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Proof step 7 from proof div-twopot (previous step)

0 ≤ i0 + j, 0 ≤ j ⊦ 2 ^ (i0 + j) / 2 ^ j = 2 ^ i0, 0 ≤ i0 ∧ 0 ≤ j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: two-minus: ⊦ i < 0 → 2 ^ i = 0;% ⊦ i +1 = i + 1; (s)% ⊦ 1 ≤ i ↔ 0 < i; (s)%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%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)%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs% ⊦ i ≤ j ↔ ¬ j < i; (ss)