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Proof step 4 from proof add-eq2pot-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%lep: ⊦ i ≤ i + j ↔ 0 ≤ j; from specification int-abs