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Proof step 2 from proof succ-01 (previous step)

m * n0 = m * m0 + 1 ⊦ m = 1

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, m, 1, n0, m0] flags: current, []

Additional information

Used simplifier rules: mult-c: ⊦ m * n = n * m;%zero: ⊦ 0 * n = 0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1%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 Used terms: 0, m, 1, n0, m0