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Proof step 1 from proof n2i-mult (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: nat freely generated by 0, +1;;% ⊦ m * 0 = 0; (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ i * 0 = 0; (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult