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Proof step 5 from proof le-same (previous step)

m * n + n0 ≤ m, n0 ≤ m ⊦ m = 0, n = 0, n = 1 ∧ n0 = 0

User applied rule cut formula

And obtained the following new goals

Proof step 6:
m * n = m + m * n -1, m * n + n0 ≤ m, n0 ≤ m ⊦ m = 0, n = 0, n = 1 ∧ n0 = 0

Rule argument

fma: m * n = m + m * n -1

Additional information

Used simplifier rules: coef-17: ⊦ m * n + n = m1 * n ↔ ¬ ¬(m + 1 = m1 ∨ n = 0);%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%pred-03: ⊦ n -1 = n - 1; from specification nat%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%eq-02: n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2 Used formulas: m * n = m + m * n -1