Metamath Proof Explorer


Theorem djuun

Description: The disjoint union of two classes is the union of the images of those two classes under right and left injection. (Contributed by Jim Kingdon, 22-Jun-2022)

Ref Expression
Assertion djuun ⊢ inl A ∪ inr B = A ⊔︀ B

Proof

Step Hyp Ref Expression
1 elun ⊢ x ∈ inl A ∪ inr B ↔ x ∈ inl A ∨ x ∈ inr B
2 djulf1o ⊢ inl : V ⟶ 1-1 onto ∅ × V
3 f1ofn ⊢ inl : V ⟶ 1-1 onto ∅ × V → inl Fn V
4 2 3 ax-mp ⊢ inl Fn V
5 ssv ⊢ A ⊆ V
6 fvelimab ⊢ inl Fn V ∧ A ⊆ V → x ∈ inl A ↔ ∃ u ∈ A inl ⁡ u = x
7 4 5 6 mp2an ⊢ x ∈ inl A ↔ ∃ u ∈ A inl ⁡ u = x
8 7 biimpi ⊢ x ∈ inl A → ∃ u ∈ A inl ⁡ u = x
9 simprr ⊢ x ∈ inl A ∧ u ∈ A ∧ inl ⁡ u = x → inl ⁡ u = x
10 vex ⊢ u ∈ V
11 opex ⊢ ∅ u ∈ V
12 opeq2 ⊢ z = u → ∅ z = ∅ u
13 df-inl ⊢ inl = z ∈ V ⟼ ∅ z
14 12 13 fvmptg ⊢ u ∈ V ∧ ∅ u ∈ V → inl ⁡ u = ∅ u
15 10 11 14 mp2an ⊢ inl ⁡ u = ∅ u
16 0ex ⊢ ∅ ∈ V
17 16 snid ⊢ ∅ ∈ ∅
18 opelxpi ⊢ ∅ ∈ ∅ ∧ u ∈ A → ∅ u ∈ ∅ × A
19 17 18 mpan ⊢ u ∈ A → ∅ u ∈ ∅ × A
20 19 ad2antrl ⊢ x ∈ inl A ∧ u ∈ A ∧ inl ⁡ u = x → ∅ u ∈ ∅ × A
21 15 20 eqeltrid ⊢ x ∈ inl A ∧ u ∈ A ∧ inl ⁡ u = x → inl ⁡ u ∈ ∅ × A
22 9 21 eqeltrrd ⊢ x ∈ inl A ∧ u ∈ A ∧ inl ⁡ u = x → x ∈ ∅ × A
23 8 22 rexlimddv ⊢ x ∈ inl A → x ∈ ∅ × A
24 elun1 ⊢ x ∈ ∅ × A → x ∈ ∅ × A ∪ 1 𝑜 × B
25 23 24 syl ⊢ x ∈ inl A → x ∈ ∅ × A ∪ 1 𝑜 × B
26 df-dju ⊢ A ⊔︀ B = ∅ × A ∪ 1 𝑜 × B
27 25 26 eleqtrrdi ⊢ x ∈ inl A → x ∈ A ⊔︀ B
28 djurf1o ⊢ inr : V ⟶ 1-1 onto 1 𝑜 × V
29 f1ofn ⊢ inr : V ⟶ 1-1 onto 1 𝑜 × V → inr Fn V
30 28 29 ax-mp ⊢ inr Fn V
31 ssv ⊢ B ⊆ V
32 fvelimab ⊢ inr Fn V ∧ B ⊆ V → x ∈ inr B ↔ ∃ u ∈ B inr ⁡ u = x
33 30 31 32 mp2an ⊢ x ∈ inr B ↔ ∃ u ∈ B inr ⁡ u = x
34 33 biimpi ⊢ x ∈ inr B → ∃ u ∈ B inr ⁡ u = x
35 simprr ⊢ x ∈ inr B ∧ u ∈ B ∧ inr ⁡ u = x → inr ⁡ u = x
36 opex ⊢ 1 𝑜 u ∈ V
37 opeq2 ⊢ z = u → 1 𝑜 z = 1 𝑜 u
38 df-inr ⊢ inr = z ∈ V ⟼ 1 𝑜 z
39 37 38 fvmptg ⊢ u ∈ V ∧ 1 𝑜 u ∈ V → inr ⁡ u = 1 𝑜 u
40 10 36 39 mp2an ⊢ inr ⁡ u = 1 𝑜 u
41 1oex ⊢ 1 𝑜 ∈ V
42 41 snid ⊢ 1 𝑜 ∈ 1 𝑜
43 opelxpi ⊢ 1 𝑜 ∈ 1 𝑜 ∧ u ∈ B → 1 𝑜 u ∈ 1 𝑜 × B
44 42 43 mpan ⊢ u ∈ B → 1 𝑜 u ∈ 1 𝑜 × B
