Metamath Proof Explorer

Theorem cgrdegen

Description: Two congruent segments are either both degenerate or both nondegenerate. (Contributed by Scott Fenton, 12-Jun-2013)

Ref Expression
Assertion cgrdegen ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N → A B Cgr C D → A = B ↔ C = D

Proof

Step Hyp Ref Expression
1 opeq1 ⊢ A = B → A B = B B
2 1 breq1d ⊢ A = B → A B Cgr C D ↔ B B Cgr C D
3 2 biimpac ⊢ A B Cgr C D ∧ A = B → B B Cgr C D
4 simp1 ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N → N ∈ ℕ
5 simp2r ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N → B ∈ 𝔼 ⁡ N
6 simp3l ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N → C ∈ 𝔼 ⁡ N
7 simp3r ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N → D ∈ 𝔼 ⁡ N
8 cgrid2 ⊢ N ∈ ℕ ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N → B B Cgr C D → C = D
9 4 5 6 7 8 syl13anc ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N → B B Cgr C D → C = D
10 3 9 syl5 ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N → A B Cgr C D ∧ A = B → C = D
11 10 expdimp ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N ∧ A B Cgr C D → A = B → C = D
12 opeq1 ⊢ C = D → C D = D D
13 12 breq2d ⊢ C = D → A B Cgr C D ↔ A B Cgr D D
14 13 biimpac ⊢ A B Cgr C D ∧ C = D → A B Cgr D D
15 simp2l ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N → A ∈ 𝔼 ⁡ N
16 axcgrid ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N → A B Cgr D D → A = B
17 4 15 5 7 16 syl13anc ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N → A B Cgr D D → A = B
18 14 17 syl5 ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N → A B Cgr C D ∧ C = D → A = B
19 18 expdimp ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N ∧ A B Cgr C D → C = D → A = B
20 11 19 impbid ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N ∧ A B Cgr C D → A = B ↔ C = D
21 20 ex ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N → A B Cgr C D → A = B ↔ C = D