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Proof step 9 from proof mod-two-ge-01 (previous step)

j < i, 0 ≤ j, 0 < i ⊦ 2 ^ j * 2 ^ (i - j) = 2 ^ j * ((2 ^ j * 2 ^ (i - j) - 1) / 2 ^ j + 1), 1 < 2 ^ j ∧ 0 < 2 ^ (i - j)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
i = i0 + j, j < i, 0 ≤ j, 0 < i ⊦ 2 ^ j * 2 ^ i0 = 2 ^ j * ((2 ^ j * 2 ^ i0 - 1) / 2 ^ j + 1), 1 < 2 ^ j ∧ 0 < 2 ^ i0

Rule argument

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

Additional information

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