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Proof step 6 from proof div-nearer-zero (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ j < 0 → abs(j) = 0 - j
Proof step 8:
0 < i / abs(j) * abs(j), i < i / abs(j) * abs(j), 0 ≤ i ⊦ j = 0, j < 0
Proof step 11:
0 < i / (0 - j) * (0 - j), i < i / (0 - j) * (0 - j), j < 0, 0 ≤ i ⊦ j = 0

Rule argument

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

Additional information

Used terms: j