Metamath Proof Explorer


Theorem precofval2

Description: Value of the pre-composition functor as a transposed curry of the functor composition bifunctor. (Contributed by Zhi Wang, 11-Oct-2025)

Ref Expression
Hypotheses precofval.q Q = C FuncCat D
precofval.r R = D FuncCat E
precofval.o No typesetting found for |- ( ph -> .o. = ( <. Q , R >. curryF ( ( <. C , D >. o.F E ) o.func ( Q swapF R ) ) ) ) with typecode |-
precofval.f φ F C Func D
precofval.e φ E Cat
precofval.k No typesetting found for |- ( ph -> K = ( ( 1st ` .o. ) ` F ) ) with typecode |-
Assertion precofval2 φ K = g D Func E g func F g D Func E , h D Func E a g D Nat E h a 1 st F

Proof

Step Hyp Ref Expression
1 precofval.q Q = C FuncCat D
2 precofval.r R = D FuncCat E
3 precofval.o Could not format ( ph -> .o. = ( <. Q , R >. curryF ( ( <. C , D >. o.F E ) o.func ( Q swapF R ) ) ) ) : No typesetting found for |- ( ph -> .o. = ( <. Q , R >. curryF ( ( <. C , D >. o.F E ) o.func ( Q swapF R ) ) ) ) with typecode |-
4 precofval.f φ F C Func D
5 precofval.e φ E Cat
6 precofval.k Could not format ( ph -> K = ( ( 1st ` .o. ) ` F ) ) : No typesetting found for |- ( ph -> K = ( ( 1st ` .o. ) ` F ) ) with typecode |-
7 1 2 3 4 5 6 precofval φ K = g D Func E g func F g D Func E , h D Func E a g D Nat E h x Base C a 1 st F x
8 eqid D Nat E = D Nat E
9 id a g D Nat E h a g D Nat E h
10 8 9 nat1st2nd a g D Nat E h a 1 st g 2 nd g D Nat E 1 st h 2 nd h
11 eqid Base D = Base D
12 8 10 11 natfn a g D Nat E h a Fn Base D
13 dffn2 a Fn Base D a : Base D V
14 12 13 sylib a g D Nat E h a : Base D V
15 eqid Base C = Base C
16 relfunc Rel C Func D
17 1st2ndbr Rel C Func D F C Func D 1 st F C Func D 2 nd F
18 16 4 17 sylancr φ 1 st F C Func D 2 nd F
19 15 11 18 funcf1 φ 1 st F : Base C Base D
20 fcompt a : Base D V 1 st F : Base C Base D a 1 st F = x Base C a 1 st F x
21 14 19 20 syl2anr φ a g D Nat E h a 1 st F = x Base C a 1 st F x
22 21 mpteq2dva φ a g D Nat E h a 1 st F = a g D Nat E h x Base C a 1 st F x
23 22 mpoeq3dv φ g D Func E , h D Func E a g D Nat E h a 1 st F = g D Func E , h D Func E a g D Nat E h x Base C a 1 st F x
24 23 opeq2d φ g D Func E g func F g D Func E , h D Func E a g D Nat E h a 1 st F = g D Func E g func F g D Func E , h D Func E a g D Nat E h x Base C a 1 st F x
25 7 24 eqtr4d φ K = g D Func E g func F g D Func E , h D Func E a g D Nat E h a 1 st F