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Proof step 14 from proof mod-two-ge-01 (previous step)

0 ≤ j, 0 < i0, 0 < i0 + j ⊦ 2 ^ (i0 + j) = 2 ^ j * ((2 ^ (i0 + j) - 1) / 2 ^ j + 1), 2 ^ (i0 + j) = 2 ^ j * 2 ^ i0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 15:
2 ^ (i0 + j) = i1 + 1, 0 ≤ j, 0 < i0, 0 < i0 + j ⊦ 2 ^ (i0 + j) = 2 ^ j * (i1 / 2 ^ j + 1), 2 ^ (i0 + j) = 2 ^ j * 2 ^ i0

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [2 ^ (i0 + j), 1, i] flags:

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: 2 ^ (i0 + j), 1, i