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I have been wondering if the following equivalence holds:

$(\forall i \, | \, 0 \leq i \leq n : f.i \leq f.n) \, \land \, f.n \leq f.(n+1) \equiv (\forall i \, | \, 0 \leq i \leq n + 1 : f.i \leq f.(n+1))$

So, I try to prove the $\Rightarrow$ direction by distributivity of $\land$ over $\forall$, and then by transitivity of $\leq$, but this gets me to:

$(\forall i \, | \, 0 \leq i \leq n : f.i \leq f.(n+1))$

I believe that this is equivalent to the consequent I am trying to establish, however, I don't I know of a rule that allows me to expand the range of the quantification to include $n+1$.

By the way, I am using the rules of the predicate calculus according to the book "A Logical Approach to Discrete Math", by Gries and Schnedier.

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    If $i$ is restricted to the natural numbers, your first equivalence is the principle of strong induction on the natural numbers. The final formula you showed is a consequence, but it says a lot less than the formula you started with, and it won't get you where you need to go. – 2017-02-02

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You need your logic to somehow recognize the equivalence $$ ((0 \leq i \land i \leq n) \lor i = n + 1) \iff (0 \leq i \land i \leq n + 1). $$ Without some kind of rules about how natural numbers function, this will not work. Just the rules of predicate logic, together with, say, the axioms of a partial/total ordering, are not enough. For instance, the sentence does not hold if the letter $i$ ranges over all real numbers.

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    Exactly! So maybe add the Peano axioms together with a definition for $<$. – 2017-02-03