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Proof step 2 from proof psh-05 (previous step)

⊦ m < n -1 -1 ↔ m +1 +1 < n, n = 1, n = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: nat-basic1 inst: name: elim-1 seq: ⊦ n ≠ 0 → (n0 = n -1 ↔ n = n0 +1) subst: [n, n0] → [n, n0] flags: rewrite, current

Additional information

Used simplifier rules: lrec: ⊦ m +1 < n +1 ↔ m < n;%psh-07: ⊦ ¬ n < m -1 → ¬ n +1 < m;%numeral-1: m ≠ 0 ⊦ n +1 = (* m) ↔ n = m -1;%disj-1: ⊦ 0 ≠ n +1;%psh-04: ⊦ n ≠ 0 → (m < n -1 ↔ m +1 < n); Used terms: n, n0