Metamath Proof Explorer


Theorem ringccoALTV

Description: Composition in the category of rings. (Contributed by AV, 14-Feb-2020) (New usage is discouraged.)

Ref Expression
Hypotheses ringcbasALTV.c C = RingCatALTV U
ringcbasALTV.b B = Base C
ringcbasALTV.u φ U V
ringccoALTV.o · ˙ = comp C
ringccoALTV.x φ X B
ringccoALTV.y φ Y B
ringccoALTV.z φ Z B
ringccoALTV.f φ F X RingHom Y
ringccoALTV.g φ G Y RingHom Z
Assertion ringccoALTV φ G X Y · ˙ Z F = G F

Proof

Step Hyp Ref Expression
1 ringcbasALTV.c C = RingCatALTV U
2 ringcbasALTV.b B = Base C
3 ringcbasALTV.u φ U V
4 ringccoALTV.o · ˙ = comp C
5 ringccoALTV.x φ X B
6 ringccoALTV.y φ Y B
7 ringccoALTV.z φ Z B
8 ringccoALTV.f φ F X RingHom Y
9 ringccoALTV.g φ G Y RingHom Z
10 1 2 3 4 ringccofvalALTV φ · ˙ = v B × B , z B g 2 nd v RingHom z , f 1 st v RingHom 2 nd v g f
11 simprl φ v = X Y z = Z v = X Y
12 11 fveq2d φ v = X Y z = Z 2 nd v = 2 nd X Y
13 op2ndg X B Y B 2 nd X Y = Y
14 5 6 13 syl2anc φ 2 nd X Y = Y
15 14 adantr φ v = X Y z = Z 2 nd X Y = Y
16 12 15 eqtrd φ v = X Y z = Z 2 nd v = Y
17 simprr φ v = X Y z = Z z = Z
18 16 17 oveq12d φ v = X Y z = Z 2 nd v RingHom z = Y RingHom Z
19 11 fveq2d φ v = X Y z = Z 1 st v = 1 st X Y
20 op1stg X B Y B 1 st X Y = X
21 5 6 20 syl2anc φ 1 st X Y = X
22 21 adantr φ v = X Y z = Z 1 st X Y = X
23 19 22 eqtrd φ v = X Y z = Z 1 st v = X
24 23 16 oveq12d φ v = X Y z = Z 1 st v RingHom 2 nd v = X RingHom Y
25 eqidd φ v = X Y z = Z g f = g f
26 18 24 25 mpoeq123dv φ v = X Y z = Z g 2 nd v RingHom z , f 1 st v RingHom 2 nd v g f = g Y RingHom Z , f X RingHom Y g f
27 opelxpi X B Y B X Y B × B
28 5 6 27 syl2anc φ X Y B × B
29 ovex Y RingHom Z V
30 ovex X RingHom Y V
31 29 30 mpoex g Y RingHom Z , f X RingHom Y g f V
32 31 a1i φ g Y RingHom Z , f X RingHom Y g f V
33 10 26 28 7 32 ovmpod φ X Y · ˙ Z = g Y RingHom Z , f X RingHom Y g f
34 simprl φ g = G f = F g = G
35 simprr φ g = G f = F f = F
36 34 35 coeq12d φ g = G f = F g f = G F
37 coexg G Y RingHom Z F X RingHom Y G F V
38 9 8 37 syl2anc φ G F V
39 33 36 9 8 38 ovmpod φ G X Y · ˙ Z F = G F