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

¬((2 * i) % 2 = 0 ∧ 0 ≤ 2 * i ∧ 0 ≤ 1 ∧ 0 < 2), (2 * i + 1) % 2 = 0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
(2 * i + 1) % 2 = 0, i < 0 ⊦

Rule argument

Additional information

Used simplifier rules: mod-two-02: ⊦ (i * 2) % 2 = 0;%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%zero-ge: ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult