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Proof step 4 from proof even-not-ex (previous step)

⊦ i % 2 = 0 ↔ ¬(∃ i0. i = 2 * i0 + 1), 0 ≤ i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
i < 0 ⊦ i % 2 = 0 ↔ ¬(∃ i0. i = 2 * i0 + 1)

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)