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Proof step 8 from proof square-not-neg-05 (previous step)

sqrt(i) = 0, 0 < (j + 1) * (j + 1), 0 < i, 0 < j, i < (j + 1) * (j + 1), 1 < (j + 1) * (j + 1), j * j ≤ i, 0 ≤ i → sqrt(i) ^2 ≤ i ∧ i < sqrt(i) +1 ^2 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
sqrt(i) = 0, 0 < (j + 1) * (j + 1), 0 < i, 0 < j, i < (j + 1) * (j + 1), 1 < (j + 1) * (j + 1), i < 1 ^2, j * j < 1 ^2, 0 ^2 < (j + 1) * (j + 1), 0 ^2 < 1 ^2, 0 < 1 ^2, 1 < 1 ^2, j * j ≤ i, 0 ^2 ≤ i, 0 ≤ i ⊦

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%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 → ¬ j < i; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i +1 = i + 1; (s)