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Proof step 3 from proof square-ex (previous step)

0 < i0 → (∃ j. 0 < j ∧ j * j ≤ i0 ∧ i0 < (j + 1) * (j + 1)), 0 < i0 +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 4:
0 ≤ i0, 0 < i0 → 0 < j ∧ j * j ≤ i0 ∧ i0 < (j + 1) * (j + 1) ⊦ ∃ j0. 0 < j0 ∧ j0 * j0 ≤ i0 + 1 ∧ i0 + 1 < (j0 + 1) * (j0 + 1)

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) 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%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%succ-01: ⊦ i +1 = i + 1; from specification int-basic2