Metamath Proof Explorer


Theorem brfs

Description: Binary relation form of the general five segment predicate. (Contributed by Scott Fenton, 5-Oct-2013)

Ref Expression
Assertion brfs ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N ∧ E ∈ 𝔼 ⁡ N ∧ F ∈ 𝔼 ⁡ N ∧ G ∈ 𝔼 ⁡ N ∧ H ∈ 𝔼 ⁡ N → A B C D FiveSeg E F G H ↔ A Colinear B C ∧ A B C Cgr3 E F G ∧ A D Cgr E H ∧ B D Cgr F H

Proof

Step Hyp Ref Expression
1 breq1 ⊢ a = A → a Colinear b c ↔ A Colinear b c
2 opeq1 ⊢ a = A → a b c = A b c
3 2 breq1d ⊢ a = A → a b c Cgr3 e f g ↔ A b c Cgr3 e f g
4 opeq1 ⊢ a = A → a d = A d
5 4 breq1d ⊢ a = A → a d Cgr e h ↔ A d Cgr e h
6 5 anbi1d ⊢ a = A → a d Cgr e h ∧ b d Cgr f h ↔ A d Cgr e h ∧ b d Cgr f h
7 1 3 6 3anbi123d ⊢ a = A → a Colinear b c ∧ a b c Cgr3 e f g ∧ a d Cgr e h ∧ b d Cgr f h ↔ A Colinear b c ∧ A b c Cgr3 e f g ∧ A d Cgr e h ∧ b d Cgr f h
8 opeq1 ⊢ b = B → b c = B c
9 8 breq2d ⊢ b = B → A Colinear b c ↔ A Colinear B c
10 8 opeq2d ⊢ b = B → A b c = A B c
11 10 breq1d ⊢ b = B → A b c Cgr3 e f g ↔ A B c Cgr3 e f g
12 opeq1 ⊢ b = B → b d = B d
13 12 breq1d ⊢ b = B → b d Cgr f h ↔ B d Cgr f h
14 13 anbi2d ⊢ b = B → A d Cgr e h ∧ b d Cgr f h ↔ A d Cgr e h ∧ B d Cgr f h
15 9 11 14 3anbi123d ⊢ b = B → A Colinear b c ∧ A b c Cgr3 e f g ∧ A d Cgr e h ∧ b d Cgr f h ↔ A Colinear B c ∧ A B c Cgr3 e f g ∧ A d Cgr e h ∧ B d Cgr f h
16 opeq2 ⊢ c = C → B c = B C
17 16 breq2d ⊢ c = C → A Colinear B c ↔ A Colinear B C
18 16 opeq2d ⊢ c = C → A B c = A B C
19 18 breq1d ⊢ c = C → A B c Cgr3 e f g ↔ A B C Cgr3 e f g
20 17 19 3anbi12d ⊢ c = C → A Colinear B c ∧ A B c Cgr3 e f g ∧ A d Cgr e h ∧ B d Cgr f h ↔ A Colinear B C ∧ A B C Cgr3 e f g ∧ A d Cgr e h ∧ B d Cgr f h
21 opeq2 ⊢ d = D → A d = A D
22 21 breq1d ⊢ d = D → A d Cgr e h ↔ A D Cgr e h
23 opeq2 ⊢ d = D → B d = B D
24 23 breq1d ⊢ d = D → B d Cgr f h ↔ B D Cgr f h
25 22 24 anbi12d ⊢ d = D → A d Cgr e h ∧ B d Cgr f h ↔ A D Cgr e h ∧ B D Cgr f h
26 25 3anbi3d ⊢ d = D → A Colinear B C ∧ A B C Cgr3 e f g ∧ A d Cgr e h ∧ B d Cgr f h ↔ A Colinear B C ∧ A B C Cgr3 e f g ∧ A D Cgr e h ∧ B D Cgr f h
