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:
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?