Value of an expression with an impossible predicate

Question

let $a = (a_1, a_2, a_3, \ldots , a_n)^T$ for some arbitrary large (irrelevant to the question) value of $n$.

and let $i,k \in \mathbb{Z}$

What would the value of the following expression be when $i = 0$?

$$\forall k : 0 < k < i: a_{k-1} \le a_k$$

As $i$ and $k$ are integers, there is no k such that $0 < k < i$. Does that mean that the entire expression is true because it holds "for all" (zero) admissible values of $k$?

Answer 1

(Terminlogy alert: it is not an "expression" but a "formula" -- it can be true or false, but doesn't have a "value").

It is an abbreviation for

$$ \forall k :\quad 0<k<i \implies a_{k-1} \le a_k $$

When $i=0$, then no matter what you choose for $k$, the antecedent $0<k<i$ will be false, and "$\mathsf{false}\Rightarrow\mathit{whatever}$" is always true, so the entire thing behind the $\forall k$ is always true. And that makes the entire formula true.

Answer 2

Yes, you are correct. $$\forall k\in S: P (k)$$

is an abbreviation for

$$\forall k:k\in S \implies P (k)$$

so when $S$ is empty, as it is here, the left side of the implication is false and the entire implication is true.