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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
i % 2 = 1, i < 0 ⊦ ∃ j. i = 2 * j + 1 ∧ 0 ≤ j

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs