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Proof step 2 from proof mult (previous step)

prime(m * n) ⊦ n = 1, m = 1

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 * n ∧ (∀ m0. m0 | m * n → m0 = 1 ∨ m0 = m * n) ⊦ n = 1, m = 1

Rule argument

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

Additional information

Used terms: m * n