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Proof step 23 from proof divi-sub (previous step)

n < n2, m * n = n1 + m * n2 ⊦ ∃ n0. m * n0 = n1

User applied rule cut formula

And closed the goal!

Rule argument

fma: m * n < m * n2

Additional information

Used simplifier rules: ls: ⊦ m0 * m < m0 * n ↔ m0 ≠ 0 ∧ m < n; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%zero: ⊦ 0 * n = 0; from specification nat-mult%add-base: ⊦ n + 0 = n; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%addle-01: ⊦ m * n1 = m * m0 + n0 ∧ m ≠ 0 → ¬ n1 < m0; from specification nat-mult Used formulas: m * n < m * n2