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Proof step 14 from proof mod-by6 (previous step)

0 ≤ i0 -1, 0 ≤ i0 → (i0 * (i0 + 1) * (2 * i0 + 1)) % 6 = 0 ⊦ (i0 -1 * (i0 -1 + 1) * (2 * i0 -1 + 1)) % 6 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
(i0 * (i0 + 1) * (2 * i0 + 1)) % 6 = 0, 0 < i0, 0 ≤ i0 ⊦ ((i0 - 1) * i0 * (2 * (i0 - 1) + 1)) % 6 = 0

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs% ⊦ i < j → ¬ j < i; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs% ⊦ 1 ≤ i ↔ 0 < i; (s)% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + (j - i) = j; (s)% ⊦ i -1 = i - 1; (s)