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Proof step 4 from proof square-gez-02 (previous step)

abs(i) ≤ i * i, i + i * i < 0 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-abs inst: name: Abs-neg seq: ⊦ j < 0 → abs(j) = 0 - j subst: [j] → [i] flags: rewrite

Additional information

Used simplifier rules: square-le-01: ⊦ i ≤ i * i;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%zero-ls-02: ⊦ 0 ≤ i → (i * j < 0 ↔ ¬ ¬(j < 0 ∧ i ≠ 0));%neg-06: ⊦ 0 - i = ~ i; from specification int-abs%negle-13: ⊦ ~ i ≤ j ↔ 0 ≤ i + j; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs%Abs-pos: ⊦ ¬ j < 0 → abs(j) = j; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2%lenls: ⊦ i ≤ j → ¬ j < 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 Used terms: i