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Proof step 2 from proof add-nonneg-int-to-nat (previous step)

0 ≤ i ⊦ i→n(i + n→i(n)) = i→n(i) + n

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ 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

Rule argument

indvar: n

Additional information

Used simplifier rules: nat freely generated by 0, +1;;%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%add-base: ⊦ n + 0 = n; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2