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Metamath Proof Explorer

Theorem grpnpncan0

Description: Cancellation law for group subtraction ( npncan2 analog). (Contributed by AV, 24-Nov-2019)

Ref Expression
Hypotheses grpsubadd.b 
|- B = ( Base ` G )
grpsubadd.p 
|- .+ = ( +g ` G )
grpsubadd.m 
|- .- = ( -g ` G )
grpnpncan0.0 
|- .0. = ( 0g ` G )
Assertion grpnpncan0 
|- ( ( G e. Grp /\ ( X e. B /\ Y e. B ) ) -> ( ( X .- Y ) .+ ( Y .- X ) ) = .0. )

Proof

Step Hyp Ref Expression
1 grpsubadd.b 
 |-  B = ( Base ` G )
2 grpsubadd.p 
 |-  .+ = ( +g ` G )
3 grpsubadd.m 
 |-  .- = ( -g ` G )
4 grpnpncan0.0 
 |-  .0. = ( 0g ` G )
5 simpl 
 |-  ( ( G e. Grp /\ ( X e. B /\ Y e. B ) ) -> G e. Grp )
6 simprl 
 |-  ( ( G e. Grp /\ ( X e. B /\ Y e. B ) ) -> X e. B )
7 simprr 
 |-  ( ( G e. Grp /\ ( X e. B /\ Y e. B ) ) -> Y e. B )
8 1 2 3 grpnpncan 
 |-  ( ( G e. Grp /\ ( X e. B /\ Y e. B /\ X e. B ) ) -> ( ( X .- Y ) .+ ( Y .- X ) ) = ( X .- X ) )
9 5 6 7 6 8 syl13anc 
 |-  ( ( G e. Grp /\ ( X e. B /\ Y e. B ) ) -> ( ( X .- Y ) .+ ( Y .- X ) ) = ( X .- X ) )
10 1 4 3 grpsubid 
 |-  ( ( G e. Grp /\ X e. B ) -> ( X .- X ) = .0. )
11 10 adantrr 
 |-  ( ( G e. Grp /\ ( X e. B /\ Y e. B ) ) -> ( X .- X ) = .0. )
12 9 11 eqtrd 
 |-  ( ( G e. Grp /\ ( X e. B /\ Y e. B ) ) -> ( ( X .- Y ) .+ ( Y .- X ) ) = .0. )