Metamath Proof Explorer


Theorem psgn0fv0

Description: The permutation sign function for an empty set at an empty set is 1. (Contributed by AV, 27-Feb-2019)

Ref Expression
Assertion psgn0fv0 pmSgn = 1

Proof

Step Hyp Ref Expression
1 0ex V
2 wrd0 Word ran pmTrsp
3 eqid 0 SymGrp = 0 SymGrp
4 3 gsum0 SymGrp = 0 SymGrp
5 eqid SymGrp = SymGrp
6 5 symgid V I = 0 SymGrp
7 1 6 ax-mp I = 0 SymGrp
8 res0 I =
9 7 8 eqtr3i 0 SymGrp =
10 9 a1i V Word ran pmTrsp 0 SymGrp =
11 4 10 eqtr2id V Word ran pmTrsp = SymGrp
12 11 fveq2d V Word ran pmTrsp pmSgn = pmSgn SymGrp
13 eqid ran pmTrsp = ran pmTrsp
14 eqid pmSgn = pmSgn
15 5 13 14 psgnvalii V Word ran pmTrsp pmSgn SymGrp = 1
16 hash0 = 0
17 16 oveq2i 1 = 1 0
18 neg1cn 1
19 exp0 1 1 0 = 1
20 18 19 ax-mp 1 0 = 1
21 17 20 eqtri 1 = 1
22 21 a1i V Word ran pmTrsp 1 = 1
23 12 15 22 3eqtrd V Word ran pmTrsp pmSgn = 1
24 1 2 23 mp2an pmSgn = 1