Metamath Proof Explorer


Theorem m2detleiblem5

Description: Lemma 5 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
Assertion m2detleiblem5 R Ring Q = 1 1 2 2 Y S Q = 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 1ex 1 V
7 2nn 2
8 prex 1 1 2 2 V
9 8 prid1 1 1 2 2 1 1 2 2 1 2 2 1
10 eqid SymGrp N = SymGrp N
11 10 2 1 symg2bas 1 V 2 P = 1 1 2 2 1 2 2 1
12 9 11 eleqtrrid 1 V 2 1 1 2 2 P
13 6 7 12 mp2an 1 1 2 2 P
14 eleq1 Q = 1 1 2 2 Q P 1 1 2 2 P
15 13 14 mpbiri Q = 1 1 2 2 Q P
16 1 2 3 4 5 m2detleiblem1 R Ring Q P Y S Q = pmSgn N Q R 1 ˙
17 15 16 sylan2 R Ring Q = 1 1 2 2 Y S Q = pmSgn N Q R 1 ˙
18 fveq2 Q = 1 1 2 2 pmSgn N Q = pmSgn N 1 1 2 2
19 18 adantl R Ring Q = 1 1 2 2 pmSgn N Q = pmSgn N 1 1 2 2
20 eqid ran pmTrsp N = ran pmTrsp N
21 eqid pmSgn N = pmSgn N
22 1 10 2 20 21 psgnprfval1 pmSgn N 1 1 2 2 = 1
23 19 22 eqtrdi R Ring Q = 1 1 2 2 pmSgn N Q = 1
24 23 oveq1d R Ring Q = 1 1 2 2 pmSgn N Q R 1 ˙ = 1 R 1 ˙
25 eqid Base R = Base R
26 25 5 ringidcl R Ring 1 ˙ Base R
27 26 adantr R Ring Q = 1 1 2 2 1 ˙ Base R
28 eqid R = R
29 25 28 mulg1 1 ˙ Base R 1 R 1 ˙ = 1 ˙
30 27 29 syl R Ring Q = 1 1 2 2 1 R 1 ˙ = 1 ˙
31 17 24 30 3eqtrd R Ring Q = 1 1 2 2 Y S Q = 1 ˙