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Metamath Proof Explorer

Theorem shftcan1

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 shftcan1 
|- ( ( A e. CC /\ B e. CC ) -> ( ( ( F shift A ) shift -u A ) ` B ) = ( F ` B ) )

Proof

Step Hyp Ref Expression
1 shftfval.1 
 |-  F e. _V
2 negcl 
 |-  ( A e. CC -> -u A e. CC )
3 1 2shfti 
 |-  ( ( A e. CC /\ -u A e. CC ) -> ( ( F shift A ) shift -u A ) = ( F shift ( A + -u A ) ) )
4 2 3 mpdan 
 |-  ( A e. CC -> ( ( F shift A ) shift -u A ) = ( F shift ( A + -u A ) ) )
5 negid 
 |-  ( A e. CC -> ( A + -u A ) = 0 )
6 5 oveq2d 
 |-  ( A e. CC -> ( F shift ( A + -u A ) ) = ( F shift 0 ) )
7 4 6 eqtrd 
 |-  ( A e. CC -> ( ( F shift A ) shift -u A ) = ( F shift 0 ) )
8 7 fveq1d 
 |-  ( A e. CC -> ( ( ( F shift A ) shift -u A ) ` B ) = ( ( F shift 0 ) ` B ) )
9 1 shftidt 
 |-  ( B e. CC -> ( ( F shift 0 ) ` B ) = ( F ` B ) )
10 8 9 sylan9eq 
 |-  ( ( A e. CC /\ B e. CC ) -> ( ( ( F shift A ) shift -u A ) ` B ) = ( F ` B ) )