Metamath Proof Explorer


Theorem m2detleiblem5

Description: Lemma 5 for m2detleib . (Contributed by AV, 20-Dec-2018)

Ref Expression
Hypotheses m2detleiblem1.n N=12
m2detleiblem1.p P=BaseSymGrpN
m2detleiblem1.y Y=ℤRHomR
m2detleiblem1.s S=pmSgnN
m2detleiblem1.o 1˙=1R
Assertion m2detleiblem5 RRingQ=1122YSQ=1˙

Proof

Step Hyp Ref Expression
1 m2detleiblem1.n N=12
2 m2detleiblem1.p P=BaseSymGrpN
3 m2detleiblem1.y Y=ℤRHomR
4 m2detleiblem1.s S=pmSgnN
5 m2detleiblem1.o 1˙=1R
6 1ex 1V
7 2nn 2
8 prex 1122V
9 8 prid1 112211221221
10 eqid SymGrpN=SymGrpN
11 10 2 1 symg2bas 1V2P=11221221
12 9 11 eleqtrrid 1V21122P
13 6 7 12 mp2an 1122P
14 eleq1 Q=1122QP1122P
15 13 14 mpbiri Q=1122QP
16 1 2 3 4 5 m2detleiblem1 RRingQPYSQ=pmSgnNQR1˙
17 15 16 sylan2 RRingQ=1122YSQ=pmSgnNQR1˙
18 fveq2 Q=1122pmSgnNQ=pmSgnN1122
19 18 adantl RRingQ=1122pmSgnNQ=pmSgnN1122
20 eqid ranpmTrspN=ranpmTrspN
21 eqid pmSgnN=pmSgnN
22 1 10 2 20 21 psgnprfval1 pmSgnN1122=1
23 19 22 eqtrdi RRingQ=1122pmSgnNQ=1
24 23 oveq1d RRingQ=1122pmSgnNQR1˙=1R1˙
25 eqid BaseR=BaseR
26 25 5 ringidcl RRing1˙BaseR
27 26 adantr RRingQ=11221˙BaseR
28 eqid R=R
29 25 28 mulg1 1˙BaseR1R1˙=1˙
30 27 29 syl RRingQ=11221R1˙=1˙
31 17 24 30 3eqtrd RRingQ=1122YSQ=1˙