27 opeq1 ⊢ e = E → e f g = E f g
28 27 breq2d ⊢ e = E → A B C Cgr3 e f g ↔ A B C Cgr3 E f g
29 opeq1 ⊢ e = E → e h = E h
30 29 breq2d ⊢ e = E → A D Cgr e h ↔ A D Cgr E h
31 30 anbi1d ⊢ e = E → A D Cgr e h ∧ B D Cgr f h ↔ A D Cgr E h ∧ B D Cgr f h
32 28 31 3anbi23d ⊢ e = E → A Colinear B C ∧ A B C Cgr3 e f g ∧ A D Cgr e h ∧ B D Cgr f h ↔ A Colinear B C ∧ A B C Cgr3 E f g ∧ A D Cgr E h ∧ B D Cgr f h
33 opeq1 ⊢ f = F → f g = F g
34 33 opeq2d ⊢ f = F → E f g = E F g
35 34 breq2d ⊢ f = F → A B C Cgr3 E f g ↔ A B C Cgr3 E F g
36 opeq1 ⊢ f = F → f h = F h
37 36 breq2d ⊢ f = F → B D Cgr f h ↔ B D Cgr F h
38 37 anbi2d ⊢ f = F → A D Cgr E h ∧ B D Cgr f h ↔ A D Cgr E h ∧ B D Cgr F h
39 35 38 3anbi23d ⊢ f = F → A Colinear B C ∧ A B C Cgr3 E f g ∧ A D Cgr E h ∧ B D Cgr f h ↔ A Colinear B C ∧ A B C Cgr3 E F g ∧ A D Cgr E h ∧ B D Cgr F h
40 opeq2 ⊢ g = G → F g = F G
41 40 opeq2d ⊢ g = G → E F g = E F G
42 41 breq2d ⊢ g = G → A B C Cgr3 E F g ↔ A B C Cgr3 E F G
43 42 3anbi2d ⊢ g = G → A Colinear B C ∧ A B C Cgr3 E F g ∧ A D Cgr E h ∧ B D Cgr F h ↔ A Colinear B C ∧ A B C Cgr3 E F G ∧ A D Cgr E h ∧ B D Cgr F h
44 opeq2 ⊢ h = H → E h = E H
45 44 breq2d ⊢ h = H → A D Cgr E h ↔ A D Cgr E H
46 opeq2 ⊢ h = H → F h = F H
47 46 breq2d ⊢ h = H → B D Cgr F h ↔ B D Cgr F H
48 45 47 anbi12d ⊢ h = H → A D Cgr E h ∧ B D Cgr F h ↔ A D Cgr E H ∧ B D Cgr F H
49 48 3anbi3d ⊢ h = H → A Colinear B C ∧ A B C Cgr3 E F G ∧ A D Cgr E h ∧ B D Cgr F h ↔ A Colinear B C ∧ A B C Cgr3 E F G ∧ A D Cgr E H ∧ B D Cgr F H
50 fveq2 ⊢ n = N → 𝔼 ⁡ n = 𝔼 ⁡ N
51 df-fs ⊢ FiveSeg = p q | ∃ n ∈ ℕ ∃ a ∈ 𝔼 ⁡ n ∃ b ∈ 𝔼 ⁡ n ∃ c ∈ 𝔼 ⁡ n ∃ d ∈ 𝔼 ⁡ n ∃ e ∈ 𝔼 ⁡ n ∃ f ∈ 𝔼 ⁡ n ∃ g ∈ 𝔼 ⁡ n ∃ h ∈ 𝔼 ⁡ n p = a b c d ∧ q = e f g h ∧ a Colinear b c ∧ a b c Cgr3 e f g ∧ a d Cgr e h ∧ b d Cgr f h
52 7 15 20 26 32 39 43 49 50 51 br8 ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ C ∈ 𝔼 ⁡ N ∧ D ∈ 𝔼 ⁡ N ∧ E ∈ 𝔼 ⁡ N ∧ F ∈ 𝔼 ⁡ N ∧ G ∈ 𝔼 ⁡ N ∧ H ∈ 𝔼 ⁡ N → A B C D FiveSeg E F G H ↔ A Colinear B C ∧ A B C Cgr3 E F G ∧ A D Cgr E H ∧ B D Cgr F H