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Proof step 9 from proof mod-next-square (previous step)

i % 2 = 1, 0 < i ⊦ ((i * i) * (i + 1) * (i + 1)) % 4 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: mod-mult-04: j ≠ 0 ⊦ (i * j) % j = 0;%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%succ-rmerge: ⊦ i * j + i = i * (j + 1); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: i