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Proof step 2 from proof n2i-mult (previous step)

∀ m. n→i(m * n0) = n→i(m) * n→i(n0) ⊦ n→i(m * n0 +1) = n→i(m) * n→i(n0 +1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ m. n→i(m * n0) = n→i(m) * n→i(n0) ⊦ n→i(m * n0) + n→i(m) = n→i(m) * (n→i(n0) + 1)

Rule argument

Additional information

Used simplifier rules: n2i-add: ⊦ n→i(m + n) = n→i(m) + n→i(n);%succ-01: ⊦ n→i(m + 1) = n→i(m) + 1;%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%rsucc-02: ⊦ m * (n + 1) = m * n + m; from specification nat-mult%succ: ⊦ m +1 = m + 1; from specification nat-basic2