Metamath Proof Explorer


Theorem psgneu

Description: A finitary permutation has exactly one parity. (Contributed by Stefan O'Rear, 28-Aug-2015)

Ref Expression
Hypotheses psgnval.g ⊢ G = SymGrp ⁡ D
psgnval.t ⊢ T = ran ⁡ pmTrsp ⁡ D
psgnval.n ⊢ N = pmSgn ⁡ D
Assertion psgneu ⊢ P ∈ dom ⁡ N → ∃! s ∃ w ∈ Word T P = ∑ G w ∧ s = − 1 w

Proof

Step Hyp Ref Expression
1 psgnval.g ⊢ G = SymGrp ⁡ D
2 psgnval.t ⊢ T = ran ⁡ pmTrsp ⁡ D
3 psgnval.n ⊢ N = pmSgn ⁡ D
4 eqid ⊢ Base G = Base G
5 1 3 4 psgneldm ⊢ P ∈ dom ⁡ N ↔ P ∈ Base G ∧ dom ⁡ P ∖ I ∈ Fin
6 5 simplbi ⊢ P ∈ dom ⁡ N → P ∈ Base G
7 1 4 elbasfv ⊢ P ∈ Base G → D ∈ V
8 6 7 syl ⊢ P ∈ dom ⁡ N → D ∈ V
9 1 2 3 psgneldm2 ⊢ D ∈ V → P ∈ dom ⁡ N ↔ ∃ w ∈ Word T P = ∑ G w
10 8 9 syl ⊢ P ∈ dom ⁡ N → P ∈ dom ⁡ N ↔ ∃ w ∈ Word T P = ∑ G w
11 10 ibi ⊢ P ∈ dom ⁡ N → ∃ w ∈ Word T P = ∑ G w
12 simpr ⊢ P ∈ dom ⁡ N ∧ w ∈ Word T ∧ P = ∑ G w → P = ∑ G w
13 eqid ⊢ − 1 w = − 1 w
14 ovex ⊢ − 1 w ∈ V
15 eqeq1 ⊢ s = − 1 w → s = − 1 w ↔ − 1 w = − 1 w
16 15 anbi2d ⊢ s = − 1 w → P = ∑ G w ∧ s = − 1 w ↔ P = ∑ G w ∧ − 1 w = − 1 w
17 14 16 spcev ⊢ P = ∑ G w ∧ − 1 w = − 1 w → ∃ s P = ∑ G w ∧ s = − 1 w
18 12 13 17 sylancl ⊢ P ∈ dom ⁡ N ∧ w ∈ Word T ∧ P = ∑ G w → ∃ s P = ∑ G w ∧ s = − 1 w
19 18 ex ⊢ P ∈ dom ⁡ N ∧ w ∈ Word T → P = ∑ G w → ∃ s P = ∑ G w ∧ s = − 1 w
20 19 reximdva ⊢ P ∈ dom ⁡ N → ∃ w ∈ Word T P = ∑ G w → ∃ w ∈ Word T ∃ s P = ∑ G w ∧ s = − 1 w
21 11 20 mpd ⊢ P ∈ dom ⁡ N → ∃ w ∈ Word T ∃ s P = ∑ G w ∧ s = − 1 w
22 rexcom4 ⊢ ∃ w ∈ Word T ∃ s P = ∑ G w ∧ s = − 1 w ↔ ∃ s ∃ w ∈ Word T P = ∑ G w ∧ s = − 1 w
23 21 22 sylib ⊢ P ∈ dom ⁡ N → ∃ s ∃ w ∈ Word T P = ∑ G w ∧ s = − 1 w
24 reeanv ⊢ ∃ w ∈ Word T ∃ x ∈ Word T P = ∑ G w ∧ s = − 1 w ∧ P = ∑ G x ∧ t = − 1 x ↔ ∃ w ∈ Word T P = ∑ G w ∧ s = − 1 w ∧ ∃ x ∈ Word T P = ∑ G x ∧ t = − 1 x
25 8 ad2antrr ⊢ P ∈ dom ⁡ N ∧ w ∈ Word T ∧ x ∈ Word T ∧ P = ∑ G w ∧ s = − 1 w ∧ P = ∑ G x ∧ t = − 1 x → D ∈ V
26 simplrl ⊢ P ∈ dom ⁡ N ∧ w ∈ Word T ∧ x ∈ Word T ∧ P = ∑ G w ∧ s = − 1 w ∧ P = ∑ G x ∧ t = − 1 x → w ∈ Word T
27 simplrr ⊢ P ∈ dom ⁡ N ∧ w ∈ Word T ∧ x ∈ Word T ∧ P = ∑ G w ∧ s = − 1 w ∧ P = ∑ G x ∧ t = − 1 x → x ∈ Word T
