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 ) )