Metamath Proof Explorer


Theorem mapex

Description: The class of all functions mapping one set to another is a set. Remark after Definition 10.24 of Kunen p. 31. (Contributed by Raph Levien, 4-Dec-2003) (Proof shortened by AV, 16-Jun-2025)

Ref Expression
Assertion mapex
|- ( ( A e. C /\ B e. D ) -> { f | f : A --> B } e. _V )

Proof

Step Hyp Ref Expression
1 eqid
 |-  { f | ( f : A --> B /\ f : A --> B ) } = { f | ( f : A --> B /\ f : A --> B ) }
2 1 fabexg
 |-  ( ( A e. C /\ B e. D ) -> { f | ( f : A --> B /\ f : A --> B ) } e. _V )
3 id
 |-  ( f : A --> B -> f : A --> B )
4 3 ancli
 |-  ( f : A --> B -> ( f : A --> B /\ f : A --> B ) )
5 4 ss2abi
 |-  { f | f : A --> B } C_ { f | ( f : A --> B /\ f : A --> B ) }
6 5 a1i
 |-  ( ( A e. C /\ B e. D ) -> { f | f : A --> B } C_ { f | ( f : A --> B /\ f : A --> B ) } )
7 2 6 ssexd
 |-  ( ( A e. C /\ B e. D ) -> { f | f : A --> B } e. _V )