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

m ≤ m1 ⊦ m + 1 = m1 * m0 ↔ m0 = 1 ∧ m1 = m + 1 ∨ m = 1 ∧ m0 = 2 ∧ m1 = 1

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: nat-mult inst: name: mult-eq-succ seq: ⊦ m < m1 → (m + 1 = m1 * m0 ↔ m0 = 1 ∧ m1 = m + 1) subst: [m0, m1, m] → [m0, m1, m] flags: rewrite, current

Additional information

Used simplifier rules: mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%mult-eq-succ-02: ⊦ m1 + 1 = m1 * m0 ↔ m1 = 1 ∧ m0 = 2;%lels-03: ⊦ m < n → m ≤ n; from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%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%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%anti: ⊦ m ≤ n → (n ≤ m ↔ m = n); from specification nat%eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%ler: ⊦ m ≤ m; from specification nat Used terms: m0, m1, m