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Proof step 2 from proof exin2pot-neg (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
j < 0 ⊦ ∃ i. ~1 ≤ i ∧ j + 2 ^ i < 0 ∧ 0 ≤ 2 ^ (i + 1) + j

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs% ⊦ ~ i ≤ j ↔ 0 ≤ i + j; (s)% ⊦ i < ~ j ↔ i + j < 0; (s)