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Consider the following implication.

Let $k\in \mathbb{Z}$. If $k^{2} + 5k$ is odd, then $k^{2}+5k+1$ is odd.

At first it seems to be false, and one could proceed easily to prove it false directly by assuming that the premise ($k^{2} + 5k$ is odd) is true. However, there is no integer $k$ such that $k^{2} + 5k = k(k+5)$ is odd, and the premise is false in all posible cases, and so the implication follows vacuously (true).

I do have a lot of questions about this issue. But I want to be concise. Does the statement is true or false? and why?

ryang
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Kr'aamkh
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    It is vacuously true, exactly as you suggest. – lulu Apr 21 '22 at 16:19
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    Thanks for your answer. Yes! That was what made sense to me. But the process of direct proof (which proves it wrong) is correct as far as I'm aware, That's the problem I have. Does this has to do with the premise being false?? – Kr'aamkh Apr 21 '22 at 16:25
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    False $\Rightarrow$ True is always True, by definition. – dsillman2000 Apr 21 '22 at 16:27
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    Of course. The statement admits no counterexamples, because the premise is impossible. That makes the statement true, though not terribly interesting because there are no examples either. – lulu Apr 21 '22 at 16:27
  • Oh I see. So the process of direct proof is correct but it assumes the premise to be true and since it is impossible for the premise to be true, it follows that the implication is false. So this has to do with proof methods focusing more on the processes of deduction than in the truth of the premises. – Kr'aamkh Apr 21 '22 at 16:40
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    Usually $A \Longrightarrow B$ simply means, that if $A$ is true, so is $B$. It doesn't say anything if $A$ doesn't hold. Then $B$ may or may not be true. – Dietrich Burde Apr 21 '22 at 16:54
  • @Kr'aamkh The question asks for the truth value of the implication, which is (vacuously) true, as already explained. Note, however, that the implications "if $k^{2} + 5k$ is odd, then $k^{2}+5k+1$ is *even" and "if $k^{2} + 5k$ is odd, then* $1+1=3$" are also true for the same reason. Also note that if you slightly change the question to "what is the parity of $k^{2}+5k+1$ if $k^{2} + 5k$ is odd" then the answer "even" is wrong. – dxiv Apr 22 '22 at 06:37

1 Answers1

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Let $k\in \mathbb{Z}$. If $k^{2} + 5k$ is odd, then $k^{2}+5k+1$ is odd.

So the process of direct proof is correct but it assumes the premise to be true and since it is impossible for the premise to be true, it follows that the implication is false.

However, there is no integer $k$ such that $k^{2} + 5k = > k(k+5)$ is odd, and the premise is false in all possible cases, and so the implication follows vacuously (true).

But the process of direct proof (which proves it wrong) is correct

To be clear, formally, the implication here is

  • if $\bigg(k\in\mathbb Z\:\:$ and $\:\:k^{2} + 5k$ is odd$\bigg),$ then $\bigg[k^{2}+5k+1$ is odd$\bigg]$

rather than

  • $\bigg[k^{2}+5k+1$ is odd$\bigg],$

which is just its consequent/conclusion. And the statement to be proved is this:

  • for each $k$, if $\bigg(k\in\mathbb Z\:\:$ and $\:\:k^{2} + 5k$ is odd$\bigg),$ then $\bigg[k^{2}+5k+1$ is odd$\bigg].$

(Informally, “implication” has a second meaning as a synonym of consequent, and conflating these two meanings is causing confusion, I think.)

So this has to do with proof methods focusing more on the processes of deduction than in the truth of the premises.

Yes, proving an implication is typically a sequence of derivations that begins by assuming that its premises are true and ends by deducing its conclusion. In other words, the process is showing that whenever its premises are true, its conclusion must be true. There are two exceptions:

  • showing that its conclusion is true immediately (logically) derives the implication, even if none or only some of its premises are assumed: \begin{align}∀xCx &⊨ ∀x\big(Px → Cx\big)\\C&⊨P→C ;\end{align}
  • showing that its premise is false immediately (logically) derives the implication as being vacuously true: \begin{align}¬∃xPx &⊨ ∀x\big(Px → Cx\big)\\¬P &⊨ P→C.\end{align}

Thus, when we know that an implication's premise is false, we can just show this, even when it's possible to assume (against the facts) its premise as true then using axioms to deduce its conclusion, that is, even when we can, for example, exhibit the (vacuously true) chain $$-2=2\implies(-2)^2=2^2\implies4=4$$ or the (vacuously true) chain $$-2=2\implies-2+5=2+5\implies3=7.$$

the statement is true or false? and why?

Your given statement is vacuously true.

ryang
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