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Proof step 6 from proof exin2pot-neg (previous step)

0 < 2 ^ i * 2 + j, 2 ^ i + j ≤ 0, 0 ≤ i, j < 0 ⊦ ~1 ≤ i ∧ j + 2 ^ i < 0 ∧ 0 ≤ 2 ^ (i + 1) + j, ∃ i. ~1 ≤ i ∧ j + 2 ^ i < 0 ∧ 0 ≤ 2 ^ (i + 1) + j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
j + 2 ^ i = 0, 0 ≤ 2 ^ i * 2 + j, 0 < 2 ^ i * 2 + j, 0 ≤ i, j < 0 ⊦ ∃ i. ~1 ≤ i ∧ j + 2 ^ i < 0 ∧ 0 ≤ 2 ^ (i + 1) + j

Rule argument

Additional information

Used simplifier rules: succ: ⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j;%pot2-pos: ⊦ 0 < 2 ^ i ↔ 0 ≤ i;%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%ls-02: ⊦ i < i + j ↔ 0 < j; from specification int-basic2%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%lels-01: ⊦ ¬ i ≤ j ↔ j < i; 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%numleledel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; from specification int-abs