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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 $k$?

thwd
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2 Answers2

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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.

MJD
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  • This might be a stupid detail but in my case it's a colon, not a period as in your answer. Does that make any difference? Also, I'm sorry for the wrong terminology used in the question. – thwd Dec 29 '14 at 23:06
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    It doesn't make a difference. Sorry for my needlessly confusing variation! – MJD Dec 29 '14 at 23:29
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    @thwd: Each author seems to have his own preferences for which punctuation to use with quantifiers, and in general it doesn't make any difference. See here for an attempt at surveying the options. – hmakholm left over Monica Dec 29 '14 at 23:31
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(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.

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    Thanks for the clarification on terminology. I'm (evidently) a CS-student. – thwd Dec 29 '14 at 23:08
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    @thwd: I guessed as much -- programming languages rarely bother to make a distinction between Boolean expressions and other ones, and seem to do fine without one -- but for better or worse, the tradition in mathematics is to distinguish them syntactically. – hmakholm left over Monica Dec 29 '14 at 23:15