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Proof step 3 from proof div-nearer-zero (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ j < 0 → abs(j) = 0 - j
Proof step 5:
abs(i) < abs(i) / abs(j) * abs(j) ⊦ j = 0, i < 0
Proof step 14:
0 - i < (0 - i) / abs(j) * abs(j), i < 0 ⊦ j = 0

Rule argument

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

Additional information

Used terms: i