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Proof step 21 from proof mod-by6 (previous step)

(6 * (i1 + 1) * (i1 + 1) + i1 * (i1 + 1) * (2 * i1 + 1)) % 6 = 0, 0 ≤ i1 ⊦ (i1 * (i1 + 1) * (2 * i1 + 1)) % 6 = 0, (6 * (i1 + 1) * (i1 + 1)) % 6 = 0 ∧ 0 ≤ 6 * (i1 + 1) * (i1 + 1) ∧ 0 ≤ i1 * (i1 + 1) * (2 * i1 + 1) ∧ 0 < 6

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: mod-zero-num: i0 % j = 0, j ≠ 0 ⊦ (i * (* i0)) % (* j) = 0;%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)% ⊦ 0 ≤ i → (0 ≤ i * j ↔ i = 0 ∨ 0 ≤ j); (s)% ⊦ 0 ≤ i * i; (s)% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult