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Proof step 7 from proof div2-2pot-03 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
0 ≤ j, 0 ≤ i0 ⊦ (j * 2 ^ i0 * 2 + 1) / 2 = j * 2 ^ i0

Rule argument

Additional information

Used simplifier rules: ⊦ ~1 < i ↔ 0 ≤ i; (s)% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ (i + 1)-1 = i; (s)