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Proof step 3 from proof psh-05 (previous step)

⊦ m < n -1 -1 ↔ m +1 +1 < n, 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: disj: ⊦ 0 ≠ n +1;%numeral-1: m ≠ 0 ⊦ n +1 = (* m) ↔ n = m -1;%psh-04: ⊦ n ≠ 0 → (m < n -1 ↔ m +1 < n);%lrec: ⊦ m +1 < n +1 ↔ m < n;%psh-01: ⊦ m < n0 → (n < n0 -1 ↔ n +1 < n0); Used terms: n, n0