If the following statement is vacuously true, could it also be trivially true?

Question

Let $n∈ℕ$. If $n+n<1$ then $n+n∕2>1$.

Let $P(n)=n+n<1$

Let $Q(n)=n+n/2>1$

$P(n)$ is false $∀n∈ℕ$, so is $Q(n)$ is vacuously true?

But also, $Q(n)$ is true $∀n∈ℕ$, independent of $P(n)$, so is the overall open sentence trivially true, vacuously true, or both?

Answer 1

$\forall n\in\Bbb N^+~(n+n<1\to n+n/2>1)$ is vacuously true, because $\{n\in\Bbb N^+: n+n<1\}$ is indeed an empty set.   [Note: That is $\Bbb N^+$, to explicitly exclude $0$ from the domain .]

$\forall n\in\Bbb N^+~(n+n/2>1)$ is not vacuously true.   It is true but not vacuously so, because $\Bbb N^+$ is not an empty set.   Further, its truth is not derived from the above statement.†   It may be considered trivially true, because an inductive proof seems obvious and unnecessary.

$\dagger$ obviously: $\{\forall x~\neg P(x), \forall x~(P(x)\to Q(x))\}\nvDash \forall x~Q(x)$

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Well, $1+1/2>1$ is trivially true, and if, for any positive integer $n$, we assume that $n+n/2>1$, then clearly $n+1+(n+1)/2>1$ because: $$n+1+(n+1)/2~{=(n+n/2)+(1+1/2) \\> 1+1\\>1}$$ Thus it is proven that $\forall n\in\Bbb N^+~(n+n/2>1)$.

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Answer 2

The statement $$If\text { } n+n<1, \text { then } n+n∕2>1$$

Does not prove nor disprove $n+n∕2>1$

You might as well say $$If\text { } n+n<1\text { then } n+n∕2<1$$ which does not prove nor disprove $n+n∕2<1$

What we have here is the truth of $$If\text { } n+n<1, \text { then } n+n∕2>1$$based on the falsehood of $n+n<1.$

Since $ n+n∕2>1$ is true independent of $n+n<1$ ,$n+n∕2>1$ it is true.