45 44 ad2antrl ⊢ x ∈ inr B ∧ u ∈ B ∧ inr ⁡ u = x → 1 𝑜 u ∈ 1 𝑜 × B
46 40 45 eqeltrid ⊢ x ∈ inr B ∧ u ∈ B ∧ inr ⁡ u = x → inr ⁡ u ∈ 1 𝑜 × B
47 35 46 eqeltrrd ⊢ x ∈ inr B ∧ u ∈ B ∧ inr ⁡ u = x → x ∈ 1 𝑜 × B
48 34 47 rexlimddv ⊢ x ∈ inr B → x ∈ 1 𝑜 × B
49 elun2 ⊢ x ∈ 1 𝑜 × B → x ∈ ∅ × A ∪ 1 𝑜 × B
50 48 49 syl ⊢ x ∈ inr B → x ∈ ∅ × A ∪ 1 𝑜 × B
51 50 26 eleqtrrdi ⊢ x ∈ inr B → x ∈ A ⊔︀ B
52 27 51 jaoi ⊢ x ∈ inl A ∨ x ∈ inr B → x ∈ A ⊔︀ B
53 1 52 sylbi ⊢ x ∈ inl A ∪ inr B → x ∈ A ⊔︀ B
54 53 ssriv ⊢ inl A ∪ inr B ⊆ A ⊔︀ B
55 djur ⊢ x ∈ A ⊔︀ B → ∃ y ∈ A x = inl ⁡ y ∨ ∃ y ∈ B x = inr ⁡ y
56 vex ⊢ y ∈ V
57 f1odm ⊢ inl : V ⟶ 1-1 onto ∅ × V → dom ⁡ inl = V
58 2 57 ax-mp ⊢ dom ⁡ inl = V
59 56 58 eleqtrri ⊢ y ∈ dom ⁡ inl
60 simpl ⊢ y ∈ A ∧ x = inl ⁡ y → y ∈ A
61 13 funmpt2 ⊢ Fun ⁡ inl
62 funfvima ⊢ Fun ⁡ inl ∧ y ∈ dom ⁡ inl → y ∈ A → inl ⁡ y ∈ inl A
63 61 62 mpan ⊢ y ∈ dom ⁡ inl → y ∈ A → inl ⁡ y ∈ inl A
64 59 60 63 mpsyl ⊢ y ∈ A ∧ x = inl ⁡ y → inl ⁡ y ∈ inl A
65 eleq1 ⊢ x = inl ⁡ y → x ∈ inl A ↔ inl ⁡ y ∈ inl A
66 65 adantl ⊢ y ∈ A ∧ x = inl ⁡ y → x ∈ inl A ↔ inl ⁡ y ∈ inl A
67 64 66 mpbird ⊢ y ∈ A ∧ x = inl ⁡ y → x ∈ inl A
68 67 rexlimiva ⊢ ∃ y ∈ A x = inl ⁡ y → x ∈ inl A
69 f1odm ⊢ inr : V ⟶ 1-1 onto 1 𝑜 × V → dom ⁡ inr = V
70 28 69 ax-mp ⊢ dom ⁡ inr = V
71 56 70 eleqtrri ⊢ y ∈ dom ⁡ inr
72 simpl ⊢ y ∈ B ∧ x = inr ⁡ y → y ∈ B
73 f1ofun ⊢ inr : V ⟶ 1-1 onto 1 𝑜 × V → Fun ⁡ inr
74 28 73 ax-mp ⊢ Fun ⁡ inr
75 funfvima ⊢ Fun ⁡ inr ∧ y ∈ dom ⁡ inr → y ∈ B → inr ⁡ y ∈ inr B
76 74 75 mpan ⊢ y ∈ dom ⁡ inr → y ∈ B → inr ⁡ y ∈ inr B
77 71 72 76 mpsyl ⊢ y ∈ B ∧ x = inr ⁡ y → inr ⁡ y ∈ inr B
78 eleq1 ⊢ x = inr ⁡ y → x ∈ inr B ↔ inr ⁡ y ∈ inr B
79 78 adantl ⊢ y ∈ B ∧ x = inr ⁡ y → x ∈ inr B ↔ inr ⁡ y ∈ inr B
80 77 79 mpbird ⊢ y ∈ B ∧ x = inr ⁡ y → x ∈ inr B
81 80 rexlimiva ⊢ ∃ y ∈ B x = inr ⁡ y → x ∈ inr B
82 68 81 orim12i ⊢ ∃ y ∈ A x = inl ⁡ y ∨ ∃ y ∈ B x = inr ⁡ y → x ∈ inl A ∨ x ∈ inr B
83 55 82 syl ⊢ x ∈ A ⊔︀ B → x ∈ inl A ∨ x ∈ inr B
84 83 1 sylibr ⊢ x ∈ A ⊔︀ B → x ∈ inl A ∪ inr B
85 84 ssriv ⊢ A ⊔︀ B ⊆ inl A ∪ inr B
86 54 85 eqssi ⊢ inl A ∪ inr B = A ⊔︀ B