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Proof step 1 from proof square-zero-01 (conclusion)

⊦ 0 ≤ i → ¬ (i + 1) * (i + 2) ≤ 0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: dist-lsplit: ⊦ (i + i0) * j = i * j + i0 * j;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%one: ⊦ 1 * i = i;%numsucc: ⊦ (* i0) * i + i = (i0 + 1) * i;%numls-13: ⊦ (* i1) * j + i < j ↔ (i1 - 1) * j + i < 0;%square-lez: ⊦ ¬ i * i < 0;%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%zero-gels-03: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%forward-le-03: ⊦ j ≤ i ∧ i < 0 → j < 0; (forward) from specification int-abs%numle-03: ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2