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Proof step 5 from proof even-not (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
(2 * i + 1) % 2 = 0, i < 0 ⊦

Rule argument

Additional information

Used simplifier rules: mod-two-02: ⊦ (i * 2) % 2 = 0;%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult