Step |
Hyp |
Ref |
Expression |
1 |
|
brcup.1 |
⊢ 𝐴 ∈ V |
2 |
|
brcup.2 |
⊢ 𝐵 ∈ V |
3 |
|
brcup.3 |
⊢ 𝐶 ∈ V |
4 |
|
opex |
⊢ 〈 𝐴 , 𝐵 〉 ∈ V |
5 |
|
df-cup |
⊢ Cup = ( ( ( V × V ) × V ) ∖ ran ( ( V ⊗ E ) △ ( ( ( ◡ 1st ∘ E ) ∪ ( ◡ 2nd ∘ E ) ) ⊗ V ) ) ) |
6 |
1 2
|
opelvv |
⊢ 〈 𝐴 , 𝐵 〉 ∈ ( V × V ) |
7 |
|
brxp |
⊢ ( 〈 𝐴 , 𝐵 〉 ( ( V × V ) × V ) 𝐶 ↔ ( 〈 𝐴 , 𝐵 〉 ∈ ( V × V ) ∧ 𝐶 ∈ V ) ) |
8 |
6 3 7
|
mpbir2an |
⊢ 〈 𝐴 , 𝐵 〉 ( ( V × V ) × V ) 𝐶 |
9 |
|
epel |
⊢ ( 𝑥 E 𝑦 ↔ 𝑥 ∈ 𝑦 ) |
10 |
|
vex |
⊢ 𝑦 ∈ V |
11 |
10 4
|
brcnv |
⊢ ( 𝑦 ◡ 1st 〈 𝐴 , 𝐵 〉 ↔ 〈 𝐴 , 𝐵 〉 1st 𝑦 ) |
12 |
1 2
|
br1steq |
⊢ ( 〈 𝐴 , 𝐵 〉 1st 𝑦 ↔ 𝑦 = 𝐴 ) |
13 |
11 12
|
bitri |
⊢ ( 𝑦 ◡ 1st 〈 𝐴 , 𝐵 〉 ↔ 𝑦 = 𝐴 ) |
14 |
9 13
|
anbi12ci |
⊢ ( ( 𝑥 E 𝑦 ∧ 𝑦 ◡ 1st 〈 𝐴 , 𝐵 〉 ) ↔ ( 𝑦 = 𝐴 ∧ 𝑥 ∈ 𝑦 ) ) |
15 |
14
|
exbii |
⊢ ( ∃ 𝑦 ( 𝑥 E 𝑦 ∧ 𝑦 ◡ 1st 〈 𝐴 , 𝐵 〉 ) ↔ ∃ 𝑦 ( 𝑦 = 𝐴 ∧ 𝑥 ∈ 𝑦 ) ) |
16 |
|
vex |
⊢ 𝑥 ∈ V |
17 |
16 4
|
brco |
⊢ ( 𝑥 ( ◡ 1st ∘ E ) 〈 𝐴 , 𝐵 〉 ↔ ∃ 𝑦 ( 𝑥 E 𝑦 ∧ 𝑦 ◡ 1st 〈 𝐴 , 𝐵 〉 ) ) |
18 |
1
|
clel3 |
⊢ ( 𝑥 ∈ 𝐴 ↔ ∃ 𝑦 ( 𝑦 = 𝐴 ∧ 𝑥 ∈ 𝑦 ) ) |
19 |
15 17 18
|
3bitr4i |
⊢ ( 𝑥 ( ◡ 1st ∘ E ) 〈 𝐴 , 𝐵 〉 ↔ 𝑥 ∈ 𝐴 ) |
20 |
10 4
|
brcnv |
⊢ ( 𝑦 ◡ 2nd 〈 𝐴 , 𝐵 〉 ↔ 〈 𝐴 , 𝐵 〉 2nd 𝑦 ) |
21 |
1 2
|
br2ndeq |
⊢ ( 〈 𝐴 , 𝐵 〉 2nd 𝑦 ↔ 𝑦 = 𝐵 ) |
22 |
20 21
|
bitri |
⊢ ( 𝑦 ◡ 2nd 〈 𝐴 , 𝐵 〉 ↔ 𝑦 = 𝐵 ) |
23 |
9 22
|
anbi12ci |
⊢ ( ( 𝑥 E 𝑦 ∧ 𝑦 ◡ 2nd 〈 𝐴 , 𝐵 〉 ) ↔ ( 𝑦 = 𝐵 ∧ 𝑥 ∈ 𝑦 ) ) |
24 |
23
|
exbii |
⊢ ( ∃ 𝑦 ( 𝑥 E 𝑦 ∧ 𝑦 ◡ 2nd 〈 𝐴 , 𝐵 〉 ) ↔ ∃ 𝑦 ( 𝑦 = 𝐵 ∧ 𝑥 ∈ 𝑦 ) ) |
25 |
16 4
|
brco |
⊢ ( 𝑥 ( ◡ 2nd ∘ E ) 〈 𝐴 , 𝐵 〉 ↔ ∃ 𝑦 ( 𝑥 E 𝑦 ∧ 𝑦 ◡ 2nd 〈 𝐴 , 𝐵 〉 ) ) |
26 |
2
|
clel3 |
⊢ ( 𝑥 ∈ 𝐵 ↔ ∃ 𝑦 ( 𝑦 = 𝐵 ∧ 𝑥 ∈ 𝑦 ) ) |
27 |
24 25 26
|
3bitr4i |
⊢ ( 𝑥 ( ◡ 2nd ∘ E ) 〈 𝐴 , 𝐵 〉 ↔ 𝑥 ∈ 𝐵 ) |
28 |
19 27
|
orbi12i |
⊢ ( ( 𝑥 ( ◡ 1st ∘ E ) 〈 𝐴 , 𝐵 〉 ∨ 𝑥 ( ◡ 2nd ∘ E ) 〈 𝐴 , 𝐵 〉 ) ↔ ( 𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵 ) ) |
29 |
|
brun |
⊢ ( 𝑥 ( ( ◡ 1st ∘ E ) ∪ ( ◡ 2nd ∘ E ) ) 〈 𝐴 , 𝐵 〉 ↔ ( 𝑥 ( ◡ 1st ∘ E ) 〈 𝐴 , 𝐵 〉 ∨ 𝑥 ( ◡ 2nd ∘ E ) 〈 𝐴 , 𝐵 〉 ) ) |
30 |
|
elun |
⊢ ( 𝑥 ∈ ( 𝐴 ∪ 𝐵 ) ↔ ( 𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵 ) ) |
31 |
28 29 30
|
3bitr4ri |
⊢ ( 𝑥 ∈ ( 𝐴 ∪ 𝐵 ) ↔ 𝑥 ( ( ◡ 1st ∘ E ) ∪ ( ◡ 2nd ∘ E ) ) 〈 𝐴 , 𝐵 〉 ) |
32 |
4 3 5 8 31
|
brtxpsd3 |
⊢ ( 〈 𝐴 , 𝐵 〉 Cup 𝐶 ↔ 𝐶 = ( 𝐴 ∪ 𝐵 ) ) |