Metamath Proof Explorer


Theorem psgnid

Description: Permutation sign of the identity. (Contributed by Thierry Arnoux, 21-Aug-2020)

Ref Expression
Hypothesis psgnid.s ⊢ S = pmSgn ⁡ D
Assertion psgnid ⊢ D ∈ Fin → S ⁡ I ↾ D = 1

Proof

Step Hyp Ref Expression
1 psgnid.s ⊢ S = pmSgn ⁡ D
2 eqid ⊢ SymGrp ⁡ D = SymGrp ⁡ D
3 2 symgid ⊢ D ∈ Fin → I ↾ D = 0 SymGrp ⁡ D
4 3 fveq2d ⊢ D ∈ Fin → S ⁡ I ↾ D = S ⁡ 0 SymGrp ⁡ D
5 eqid ⊢ mulGrp ℂ fld ↾ 𝑠 1 − 1 = mulGrp ℂ fld ↾ 𝑠 1 − 1
6 2 1 5 psgnghm2 ⊢ D ∈ Fin → S ∈ SymGrp ⁡ D GrpHom mulGrp ℂ fld ↾ 𝑠 1 − 1
7 eqid ⊢ 0 SymGrp ⁡ D = 0 SymGrp ⁡ D
8 cnring ⊢ ℂ fld ∈ Ring
9 eqid ⊢ mulGrp ℂ fld = mulGrp ℂ fld
10 9 ringmgp ⊢ ℂ fld ∈ Ring → mulGrp ℂ fld ∈ Mnd
11 8 10 ax-mp ⊢ mulGrp ℂ fld ∈ Mnd
12 1ex ⊢ 1 ∈ V
13 12 prid1 ⊢ 1 ∈ 1 − 1
14 ax-1cn ⊢ 1 ∈ ℂ
15 neg1cn ⊢ − 1 ∈ ℂ
16 prssi ⊢ 1 ∈ ℂ ∧ − 1 ∈ ℂ → 1 − 1 ⊆ ℂ
17 14 15 16 mp2an ⊢ 1 − 1 ⊆ ℂ
18 cnfldbas ⊢ ℂ = Base ℂ fld
19 9 18 mgpbas ⊢ ℂ = Base mulGrp ℂ fld
20 cnfld1 ⊢ 1 = 1 ℂ fld
21 9 20 ringidval ⊢ 1 = 0 mulGrp ℂ fld
22 5 19 21 ress0g ⊢ mulGrp ℂ fld ∈ Mnd ∧ 1 ∈ 1 − 1 ∧ 1 − 1 ⊆ ℂ → 1 = 0 mulGrp ℂ fld ↾ 𝑠 1 − 1
23 11 13 17 22 mp3an ⊢ 1 = 0 mulGrp ℂ fld ↾ 𝑠 1 − 1
24 7 23 ghmid ⊢ S ∈ SymGrp ⁡ D GrpHom mulGrp ℂ fld ↾ 𝑠 1 − 1 → S ⁡ 0 SymGrp ⁡ D = 1
25 6 24 syl ⊢ D ∈ Fin → S ⁡ 0 SymGrp ⁡ D = 1
26 4 25 eqtrd ⊢ D ∈ Fin → S ⁡ I ↾ D = 1