Metamath Proof Explorer


Theorem fun2dmnop

Description: A function with a domain containing (at least) two different elements is not an ordered pair. (Contributed by AV, 21-Sep-2020) (Proof shortened by AV, 9-Jun-2021)

Ref Expression
Hypotheses fun2dmnop.a
|- A e. _V
fun2dmnop.b
|- B e. _V
Assertion fun2dmnop
|- ( ( Fun G /\ A =/= B /\ { A , B } C_ dom G ) -> -. G e. ( _V X. _V ) )

Proof

Step Hyp Ref Expression
1 fun2dmnop.a
 |-  A e. _V
2 fun2dmnop.b
 |-  B e. _V
3 fundif
 |-  ( Fun G -> Fun ( G \ { (/) } ) )
4 1 2 fun2dmnop0
 |-  ( ( Fun ( G \ { (/) } ) /\ A =/= B /\ { A , B } C_ dom G ) -> -. G e. ( _V X. _V ) )
5 3 4 syl3an1
 |-  ( ( Fun G /\ A =/= B /\ { A , B } C_ dom G ) -> -. G e. ( _V X. _V ) )