Metamath Proof Explorer


Theorem wwlktovf

Description: Lemma 1 for wrd2f1tovbij . (Contributed by Alexander van der Vekens, 27-Jul-2018)

Ref Expression
Hypotheses wwlktovf1o.d D = w Word V | w = 2 w 0 = P w 0 w 1 X
wwlktovf1o.r R = n V | P n X
wwlktovf1o.f F = t D t 1
Assertion wwlktovf F : D R

Proof

Step Hyp Ref Expression
1 wwlktovf1o.d D = w Word V | w = 2 w 0 = P w 0 w 1 X
2 wwlktovf1o.r R = n V | P n X
3 wwlktovf1o.f F = t D t 1
4 wrdf t Word V t : 0 ..^ t V
5 oveq2 t = 2 0 ..^ t = 0 ..^ 2
6 5 feq2d t = 2 t : 0 ..^ t V t : 0 ..^ 2 V
7 1nn0 1 0
8 2nn 2
9 1lt2 1 < 2
10 elfzo0 1 0 ..^ 2 1 0 2 1 < 2
11 7 8 9 10 mpbir3an 1 0 ..^ 2
12 ffvelrn t : 0 ..^ 2 V 1 0 ..^ 2 t 1 V
13 11 12 mpan2 t : 0 ..^ 2 V t 1 V
14 6 13 syl6bi t = 2 t : 0 ..^ t V t 1 V
15 14 3ad2ant1 t = 2 t 0 = P t 0 t 1 X t : 0 ..^ t V t 1 V
16 4 15 mpan9 t Word V t = 2 t 0 = P t 0 t 1 X t 1 V
17 preq1 t 0 = P t 0 t 1 = P t 1
18 17 eleq1d t 0 = P t 0 t 1 X P t 1 X
19 18 biimpa t 0 = P t 0 t 1 X P t 1 X
20 19 3adant1 t = 2 t 0 = P t 0 t 1 X P t 1 X
21 20 adantl t Word V t = 2 t 0 = P t 0 t 1 X P t 1 X
22 16 21 jca t Word V t = 2 t 0 = P t 0 t 1 X t 1 V P t 1 X
23 fveqeq2 w = t w = 2 t = 2
24 fveq1 w = t w 0 = t 0
25 24 eqeq1d w = t w 0 = P t 0 = P
26 fveq1 w = t w 1 = t 1
27 24 26 preq12d w = t w 0 w 1 = t 0 t 1
28 27 eleq1d w = t w 0 w 1 X t 0 t 1 X
29 23 25 28 3anbi123d w = t w = 2 w 0 = P w 0 w 1 X t = 2 t 0 = P t 0 t 1 X
30 29 1 elrab2 t D t Word V t = 2 t 0 = P t 0 t 1 X
31 preq2 n = t 1 P n = P t 1
32 31 eleq1d n = t 1 P n X P t 1 X
33 32 2 elrab2 t 1 R t 1 V P t 1 X
34 22 30 33 3imtr4i t D t 1 R
35 3 34 fmpti F : D R