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Proof step 5 from proof cast-b2i-01 (previous step)

8 ≤ i, 0 < i ⊦ cast(i, b2i(by)) = b2i(by), 0 = i, ¬ 2 ^ (i - 1) ≤ b2i(by) ∧ ¬ 2 ^ (i - 1) + b2i(by) < 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
i = i0 + 1, 8 ≤ i, 0 < i ⊦ cast(i, b2i(by)) = b2i(by), 0 = i, ¬ 2 ^ i0 ≤ b2i(by) ∧ ¬ 2 ^ i0 + b2i(by) < 0

Rule argument

seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [i, 1, i0] rotate: T spec: int-abs inst: name: elim-sub

Additional information

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