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Proof step 7 from proof square-ex (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%ler: ⊦ i ≤ i; from specification int-abs%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%int-add-pos: ⊦ 0 ≤ j ∧ j < j1 ∧ j + i = j1 → 0 < i; from specification int-abs%les: ⊦ i < j → (j ≤ i + 1 ↔ j = i + 1); from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; 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%le-add-03: ⊦ i ≤ i0 ∧ j0 ≥ 0 → i ≤ i0 + j0; from specification int-abs