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Proof step 5 from proof even-02 (previous step)

2 + m0 + m0 < n + n, m0 + m0 < n -1 + n -1 → n -1 + n -1 = 0 ∨ m0 < n -1 ⊦ n + n = 0, m0 + 1 < n

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat-basic1 inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n -1 ↔ n = m +1 subst: [n, m] → [n, m] flags:

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%succ: ⊦ m +1 = m + 1;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%oz: ⊦ m + 1 ≠ 0;%l: ⊦ m + n < m + n0 ↔ n < n0;%l-04: ⊦ m + m0 < n + n0 → (¬ m + m0 < n + n0 ↔ false);%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0;%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lf-05: ⊦ ¬ n < m ∧ m = n0 → ¬ n < n0; (forward) from specification nat-basic1%n ≠ 0 ⊦ m = n -1 ↔ n = m +1; (elim) Used terms: n, m