Metamath Proof Explorer

Theorem endofsegid

Description: If A , B , and C fall in order on a line, and A B and A C are congruent, then C = B . (Contributed by Scott Fenton, 7-Oct-2013)

Ref Expression
Assertion endofsegid ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N → B Btwn A C ∧ A C Cgr A B → C = B

Proof

Step Hyp Ref Expression
1 simpl ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N → N ∈ ℕ
2 simpr1 ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N → A ∈ 𝔼 ⁡ N
3 simpr3 ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N → C ∈ 𝔼 ⁡ N
4 simpr2 ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N → B ∈ 𝔼 ⁡ N
5 cgrcom ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N → A C Cgr A B ↔ A B Cgr A C
6 1 2 3 2 4 5 syl122anc ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N → A C Cgr A B ↔ A B Cgr A C
7 6 biimpd ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N → A C Cgr A B → A B Cgr A C
8 idd ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N → A C Cgr A B → A C Cgr A B
9 axcgrrflx ⊢ N ∈ ℕ ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N → B C Cgr C B
10 9 3adant3r1 ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N → B C Cgr C B
11 10 a1d ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N → A C Cgr A B → B C Cgr C B
12 7 8 11 3jcad ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N → A C Cgr A B → A B Cgr A C ∧ A C Cgr A B ∧ B C Cgr C B
13 3ancomb ⊢ A ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ↔ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N
14 brcgr3 ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ A ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N → A B C Cgr3 A C B ↔ A B Cgr A C ∧ A C Cgr A B ∧ B C Cgr C B
15 13 14 syl3an3br ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N → A B C Cgr3 A C B ↔ A B Cgr A C ∧ A C Cgr A B ∧ B C Cgr C B
16 15 3anidm23 ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N → A B C Cgr3 A C B ↔ A B Cgr A C ∧ A C Cgr A B ∧ B C Cgr C B
17 12 16 sylibrd ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N → A C Cgr A B → A B C Cgr3 A C B
18 btwnxfr ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ A ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N → B Btwn A C ∧ A B C Cgr3 A C B → C Btwn A B
19 13 18 syl3an3br ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N → B Btwn A C ∧ A B C Cgr3 A C B → C Btwn A B
20 19 3anidm23 ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N → B Btwn A C ∧ A B C Cgr3 A C B → C Btwn A B
21 17 20 sylan2d ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N → B Btwn A C ∧ A C Cgr A B → C Btwn A B
22 simpl ⊢ B Btwn A C ∧ A C Cgr A B → B Btwn A C
23 22 a1i ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N → B Btwn A C ∧ A C Cgr A B → B Btwn A C
24 21 23 jcad ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N → B Btwn A C ∧ A C Cgr A B → C Btwn A B ∧ B Btwn A C
25 3anrot ⊢ C ∈ 𝔼 ⁡ N ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ↔ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N
26 btwnswapid2 ⊢ N ∈ ℕ ∧ C ∈ 𝔼 ⁡ N ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N → C Btwn A B ∧ B Btwn A C → C = B
27 25 26 sylan2br ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N → C Btwn A B ∧ B Btwn A C → C = B
28 24 27 syld ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N → B Btwn A C ∧ A C Cgr A B → C = B