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Proof step 3 from proof divi-two-01 (
previous step
)
⊦ (∃ n0. 2 * n0 = m + 2) ↔ 2 | m
User applied rule insert rewrite lemma
And obtained the following new goals
Proof step 4
:
⊦ m | n ↔ (∃ n0. m * n0 = n)
Proof step 5
:
⊦ (∃ n0. 2 * n0 = m + 2) ↔ (∃ n0. 2 * n0 = m)
Rule argument
spec: nat-prime inst: name: Divides seq: ⊦ m | n ↔ (∃ n0. m * n0 = n) subst: [n, m] → [m, 2] flags: rewrite, current
Additional information
Used terms: m, 2