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Proof step 4 from proof modex-odd-neg (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
i < 0 ⊦ i % 2 = ~1 ↔ (∃ j. 2 * j = i + 1 ∧ j ≤ 0)

Rule argument

Additional information

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