Metamath Proof Explorer


Theorem fnxpdmdm

Description: The domain of the domain of a function over a Cartesian square. (Contributed by AV, 13-Jan-2020)

Ref Expression
Assertion fnxpdmdm FFnA×AdomdomF=A

Proof

Step Hyp Ref Expression
1 fndm FFnA×AdomF=A×A
2 dmeq domF=A×AdomdomF=domA×A
3 dmxpid domA×A=A
4 2 3 eqtrdi domF=A×AdomdomF=A
5 1 4 syl FFnA×AdomdomF=A