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Metamath Proof Explorer

Theorem fcof1o

Description: Show that two functions are inverse to each other by computing their compositions. (Contributed by Mario Carneiro, 21-Mar-2015) (Proof shortened by AV, 15-Dec-2019)

Ref Expression
Assertion fcof1o 
|- ( ( ( F : A --> B /\ G : B --> A ) /\ ( ( F o. G ) = ( _I |` B ) /\ ( G o. F ) = ( _I |` A ) ) ) -> ( F : A -1-1-onto-> B /\ `' F = G ) )

Proof

Step Hyp Ref Expression
1 simpll 
 |-  ( ( ( F : A --> B /\ G : B --> A ) /\ ( ( F o. G ) = ( _I |` B ) /\ ( G o. F ) = ( _I |` A ) ) ) -> F : A --> B )
2 simplr 
 |-  ( ( ( F : A --> B /\ G : B --> A ) /\ ( ( F o. G ) = ( _I |` B ) /\ ( G o. F ) = ( _I |` A ) ) ) -> G : B --> A )
3 simprr 
 |-  ( ( ( F : A --> B /\ G : B --> A ) /\ ( ( F o. G ) = ( _I |` B ) /\ ( G o. F ) = ( _I |` A ) ) ) -> ( G o. F ) = ( _I |` A ) )
4 simprl 
 |-  ( ( ( F : A --> B /\ G : B --> A ) /\ ( ( F o. G ) = ( _I |` B ) /\ ( G o. F ) = ( _I |` A ) ) ) -> ( F o. G ) = ( _I |` B ) )
5 1 2 3 4 fcof1od 
 |-  ( ( ( F : A --> B /\ G : B --> A ) /\ ( ( F o. G ) = ( _I |` B ) /\ ( G o. F ) = ( _I |` A ) ) ) -> F : A -1-1-onto-> B )
6 1 2 3 4 2fcoidinvd 
 |-  ( ( ( F : A --> B /\ G : B --> A ) /\ ( ( F o. G ) = ( _I |` B ) /\ ( G o. F ) = ( _I |` A ) ) ) -> `' F = G )
7 5 6 jca 
 |-  ( ( ( F : A --> B /\ G : B --> A ) /\ ( ( F o. G ) = ( _I |` B ) /\ ( G o. F ) = ( _I |` A ) ) ) -> ( F : A -1-1-onto-> B /\ `' F = G ) )