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Proof step 10 from proof mod-two-ge (previous step)

i = i0 + j, j ≤ i, 0 ≤ i, 0 ≤ j ⊦ 2 ^ i = 2 ^ i0 * 2 ^ j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
0 ≤ j, 0 ≤ i0, 0 ≤ i0 + j ⊦ 2 ^ (i0 + j) = 2 ^ i0 * 2 ^ j

Rule argument

Additional information

Used simplifier rules: ⊦ ~1 < i ↔ 0 ≤ i; (s)%numlef-01: ⊦ (* 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)