Theorem elpr2elpr

Description: For an element A of an unordered pair which is a subset of a given set V , there is another (maybe the same) element b of the given set V being an element of the unordered pair. (Contributed by AV, 5-Dec-2020)

	Ref	Expression
Assertion	elpr2elpr	⊢ ( ( 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉 ∧ 𝐴 ∈ { 𝑋 , 𝑌 } ) → ∃ 𝑏 ∈ 𝑉 { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } )

Proof

Step	Hyp	Ref	Expression
1		simprr	⊢ ( ( 𝐴 = 𝑋 ∧ ( 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉 ) ) → 𝑌 ∈ 𝑉 )
2		preq12	⊢ ( ( 𝐴 = 𝑋 ∧ 𝑏 = 𝑌 ) → { 𝐴 , 𝑏 } = { 𝑋 , 𝑌 } )
3	2	eqcomd	⊢ ( ( 𝐴 = 𝑋 ∧ 𝑏 = 𝑌 ) → { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } )
4	3	adantlr	⊢ ( ( ( 𝐴 = 𝑋 ∧ ( 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉 ) ) ∧ 𝑏 = 𝑌 ) → { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } )
5	1 4	rspcedeqvd	⊢ ( ( 𝐴 = 𝑋 ∧ ( 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉 ) ) → ∃ 𝑏 ∈ 𝑉 { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } )
6	5	ex	⊢ ( 𝐴 = 𝑋 → ( ( 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉 ) → ∃ 𝑏 ∈ 𝑉 { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } ) )
7		simprl	⊢ ( ( 𝐴 = 𝑌 ∧ ( 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉 ) ) → 𝑋 ∈ 𝑉 )
8		preq12	⊢ ( ( 𝐴 = 𝑌 ∧ 𝑏 = 𝑋 ) → { 𝐴 , 𝑏 } = { 𝑌 , 𝑋 } )
9		prcom	⊢ { 𝑌 , 𝑋 } = { 𝑋 , 𝑌 }
10	8 9	eqtr2di	⊢ ( ( 𝐴 = 𝑌 ∧ 𝑏 = 𝑋 ) → { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } )
11	10	adantlr	⊢ ( ( ( 𝐴 = 𝑌 ∧ ( 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉 ) ) ∧ 𝑏 = 𝑋 ) → { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } )
12	7 11	rspcedeqvd	⊢ ( ( 𝐴 = 𝑌 ∧ ( 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉 ) ) → ∃ 𝑏 ∈ 𝑉 { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } )
13	12	ex	⊢ ( 𝐴 = 𝑌 → ( ( 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉 ) → ∃ 𝑏 ∈ 𝑉 { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } ) )
14	6 13	jaoi	⊢ ( ( 𝐴 = 𝑋 ∨ 𝐴 = 𝑌 ) → ( ( 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉 ) → ∃ 𝑏 ∈ 𝑉 { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } ) )
15		elpri	⊢ ( 𝐴 ∈ { 𝑋 , 𝑌 } → ( 𝐴 = 𝑋 ∨ 𝐴 = 𝑌 ) )
16	14 15	syl11	⊢ ( ( 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉 ) → ( 𝐴 ∈ { 𝑋 , 𝑌 } → ∃ 𝑏 ∈ 𝑉 { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } ) )
17	16	3impia	⊢ ( ( 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉 ∧ 𝐴 ∈ { 𝑋 , 𝑌 } ) → ∃ 𝑏 ∈ 𝑉 { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } )