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Proof step 24 from proof divides-prime (previous step)

abs(i) = 1, prime(j) ⊦ 1 = j, i | j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 25:
⊦ ¬ j < 0 → abs(j) = j
Proof step 26:
¬ ¬ i < 0, abs(i) = 1, prime(j) ⊦ 1 = j, i | j

Rule argument

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

Additional information

Used simplifier rules: div-one: ⊦ 1 | i; Used terms: i