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Proof step 1 from proof add-nonneg-int-to-nat (conclusion)

⊦ 0 ≤ i → i→n(i + n→i(n)) = i→n(i) + n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 ≤ i ⊦ i→n(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