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Proof step 13 from proof mod-by6 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs%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%Sub-def: ⊦ i + (j - i) = j; from specification int-abs%sub-pred: ⊦ i -1 = i - 1; from specification int-abs