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Proof step 6 from proof abs (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ ¬ j < 0 → abs(j) = j
Proof step 8:
j < 0, i < 0 ⊦ i / abs(j) + abs(i / j) = 0, j = 0

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-lezero: ⊦ 0 < j → (i / j ≤ 0 ↔ i < j);% ⊦ ¬ i < j → (i = j ↔ ¬ j < i); (ss)% ⊦ abs(i) + i = 0 ↔ i ≤ 0; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i < 0 ∧ 0 < j → ¬ j ≤ i; (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