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Proof step 2 from proof num-n2i-ls-06 (previous step)

0 ≤ i ⊦ i ≤ n→i(n) ↔ i→n(i) ≤ n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
0 ≤ i ⊦ i ≤ n→i(n) ↔ i→n(i) ≤ n

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs