Metamath Proof Explorer


Theorem nfcvb

Description: The "distinctor" expression -. A. x x = y , stating that x and y are not the same variable, can be written in terms of F/ in the obvious way. This theorem is not true in a one-element domain, because then F/_ x y and A. x x = y will both be true. (Contributed by Mario Carneiro, 8-Oct-2016) Usage of this theorem is discouraged because it depends on ax-13 . (New usage is discouraged.)

Ref Expression
Assertion nfcvb ( 𝑥 𝑦 ↔ ¬ ∀ 𝑥 𝑥 = 𝑦 )

Proof

Step Hyp Ref Expression
1 nfnid ¬ 𝑦 𝑦
2 eqidd ( ∀ 𝑥 𝑥 = 𝑦𝑦 = 𝑦 )
3 2 drnfc1 ( ∀ 𝑥 𝑥 = 𝑦 → ( 𝑥 𝑦 𝑦 𝑦 ) )
4 1 3 mtbiri ( ∀ 𝑥 𝑥 = 𝑦 → ¬ 𝑥 𝑦 )
5 4 con2i ( 𝑥 𝑦 → ¬ ∀ 𝑥 𝑥 = 𝑦 )
6 nfcvf ( ¬ ∀ 𝑥 𝑥 = 𝑦 𝑥 𝑦 )
7 5 6 impbii ( 𝑥 𝑦 ↔ ¬ ∀ 𝑥 𝑥 = 𝑦 )