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So contrapositives are logically equivalent statements, which can be useful when proving things. But I'm not convinced that this is true.

Take this example: $x \in \mathbb{Z}$. Then $$x \neq 2 \Rightarrow x+1 \neq 0.5$$ is true, but the contrapositive $$x+1 = 0.5 \Rightarrow x=2$$ is false.

If one statement is always false/true, then is it not logically equivalent to its contrapositive?

Matthew Towers
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thedanktouch
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5 Answers5

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You've restricted yourself to $x \in \mathbb Z$. The hypothesis $x+1=0.5$ is always false, so the implication $$x+1 = 0.5 \implies x=2$$ is vacuously true.

Umberto P.
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    Is this a matter of convention in what we call true? You can't provide me an x as a counterexample, therefore its true? – thedanktouch Oct 01 '24 at 18:55
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    @thedanktouch https://en.wikipedia.org/wiki/Vacuous_truth#Examples You can also find many question on vacuous truth on this site. – Adam Rubinson Oct 01 '24 at 20:27
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    And this is symmetric with the original proposition being "trivially true", i.e. $x+1 \neq 0.5$ is true for all integers, so $P \implies x + 1 \neq 0.5$ is true by convention regardless of the truth of $P$. – usul Oct 03 '24 at 10:34
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$x + 1 = \frac{1}{2} \implies x = 2$ is true because the premise is that $x$ is an integer and there are no integer solutions to $x + 1 = \frac{1}{2}$. Thus for all the non-existent integer solutions $x$ to that equation, we have $x = 2$.

mathination
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4

Take this example: $x \in \mathbb{Z}$. Then $$x \neq 2 \Rightarrow x+1 \neq 0.5$$ is true, but the contrapositive $$x+1 = 0.5 \Rightarrow x=2$$ is false.

These statements are logically equivalent and all true (the last one vacuously true): $$x \in \mathbb{Z} \implies (x \neq 2 \implies x+1 \neq 0.5)\\x \in \mathbb{Z} \implies (\boldsymbol{x+1 = 0.5 \implies x = 2 })\\x \in \mathbb{Z} \implies (\boldsymbol{x+1 \ne 0.5 \,\text{ or }\, x = 2} )\\(x \in \mathbb{Z} \,\text{ and }\, x +1= 0.5) \implies x = 2.$$

These statements are logically equivalent and both false (the counterexample is $x=-0.5):$ $$x \in \mathbb{R} \implies (x \neq 2 \implies x+1 \neq 0.5)\\x \in \mathbb{R} \implies (x+1 = 0.5 \implies x = 2 ).$$

ryang
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  • A nitpick (but important here) — your last two statements are not necessarily false as currently written. Their universal quantifications are false — e.g. “for all $x \in \mathbb{R}$, if $x \neq 2$ then $x + 1 \neq 0.5$” — as you presumably meant. But with $x$ free, their truth depends on $x$: e.g. “if $x \in \mathbb{R}$, then $x \neq 2$ implies $x+1 \neq 0.5$” is false for $x = -0.5$, but true for any other value of $x$. – Peter LeFanu Lumsdaine Oct 02 '24 at 12:46
  • @PeterLeFanuLumsdaine All six statements (as well as the OP's two given statements) are indeed tacitly universally quantified. Calling them statements to signal this (i.e., they aren't to be read as unquantified predicates), rather than writing out "for every $x,$" was just to avoid distracting from the discussion's main point. – ryang Oct 02 '24 at 13:11
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The reason the concept of "vacuous truth" arises here is that your statement is not fully explicit. Every condition that it is constructed under needs to be included in it before you take the contrapositive. The complete statement is $$x \in \Bbb Z \text{ and }x\ne 2 \implies x + 1 \ne 0.5$$

So the actual contrapositive is $$\text{not }(x+1 \ne 0.5) \implies \text{ not }(x \in \Bbb Z \text{ and }x\ne 2)$$ By applying some logic operations, $$x + 1 = 0.5 \implies x \notin \Bbb Z \text{ or }x = 2$$

And here it is clear that if the left side is true, then $x \notin \Bbb Z$, which makes the right side true. "Vacuous truth" is just a concept used to allows us to leave the implicit part of your original statement still implicit in the contrapositive.

Paul Sinclair
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Remember that $$P \implies Q$$ is equivalent to $$\neg P \vee Q$$ The root of your confusion comes from this, because this is slightly different than what an "if...then" statement means in ordinary language. So $$x \neq 2 \implies x+1 \neq 0.5$$ Means the same thing as $$x=2 \vee x \neq -0.5$$ Now, this makes it clear that this is sort of "vacuously true," because $x \neq -0.5$ is always true. Now the contrapositive $$x+1 = 0.5 \implies x=2$$ is equivalent to $$x \neq -0.5 \vee x=2$$ This is exactly the same as the above statement.

Jbag1212
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