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Proof step 1 from proof divi-two-01 (conclusion)
⊦ 2 | m + 2 ↔ 2 | m
User applied rule insert rewrite lemma
And obtained the following new goals
Proof step 2
:
⊦ m | n ↔ (∃ n0. m * n0 = n)
Proof step 3
:
⊦ (∃ n0. 2 * n0 = m + 2) ↔ 2 | m
Rule argument
spec: nat-prime inst: name: Divides seq: ⊦ m | n ↔ (∃ n0. m * n0 = n) subst: [n, m] → [m + 2, 2] flags: rewrite, current
Additional information
Used terms: m + 2, 2