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Proof step 17 from proof square-ex (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
1 < (j + 1) * (j + 1), i0 < (j + 1) * (j + 1), 0 < j, 0 < i0, 1 < i0, 0 < (j + 1) * (j + 1), j * j ≤ i0 ⊦ ∃ j0. 0 < j0 ∧ j0 * j0 < i0 ∧ i0 ≤ (j0 + 1) * (j0 + 1)

Rule argument

Additional information

Used simplifier rules: square-ls: ⊦ i * i < j * j ↔ abs(i) < abs(j);%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;% ⊦ i ≤ i; (s)% ⊦ 0 < j → abs(j + 1) = j + 1; (s)% ⊦ 0 < j → abs(j) = j; (s)%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1% ⊦ i < j + 1 ↔ i ≤ j; (s)% ⊦ i - j < i0 ↔ i < j + i0; (s)% ⊦ i + 1 ≤ j ↔ i < j; (s)% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i < j - i0 ↔ i + i0 < j; (s)% ⊦ i -1 = i - 1; (s)