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Proof step 4 from proof square-inject (previous step)

i0 ≤ i, j0 ≤ j, 0 ≤ i0, 0 ≤ j, 0 ≤ j0 ⊦ i * (i + 1) ≤ i * (i + 1) + 2 * i0 ∧ i * (i + 1) + 2 * i0 < (i + 1) * (i + 2) ∧ 0 ≤ i, i = j ∧ i0 = j0

User applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: dist-lsplit: ⊦ (i + i0) * j = i * j + i0 * j;%one: ⊦ 1 * i = i;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%numsucc: ⊦ (* i0) * i + i = (i0 + 1) * i;%numls-04: ⊦ i + j < (* i1) * i + j0 ↔ j < (i1 - 1) * i + j0;%zero-ls: ⊦ 0 < i → (i * j < 0 ↔ j < 0);%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-04: ⊦ i ≤ j → (¬ ¬ j < i ↔ false); from specification int-abs%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%ls-01: ⊦ i + j < i + i0 ↔ j < i0; from specification int-basic2%lep-02: ⊦ i + j ≤ i + j0 ↔ j ≤ j0; 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