Metamath Proof Explorer


Theorem m2detleiblem6

Description: Lemma 6 for m2detleib . (Contributed by AV, 20-Dec-2018)

Ref Expression
Hypotheses m2detleiblem1.n N = 1 2
m2detleiblem1.p P = Base SymGrp N
m2detleiblem1.y Y = ℤRHom R
m2detleiblem1.s S = pmSgn N
m2detleiblem1.o 1 ˙ = 1 R
m2detleiblem1.i I = inv g R
Assertion m2detleiblem6 R Ring Q = 1 2 2 1 Y S Q = I 1 ˙

Proof

Step Hyp Ref Expression
1 m2detleiblem1.n N = 1 2
2 m2detleiblem1.p P = Base SymGrp N
3 m2detleiblem1.y Y = ℤRHom R
4 m2detleiblem1.s S = pmSgn N
5 m2detleiblem1.o 1 ˙ = 1 R
6 m2detleiblem1.i I = inv g R
7 1ex 1 V
8 2nn 2
9 prex 1 2 2 1 V
10 9 prid2 1 2 2 1 1 1 2 2 1 2 2 1
11 eqid SymGrp N = SymGrp N
12 11 2 1 symg2bas 1 V 2 P = 1 1 2 2 1 2 2 1
13 10 12 eleqtrrid 1 V 2 1 2 2 1 P
14 7 8 13 mp2an 1 2 2 1 P
15 eleq1 Q = 1 2 2 1 Q P 1 2 2 1 P
16 14 15 mpbiri Q = 1 2 2 1 Q P
17 1 2 3 4 5 m2detleiblem1 R Ring Q P Y S Q = pmSgn N Q R 1 ˙
18 16 17 sylan2 R Ring Q = 1 2 2 1 Y S Q = pmSgn N Q R 1 ˙
19 fveq2 Q = 1 2 2 1 pmSgn N Q = pmSgn N 1 2 2 1
20 19 adantl R Ring Q = 1 2 2 1 pmSgn N Q = pmSgn N 1 2 2 1
21 eqid ran pmTrsp N = ran pmTrsp N
22 eqid pmSgn N = pmSgn N
23 1 11 2 21 22 psgnprfval2 pmSgn N 1 2 2 1 = 1
24 20 23 eqtrdi R Ring Q = 1 2 2 1 pmSgn N Q = 1
25 24 oveq1d R Ring Q = 1 2 2 1 pmSgn N Q R 1 ˙ = -1 R 1 ˙
26 ringgrp R Ring R Grp
27 eqid Base R = Base R
28 27 5 ringidcl R Ring 1 ˙ Base R
29 eqid R = R
30 27 29 6 mulgm1 R Grp 1 ˙ Base R -1 R 1 ˙ = I 1 ˙
31 26 28 30 syl2anc R Ring -1 R 1 ˙ = I 1 ˙
32 31 adantr R Ring Q = 1 2 2 1 -1 R 1 ˙ = I 1 ˙
33 18 25 32 3eqtrd R Ring Q = 1 2 2 1 Y S Q = I 1 ˙