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Proof step 5 from proof sqrt-estimate (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ i ^2 < j ^2 ↔ abs(i) < abs(j)

Rule argument

spec: inst: name: two-ls seq: ⊦ i ^2 < j ^2 ↔ abs(i) < abs(j) subst: [j, i] → [j + 1, sqrt(i)] flags: rewrite, current

Additional information

Used simplifier rules: root-abs: ⊦ 0 ≤ i → abs(sqrt(i)) = sqrt(i);%square-nls: ⊦ i ^2 ≤ j → ¬ sqrt(j) < i;%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%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs Used terms: j + 1, sqrt(i)