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Can multiple contrapositives be used to disprove statements?

Suppose we have $$x \implies y \implies z,$$ and that $z$ is false. Thus, by contrapositive, $y$ cannot be true; so, $x$ cannot be true either. Is this valid or logically incoherent?

ryang
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TerribleStudent
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    That logic is just fine. – Ethan Bolker Jan 07 '25 at 22:41
  • You can verify that with a truth table. You only need to consider eight inputs. – John Douma Jan 07 '25 at 22:46
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    Your question took me a while to parse because I assumed you meant $x \implies (y \implies z)$ (equivalently $(x \land y) \implies z$), not $(x \implies y) \land (y \implies z)$. The former is more common. – Naïm Camille Favier Jan 07 '25 at 22:49
  • @NaïmFavier Sorry about the confusion, I meant two different statements, $x \implies y$ and $y \implies z $ and I want to prove/disprove x. So basically if I can disprove z, I can also infer that x must be false (?). Thanks. – TerribleStudent Jan 07 '25 at 23:02

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Yes, you've validly argued that $$\Big((x \to y) \land(y \to z)\Big)\land\lnot z\quad\to\quad\lnot x$$ is a tautology.

In case the sentence $$x \implies y \implies z$$ is being read with right associativity: the assignment $(x,y,z)=(1,0,0)$ shows that $$\Big(x \to (y \to z)\Big)\land\lnot z$$ does not entail $$\lnot x.$$

ryang
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  • Question about the right associativity answer -- doesn't (0,1,0) already imply not-x? I can see (1,0,0) being a counterexample, but can you explain why (0,1,0) is? – philo_777 Jan 15 '25 at 06:21
  • Thanks for the catch; edited! – ryang Jan 15 '25 at 06:30
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This is an example of the transitive property of the material conditional:

$(a \implies b) \wedge (b \implies c) \implies (a \implies c)$

Proof

$$(a \implies b) \wedge (b \implies c) \implies (a \implies c) \equiv \neg [(\neg a \vee b) \wedge (\neg b \vee c)] \vee (\neg a \vee c)$$ by definition of material conditional.

Applying De Morgan's Rule to the antecedent, we get:

$$ (a \wedge \neg b) \vee (b \wedge \neg c) \vee \neg a \vee c$$

Looking at each disjunct, this formula will be true if $\neg a$ or $c$ is true.

If we set them to false, we must have $(a \wedge \neg b)$ or $(b \wedge \neg c)$ be true, but this must be true under our assumptions since $a$ will be true and $\neg c$ will be true and either $b$ or $\neg b$ will be true.

So the initial formula is a tautology (always true). In fact, it is truth-functionally independent of the value of $b$ and hence makes sense why we can bypass/shortcut it to get right to $a \implies c$.

Now that we've shown we can reduce $a \implies b \implies c$ to $a \implies c$, we can apply the usual Modus Tollens (as you have done) to get your conclusion: $\neg c \implies \neg a$

philo_777
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