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Proof step 3 from proof square-zero-02 (previous step)

0 ≤ i + j - 1 → ¬ ((i + j - 1) + 1) * ((i + j - 1) + 2) ≤ 0 ⊦ 0 ≤ j ∧ 0 ≤ i → ¬ (i + j) * (i + j + 1) < 0

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: zero-ls-03: ⊦ i ≤ 0 → (i * j < 0 ↔ ¬ ¬(0 < j ∧ i ≠ 0));%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%Sub-def: ⊦ i + (j - i) = j; 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%sub-add-01: ⊦ i + (j - i0) = i + j - i0; from specification int-abs%numsub: ⊦ (* i) + i0 - (* j) = (i - j) + i0; from specification int-abs%zero-gels-03: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%lenls: ⊦ i ≤ j → ¬ j < i; from specification int-abs