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Proof step 3 from proof add-eq2pot-01 (previous step)

2 ^ (i + 1 - i0) = i1 + 1, 0 ≤ j, 0 ≤ i, 0 ≤ i0 ⊦ 2 * 2 ^ i = 2 ^ i0 + j ↔ j = 2 ^ i0 * i1 ∧ i0 ≤ i + 1

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
i + 1 = i2 + i0, 2 ^ i2 = i1 + 1, 0 ≤ j, 0 ≤ i, 0 ≤ i0 ⊦ 2 * 2 ^ i = 2 ^ i0 + j ↔ j = 2 ^ i0 * i1 ∧ i0 ≤ i + 1

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [i + 1, i0, i2] flags:

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i + 1, i0, i2