Metamath Proof Explorer


Theorem btwnswapid

Description: If you can swap the first two arguments of a betweenness statement, then those arguments are identical. Theorem 3.4 of Schwabhauser p. 30. (Contributed by Scott Fenton, 12-Jun-2013)

Ref Expression
Assertion btwnswapid N A 𝔼 N B 𝔼 N C 𝔼 N A Btwn B C B Btwn A C A = B

Proof

Step Hyp Ref Expression
1 simpl N A 𝔼 N B 𝔼 N C 𝔼 N N
2 simpr2 N A 𝔼 N B 𝔼 N C 𝔼 N B 𝔼 N
3 simpr1 N A 𝔼 N B 𝔼 N C 𝔼 N A 𝔼 N
4 simpr3 N A 𝔼 N B 𝔼 N C 𝔼 N C 𝔼 N
5 axpasch N B 𝔼 N A 𝔼 N C 𝔼 N A 𝔼 N B 𝔼 N A Btwn B C B Btwn A C x 𝔼 N x Btwn A A x Btwn B B
6 1 2 3 4 3 2 5 syl132anc N A 𝔼 N B 𝔼 N C 𝔼 N A Btwn B C B Btwn A C x 𝔼 N x Btwn A A x Btwn B B
7 simpll N A 𝔼 N B 𝔼 N C 𝔼 N x 𝔼 N N
8 simpr N A 𝔼 N B 𝔼 N C 𝔼 N x 𝔼 N x 𝔼 N
9 simplr1 N A 𝔼 N B 𝔼 N C 𝔼 N x 𝔼 N A 𝔼 N
10 axbtwnid N x 𝔼 N A 𝔼 N x Btwn A A x = A
11 7 8 9 10 syl3anc N A 𝔼 N B 𝔼 N C 𝔼 N x 𝔼 N x Btwn A A x = A
12 simplr2 N A 𝔼 N B 𝔼 N C 𝔼 N x 𝔼 N B 𝔼 N
13 axbtwnid N x 𝔼 N B 𝔼 N x Btwn B B x = B
14 7 8 12 13 syl3anc N A 𝔼 N B 𝔼 N C 𝔼 N x 𝔼 N x Btwn B B x = B
15 11 14 anim12d N A 𝔼 N B 𝔼 N C 𝔼 N x 𝔼 N x Btwn A A x Btwn B B x = A x = B
16 eqtr2 x = A x = B A = B
17 15 16 syl6 N A 𝔼 N B 𝔼 N C 𝔼 N x 𝔼 N x Btwn A A x Btwn B B A = B
18 17 rexlimdva N A 𝔼 N B 𝔼 N C 𝔼 N x 𝔼 N x Btwn A A x Btwn B B A = B
19 6 18 syld N A 𝔼 N B 𝔼 N C 𝔼 N A Btwn B C B Btwn A C A = B