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Proof step 46 from proof divi-same (previous step)

m * n2 + n3 = m * n1, n1 < n2 ⊦ m * (n1 - n2) = n3, n3 = 0

User applied rule cut formula

And closed the goal!

Rule argument

fma: m * n1 < m * n2

Additional information

Used simplifier rules: ⊦ m0 * m < m0 * n ↔ m0 ≠ 0 ∧ m < n; (ws)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ m * 0 = 0; (s)% ⊦ 0 + m = m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m * n1 = m * m0 + n0 ∧ m ≠ 0 → ¬ n1 < m0; (s) Used formulas: m * n1 < m * n2