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 ∈ ℕ
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 ∈ ℕ → P = 1 1 2 2 1 2 2 1
12 9 11 eleqtrrid ⊢ 1 ∈ V ∧ 2 ∈ ℕ → 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 ˙