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Proof step 2 from proof twodiv-01 (previous step)

prime(m + 1) ⊦ m = 1, 2 | 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 + 1 ∧ (∀ m0. m0 | m + 1 → m0 = 1 ∨ m0 = m + 1) ⊦ m = 1, 2 | m

Rule argument

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

Additional information

Used terms: m + 1