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 ) )