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Proof step 4 from proof square-not-neg-05 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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%ler: ⊦ i ≤ i; from specification int-abs%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%abs-16: ⊦ 0 < j → abs(j) = j; 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%abs-17: ⊦ 0 < j → abs(j + 1) = j + 1; from specification int-abs%square-ls: ⊦ i * i < j * j ↔ abs(i) < abs(j); from specification int-mult%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%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