Proving (disproving?) a statement when its condition is not satisfied

Question

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?

Answer 1

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.