Metamath Proof Explorer


Theorem sxbrsigalem5

Description: First direction for sxbrsiga . (Contributed by Thierry Arnoux, 22-Sep-2017) (Revised by Thierry Arnoux, 11-Oct-2017)

Ref Expression
Hypotheses sxbrsiga.0 J = topGen ran .
dya2ioc.1 I = x , n x 2 n x + 1 2 n
dya2ioc.2 R = u ran I , v ran I u × v
Assertion sxbrsigalem5 𝛔 J × t J 𝔅 × s 𝔅

Proof

Step Hyp Ref Expression
1 sxbrsiga.0 J = topGen ran .
2 dya2ioc.1 I = x , n x 2 n x + 1 2 n
3 dya2ioc.2 R = u ran I , v ran I u × v
4 1 2 3 dya2iocucvr ran R = 2
5 br2base ran e 𝔅 , f 𝔅 e × f = 2
6 4 5 eqtr4i ran R = ran e 𝔅 , f 𝔅 e × f
7 brsigarn 𝔅 sigAlgebra
8 7 elexi 𝔅 V
9 8 8 mpoex e 𝔅 , f 𝔅 e × f V
10 9 rnex ran e 𝔅 , f 𝔅 e × f V
11 1 2 dya2icobrsiga ran I 𝔅
12 11 sseli u ran I u 𝔅
13 11 sseli v ran I v 𝔅
14 12 13 anim12i u ran I v ran I u 𝔅 v 𝔅
15 14 anim1i u ran I v ran I g = u × v u 𝔅 v 𝔅 g = u × v
16 15 ssoprab2i u v g | u ran I v ran I g = u × v u v g | u 𝔅 v 𝔅 g = u × v
17 df-mpo u ran I , v ran I u × v = u v g | u ran I v ran I g = u × v
18 3 17 eqtri R = u v g | u ran I v ran I g = u × v
19 xpeq1 e = u e × f = u × f
20 xpeq2 f = v u × f = u × v
21 19 20 cbvmpov e 𝔅 , f 𝔅 e × f = u 𝔅 , v 𝔅 u × v
22 df-mpo u 𝔅 , v 𝔅 u × v = u v g | u 𝔅 v 𝔅 g = u × v
23 21 22 eqtri e 𝔅 , f 𝔅 e × f = u v g | u 𝔅 v 𝔅 g = u × v
24 16 18 23 3sstr4i R e 𝔅 , f 𝔅 e × f
25 rnss R e 𝔅 , f 𝔅 e × f ran R ran e 𝔅 , f 𝔅 e × f
26 24 25 ax-mp ran R ran e 𝔅 , f 𝔅 e × f
27 sssigagen2 ran e 𝔅 , f 𝔅 e × f V ran R ran e 𝔅 , f 𝔅 e × f ran R 𝛔 ran e 𝔅 , f 𝔅 e × f
28 10 26 27 mp2an ran R 𝛔 ran e 𝔅 , f 𝔅 e × f
29 sigagenss2 ran R = ran e 𝔅 , f 𝔅 e × f ran R 𝛔 ran e 𝔅 , f 𝔅 e × f ran e 𝔅 , f 𝔅 e × f V 𝛔 ran R 𝛔 ran e 𝔅 , f 𝔅 e × f
30 6 28 10 29 mp3an 𝛔 ran R 𝛔 ran e 𝔅 , f 𝔅 e × f
31 1 2 3 sxbrsigalem4 𝛔 J × t J = 𝛔 ran R
32 eqid ran e 𝔅 , f 𝔅 e × f = ran e 𝔅 , f 𝔅 e × f
33 32 sxval 𝔅 sigAlgebra 𝔅 sigAlgebra 𝔅 × s 𝔅 = 𝛔 ran e 𝔅 , f 𝔅 e × f
34 7 7 33 mp2an 𝔅 × s 𝔅 = 𝛔 ran e 𝔅 , f 𝔅 e × f
35 30 31 34 3sstr4i 𝛔 J × t J 𝔅 × s 𝔅