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Proof step 1 from proof mod-ngez-01 (conclusion)

⊦ j < 0 ∧ 0 < i → (0 ≤ j % 2 ^ i ↔ j % 2 ^ i = 0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
j < i, j < 0, 0 < i, j + 1 < i ⊦ 0 ≤ j % 2 ^ i ↔ j % 2 ^ i = 0

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ i +1 = i + 1; from specification int-basic2%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs