Metamath Proof Explorer


Theorem brxrn

Description: Characterize a ternary relation over a range Cartesian product. Together with xrnss3v , this characterizes elementhood in a range cross. (Contributed by Peter Mazsa, 27-Jun-2021)

Ref Expression
Assertion brxrn ⊢ A ∈ V ∧ B ∈ W ∧ C ∈ X → A R ⋉ S B C ↔ A R B ∧ A S C

Proof

Step Hyp Ref Expression
1 df-xrn ⊢ R ⋉ S = 1 st ↾ V × V -1 ∘ R ∩ 2 nd ↾ V × V -1 ∘ S
2 1 breqi ⊢ A R ⋉ S B C ↔ A 1 st ↾ V × V -1 ∘ R ∩ 2 nd ↾ V × V -1 ∘ S B C
3 2 a1i ⊢ A ∈ V ∧ B ∈ W ∧ C ∈ X → A R ⋉ S B C ↔ A 1 st ↾ V × V -1 ∘ R ∩ 2 nd ↾ V × V -1 ∘ S B C
4 brin ⊢ A 1 st ↾ V × V -1 ∘ R ∩ 2 nd ↾ V × V -1 ∘ S B C ↔ A 1 st ↾ V × V -1 ∘ R B C ∧ A 2 nd ↾ V × V -1 ∘ S B C
5 4 a1i ⊢ A ∈ V ∧ B ∈ W ∧ C ∈ X → A 1 st ↾ V × V -1 ∘ R ∩ 2 nd ↾ V × V -1 ∘ S B C ↔ A 1 st ↾ V × V -1 ∘ R B C ∧ A 2 nd ↾ V × V -1 ∘ S B C
6 opex ⊢ B C ∈ V
7 brcog ⊢ A ∈ V ∧ B C ∈ V → A 1 st ↾ V × V -1 ∘ R B C ↔ ∃ x A R x ∧ x 1 st ↾ V × V -1 B C
8 6 7 mpan2 ⊢ A ∈ V → A 1 st ↾ V × V -1 ∘ R B C ↔ ∃ x A R x ∧ x 1 st ↾ V × V -1 B C
9 8 3ad2ant1 ⊢ A ∈ V ∧ B ∈ W ∧ C ∈ X → A 1 st ↾ V × V -1 ∘ R B C ↔ ∃ x A R x ∧ x 1 st ↾ V × V -1 B C
10 brcnvg ⊢ x ∈ V ∧ B C ∈ V → x 1 st ↾ V × V -1 B C ↔ B C 1 st ↾ V × V x
11 6 10 mpan2 ⊢ x ∈ V → x 1 st ↾ V × V -1 B C ↔ B C 1 st ↾ V × V x
12 11 elv ⊢ x 1 st ↾ V × V -1 B C ↔ B C 1 st ↾ V × V x
13 brres ⊢ x ∈ V → B C 1 st ↾ V × V x ↔ B C ∈ V × V ∧ B C 1 st x
14 13 elv ⊢ B C 1 st ↾ V × V x ↔ B C ∈ V × V ∧ B C 1 st x
15 opelvvg ⊢ B ∈ W ∧ C ∈ X → B C ∈ V × V
16 15 biantrurd ⊢ B ∈ W ∧ C ∈ X → B C 1 st x ↔ B C ∈ V × V ∧ B C 1 st x
17 14 16 bitr4id ⊢ B ∈ W ∧ C ∈ X → B C 1 st ↾ V × V x ↔ B C 1 st x
18 br1steqg ⊢ B ∈ W ∧ C ∈ X → B C 1 st x ↔ x = B
19 17 18 bitrd ⊢ B ∈ W ∧ C ∈ X → B C 1 st ↾ V × V x ↔ x = B
20 19 3adant1 ⊢ A ∈ V ∧ B ∈ W ∧ C ∈ X → B C 1 st ↾ V × V x ↔ x = B
21 12 20 bitrid ⊢ A ∈ V ∧ B ∈ W ∧ C ∈ X → x 1 st ↾ V × V -1 B C ↔ x = B
22 21 anbi1cd ⊢ A ∈ V ∧ B ∈ W ∧ C ∈ X → A R x ∧ x 1 st ↾ V × V -1 B C ↔ x = B ∧ A R x
23 22 exbidv ⊢ A ∈ V ∧ B ∈ W ∧ C ∈ X → ∃ x A R x ∧ x 1 st ↾ V × V -1 B C ↔ ∃ x x = B ∧ A R x
24 breq2 ⊢ x = B → A R x ↔ A R B
25 24 ceqsexgv ⊢ B ∈ W → ∃ x x = B ∧ A R x ↔ A R B
26 25 3ad2ant2 ⊢ A ∈ V ∧ B ∈ W ∧ C ∈ X → ∃ x x = B ∧ A R x ↔ A R B
27 9 23 26 3bitrd ⊢ A ∈ V ∧ B ∈ W ∧ C ∈ X → A 1 st ↾ V × V -1 ∘ R B C ↔ A R B
28 brcog ⊢ A ∈ V ∧ B C ∈ V → A 2 nd ↾ V × V -1 ∘ S B C ↔ ∃ y A S y ∧ y 2 nd ↾ V × V -1 B C
29 6 28 mpan2 ⊢ A ∈ V → A 2 nd ↾ V × V -1 ∘ S B C ↔ ∃ y A S y ∧ y 2 nd ↾ V × V -1 B C
30 29 3ad2ant1 ⊢ A ∈ V ∧ B ∈ W ∧ C ∈ X → A 2 nd ↾ V × V -1 ∘ S B C ↔ ∃ y A S y ∧ y 2 nd ↾ V × V -1 B C
31 brcnvg ⊢ y ∈ V ∧ B C ∈ V → y 2 nd ↾ V × V -1 B C ↔ B C 2 nd ↾ V × V y
32 6 31 mpan2 ⊢ y ∈ V → y 2 nd ↾ V × V -1 B C ↔ B C 2 nd ↾ V × V y
33 32 elv ⊢ y 2 nd ↾ V × V -1 B C ↔ B C 2 nd ↾ V × V y
34 brres ⊢ y ∈ V → B C 2 nd ↾ V × V y ↔ B C ∈ V × V ∧ B C 2 nd y
35 34 elv ⊢ B C 2 nd ↾ V × V y ↔ B C ∈ V × V ∧ B C 2 nd y
36 15 biantrurd ⊢ B ∈ W ∧ C ∈ X → B C 2 nd y ↔ B C ∈ V × V ∧ B C 2 nd y
37 35 36 bitr4id ⊢ B ∈ W ∧ C ∈ X → B C 2 nd ↾ V × V y ↔ B C 2 nd y
38 br2ndeqg ⊢ B ∈ W ∧ C ∈ X → B C 2 nd y ↔ y = C
39 37 38 bitrd ⊢ B ∈ W ∧ C ∈ X → B C 2 nd ↾ V × V y ↔ y = C
40 39 3adant1 ⊢ A ∈ V ∧ B ∈ W ∧ C ∈ X → B C 2 nd ↾ V × V y ↔ y = C
41 33 40 bitrid ⊢ A ∈ V ∧ B ∈ W ∧ C ∈ X → y 2 nd ↾ V × V -1 B C ↔ y = C
42 41 anbi1cd ⊢ A ∈ V ∧ B ∈ W ∧ C ∈ X → A S y ∧ y 2 nd ↾ V × V -1 B C ↔ y = C ∧ A S y
43 42 exbidv ⊢ A ∈ V ∧ B ∈ W ∧ C ∈ X → ∃ y A S y ∧ y 2 nd ↾ V × V -1 B C ↔ ∃ y y = C ∧ A S y
44 breq2 ⊢ y = C → A S y ↔ A S C
45 44 ceqsexgv ⊢ C ∈ X → ∃ y y = C ∧ A S y ↔ A S C
46 45 3ad2ant3 ⊢ A ∈ V ∧ B ∈ W ∧ C ∈ X → ∃ y y = C ∧ A S y ↔ A S C
47 30 43 46 3bitrd ⊢ A ∈ V ∧ B ∈ W ∧ C ∈ X → A 2 nd ↾ V × V -1 ∘ S B C ↔ A S C
48 27 47 anbi12d ⊢ A ∈ V ∧ B ∈ W ∧ C ∈ X → A 1 st ↾ V × V -1 ∘ R B C ∧ A 2 nd ↾ V × V -1 ∘ S B C ↔ A R B ∧ A S C
49 3 5 48 3bitrd ⊢ A ∈ V ∧ B ∈ W ∧ C ∈ X → A R ⋉ S B C ↔ A R B ∧ A S C