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