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Proof step 2 from proof root-pos (previous step)

sqrt(i) < 0, sqrt(i) < i, 0 ≤ i ⊦

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ 0 ≤ i → sqrt(i) ^2 ≤ i ∧ i < sqrt(i) +1 ^2

Rule argument

spec: int-square inst: name: Sqrt-def seq: ⊦ 0 ≤ i → sqrt(i) ^2 ≤ i ∧ i < sqrt(i) +1 ^2 subst: [i] → [i] flags: current, []

Additional information

Used simplifier rules: Square-def: ⊦ i ^2 = i * i;%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%square-ls: ⊦ i * i < j * j ↔ abs(i) < abs(j); from specification int-mult%abs-05: ⊦ i < 0 → abs(i + 1) = abs(i) - 1; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j; from specification int-abs%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: i