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Proof step 3 from proof psh-01 (previous step)

m < n0 ⊦ n < n0 -1 ↔ n + 1 < n0, n0 = 0

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] → [n0, n1] flags: <rotated> <all paths>

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1;%l: ⊦ m + n < m + n0 ↔ n < n0;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m; Used terms: n0, n1