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Proof step 1 from proof modex-odd-neg (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs