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Proof step 12 from proof abs (previous step)

0 ≤ i ⊦ abs(i / j) = i / abs(j), j = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: Div-right: ⊦ j ≠ 0 → i / ~ j = ~(i / j);%div-zero-ngr-01: ⊦ 0 ≤ i ∧ j < 0 → ¬ 0 < i / j;%div-zero-nls: ⊦ 0 ≤ i ∧ 0 < j → ¬ i / j < 0;% ⊦ ¬ i < j → (i = j ↔ ¬ j < i); (ss)% ⊦ abs(i) = i ↔ 0 ≤ i; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ ~ i = j ↔ i + j = 0; (s)% ⊦ abs(i) + i = 0 ↔ 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: j