Metamath Proof Explorer


Theorem tgcgreq

Description: Congruence and equality. (Contributed by Thierry Arnoux, 27-Aug-2019)

Ref Expression
Hypotheses tkgeom.p
|- P = ( Base ` G )
tkgeom.d
|- .- = ( dist ` G )
tkgeom.i
|- I = ( Itv ` G )
tkgeom.g
|- ( ph -> G e. TarskiG )
tgcgrcomlr.a
|- ( ph -> A e. P )
tgcgrcomlr.b
|- ( ph -> B e. P )
tgcgrcomlr.c
|- ( ph -> C e. P )
tgcgrcomlr.d
|- ( ph -> D e. P )
tgcgrcomlr.6
|- ( ph -> ( A .- B ) = ( C .- D ) )
tgcgreq.1
|- ( ph -> A = B )
Assertion tgcgreq
|- ( ph -> C = D )

Proof

Step Hyp Ref Expression
1 tkgeom.p
 |-  P = ( Base ` G )
2 tkgeom.d
 |-  .- = ( dist ` G )
3 tkgeom.i
 |-  I = ( Itv ` G )
4 tkgeom.g
 |-  ( ph -> G e. TarskiG )
5 tgcgrcomlr.a
 |-  ( ph -> A e. P )
6 tgcgrcomlr.b
 |-  ( ph -> B e. P )
7 tgcgrcomlr.c
 |-  ( ph -> C e. P )
8 tgcgrcomlr.d
 |-  ( ph -> D e. P )
9 tgcgrcomlr.6
 |-  ( ph -> ( A .- B ) = ( C .- D ) )
10 tgcgreq.1
 |-  ( ph -> A = B )
11 1 2 3 4 5 6 7 8 9 tgcgreqb
 |-  ( ph -> ( A = B <-> C = D ) )
12 10 11 mpbid
 |-  ( ph -> C = D )