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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 ( ( 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 rspcedeq2vd ⊢ ( ( 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 rspcedeq2vd ⊢ ( ( 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 } )