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Proof step 18 from proof mod-next-square (previous step)

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

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 3

Additional information

Used simplifier rules: mod-zero-num: i0 % j = 0, j ≠ 0 ⊦ (i * (* i0)) % (* j) = 0;%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%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