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Proof step 2 from proof pse (previous step)

⊦ m + n -1 -1 = (m + n)-1 -1, n = 1, n = 0

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, m0] flags:

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%pred: ⊦ (m + 1)-1 = m;%c: ⊦ m + n = n + m;%one-03: ⊦ 1 = m + 1 ↔ m = 0;%oz: ⊦ m + 1 ≠ 0;%pe: ⊦ n ≠ 0 → m + n -1 = (m + n)-1;%n ≠ 0 ⊦ m = n -1 ↔ n = m +1; (elim) Used terms: n, m0