Metamath Proof Explorer


Theorem f2ndres

Description: Mapping of a restriction of the 2nd (second member of an ordered pair) function. (Contributed by NM, 7-Aug-2006) (Revised by Mario Carneiro, 8-Sep-2013)

Ref Expression
Assertion f2ndres ( 2nd ↾ ( 𝐴 × 𝐵 ) ) : ( 𝐴 × 𝐵 ) ⟶ 𝐵

Proof

Step Hyp Ref Expression
1 vex ⊢ 𝑦 ∈ V
2 vex ⊢ 𝑧 ∈ V
3 1 2 op2nda ⊢ ∪ ran { ⟨ 𝑦 , 𝑧 ⟩ } = 𝑧
4 3 eleq1i ⊢ ( ∪ ran { ⟨ 𝑦 , 𝑧 ⟩ } ∈ 𝐵 ↔ 𝑧 ∈ 𝐵 )
5 4 bilanri ⊢ ( ( 𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵 ) → ∪ ran { ⟨ 𝑦 , 𝑧 ⟩ } ∈ 𝐵 )
6 5 rgen2 ⊢ ∀ 𝑦 ∈ 𝐴 ∀ 𝑧 ∈ 𝐵 ∪ ran { ⟨ 𝑦 , 𝑧 ⟩ } ∈ 𝐵
7 sneq ⊢ ( 𝑥 = ⟨ 𝑦 , 𝑧 ⟩ → { 𝑥 } = { ⟨ 𝑦 , 𝑧 ⟩ } )
8 7 rneqd ⊢ ( 𝑥 = ⟨ 𝑦 , 𝑧 ⟩ → ran { 𝑥 } = ran { ⟨ 𝑦 , 𝑧 ⟩ } )
9 8 unieqd ⊢ ( 𝑥 = ⟨ 𝑦 , 𝑧 ⟩ → ∪ ran { 𝑥 } = ∪ ran { ⟨ 𝑦 , 𝑧 ⟩ } )
10 9 eleq1d ⊢ ( 𝑥 = ⟨ 𝑦 , 𝑧 ⟩ → ( ∪ ran { 𝑥 } ∈ 𝐵 ↔ ∪ ran { ⟨ 𝑦 , 𝑧 ⟩ } ∈ 𝐵 ) )
11 10 ralxp ⊢ ( ∀ 𝑥 ∈ ( 𝐴 × 𝐵 ) ∪ ran { 𝑥 } ∈ 𝐵 ↔ ∀ 𝑦 ∈ 𝐴 ∀ 𝑧 ∈ 𝐵 ∪ ran { ⟨ 𝑦 , 𝑧 ⟩ } ∈ 𝐵 )
12 6 11 mpbir ⊢ ∀ 𝑥 ∈ ( 𝐴 × 𝐵 ) ∪ ran { 𝑥 } ∈ 𝐵
13 df-2nd ⊢ 2nd = ( 𝑥 ∈ V ↦ ∪ ran { 𝑥 } )
14 13 reseq1i ⊢ ( 2nd ↾ ( 𝐴 × 𝐵 ) ) = ( ( 𝑥 ∈ V ↦ ∪ ran { 𝑥 } ) ↾ ( 𝐴 × 𝐵 ) )
15 ssv ⊢ ( 𝐴 × 𝐵 ) ⊆ V
16 resmpt ⊢ ( ( 𝐴 × 𝐵 ) ⊆ V → ( ( 𝑥 ∈ V ↦ ∪ ran { 𝑥 } ) ↾ ( 𝐴 × 𝐵 ) ) = ( 𝑥 ∈ ( 𝐴 × 𝐵 ) ↦ ∪ ran { 𝑥 } ) )
17 15 16 ax-mp ⊢ ( ( 𝑥 ∈ V ↦ ∪ ran { 𝑥 } ) ↾ ( 𝐴 × 𝐵 ) ) = ( 𝑥 ∈ ( 𝐴 × 𝐵 ) ↦ ∪ ran { 𝑥 } )
18 14 17 eqtri ⊢ ( 2nd ↾ ( 𝐴 × 𝐵 ) ) = ( 𝑥 ∈ ( 𝐴 × 𝐵 ) ↦ ∪ ran { 𝑥 } )
19 18 fmpt ⊢ ( ∀ 𝑥 ∈ ( 𝐴 × 𝐵 ) ∪ ran { 𝑥 } ∈ 𝐵 ↔ ( 2nd ↾ ( 𝐴 × 𝐵 ) ) : ( 𝐴 × 𝐵 ) ⟶ 𝐵 )
20 12 19 mpbi ⊢ ( 2nd ↾ ( 𝐴 × 𝐵 ) ) : ( 𝐴 × 𝐵 ) ⟶ 𝐵