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Is the statement “Every implication or its converse must be true” correct, or is it possible to have both implication and converse be false?

If $A$ stands for "$x$ is a prime number" and $B$ stands for "$x$ is greater than $100$", then both $(A → B)$ and $(B → A)$ are false. Is this a counterexample to the quoted statement? How does this make sense with respect to the truth table of $(A → B) ∨ (B → A)$ showing that it is always true?

ryang
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sarah003
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    Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – CrSb0001 Oct 05 '23 at 17:11
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    Because "$x$ is a prime number" is not an absolute statement until you tell me what $x$ is. Basically, your statement has a quantifier in it. – Randall Oct 05 '23 at 17:13
  • Note that your example is implicitly quantified, which changes how it would be analyzed in this context.

    Either way, $(A \to B) \lor (B \to A)$ is a Classical tautology as you mention, and even Intuitionistic Logic can’t have both an implication and its converse as false. However, an implication and its converse can both fail to be true in Intuitionistic Logic.

    There are weaker logics where both an implication and its converse can be false.

     – PW_246 Oct 05 '23 at 17:21
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    @CrSb0001 that’s clearly false. The post is more than intelligible and asks a specific question. – PW_246 Oct 05 '23 at 17:23
  • Related: Proving an implication by disproving its converse – ryang Oct 06 '23 at 23:56

1 Answers1

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Put $Px$ for $x$ is prime, and $Gx$ for $x > 100$.

Now distinguish

  1. $\forall x(Px \to Gx) \lor \forall x(Gx \to Px)$

from

  1. $\forall x((Px \to Gx) \lor (Gx \to Px))$

The first I think properly captures the thought that you say is false (for yes, the generalization $\forall x(Px \to Gx)$ is false, and likewise $\forall x(Gx \to Px)$ is false). But (1) is not a logical theorem.

On the other had (2) is a classical theorem, but is not obviously false, once distinguished from (1). Take any $n$ you like. Then either $Pn$ is false, in which case the supposition that $Pn$ is true leads to contradiction, and anything goes, so we get in particular $(Pn \to Gn)$. Or $Pn$ is true, in which case $(Gn \to Pn)$. So either way we get $((Pn \to Gn) \lor (Gn \to Pn))$. But $n$ was arbitrary so we can generalize.

Peter Smith
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