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Proof step 6 from proof div-twopot (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
i0 < 0, 0 ≤ i0 + j, 0 ≤ j ⊦ 2 ^ (i0 + j) / 2 ^ j = 0

Rule argument

Additional information

Used simplifier rules: two-minus: ⊦ i < 0 → 2 ^ i = 0;%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs