Metamath Proof Explorer


Theorem f1resfz0f1d

Description: If a function with a sequence of nonnegative integers (starting at 0) as its domain is one-to-one when 0 is removed, and if the range of that restriction does not contain the function's value at the removed integer, then the function is itself one-to-one. (Contributed by BTernaryTau, 4-Oct-2023)

Ref Expression
Hypotheses f1resfz0f1d.1 φ K 0
f1resfz0f1d.2 φ F : 0 K V
f1resfz0f1d.3 φ F 1 K : 1 K 1-1 V
f1resfz0f1d.4 φ F 0 F 1 K =
Assertion f1resfz0f1d φ F : 0 K 1-1 V

Proof

Step Hyp Ref Expression
1 f1resfz0f1d.1 φ K 0
2 f1resfz0f1d.2 φ F : 0 K V
3 f1resfz0f1d.3 φ F 1 K : 1 K 1-1 V
4 f1resfz0f1d.4 φ F 0 F 1 K =
5 fz1ssfz0 1 K 0 K
6 5 a1i φ 1 K 0 K
7 0elfz K 0 0 0 K
8 snssi 0 0 K 0 0 K
9 1 7 8 3syl φ 0 0 K
10 2 9 fssresd φ F 0 : 0 V
11 eqidd F 0 0 = F 0 0 0 = 0
12 0nn0 0 0
13 fveqeq2 x = 0 F 0 x = F 0 y F 0 0 = F 0 y
14 eqeq1 x = 0 x = y 0 = y
15 13 14 imbi12d x = 0 F 0 x = F 0 y x = y F 0 0 = F 0 y 0 = y
16 fveq2 y = 0 F 0 y = F 0 0
17 16 eqeq2d y = 0 F 0 0 = F 0 y F 0 0 = F 0 0
18 eqeq2 y = 0 0 = y 0 = 0
19 17 18 imbi12d y = 0 F 0 0 = F 0 y 0 = y F 0 0 = F 0 0 0 = 0
20 15 19 2ralsng 0 0 0 0 x 0 y 0 F 0 x = F 0 y x = y F 0 0 = F 0 0 0 = 0
21 12 12 20 mp2an x 0 y 0 F 0 x = F 0 y x = y F 0 0 = F 0 0 0 = 0
22 11 21 mpbir x 0 y 0 F 0 x = F 0 y x = y
23 dff13 F 0 : 0 1-1 V F 0 : 0 V x 0 y 0 F 0 x = F 0 y x = y
24 10 22 23 sylanblrc φ F 0 : 0 1-1 V
25 uncom 1 K 0 = 0 1 K
26 fz0sn0fz1 K 0 0 K = 0 1 K
27 1 26 syl φ 0 K = 0 1 K
28 25 27 eqtr4id φ 1 K 0 = 0 K
29 0nelfz1 0 1 K
30 29 neli ¬ 0 1 K
31 disjsn 1 K 0 = ¬ 0 1 K
32 30 31 mpbir 1 K 0 =
33 uneqdifeq 1 K 0 K 1 K 0 = 1 K 0 = 0 K 0 K 1 K = 0
34 5 32 33 mp2an 1 K 0 = 0 K 0 K 1 K = 0
35 28 34 sylib φ 0 K 1 K = 0
36 35 eqcomd φ 0 = 0 K 1 K
37 36 reseq2d φ F 0 = F 0 K 1 K
38 eqidd φ V = V
39 37 36 38 f1eq123d φ F 0 : 0 1-1 V F 0 K 1 K : 0 K 1 K 1-1 V
40 24 39 mpbid φ F 0 K 1 K : 0 K 1 K 1-1 V
41 36 imaeq2d φ F 0 = F 0 K 1 K
42 41 ineq2d φ F 1 K F 0 = F 1 K F 0 K 1 K
43 incom F 0 F 1 K = F 1 K F 0
44 43 4 syl5eqr φ F 1 K F 0 =
45 42 44 eqtr3d φ F 1 K F 0 K 1 K =
46 6 2 3 40 45 f1resrcmplf1d φ F : 0 K 1-1 V