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Proof step 7 from proof divi-two-01 (previous step)

2 * n = m + 2 ⊦ 2 * (n - 1) = m, ∃ n0. 2 * n0 = m

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m0] flags:

Additional information

Used simplifier rules: rsucc-02: ⊦ m * (n + 1) = m * n + m; from specification nat-mult%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m0