Metamath Proof Explorer


Theorem up1st2nd

Description: Rewrite the universal property predicate with separated parts. (Contributed by Zhi Wang, 23-Oct-2025)

Ref Expression
Hypothesis up1st2nd.1 ( 𝜑𝑋 ( 𝐹 ( 𝐷 UP 𝐸 ) 𝑊 ) 𝑀 )
Assertion up1st2nd ( 𝜑𝑋 ( ⟨ ( 1st𝐹 ) , ( 2nd𝐹 ) ⟩ ( 𝐷 UP 𝐸 ) 𝑊 ) 𝑀 )

Proof

Step Hyp Ref Expression
1 up1st2nd.1 ( 𝜑𝑋 ( 𝐹 ( 𝐷 UP 𝐸 ) 𝑊 ) 𝑀 )
2 relfunc Rel ( 𝐷 Func 𝐸 )
3 df-br ( 𝑋 ( 𝐹 ( 𝐷 UP 𝐸 ) 𝑊 ) 𝑀 ↔ ⟨ 𝑋 , 𝑀 ⟩ ∈ ( 𝐹 ( 𝐷 UP 𝐸 ) 𝑊 ) )
4 1 3 sylib ( 𝜑 → ⟨ 𝑋 , 𝑀 ⟩ ∈ ( 𝐹 ( 𝐷 UP 𝐸 ) 𝑊 ) )
5 eqid ( Base ‘ 𝐸 ) = ( Base ‘ 𝐸 )
6 5 uprcl ( ⟨ 𝑋 , 𝑀 ⟩ ∈ ( 𝐹 ( 𝐷 UP 𝐸 ) 𝑊 ) → ( 𝐹 ∈ ( 𝐷 Func 𝐸 ) ∧ 𝑊 ∈ ( Base ‘ 𝐸 ) ) )
7 4 6 syl ( 𝜑 → ( 𝐹 ∈ ( 𝐷 Func 𝐸 ) ∧ 𝑊 ∈ ( Base ‘ 𝐸 ) ) )
8 7 simpld ( 𝜑𝐹 ∈ ( 𝐷 Func 𝐸 ) )
9 1st2nd ( ( Rel ( 𝐷 Func 𝐸 ) ∧ 𝐹 ∈ ( 𝐷 Func 𝐸 ) ) → 𝐹 = ⟨ ( 1st𝐹 ) , ( 2nd𝐹 ) ⟩ )
10 2 8 9 sylancr ( 𝜑𝐹 = ⟨ ( 1st𝐹 ) , ( 2nd𝐹 ) ⟩ )
11 10 oveq1d ( 𝜑 → ( 𝐹 ( 𝐷 UP 𝐸 ) 𝑊 ) = ( ⟨ ( 1st𝐹 ) , ( 2nd𝐹 ) ⟩ ( 𝐷 UP 𝐸 ) 𝑊 ) )
12 11 1 breqdi ( 𝜑𝑋 ( ⟨ ( 1st𝐹 ) , ( 2nd𝐹 ) ⟩ ( 𝐷 UP 𝐸 ) 𝑊 ) 𝑀 )