Step |
Hyp |
Ref |
Expression |
1 |
|
1st2nd2 |
⊢ ( 𝑥 ∈ ( 𝐵 × { 𝒫 ∪ ran 𝐴 } ) → 𝑥 = 〈 ( 1st ‘ 𝑥 ) , ( 2nd ‘ 𝑥 ) 〉 ) |
2 |
1
|
ad2antll |
⊢ ( ( ( 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ) ∧ ( 𝑥 ∈ 𝐴 ∧ 𝑥 ∈ ( 𝐵 × { 𝒫 ∪ ran 𝐴 } ) ) ) → 𝑥 = 〈 ( 1st ‘ 𝑥 ) , ( 2nd ‘ 𝑥 ) 〉 ) |
3 |
|
simprl |
⊢ ( ( ( 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ) ∧ ( 𝑥 ∈ 𝐴 ∧ 𝑥 ∈ ( 𝐵 × { 𝒫 ∪ ran 𝐴 } ) ) ) → 𝑥 ∈ 𝐴 ) |
4 |
2 3
|
eqeltrrd |
⊢ ( ( ( 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ) ∧ ( 𝑥 ∈ 𝐴 ∧ 𝑥 ∈ ( 𝐵 × { 𝒫 ∪ ran 𝐴 } ) ) ) → 〈 ( 1st ‘ 𝑥 ) , ( 2nd ‘ 𝑥 ) 〉 ∈ 𝐴 ) |
5 |
|
fvex |
⊢ ( 1st ‘ 𝑥 ) ∈ V |
6 |
|
fvex |
⊢ ( 2nd ‘ 𝑥 ) ∈ V |
7 |
5 6
|
opelrn |
⊢ ( 〈 ( 1st ‘ 𝑥 ) , ( 2nd ‘ 𝑥 ) 〉 ∈ 𝐴 → ( 2nd ‘ 𝑥 ) ∈ ran 𝐴 ) |
8 |
4 7
|
syl |
⊢ ( ( ( 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ) ∧ ( 𝑥 ∈ 𝐴 ∧ 𝑥 ∈ ( 𝐵 × { 𝒫 ∪ ran 𝐴 } ) ) ) → ( 2nd ‘ 𝑥 ) ∈ ran 𝐴 ) |
9 |
|
pwuninel |
⊢ ¬ 𝒫 ∪ ran 𝐴 ∈ ran 𝐴 |
10 |
|
xp2nd |
⊢ ( 𝑥 ∈ ( 𝐵 × { 𝒫 ∪ ran 𝐴 } ) → ( 2nd ‘ 𝑥 ) ∈ { 𝒫 ∪ ran 𝐴 } ) |
11 |
10
|
ad2antll |
⊢ ( ( ( 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ) ∧ ( 𝑥 ∈ 𝐴 ∧ 𝑥 ∈ ( 𝐵 × { 𝒫 ∪ ran 𝐴 } ) ) ) → ( 2nd ‘ 𝑥 ) ∈ { 𝒫 ∪ ran 𝐴 } ) |
12 |
|
elsni |
⊢ ( ( 2nd ‘ 𝑥 ) ∈ { 𝒫 ∪ ran 𝐴 } → ( 2nd ‘ 𝑥 ) = 𝒫 ∪ ran 𝐴 ) |
13 |
11 12
|
syl |
⊢ ( ( ( 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ) ∧ ( 𝑥 ∈ 𝐴 ∧ 𝑥 ∈ ( 𝐵 × { 𝒫 ∪ ran 𝐴 } ) ) ) → ( 2nd ‘ 𝑥 ) = 𝒫 ∪ ran 𝐴 ) |
14 |
13
|
eleq1d |
⊢ ( ( ( 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ) ∧ ( 𝑥 ∈ 𝐴 ∧ 𝑥 ∈ ( 𝐵 × { 𝒫 ∪ ran 𝐴 } ) ) ) → ( ( 2nd ‘ 𝑥 ) ∈ ran 𝐴 ↔ 𝒫 ∪ ran 𝐴 ∈ ran 𝐴 ) ) |
15 |
9 14
|
mtbiri |
⊢ ( ( ( 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ) ∧ ( 𝑥 ∈ 𝐴 ∧ 𝑥 ∈ ( 𝐵 × { 𝒫 ∪ ran 𝐴 } ) ) ) → ¬ ( 2nd ‘ 𝑥 ) ∈ ran 𝐴 ) |
16 |
8 15
|
pm2.65da |
⊢ ( ( 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ) → ¬ ( 𝑥 ∈ 𝐴 ∧ 𝑥 ∈ ( 𝐵 × { 𝒫 ∪ ran 𝐴 } ) ) ) |
17 |
|
elin |
⊢ ( 𝑥 ∈ ( 𝐴 ∩ ( 𝐵 × { 𝒫 ∪ ran 𝐴 } ) ) ↔ ( 𝑥 ∈ 𝐴 ∧ 𝑥 ∈ ( 𝐵 × { 𝒫 ∪ ran 𝐴 } ) ) ) |
18 |
16 17
|
sylnibr |
⊢ ( ( 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ) → ¬ 𝑥 ∈ ( 𝐴 ∩ ( 𝐵 × { 𝒫 ∪ ran 𝐴 } ) ) ) |
19 |
18
|
eq0rdv |
⊢ ( ( 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ) → ( 𝐴 ∩ ( 𝐵 × { 𝒫 ∪ ran 𝐴 } ) ) = ∅ ) |
20 |
|
simpr |
⊢ ( ( 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ) → 𝐵 ∈ 𝑊 ) |
21 |
|
rnexg |
⊢ ( 𝐴 ∈ 𝑉 → ran 𝐴 ∈ V ) |
22 |
21
|
adantr |
⊢ ( ( 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ) → ran 𝐴 ∈ V ) |
23 |
|
uniexg |
⊢ ( ran 𝐴 ∈ V → ∪ ran 𝐴 ∈ V ) |
24 |
|
pwexg |
⊢ ( ∪ ran 𝐴 ∈ V → 𝒫 ∪ ran 𝐴 ∈ V ) |
25 |
22 23 24
|
3syl |
⊢ ( ( 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ) → 𝒫 ∪ ran 𝐴 ∈ V ) |
26 |
|
xpsneng |
⊢ ( ( 𝐵 ∈ 𝑊 ∧ 𝒫 ∪ ran 𝐴 ∈ V ) → ( 𝐵 × { 𝒫 ∪ ran 𝐴 } ) ≈ 𝐵 ) |
27 |
20 25 26
|
syl2anc |
⊢ ( ( 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ) → ( 𝐵 × { 𝒫 ∪ ran 𝐴 } ) ≈ 𝐵 ) |
28 |
19 27
|
jca |
⊢ ( ( 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ) → ( ( 𝐴 ∩ ( 𝐵 × { 𝒫 ∪ ran 𝐴 } ) ) = ∅ ∧ ( 𝐵 × { 𝒫 ∪ ran 𝐴 } ) ≈ 𝐵 ) ) |