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Proof step 3 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: numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; 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%succ-01: ⊦ i +1 = i + 1; from specification int-basic2