Metamath Proof Explorer


Theorem cchhllem

Description: Lemma for chlbas and chlvsca . (Contributed by Thierry Arnoux, 15-Apr-2019)

Ref Expression
Hypotheses cchhl.c C = subringAlg fld sSet 𝑖 ndx x , y x y
cchhllem.2 E = Slot N
cchhllem.3 N
cchhllem.4 N < 5 8 < N
Assertion cchhllem E fld = E C

Proof

Step Hyp Ref Expression
1 cchhl.c C = subringAlg fld sSet 𝑖 ndx x , y x y
2 cchhllem.2 E = Slot N
3 cchhllem.3 N
4 cchhllem.4 N < 5 8 < N
5 2 3 ndxid E = Slot E ndx
6 5lt8 5 < 8
7 3 nnrei N
8 5re 5
9 8re 8
10 7 8 9 lttri N < 5 5 < 8 N < 8
11 6 10 mpan2 N < 5 N < 8
12 7 9 ltnei N < 8 8 N
13 11 12 syl N < 5 8 N
14 13 necomd N < 5 N 8
15 9 7 ltnei 8 < N N 8
16 14 15 jaoi N < 5 8 < N N 8
17 4 16 ax-mp N 8
18 2 3 ndxarg E ndx = N
19 ipndx 𝑖 ndx = 8
20 18 19 neeq12i E ndx 𝑖 ndx N 8
21 17 20 mpbir E ndx 𝑖 ndx
22 5 21 setsnid E subringAlg fld = E subringAlg fld sSet 𝑖 ndx x , y x y
23 eqidd subringAlg fld = subringAlg fld
24 ax-resscn
25 cnfldbas = Base fld
26 24 25 sseqtri Base fld
27 26 a1i Base fld
28 23 27 2 3 4 sralem E fld = E subringAlg fld
29 28 mptru E fld = E subringAlg fld
30 1 fveq2i E C = E subringAlg fld sSet 𝑖 ndx x , y x y
31 22 29 30 3eqtr4i E fld = E C