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Proof step 3 from proof eo-02 (previous step)

2 * m = n + 1 ⊦ n = 2 * (m - 1) + 1, ∃ m0. n = 2 * m0 + 1

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] → [m, m1] flags:

Additional information

Used simplifier rules: dist: ⊦ m0 * (m + n) = m0 * m + m0 * n; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = 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%add-num-01: n < n0 ⊦ m + (* n) = m0 + (* n0) ↔ m = m0 + (n0 - n); from specification nat%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m, m1