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Proof step 6 from proof gcd-one-prime (previous step)

0 < i0, 1 < i0, prime(i), i0 | i, i0 | j ⊦ i | j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ prime(j) → (i | j ↔ abs(i) = j ∨ abs(i) = 1)

Rule argument

spec: <toplevel> name: divides-prime seq: ⊦ prime(j) → (i | j ↔ abs(i) = j ∨ abs(i) = 1) subst: [j, i] → [i, i0] flags: <all paths>

Additional information

Used simplifier rules: ⊦ 1 < i → abs(i) = i; (s) Used terms: i, i0