Metamath Proof Explorer


Theorem isopropdlem

Description: Lemma for isopropd . (Contributed by Zhi Wang, 27-Oct-2025)

Ref Expression
Hypotheses sectpropd.1 φ Hom 𝑓 C = Hom 𝑓 D
sectpropd.2 φ comp 𝑓 C = comp 𝑓 D
Assertion isopropdlem φ P Iso C P Iso D

Proof

Step Hyp Ref Expression
1 sectpropd.1 φ Hom 𝑓 C = Hom 𝑓 D
2 sectpropd.2 φ comp 𝑓 C = comp 𝑓 D
3 simpr φ P Iso C P Iso C
4 eqid Base C = Base C
5 eqid Inv C = Inv C
6 df-iso Iso = c Cat x V dom x Inv c
7 6 mptrcl P Iso C C Cat
8 7 adantl φ P Iso C C Cat
9 eqid Iso C = Iso C
10 4 5 8 9 isofval2 φ P Iso C Iso C = x Base C , y Base C dom x Inv C y
11 df-mpo x Base C , y Base C dom x Inv C y = x y z | x Base C y Base C z = dom x Inv C y
12 10 11 eqtrdi φ P Iso C Iso C = x y z | x Base C y Base C z = dom x Inv C y
13 3 12 eleqtrd φ P Iso C P x y z | x Base C y Base C z = dom x Inv C y
14 eloprab1st2nd P x y z | x Base C y Base C z = dom x Inv C y P = 1 st 1 st P 2 nd 1 st P 2 nd P
15 13 14 syl φ P Iso C P = 1 st 1 st P 2 nd 1 st P 2 nd P
16 1 adantr φ P Iso C Hom 𝑓 C = Hom 𝑓 D
17 2 adantr φ P Iso C comp 𝑓 C = comp 𝑓 D
18 16 17 invpropd φ P Iso C Inv C = Inv D
19 18 oveqd φ P Iso C 1 st 1 st P Inv C 2 nd 1 st P = 1 st 1 st P Inv D 2 nd 1 st P
20 19 dmeqd φ P Iso C dom 1 st 1 st P Inv C 2 nd 1 st P = dom 1 st 1 st P Inv D 2 nd 1 st P
21 eleq1 x = 1 st 1 st P x Base C 1 st 1 st P Base C
22 21 anbi1d x = 1 st 1 st P x Base C y Base C 1 st 1 st P Base C y Base C
23 oveq1 x = 1 st 1 st P x Inv C y = 1 st 1 st P Inv C y
24 23 dmeqd x = 1 st 1 st P dom x Inv C y = dom 1 st 1 st P Inv C y
25 24 eqeq2d x = 1 st 1 st P z = dom x Inv C y z = dom 1 st 1 st P Inv C y
26 22 25 anbi12d x = 1 st 1 st P x Base C y Base C z = dom x Inv C y 1 st 1 st P Base C y Base C z = dom 1 st 1 st P Inv C y
27 eleq1 y = 2 nd 1 st P y Base C 2 nd 1 st P Base C
28 27 anbi2d y = 2 nd 1 st P 1 st 1 st P Base C y Base C 1 st 1 st P Base C 2 nd 1 st P Base C
29 oveq2 y = 2 nd 1 st P 1 st 1 st P Inv C y = 1 st 1 st P Inv C 2 nd 1 st P
30 29 dmeqd y = 2 nd 1 st P dom 1 st 1 st P Inv C y = dom 1 st 1 st P Inv C 2 nd 1 st P
31 30 eqeq2d y = 2 nd 1 st P z = dom 1 st 1 st P Inv C y z = dom 1 st 1 st P Inv C 2 nd 1 st P
32 28 31 anbi12d y = 2 nd 1 st P 1 st 1 st P Base C y Base C z = dom 1 st 1 st P Inv C y 1 st 1 st P Base C 2 nd 1 st P Base C z = dom 1 st 1 st P Inv C 2 nd 1 st P
33 eqeq1 z = 2 nd P z = dom 1 st 1 st P Inv C 2 nd 1 st P 2 nd P = dom 1 st 1 st P Inv C 2 nd 1 st P
34 33 anbi2d z = 2 nd P 1 st 1 st P Base C 2 nd 1 st P Base C z = dom 1 st 1 st P Inv C 2 nd 1 st P 1 st 1 st P Base C 2 nd 1 st P Base C 2 nd P = dom 1 st 1 st P Inv C 2 nd 1 st P
35 26 32 34 eloprabi P x y z | x Base C y Base C z = dom x Inv C y 1 st 1 st P Base C 2 nd 1 st P Base C 2 nd P = dom 1 st 1 st P Inv C 2 nd 1 st P
36 13 35 syl φ P Iso C 1 st 1 st P Base C 2 nd 1 st P Base C 2 nd P = dom 1 st 1 st P Inv C 2 nd 1 st P
37 36 simprd φ P Iso C 2 nd P = dom 1 st 1 st P Inv C 2 nd 1 st P
38 eqid Base D = Base D
39 eqid Inv D = Inv D
40 36 simplld φ P Iso C 1 st 1 st P Base C
41 16 homfeqbas φ P Iso C Base C = Base D
42 40 41 eleqtrd φ P Iso C 1 st 1 st P Base D
43 42 elfvexd φ P Iso C D V
44 16 17 8 43 catpropd φ P Iso C C Cat D Cat
45 8 44 mpbid φ P Iso C D Cat
46 36 simplrd φ P Iso C 2 nd 1 st P Base C
47 46 41 eleqtrd φ P Iso C 2 nd 1 st P Base D
48 eqid Iso D = Iso D
49 38 39 45 42 47 48 isoval φ P Iso C 1 st 1 st P Iso D 2 nd 1 st P = dom 1 st 1 st P Inv D 2 nd 1 st P
50 20 37 49 3eqtr4rd φ P Iso C 1 st 1 st P Iso D 2 nd 1 st P = 2 nd P
51 isofn D Cat Iso D Fn Base D × Base D
52 45 51 syl φ P Iso C Iso D Fn Base D × Base D
53 fnbrovb Iso D Fn Base D × Base D 1 st 1 st P Base D 2 nd 1 st P Base D 1 st 1 st P Iso D 2 nd 1 st P = 2 nd P 1 st 1 st P 2 nd 1 st P Iso D 2 nd P
54 52 42 47 53 syl12anc φ P Iso C 1 st 1 st P Iso D 2 nd 1 st P = 2 nd P 1 st 1 st P 2 nd 1 st P Iso D 2 nd P
55 50 54 mpbid φ P Iso C 1 st 1 st P 2 nd 1 st P Iso D 2 nd P
56 df-br 1 st 1 st P 2 nd 1 st P Iso D 2 nd P 1 st 1 st P 2 nd 1 st P 2 nd P Iso D
57 55 56 sylib φ P Iso C 1 st 1 st P 2 nd 1 st P 2 nd P Iso D
58 15 57 eqeltrd φ P Iso C P Iso D