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Proof step 4 from proof mod-by6 (previous step)

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

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: 0 ≤ i0

Additional information

Used simplifier rules: lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%mone-08: ⊦ i < 0 → (~1 ≤ i ↔ i = ~1); from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) 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%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used formulas: 0 ≤ i0