28 simprll ⊢ P ∈ dom ⁡ N ∧ w ∈ Word T ∧ x ∈ Word T ∧ P = ∑ G w ∧ s = − 1 w ∧ P = ∑ G x ∧ t = − 1 x → P = ∑ G w
29 simprrl ⊢ P ∈ dom ⁡ N ∧ w ∈ Word T ∧ x ∈ Word T ∧ P = ∑ G w ∧ s = − 1 w ∧ P = ∑ G x ∧ t = − 1 x → P = ∑ G x
30 28 29 eqtr3d ⊢ P ∈ dom ⁡ N ∧ w ∈ Word T ∧ x ∈ Word T ∧ P = ∑ G w ∧ s = − 1 w ∧ P = ∑ G x ∧ t = − 1 x → ∑ G w = ∑ G x
31 1 2 25 26 27 30 psgnuni ⊢ P ∈ dom ⁡ N ∧ w ∈ Word T ∧ x ∈ Word T ∧ P = ∑ G w ∧ s = − 1 w ∧ P = ∑ G x ∧ t = − 1 x → − 1 w = − 1 x
32 simprlr ⊢ P ∈ dom ⁡ N ∧ w ∈ Word T ∧ x ∈ Word T ∧ P = ∑ G w ∧ s = − 1 w ∧ P = ∑ G x ∧ t = − 1 x → s = − 1 w
33 simprrr ⊢ P ∈ dom ⁡ N ∧ w ∈ Word T ∧ x ∈ Word T ∧ P = ∑ G w ∧ s = − 1 w ∧ P = ∑ G x ∧ t = − 1 x → t = − 1 x
34 31 32 33 3eqtr4d ⊢ P ∈ dom ⁡ N ∧ w ∈ Word T ∧ x ∈ Word T ∧ P = ∑ G w ∧ s = − 1 w ∧ P = ∑ G x ∧ t = − 1 x → s = t
35 34 ex ⊢ P ∈ dom ⁡ N ∧ w ∈ Word T ∧ x ∈ Word T → P = ∑ G w ∧ s = − 1 w ∧ P = ∑ G x ∧ t = − 1 x → s = t
36 35 rexlimdvva ⊢ P ∈ dom ⁡ N → ∃ w ∈ Word T ∃ x ∈ Word T P = ∑ G w ∧ s = − 1 w ∧ P = ∑ G x ∧ t = − 1 x → s = t
37 24 36 biimtrrid ⊢ P ∈ dom ⁡ N → ∃ w ∈ Word T P = ∑ G w ∧ s = − 1 w ∧ ∃ x ∈ Word T P = ∑ G x ∧ t = − 1 x → s = t
38 37 alrimivv ⊢ P ∈ dom ⁡ N → ∀ s ∀ t ∃ w ∈ Word T P = ∑ G w ∧ s = − 1 w ∧ ∃ x ∈ Word T P = ∑ G x ∧ t = − 1 x → s = t
39 eqeq1 ⊢ s = t → s = − 1 w ↔ t = − 1 w
40 39 anbi2d ⊢ s = t → P = ∑ G w ∧ s = − 1 w ↔ P = ∑ G w ∧ t = − 1 w
41 40 rexbidv ⊢ s = t → ∃ w ∈ Word T P = ∑ G w ∧ s = − 1 w ↔ ∃ w ∈ Word T P = ∑ G w ∧ t = − 1 w
42 oveq2 ⊢ w = x → ∑ G w = ∑ G x
43 42 eqeq2d ⊢ w = x → P = ∑ G w ↔ P = ∑ G x
44 fveq2 ⊢ w = x → w = x
45 44 oveq2d ⊢ w = x → − 1 w = − 1 x
46 45 eqeq2d ⊢ w = x → t = − 1 w ↔ t = − 1 x
47 43 46 anbi12d ⊢ w = x → P = ∑ G w ∧ t = − 1 w ↔ P = ∑ G x ∧ t = − 1 x
48 47 cbvrexvw ⊢ ∃ w ∈ Word T P = ∑ G w ∧ t = − 1 w ↔ ∃ x ∈ Word T P = ∑ G x ∧ t = − 1 x
49 41 48 bitrdi ⊢ s = t → ∃ w ∈ Word T P = ∑ G w ∧ s = − 1 w ↔ ∃ x ∈ Word T P = ∑ G x ∧ t = − 1 x
50 49 eu4 ⊢ ∃! s ∃ w ∈ Word T P = ∑ G w ∧ s = − 1 w ↔ ∃ s ∃ w ∈ Word T P = ∑ G w ∧ s = − 1 w ∧ ∀ s ∀ t ∃ w ∈ Word T P = ∑ G w ∧ s = − 1 w ∧ ∃ x ∈ Word T P = ∑ G x ∧ t = − 1 x → s = t
51 23 38 50 sylanbrc ⊢ P ∈ dom ⁡ N → ∃! s ∃ w ∈ Word T P = ∑ G w ∧ s = − 1 w