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Proof step 5 from proof mod-two-ge (previous step)

0 ≤ i, j ≤ i, 0 ≤ j ⊦ 2 ^ i - 2 ^ i / 2 ^ j * 2 ^ j = 0, 2 ^ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
0 ≤ j, j ≤ i, 0 ≤ i ⊦ 2 ^ i = 2 ^ i / 2 ^ j * 2 ^ j

Rule argument

Additional information

Used simplifier rules: two-zero: ⊦ 0 ≤ j → 2 ^ j ≠ 0;%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + 0 = i; (s)% ⊦ i - j = i0 ↔ i = i0 + j; (s)