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Proof step 1 from proof n2i-add (conclusion)

⊦ n→i(m + n) = n→i(m) + n→i(n)

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 2:
∀ m. n→i(m + n0) = n→i(m) + n→i(n0) ⊦ n→i(m + n0 +1) = n→i(m) + n→i(n0 +1)

Rule argument

term: n

Additional information

Used simplifier rules: inj: ⊦ n→i(m) = n→i(n) ↔ m = n;%nat freely generated by 0, +1;;%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%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