Metamath Proof Explorer


Theorem prf1

Description: Value of the pairing functor on objects. (Contributed by Mario Carneiro, 12-Jan-2017)

Ref Expression
Hypotheses prfval.k ⊢ P = F ⟨,⟩ F G
prfval.b ⊢ B = Base C
prfval.h ⊢ H = Hom ⁡ C
prfval.c ⊢ φ → F ∈ C Func D
prfval.d ⊢ φ → G ∈ C Func E
prf1.x ⊢ φ → X ∈ B
Assertion prf1 ⊢ φ → 1 st ⁡ P ⁡ X = 1 st ⁡ F ⁡ X 1 st ⁡ G ⁡ X

Proof

Step Hyp Ref Expression
1 prfval.k ⊢ P = F ⟨,⟩ F G
2 prfval.b ⊢ B = Base C
3 prfval.h ⊢ H = Hom ⁡ C
4 prfval.c ⊢ φ → F ∈ C Func D
5 prfval.d ⊢ φ → G ∈ C Func E
6 prf1.x ⊢ φ → X ∈ B
7 1 2 3 4 5 prfval ⊢ φ → P = x ∈ B ⟼ 1 st ⁡ F ⁡ x 1 st ⁡ G ⁡ x x ∈ B , y ∈ B ⟼ h ∈ x H y ⟼ x 2 nd ⁡ F y ⁡ h x 2 nd ⁡ G y ⁡ h
8 2 fvexi ⊢ B ∈ V
9 8 mptex ⊢ x ∈ B ⟼ 1 st ⁡ F ⁡ x 1 st ⁡ G ⁡ x ∈ V
10 8 8 mpoex ⊢ x ∈ B , y ∈ B ⟼ h ∈ x H y ⟼ x 2 nd ⁡ F y ⁡ h x 2 nd ⁡ G y ⁡ h ∈ V
11 9 10 op1std ⊢ P = x ∈ B ⟼ 1 st ⁡ F ⁡ x 1 st ⁡ G ⁡ x x ∈ B , y ∈ B ⟼ h ∈ x H y ⟼ x 2 nd ⁡ F y ⁡ h x 2 nd ⁡ G y ⁡ h → 1 st ⁡ P = x ∈ B ⟼ 1 st ⁡ F ⁡ x 1 st ⁡ G ⁡ x
12 7 11 syl ⊢ φ → 1 st ⁡ P = x ∈ B ⟼ 1 st ⁡ F ⁡ x 1 st ⁡ G ⁡ x
13 simpr ⊢ φ ∧ x = X → x = X
14 13 fveq2d ⊢ φ ∧ x = X → 1 st ⁡ F ⁡ x = 1 st ⁡ F ⁡ X
15 13 fveq2d ⊢ φ ∧ x = X → 1 st ⁡ G ⁡ x = 1 st ⁡ G ⁡ X
16 14 15 opeq12d ⊢ φ ∧ x = X → 1 st ⁡ F ⁡ x 1 st ⁡ G ⁡ x = 1 st ⁡ F ⁡ X 1 st ⁡ G ⁡ X
17 opex ⊢ 1 st ⁡ F ⁡ X 1 st ⁡ G ⁡ X ∈ V
18 17 a1i ⊢ φ → 1 st ⁡ F ⁡ X 1 st ⁡ G ⁡ X ∈ V
19 12 16 6 18 fvmptd ⊢ φ → 1 st ⁡ P ⁡ X = 1 st ⁡ F ⁡ X 1 st ⁡ G ⁡ X