Metamath Proof Explorer


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 ( ( 𝑋𝑉𝑌𝑉𝐴 ∈ { 𝑋 , 𝑌 } ) → ∃ 𝑏𝑉 { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } )

Proof

Step Hyp Ref Expression
1 simprr ( ( 𝐴 = 𝑋 ∧ ( 𝑋𝑉𝑌𝑉 ) ) → 𝑌𝑉 )
2 preq12 ( ( 𝐴 = 𝑋𝑏 = 𝑌 ) → { 𝐴 , 𝑏 } = { 𝑋 , 𝑌 } )
3 2 eqcomd ( ( 𝐴 = 𝑋𝑏 = 𝑌 ) → { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } )
4 3 adantlr ( ( ( 𝐴 = 𝑋 ∧ ( 𝑋𝑉𝑌𝑉 ) ) ∧ 𝑏 = 𝑌 ) → { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } )
5 1 4 rspcedeq2vd ( ( 𝐴 = 𝑋 ∧ ( 𝑋𝑉𝑌𝑉 ) ) → ∃ 𝑏𝑉 { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } )
6 5 ex ( 𝐴 = 𝑋 → ( ( 𝑋𝑉𝑌𝑉 ) → ∃ 𝑏𝑉 { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } ) )
7 simprl ( ( 𝐴 = 𝑌 ∧ ( 𝑋𝑉𝑌𝑉 ) ) → 𝑋𝑉 )
8 preq12 ( ( 𝐴 = 𝑌𝑏 = 𝑋 ) → { 𝐴 , 𝑏 } = { 𝑌 , 𝑋 } )
9 prcom { 𝑌 , 𝑋 } = { 𝑋 , 𝑌 }
10 8 9 eqtr2di ( ( 𝐴 = 𝑌𝑏 = 𝑋 ) → { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } )
11 10 adantlr ( ( ( 𝐴 = 𝑌 ∧ ( 𝑋𝑉𝑌𝑉 ) ) ∧ 𝑏 = 𝑋 ) → { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } )
12 7 11 rspcedeq2vd ( ( 𝐴 = 𝑌 ∧ ( 𝑋𝑉𝑌𝑉 ) ) → ∃ 𝑏𝑉 { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } )
13 12 ex ( 𝐴 = 𝑌 → ( ( 𝑋𝑉𝑌𝑉 ) → ∃ 𝑏𝑉 { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } ) )
14 6 13 jaoi ( ( 𝐴 = 𝑋𝐴 = 𝑌 ) → ( ( 𝑋𝑉𝑌𝑉 ) → ∃ 𝑏𝑉 { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } ) )
15 elpri ( 𝐴 ∈ { 𝑋 , 𝑌 } → ( 𝐴 = 𝑋𝐴 = 𝑌 ) )
16 14 15 syl11 ( ( 𝑋𝑉𝑌𝑉 ) → ( 𝐴 ∈ { 𝑋 , 𝑌 } → ∃ 𝑏𝑉 { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } ) )
17 16 3impia ( ( 𝑋𝑉𝑌𝑉𝐴 ∈ { 𝑋 , 𝑌 } ) → ∃ 𝑏𝑉 { 𝑋 , 𝑌 } = { 𝐴 , 𝑏 } )