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Metamath Proof Explorer

Theorem shftcan2

Description: Cancellation law for the shift operation. (Contributed by NM, 4-Aug-2005) (Revised by Mario Carneiro, 5-Nov-2013)

Ref Expression
Hypothesis shftfval.1 
|- F e. _V
Assertion shftcan2 
|- ( ( A e. CC /\ B e. CC ) -> ( ( ( F shift -u A ) shift A ) ` B ) = ( F ` B ) )

Proof

Step Hyp Ref Expression
1 shftfval.1 
 |-  F e. _V
2 negneg 
 |-  ( A e. CC -> -u -u A = A )
3 2 adantr 
 |-  ( ( A e. CC /\ B e. CC ) -> -u -u A = A )
4 3 oveq2d 
 |-  ( ( A e. CC /\ B e. CC ) -> ( ( F shift -u A ) shift -u -u A ) = ( ( F shift -u A ) shift A ) )
5 4 fveq1d 
 |-  ( ( A e. CC /\ B e. CC ) -> ( ( ( F shift -u A ) shift -u -u A ) ` B ) = ( ( ( F shift -u A ) shift A ) ` B ) )
6 negcl 
 |-  ( A e. CC -> -u A e. CC )
7 1 shftcan1 
 |-  ( ( -u A e. CC /\ B e. CC ) -> ( ( ( F shift -u A ) shift -u -u A ) ` B ) = ( F ` B ) )
8 6 7 sylan 
 |-  ( ( A e. CC /\ B e. CC ) -> ( ( ( F shift -u A ) shift -u -u A ) ` B ) = ( F ` B ) )
9 5 8 eqtr3d 
 |-  ( ( A e. CC /\ B e. CC ) -> ( ( ( F shift -u A ) shift A ) ` B ) = ( F ` B ) )