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Proof step 3 from proof pse (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
⊦ n ≠ 0 → (n0 = n -1 ↔ n = n0 +1)

Rule argument

spec: <toplevel> name: elim seq: ⊦ n ≠ 0 → (n0 = n -1 ↔ n = n0 +1) subst: [n, n0] → [n, n0] flags: <rotated> <all paths>

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1;%oz: ⊦ m + 1 ≠ 0;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%one-06: ⊦ 1 = m + 1 ↔ m = 0;%pe: ⊦ n ≠ 0 → m + n -1 = (m + n)-1;%pred: ⊦ (m + 1)-1 = m; Used terms: n, n0