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 : 𝐴𝐵𝐹 ⊆ ( 𝐴 × 𝐵 ) )

Proof

Step Hyp Ref Expression
1 frel ( 𝐹 : 𝐴𝐵 → Rel 𝐹 )
2 relssdmrn ( Rel 𝐹𝐹 ⊆ ( dom 𝐹 × ran 𝐹 ) )
3 1 2 syl ( 𝐹 : 𝐴𝐵𝐹 ⊆ ( dom 𝐹 × ran 𝐹 ) )
4 fdm ( 𝐹 : 𝐴𝐵 → dom 𝐹 = 𝐴 )
5 eqimss ( dom 𝐹 = 𝐴 → dom 𝐹𝐴 )
6 4 5 syl ( 𝐹 : 𝐴𝐵 → dom 𝐹𝐴 )
7 frn ( 𝐹 : 𝐴𝐵 → ran 𝐹𝐵 )
8 xpss12 ( ( dom 𝐹𝐴 ∧ ran 𝐹𝐵 ) → ( dom 𝐹 × ran 𝐹 ) ⊆ ( 𝐴 × 𝐵 ) )
9 6 7 8 syl2anc ( 𝐹 : 𝐴𝐵 → ( dom 𝐹 × ran 𝐹 ) ⊆ ( 𝐴 × 𝐵 ) )
10 3 9 sstrd ( 𝐹 : 𝐴𝐵𝐹 ⊆ ( 𝐴 × 𝐵 ) )