Metamath Proof Explorer


Theorem m2detleiblem1

Description: Lemma 1 for m2detleib . (Contributed by AV, 12-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 m2detleiblem1 RRingQPYSQ=pmSgnNQR1˙

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 elpri Q11221221Q=1122Q=1221
7 fveq2 Q=1122SQ=S1122
8 eqid SymGrpN=SymGrpN
9 eqid ranpmTrspN=ranpmTrspN
10 1 8 2 9 4 psgnprfval1 S1122=1
11 7 10 eqtrdi Q=1122SQ=1
12 1z 1
13 11 12 eqeltrdi Q=1122SQ
14 fveq2 Q=1221SQ=S1221
15 1 8 2 9 4 psgnprfval2 S1221=1
16 14 15 eqtrdi Q=1221SQ=1
17 neg1z 1
18 16 17 eqeltrdi Q=1221SQ
19 13 18 jaoi Q=1122Q=1221SQ
20 6 19 syl Q11221221SQ
21 1ex 1V
22 2nn 2
23 8 2 1 symg2bas 1V2P=11221221
24 21 22 23 mp2an P=11221221
25 20 24 eleq2s QPSQ
26 eqid R=R
27 3 26 5 zrhmulg RRingSQYSQ=SQR1˙
28 25 27 sylan2 RRingQPYSQ=SQR1˙
29 4 a1i RRingQPS=pmSgnN
30 29 fveq1d RRingQPSQ=pmSgnNQ
31 30 oveq1d RRingQPSQR1˙=pmSgnNQR1˙
32 28 31 eqtrd RRingQPYSQ=pmSgnNQR1˙