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Proof step 18 from proof square-ex (previous step)

0 < j, 1 < (j + 1) * (j + 1), 0 < (j + 1) * (j + 1), 2 < (j + 1) * (j + 1), i0 < (j + 1) * (j + 1), j * j ≤ i0, 0 < i0, 1 < i0 ⊦ 0 < j ∧ j * j < i0 ∧ i0 ≤ (j + 1) * (j + 1), ∃ j0. 0 < j0 ∧ j0 * j0 < i0 ∧ i0 ≤ (j0 + 1) * (j0 + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: square-le: ⊦ i * i ≤ j * j ↔ abs(i) ≤ abs(j);%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%zero-gr: ⊦ 0 < i → (0 < i * j ↔ 0 < j);%square-ls: ⊦ i * i < j * j ↔ abs(i) < abs(j);%le-pos: ⊦ 0 < i → (i * j ≤ i * i0 ↔ j ≤ i0);%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%les: ⊦ i < j → (j ≤ i + 1 ↔ j = i + 1); from specification int-abs%numlelsdel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) < j → (* i0) < j; from specification int-abs%ler: ⊦ i ≤ i; from specification int-abs%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%lep: ⊦ i ≤ i + j ↔ 0 ≤ j; from specification int-abs%abs-16: ⊦ 0 < j → abs(j) = 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%abs-17: ⊦ 0 < j → abs(j + 1) = j + 1; from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs