Metamath Proof Explorer


Theorem tposfn

Description: Functionality of a transposition. (Contributed by Mario Carneiro, 4-Oct-2015)

Ref Expression
Assertion tposfn ( 𝐹 Fn ( 𝐴 × 𝐵 ) → tpos 𝐹 Fn ( 𝐵 × 𝐴 ) )

Proof

Step Hyp Ref Expression
1 tposf ( 𝐹 : ( 𝐴 × 𝐵 ) ⟶ V → tpos 𝐹 : ( 𝐵 × 𝐴 ) ⟶ V )
2 dffn2 ( 𝐹 Fn ( 𝐴 × 𝐵 ) ↔ 𝐹 : ( 𝐴 × 𝐵 ) ⟶ V )
3 dffn2 ( tpos 𝐹 Fn ( 𝐵 × 𝐴 ) ↔ tpos 𝐹 : ( 𝐵 × 𝐴 ) ⟶ V )
4 1 2 3 3imtr4i ( 𝐹 Fn ( 𝐴 × 𝐵 ) → tpos 𝐹 Fn ( 𝐵 × 𝐴 ) )