Metamath Proof Explorer


Theorem signstfveq0a

Description: Lemma for signstfveq0 . (Contributed by Thierry Arnoux, 11-Oct-2018)

Ref Expression
Hypotheses signsv.p ˙ = a 1 0 1 , b 1 0 1 if b = 0 a b
signsv.w W = Base ndx 1 0 1 + ndx ˙
signsv.t T = f Word n 0 ..^ f W i = 0 n sgn f i
signsv.v V = f Word j 1 ..^ f if T f j T f j 1 1 0
signstfveq0.1 N = F
Assertion signstfveq0a F Word F 0 0 F N 1 = 0 N 2

Proof

Step Hyp Ref Expression
1 signsv.p ˙ = a 1 0 1 , b 1 0 1 if b = 0 a b
2 signsv.w W = Base ndx 1 0 1 + ndx ˙
3 signsv.t T = f Word n 0 ..^ f W i = 0 n sgn f i
4 signsv.v V = f Word j 1 ..^ f if T f j T f j 1 1 0
5 signstfveq0.1 N = F
6 simpll F Word F 0 0 F N 1 = 0 F Word
7 6 eldifad F Word F 0 0 F N 1 = 0 F Word
8 lencl F Word F 0
9 5 8 eqeltrid F Word N 0
10 7 9 syl F Word F 0 0 F N 1 = 0 N 0
11 eldifsn F Word F Word F
12 6 11 sylib F Word F 0 0 F N 1 = 0 F Word F
13 hasheq0 F Word F = 0 F =
14 13 necon3bid F Word F 0 F
15 14 biimpar F Word F F 0
16 5 neeq1i N 0 F 0
17 15 16 sylibr F Word F N 0
18 12 17 syl F Word F 0 0 F N 1 = 0 N 0
19 elnnne0 N N 0 N 0
20 10 18 19 sylanbrc F Word F 0 0 F N 1 = 0 N
21 simplr F Word F 0 0 F N 1 = 0 F 0 0
22 simpr F Word F 0 0 F N 1 = 0 F N 1 = 0
23 21 22 neeqtrrd F Word F 0 0 F N 1 = 0 F 0 F N 1
24 23 necomd F Word F 0 0 F N 1 = 0 F N 1 F 0
25 oveq1 N = 1 N 1 = 1 1
26 1m1e0 1 1 = 0
27 25 26 eqtrdi N = 1 N 1 = 0
28 27 fveq2d N = 1 F N 1 = F 0
29 28 necon3i F N 1 F 0 N 1
30 24 29 syl F Word F 0 0 F N 1 = 0 N 1
31 eluz2b3 N 2 N N 1
32 20 30 31 sylanbrc F Word F 0 0 F N 1 = 0 N 2