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Proof step 8 from proof modls (previous step)

m0 * n = m0 * m + n0, m < n, n0 < m0 ⊦

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: nat-mult inst: name: modeq seq: ⊦ n0 < n ∧ n1 < n → (m * n + n0 = m0 * n + n1 ↔ m = m0 ∧ n0 = n1) subst: [n0, n, n1, m, m0] → [0, m0, n0, n, m] flags: current, []

Additional information

Used simplifier rules: mult-c: ⊦ m * n = n * m;%z: ⊦ 0 < n ↔ n ≠ 0; from specification nat-basic1%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%z: ⊦ 0 + m = m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1 Used terms: 0, m0, n0, n, m