Metamath Proof Explorer


Theorem djur

Description: A member of a disjoint union can be mapped from one of the classes which produced it. (Contributed by Jim Kingdon, 23-Jun-2022)

Ref Expression
Assertion djur C A ⊔︀ B x A C = inl x x B C = inr x

Proof

Step Hyp Ref Expression
1 df-dju A ⊔︀ B = × A 1 𝑜 × B
2 1 eleq2i C A ⊔︀ B C × A 1 𝑜 × B
3 elun C × A 1 𝑜 × B C × A C 1 𝑜 × B
4 2 3 sylbb C A ⊔︀ B C × A C 1 𝑜 × B
5 xp2nd C × A 2 nd C A
6 1st2nd2 C × A C = 1 st C 2 nd C
7 xp1st C × A 1 st C
8 elsni 1 st C 1 st C =
9 opeq1 1 st C = 1 st C 2 nd C = 2 nd C
10 9 eqeq2d 1 st C = C = 1 st C 2 nd C C = 2 nd C
11 7 8 10 3syl C × A C = 1 st C 2 nd C C = 2 nd C
12 6 11 mpbid C × A C = 2 nd C
13 fvexd C × A 2 nd C V
14 opex 2 nd C V
15 opeq2 y = 2 nd C y = 2 nd C
16 df-inl inl = y V y
17 15 16 fvmptg 2 nd C V 2 nd C V inl 2 nd C = 2 nd C
18 13 14 17 sylancl C × A inl 2 nd C = 2 nd C
19 12 18 eqtr4d C × A C = inl 2 nd C
20 fveq2 x = 2 nd C inl x = inl 2 nd C
21 20 rspceeqv 2 nd C A C = inl 2 nd C x A C = inl x
22 5 19 21 syl2anc C × A x A C = inl x
23 xp2nd C 1 𝑜 × B 2 nd C B
24 1st2nd2 C 1 𝑜 × B C = 1 st C 2 nd C
25 xp1st C 1 𝑜 × B 1 st C 1 𝑜
26 elsni 1 st C 1 𝑜 1 st C = 1 𝑜
27 opeq1 1 st C = 1 𝑜 1 st C 2 nd C = 1 𝑜 2 nd C
28 27 eqeq2d 1 st C = 1 𝑜 C = 1 st C 2 nd C C = 1 𝑜 2 nd C
29 25 26 28 3syl C 1 𝑜 × B C = 1 st C 2 nd C C = 1 𝑜 2 nd C
30 24 29 mpbid C 1 𝑜 × B C = 1 𝑜 2 nd C
31 fvexd C 1 𝑜 × B 2 nd C V
32 opex 1 𝑜 2 nd C V
33 opeq2 z = 2 nd C 1 𝑜 z = 1 𝑜 2 nd C
34 df-inr inr = z V 1 𝑜 z
35 33 34 fvmptg 2 nd C V 1 𝑜 2 nd C V inr 2 nd C = 1 𝑜 2 nd C
36 31 32 35 sylancl C 1 𝑜 × B inr 2 nd C = 1 𝑜 2 nd C
37 30 36 eqtr4d C 1 𝑜 × B C = inr 2 nd C
38 fveq2 x = 2 nd C inr x = inr 2 nd C
39 38 rspceeqv 2 nd C B C = inr 2 nd C x B C = inr x
40 23 37 39 syl2anc C 1 𝑜 × B x B C = inr x
41 22 40 orim12i C × A C 1 𝑜 × B x A C = inl x x B C = inr x
42 4 41 syl C A ⊔︀ B x A C = inl x x B C = inr x