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

i + i * i < 0, abs(i) ≤ i * i ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ j < 0 → abs(j) = 0 - j

Rule argument

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

Additional information

Used simplifier rules: square-le-01: ⊦ i ≤ i * i;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%square-lez: ⊦ ¬ i * i < 0;% ⊦ 0 - i = ~ i; (s)% ⊦ ~ i ≤ j ↔ 0 ≤ i + j; (s)% ⊦ i < j → (j ≤ i ↔ false); (as)% ⊦ ¬ j < 0 → abs(j) = j; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs% ⊦ i + j < j ↔ i < 0; (s)%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