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Proof step 5 from proof abs (previous step)

i < 0 ⊦ j = 0, i / abs(j) + abs(i / j) = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ ¬ j < 0 → abs(j) = j
Proof step 7:
¬ ¬ j < 0, i < 0 ⊦ j = 0, i / abs(j) + abs(i / j) = 0

Rule argument

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

Additional information

Used simplifier rules: div-lezero: ⊦ 0 < j → (i / j ≤ 0 ↔ i < j);%lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1%abseq: ⊦ abs(i) + i = 0 ↔ i ≤ 0; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%zero-gels-02: ⊦ i < 0 ∧ 0 < 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: j