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 ∈ ℕ
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 ∈ ℕ → P = 1 1 2 2 1 2 2 1
13 10 12 eleqtrrid ⊢ 1 ∈ V ∧ 2 ∈ ℕ → 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 ˙