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Proof step 5 from proof square-not-neg-05 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
sqrt(i) = 0, 1 < (j + 1) * (j + 1), i < (j + 1) * (j + 1), 0 < j, 0 < i, 0 < (j + 1) * (j + 1), j * j ≤ i ⊦

Rule argument

Additional information

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