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Proof step 9 from proof exin2pot-neg (previous step)

j = ~(2 ^ i), 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ls-2pot: ⊦ 0 ≤ i → (2 ^ i < j * 2 ^ i ↔ 1 < j);%pot2-pos: ⊦ 0 < 2 ^ i ↔ 0 ≤ i;%two-le: ⊦ 0 ≤ i → (2 ^ i ≤ 2 ^ j ↔ i ≤ j);%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%neg-05: ⊦ ~ i < 0 ↔ 0 < i; from specification int-abs%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j; from specification int-abs%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%one: ⊦ 1 * i = i; from specification int-mult%numle-09: ⊦ j ≤ (* i1) * j ↔ 0 ≤ (i1 - 1) * j; from specification int-mult%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; 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%neg-15: ⊦ i + ~ j = i - j; from specification int-abs