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Proof step 14 from proof mod-next-square (previous step)

¬(i % 2 + 1 ≠ 0 ∧ i < 0), i ≤ 0 ⊦ i % 2 = 0, ((i * i) * (i + 1) * (i + 1)) % 4 = 0, i % 2 ≠ 1 ∧ 0 < i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
i ≤ 0 ⊦ ((i * i) * (i + 1) * (i + 1)) % 4 = 0, i % 2 = 0, ~1 ≠ i % 2 ∧ i < 0

Rule argument

Additional information

Used simplifier rules: lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs