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Proof step 2 from proof odd-rec (previous step)

0 < i, 1 < i ⊦ odd(i) ↔ odd(i - 2)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
i = i0 + 2, 0 < i, 1 < i ⊦ odd(i) ↔ odd(i0)

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [i, 2, i0] flags:

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i, 2, i0