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Proof step 7 from proof div-abs-02 (previous step)

i * i0 = abs(j) ⊦ ∃ i0. i * i0 = j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ ¬ j < 0 → abs(j) = j
Proof step 9:
i * i0 = abs(j), j < 0 ⊦ ∃ i0. i * i0 = j
Proof step 11:
i * i0 = j ⊦ j < 0, ∃ i0. i * i0 = j

Rule argument

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

Additional information

Used terms: j