Metamath Proof Explorer


Theorem swrd0

Description: A subword of an empty set is always the empty set. (Contributed by AV, 31-Mar-2018) (Revised by AV, 20-Oct-2018) (Proof shortened by AV, 2-May-2020)

Ref Expression
Assertion swrd0 substrFL=

Proof

Step Hyp Ref Expression
1 opelxp FLV××VFL×
2 opelxp FL×FL
3 swrdval VFLsubstrFL=ifF..^Ldomx0..^LFx+F
4 fzonlt0 FL¬F<LF..^L=
5 4 biimprd FLF..^L=¬F<L
6 5 con2d FLF<L¬F..^L=
7 6 impcom F<LFL¬F..^L=
8 ss0 F..^LF..^L=
9 7 8 nsyl F<LFL¬F..^L
10 dm0 dom=
11 10 a1i F<LFLdom=
12 11 sseq2d F<LFLF..^LdomF..^L
13 9 12 mtbird F<LFL¬F..^Ldom
14 13 iffalsed F<LFLifF..^Ldomx0..^LFx+F=
15 ssidd ¬F<LFL
16 4 biimpac ¬F<LFLF..^L=
17 10 a1i ¬F<LFLdom=
18 15 16 17 3sstr4d ¬F<LFLF..^Ldom
19 18 iftrued ¬F<LFLifF..^Ldomx0..^LFx+F=x0..^LFx+F
20 zre LL
21 zre FF
22 lenlt LFLF¬F<L
23 22 bicomd LF¬F<LLF
24 20 21 23 syl2anr FL¬F<LLF
25 fzo0n FLLF0..^LF=
26 24 25 bitrd FL¬F<L0..^LF=
27 26 biimpac ¬F<LFL0..^LF=
28 27 mpteq1d ¬F<LFLx0..^LFx+F=xx+F
29 28 dmeqd ¬F<LFLdomx0..^LFx+F=domxx+F
30 ral0 xx+FV
31 dmmptg xx+FVdomxx+F=
32 30 31 mp1i ¬F<LFLdomxx+F=
33 29 32 eqtrd ¬F<LFLdomx0..^LFx+F=
34 mptrel Relx0..^LFx+F
35 reldm0 Relx0..^LFx+Fx0..^LFx+F=domx0..^LFx+F=
36 34 35 mp1i ¬F<LFLx0..^LFx+F=domx0..^LFx+F=
37 33 36 mpbird ¬F<LFLx0..^LFx+F=
38 19 37 eqtrd ¬F<LFLifF..^Ldomx0..^LFx+F=
39 14 38 pm2.61ian FLifF..^Ldomx0..^LFx+F=
40 39 3adant1 VFLifF..^Ldomx0..^LFx+F=
41 3 40 eqtrd VFLsubstrFL=
42 41 3expb VFLsubstrFL=
43 2 42 sylan2b VFL×substrFL=
44 1 43 sylbi FLV××substrFL=
45 df-substr substr=sV,b×if1stb..^2ndbdomsz0..^2ndb1stbsz+1stb
46 ovex 0..^2ndb1stbV
47 46 mptex z0..^2ndb1stbsz+1stbV
48 0ex V
49 47 48 ifex if1stb..^2ndbdomsz0..^2ndb1stbsz+1stbV
50 45 49 dmmpo domsubstr=V××
51 44 50 eleq2s FLdomsubstrsubstrFL=
52 df-ov substrFL=substrFL
53 ndmfv ¬FLdomsubstrsubstrFL=
54 52 53 eqtrid ¬FLdomsubstrsubstrFL=
55 51 54 pm2.61i substrFL=