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Proof step 22 from proof cong-equiv (previous step)

n1 + m1 * m = m0 * m, m0 < m1, n2 < m ⊦ m * (m0 - m1) = n1, n1 = 0

User applied rule cut formula

And obtained the following new goals

Proof step 23:
m0 * m < m1 * m, n1 + m1 * m = m0 * m, m0 < m1, n2 < m ⊦ m * (m0 - m1) = n1, n1 = 0

Rule argument

fma: m0 * m < m1 * m

Additional information

Used simplifier rules: ls: ⊦ m0 * m < m0 * n ↔ m0 ≠ 0 ∧ m < n; from specification nat-mult%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult Used formulas: m0 * m < m1 * m