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Proof step 3 from proof abs (previous step)

⊦ abs(i / j) = abs(i) / abs(j), j = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ ¬ j < 0 → abs(j) = j
Proof step 5:
i < 0 ⊦ abs(i / j) = abs(i) / abs(j), j = 0
Proof step 11:
⊦ abs(i / j) = i / abs(j), j = 0, i < 0

Rule argument

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

Additional information

Used terms: i