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Proof step 7 from proof square-not-neg-05 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
sqrt(i) = 0, 0 ^2 < (j + 1) * (j + 1), 0 ≤ i, j * j < 0 ^2 + 1 ^2, 1 < 0 ^2 + 1 ^2, i < 0 ^2 + 1 ^2, 0 ^2 ≤ i, 0 < j, 1 < (j + 1) * (j + 1), 0 < (j + 1) * (j + 1), i < (j + 1) * (j + 1), j * j ≤ i, 0 < i, 0 < 0 ^2 + 1 ^2, 0 < 1 ^2 ⊦

Rule argument

Additional information

Used simplifier rules: two-plus: ⊦ (i + j)^2 = i ^2 + 2 * i * j + j ^2;%ls-02: ⊦ i < i + j ↔ 0 < j; from specification int-basic2%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%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%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%succ-01: ⊦ i +1 = i + 1; from specification int-basic2