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Proof step 13 from proof dec-nat-01 (previous step)

⊦ n % 2 = 0, 2 * (n + 1) / 2 - 1 = n, n = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [2 * (n + 1) / 2, m]

Additional information

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