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Proof step 3 from proof mods-nls (previous step)

i * abs(j) < j, 0 < i ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ j < 0 → abs(j) = 0 - j
Proof step 5:
0 < i, i * abs(j) < j ⊦ j < 0
Proof step 8:
0 < i, i * (0 - j) < j, 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