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Proof step 3 from proof two-eq (previous step)

m = m / 2 ⊦ m = 0, m = 1

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: div-elim seq: n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n subst: [m, n, m0, n0] → [m, 2, m0, n] flags: <rotated> <all paths>

Additional information

Used simplifier rules: m1 ≠ 0 ⊦ m = (* m1) * m + n0 ↔ 0 = (m1 - 1) * m + n0; (s)% ⊦ m0 * 1 = m0; (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0); (s) Used terms: m, 2, m0, n