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Proof step 7 from proof eq (previous step)

m0 * m1 + m0 = n * m0, m1 * m0 = (n - 1) * m0 → m1 = n - 1, ∀ m0, n. m1 * m0 = n * m0 → m1 = n ∨ m0 = 0 ⊦ m1 + 1 = n, m0 = 0

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m]

Additional information

Used simplifier rules: mult-base: ⊦ m * 0 = 0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%rsucc-02: ⊦ m * (n + 1) = m * n + m;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ m + 1 ≠ 0; (s)% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m