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Proof step 2 from proof add (previous step)

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

User applied rule structural induction

And obtained the following new goals

Proof step 3:
0 ≤ i0 +1, 0 ≤ j, ∀ j. 0 ≤ j ∧ 0 ≤ i0 → i→n(i0) + i→n(j) = i→n(i0 + j) ⊦ i→n(i0 +1) + i→n(j) = i→n(i0 +1 + j)
Proof step 8:
0 ≤ i0 -1, 0 ≤ j, ∀ j. 0 ≤ j ∧ 0 ≤ i0 → i→n(i0) + i→n(j) = i→n(i0 + j) ⊦ i→n(i0 -1) + i→n(j) = i→n(i0 -1 + j)

Rule argument

indvar: i

Additional information

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