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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
0 ≤ j, 0 ≤ i0, 0 ≤ i1 + i0 ⊦ 2 ^ (i1 + i0) = 2 ^ i0 + j ↔ j = 2 ^ i0 * (2 ^ i1 - 1) ∧ 0 ≤ i1

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%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