Metamath Proof Explorer


Theorem inagne2

Description: Deduce inequality from the in-angle relation. (Contributed by Thierry Arnoux, 29-Oct-2021)

Ref Expression
Hypotheses isinag.p
|- P = ( Base ` G )
isinag.i
|- I = ( Itv ` G )
isinag.k
|- K = ( hlG ` G )
isinag.x
|- ( ph -> X e. P )
isinag.a
|- ( ph -> A e. P )
isinag.b
|- ( ph -> B e. P )
isinag.c
|- ( ph -> C e. P )
inagflat.g
|- ( ph -> G e. TarskiG )
inagswap.1
|- ( ph -> X ( inA ` G ) <" A B C "> )
Assertion inagne2
|- ( ph -> C =/= B )

Proof

Step Hyp Ref Expression
1 isinag.p
 |-  P = ( Base ` G )
2 isinag.i
 |-  I = ( Itv ` G )
3 isinag.k
 |-  K = ( hlG ` G )
4 isinag.x
 |-  ( ph -> X e. P )
5 isinag.a
 |-  ( ph -> A e. P )
6 isinag.b
 |-  ( ph -> B e. P )
7 isinag.c
 |-  ( ph -> C e. P )
8 inagflat.g
 |-  ( ph -> G e. TarskiG )
9 inagswap.1
 |-  ( ph -> X ( inA ` G ) <" A B C "> )
10 1 2 3 4 5 6 7 8 isinag
 |-  ( ph -> ( X ( inA ` G ) <" A B C "> <-> ( ( A =/= B /\ C =/= B /\ X =/= B ) /\ E. x e. P ( x e. ( A I C ) /\ ( x = B \/ x ( K ` B ) X ) ) ) ) )
11 9 10 mpbid
 |-  ( ph -> ( ( A =/= B /\ C =/= B /\ X =/= B ) /\ E. x e. P ( x e. ( A I C ) /\ ( x = B \/ x ( K ` B ) X ) ) ) )
12 11 simpld
 |-  ( ph -> ( A =/= B /\ C =/= B /\ X =/= B ) )
13 12 simp2d
 |-  ( ph -> C =/= B )