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Proof step 2 from proof div (previous step)

prime(m), n | m ⊦ n = 1, n = m

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ prime(n) ↔ 1 < n ∧ (∀ m. m | n → m = 1 ∨ m = n)
Proof step 4:
1 < m ∧ (∀ m0. m0 | m → m0 = 1 ∨ m0 = m), n | m ⊦ n = 1, n = m

Rule argument

spec: nat-prime inst: name: Prime seq: ⊦ prime(n) ↔ 1 < n ∧ (∀ m. m | n → m = 1 ∨ m = n) subst: [n] → [m] flags: rewrite, current

Additional information

Used terms: m