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Proof step 1 from proof one-less (conclusion)

⊦ 1 < i ↔ 0 < i -1

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 2:
⊦ i = j -1 ↔ j = i +1
Proof step 3:
i = i0 +1 ⊦ 1 < i ↔ 0 < i0

Rule argument

spec: int-basic1 inst: name: elim-pred seq: ⊦ i = j -1 ↔ j = i +1 subst: [j, i] → [i, i0] flags: rewrite, current

Additional information

Used terms: i, i0