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I'm reviewing my previous exams for the final, and the only two true or false questions that confuse me are:

  1. If (A⊃B)∨(A⊃C) is true, then A implies either B or C.

  2. (P⊃Q)∨(P⊃~Q) means P⊃(Q∨~Q).

The answer is that both statements are false.

For 1, ((A⊃B)∨(A⊃C))⊃(A⊃(B∨C)), either by assigning truth values or by proof using inference, I find the conditional to be a tautology.

For 2, both (P⊃Q)∨(P⊃~Q) and P⊃(Q∨~Q) are tautologies. So ((P⊃Q)∨(P⊃~Q)) is equivalent to (P⊃(Q∨~Q)).

Then, why are these two statements false?

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  • 1
    Perhaps "either B or C" implies exclusive OR, but I do not see what the second issue might be unless you were going over intuitionistic logic.
    – Conifold
    Dec 8, 2016 at 22:17
  • For the second one, the only thing I can imagine seems utterly pedantic to me, namely that P itself does not entail Q v ~Q; it entails one or the other (since both entailments could never obtain simultaneously). But this seems pedantic and trivial because Q v ~Q is always true in standard sentential logic. and because material implication means 2 = ~P v Q v v ~P v ~Q which is satisfied for any value of Q.
    – virmaior
    Dec 9, 2016 at 1:46
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    Does the source of the statement explain its answer? Both of these look tautological to me, as well.
    – Mark Andrews
    Dec 10, 2016 at 2:18

1 Answer 1

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As the OP and the comments note, both of the results are tautologies.

Here is the truth table for the first one:

enter image description here

Here is the truth table for the second one:

enter image description here

As the OP notes the two sides of the biconditional are also tautologies.

As to why these were marked false, perhaps the answer key was in error or the problem that was intended was misstated. Perhaps the original question had a conjunction in place of the right-most disjunction. What was in the answer key may have been corrected versions of the original questions.

Michael Rieppel. Truth Table Generator. Generated on May 14, 2019 at https://mrieppel.net/prog/truthtable.html

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