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Metamath Proof Explorer

Theorem fssxp

Description: A mapping is a class of ordered pairs. (Contributed by NM, 3-Aug-1994) (Proof shortened by Andrew Salmon, 17-Sep-2011)

Ref Expression
Assertion fssxp 
|- ( F : A --> B -> F C_ ( A X. B ) )

Proof

Step Hyp Ref Expression
1 frel 
 |-  ( F : A --> B -> Rel F )
2 relssdmrn 
 |-  ( Rel F -> F C_ ( dom F X. ran F ) )
3 1 2 syl 
 |-  ( F : A --> B -> F C_ ( dom F X. ran F ) )
4 fdm 
 |-  ( F : A --> B -> dom F = A )
5 eqimss 
 |-  ( dom F = A -> dom F C_ A )
6 4 5 syl 
 |-  ( F : A --> B -> dom F C_ A )
7 frn 
 |-  ( F : A --> B -> ran F C_ B )
8 xpss12 
 |-  ( ( dom F C_ A /\ ran F C_ B ) -> ( dom F X. ran F ) C_ ( A X. B ) )
9 6 7 8 syl2anc 
 |-  ( F : A --> B -> ( dom F X. ran F ) C_ ( A X. B ) )
10 3 9 sstrd 
 |-  ( F : A --> B -> F C_ ( A X. B ) )