P is true and P≤S, then S is true.

Thus P∨Q≤S is true.

For case 2 P,¬Q:

Thus P≤S is true. Thus S is true.

Thus P∨Q≤S is true.

For case 3 ¬P,Q:

Thus Q≤S is true. Thus S is true.

Thus P∨Q≤S is true.