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Proof step 7 from proof inverse-ii (previous step)

sqrt(i * i) < abs(sqrt(i * i) + 1), sqrt(i * i) < abs(i), abs(i) < abs(sqrt(i * i) + 1), sqrt(i * i) < ~1 ⊦ sqrt(i * i) = abs(i)

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: int-square inst: name: root-pos seq: ⊦ 0 ≤ i → 0 ≤ sqrt(i) subst: [i] → [i * i] flags: current, []

Additional information

Used simplifier rules: square-gez: ⊦ 0 ≤ i * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%abs-04: ⊦ 0 ≤ i → abs(i + 1) = abs(i) + 1; from specification int-abs%absle: ⊦ 0 ≤ i → abs(i) = i; 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%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%ler: ⊦ i ≤ i; from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1 Used terms: i * i