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		preq2	⊢ ( 𝑏 = 𝑌 → { 𝐴 , 𝑏 } = { 𝐴 , 𝑌 } )
3	2	eqeq2d	⊢ ( 𝑏 = 𝑌 → ( { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } ↔ { 𝑋 , 𝑌 } = { 𝐴 , 𝑌 } ) )
4	3	adantl	⊢ ( ( ( 𝐴 = 𝑋 ∧ ( 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉 ) ) ∧ 𝑏 = 𝑌 ) → ( { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } ↔ { 𝑋 , 𝑌 } = { 𝐴 , 𝑌 } ) )
5		preq1	⊢ ( 𝑋 = 𝐴 → { 𝑋 , 𝑌 } = { 𝐴 , 𝑌 } )
6	5	eqcoms	⊢ ( 𝐴 = 𝑋 → { 𝑋 , 𝑌 } = { 𝐴 , 𝑌 } )
7	6	adantr	⊢ ( ( 𝐴 = 𝑋 ∧ ( 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉 ) ) → { 𝑋 , 𝑌 } = { 𝐴 , 𝑌 } )
8	1 4 7	rspcedvd	⊢ ( ( 𝐴 = 𝑋 ∧ ( 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉 ) ) → ∃ 𝑏 ∈ 𝑉 { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } )
9	8	ex	⊢ ( 𝐴 = 𝑋 → ( ( 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉 ) → ∃ 𝑏 ∈ 𝑉 { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } ) )
10		simprl	⊢ ( ( 𝐴 = 𝑌 ∧ ( 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉 ) ) → 𝑋 ∈ 𝑉 )
11		preq2	⊢ ( 𝑏 = 𝑋 → { 𝐴 , 𝑏 } = { 𝐴 , 𝑋 } )
12	11	eqeq2d	⊢ ( 𝑏 = 𝑋 → ( { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } ↔ { 𝑋 , 𝑌 } = { 𝐴 , 𝑋 } ) )
13	12	adantl	⊢ ( ( ( 𝐴 = 𝑌 ∧ ( 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉 ) ) ∧ 𝑏 = 𝑋 ) → ( { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } ↔ { 𝑋 , 𝑌 } = { 𝐴 , 𝑋 } ) )
14		preq2	⊢ ( 𝑌 = 𝐴 → { 𝑋 , 𝑌 } = { 𝑋 , 𝐴 } )
15	14	eqcoms	⊢ ( 𝐴 = 𝑌 → { 𝑋 , 𝑌 } = { 𝑋 , 𝐴 } )
16		prcom	⊢ { 𝑋 , 𝐴 } = { 𝐴 , 𝑋 }
17	15 16	eqtrdi	⊢ ( 𝐴 = 𝑌 → { 𝑋 , 𝑌 } = { 𝐴 , 𝑋 } )
18	17	adantr	⊢ ( ( 𝐴 = 𝑌 ∧ ( 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉 ) ) → { 𝑋 , 𝑌 } = { 𝐴 , 𝑋 } )
19	10 13 18	rspcedvd	⊢ ( ( 𝐴 = 𝑌 ∧ ( 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉 ) ) → ∃ 𝑏 ∈ 𝑉 { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } )
20	19	ex	⊢ ( 𝐴 = 𝑌 → ( ( 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉 ) → ∃ 𝑏 ∈ 𝑉 { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } ) )
21	9 20	jaoi	⊢ ( ( 𝐴 = 𝑋 ∨ 𝐴 = 𝑌 ) → ( ( 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉 ) → ∃ 𝑏 ∈ 𝑉 { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } ) )
22		elpri	⊢ ( 𝐴 ∈ { 𝑋 , 𝑌 } → ( 𝐴 = 𝑋 ∨ 𝐴 = 𝑌 ) )
23	21 22	syl11	⊢ ( ( 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉 ) → ( 𝐴 ∈ { 𝑋 , 𝑌 } → ∃ 𝑏 ∈ 𝑉 { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } ) )
24	23	3impia	⊢ ( ( 𝑋 ∈ 𝑉 ∧ 𝑌 ∈ 𝑉 ∧ 𝐴 ∈ { 𝑋 , 𝑌 } ) → ∃ 𝑏 ∈ 𝑉 { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } )