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Proof step 6 from proof div-succ (previous step)

⊦ (m0 * n0 + m0) / (n0 + 1) = m0

User applied rule cut formula

And obtained the following new goals

Proof step 7:
m0 * n0 + m0 = m0 * (n0 + 1) ⊦ (m0 * n0 + m0) / (n0 + 1) = m0

Rule argument

fma: m0 * n0 + m0 = m0 * (n0 + 1)

Additional information

Used simplifier rules: ⊦ m * (n + 1) = m * n + m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ m + n = m + n0 ↔ n = n0; (s) Used formulas: m0 * n0 + m0 = m0 * (n0 + 1)