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Proof step 15 from proof mod-next-square (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ i % 2 = ~1 ↔ (∃ j. i = 2 * j + 1 ∧ j < 0)
Proof step 17:
i ≤ 0 ⊦ ((i * i) * (i + 1) * (i + 1)) % 4 = 0, i % 2 = 0, ¬(∃ j. i = 2 * j + 1 ∧ j < 0) ∧ i < 0

Rule argument

spec: int-div inst: name: modex-odd-neg-01 seq: ⊦ i % 2 = ~1 ↔ (∃ j. i = 2 * j + 1 ∧ j < 0) subst: [i] → [i] flags: rewrite, current

Additional information

Used terms: i