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Proof step 3 from proof two-mone (previous step)

1 < i, 2 ≤ i, 2 ^ i -1 ≤ 0 ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
i = i0 +1, 1 < i, 2 ^ i0 ≤ 0, 2 ≤ i ⊦

Rule argument

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

Additional information

Used simplifier rules: ⊦ i = j -1 ↔ j = i +1; (elim) Used terms: i, i0