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Proof step 15 from proof divi-add (previous step)

m * n = m * n1 + n0 ⊦ m * (n - n1) = n0, n < n1

User applied rule insert spec-lemma

And obtained the following new goals

Rule argument

spec: nat-mult name: numeral-07 seq: m1 < m2 ⊦ (* m2) * m = (* m1) * m + n0 ↔ (m2 - m1) * m = n0 subst: [m1, m2, m, n0] → [n1, n, m, n0]

Additional information

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