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Proof step 14 from proof mod-two-ge-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
0 < i0, 0 < i0 + j, 0 ≤ j ⊦ 2 ^ (i0 + j) = 2 ^ j * 2 ^ i0, 2 ^ (i0 + j) = 2 ^ j * ((2 ^ (i0 + j) - 1) / 2 ^ j + 1)

Rule argument

Additional information

Used simplifier rules: ⊦ 1 ≤ i ↔ 0 < i; (s)%numlsf-02: ⊦ (* i) < j → i + 1 ≤ 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% ⊦ i < i + j ↔ 0 < j; (s)