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Proof step 9 from proof gcd-primes (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ 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, j] flags: <all paths>

Additional information

Used simplifier rules: prime-abs: ⊦ prime(j) → abs(j) = j; Used terms: i, j