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Proof step 1 from proof i2i-pos-01 (conclusion)

⊦ 0 ≤ i → n→i(i→n(i)) = i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 ≤ i ⊦ n→i(i→n(i)) = i

Rule argument

Additional information

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