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Proof step 3 from proof add-nonneg-int-to-nat (previous step)

∀ i. 0 ≤ i → i→n(i + n→i(n0)) = i→n(i) + n0, 0 ≤ i ⊦ i→n(i + n→i(n0 +1)) = i→n(i) + n0 +1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
0 ≤ i, ∀ i. 0 ≤ i → i→n(i + n→i(n0)) = i→n(i) + n0 ⊦ i→n(i + n→i(n0) + 1) = i→n(i) + n0 + 1

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ n→i(m + 1) = n→i(m) + 1;%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2