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Proof step 5 from proof div-abs-01 (previous step)

i * i0 = abs(j) ⊦ i | j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ ¬ j < 0 → abs(j) = j
Proof step 7:
¬ ¬ j < 0, i * i0 = abs(j) ⊦ i | j

Rule argument

spec: int-abs inst: name: Abs-pos seq: ⊦ ¬ j < 0 → abs(j) = j subst: [j] → [j] flags: rewrite

Additional information

Used simplifier rules: div-mult: ⊦ i | i * j;%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: j