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Proof step 4 from proof add-eq2pot (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
2 ^ i1 = i2 + 1, 0 ≤ j, 0 ≤ i0, 0 ≤ i1 + i0 ⊦ 2 ^ (i1 + i0) = 2 ^ i0 + j ↔ j = 2 ^ i0 * i2 ∧ 0 ≤ i1

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [2 ^ i1, 1, i] flags:

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: 2 ^ i1, 1, i