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Proof step 1 from proof abs-succ-03 (conclusion)

⊦ 0 < i → i→n(i + 1) = i→n(i) + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 < i ⊦ i→n(i + 1) = i→n(i) + 1

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs