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Proof step 7 from proof exin2pot-neg (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
j + 2 ^ i = 0, 0 < 2 ^ i * 2 + j, j < 0, 0 ≤ 2 ^ i * 2 + j, 0 ≤ i ⊦ ∃ 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;% ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i < j → ¬ j < i; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; (s)