Metamath Proof Explorer


Definition df-prf

Description: Define the pairing operation for functors (which takes two functors F : C --> D and G : C --> E and produces ( F pairF G ) : C --> ( D Xc. E ) ). (Contributed by Mario Carneiro, 11-Jan-2017)

Ref Expression
Assertion df-prf ⟨,⟩ F = f V , g V dom 1 st f / b x b 1 st f x 1 st g x x b , y b h dom x 2 nd f y x 2 nd f y h x 2 nd g y h

Detailed syntax breakdown

Step Hyp Ref Expression
0 cprf class ⟨,⟩ F
1 vf setvar f
2 cvv class V
3 vg setvar g
4 c1st class 1 st
5 1 cv setvar f
6 5 4 cfv class 1 st f
7 6 cdm class dom 1 st f
8 vb setvar b
9 vx setvar x
10 8 cv setvar b
11 9 cv setvar x
12 11 6 cfv class 1 st f x
13 3 cv setvar g
14 13 4 cfv class 1 st g
15 11 14 cfv class 1 st g x
16 12 15 cop class 1 st f x 1 st g x
17 9 10 16 cmpt class x b 1 st f x 1 st g x
18 vy setvar y
19 vh setvar h
20 c2nd class 2 nd
21 5 20 cfv class 2 nd f
22 18 cv setvar y
23 11 22 21 co class x 2 nd f y
24 23 cdm class dom x 2 nd f y
25 19 cv setvar h
26 25 23 cfv class x 2 nd f y h
27 13 20 cfv class 2 nd g
28 11 22 27 co class x 2 nd g y
29 25 28 cfv class x 2 nd g y h
30 26 29 cop class x 2 nd f y h x 2 nd g y h
31 19 24 30 cmpt class h dom x 2 nd f y x 2 nd f y h x 2 nd g y h
32 9 18 10 10 31 cmpo class x b , y b h dom x 2 nd f y x 2 nd f y h x 2 nd g y h
33 17 32 cop class x b 1 st f x 1 st g x x b , y b h dom x 2 nd f y x 2 nd f y h x 2 nd g y h
34 8 7 33 csb class dom 1 st f / b x b 1 st f x 1 st g x x b , y b h dom x 2 nd f y x 2 nd f y h x 2 nd g y h
35 1 3 2 2 34 cmpo class f V , g V dom 1 st f / b x b 1 st f x 1 st g x x b , y b h dom x 2 nd f y x 2 nd f y h x 2 nd g y h
36 0 35 wceq wff ⟨,⟩ F = f V , g V dom 1 st f / b x b 1 st f x 1 st g x x b , y b h dom x 2 nd f y x 2 nd f y h x 2 nd g y h