Metamath Proof Explorer


Theorem allbutfiinf

Description: Given a "for all but finitely many" condition, the condition holds from N on. (Contributed by Glauco Siliprandi, 23-Oct-2021)

Ref Expression
Hypotheses allbutfiinf.z Z = M
allbutfiinf.a A = n Z m n B
allbutfiinf.x φ X A
allbutfiinf.n N = sup n Z | m n X B <
Assertion allbutfiinf φ N Z m N X B

Proof

Step Hyp Ref Expression
1 allbutfiinf.z Z = M
2 allbutfiinf.a A = n Z m n B
3 allbutfiinf.x φ X A
4 allbutfiinf.n N = sup n Z | m n X B <
5 ssrab2 n Z | m n X B Z
6 4 a1i φ N = sup n Z | m n X B <
7 5 1 sseqtri n Z | m n X B M
8 7 a1i φ n Z | m n X B M
9 1 2 allbutfi X A n Z m n X B
10 3 9 sylib φ n Z m n X B
11 nfrab1 _ n n Z | m n X B
12 nfcv _ n
13 11 12 nfne n n Z | m n X B
14 rabid n n Z | m n X B n Z m n X B
15 14 bicomi n Z m n X B n n Z | m n X B
16 15 biimpi n Z m n X B n n Z | m n X B
17 16 ne0d n Z m n X B n Z | m n X B
18 17 ex n Z m n X B n Z | m n X B
19 13 18 rexlimi n Z m n X B n Z | m n X B
20 19 a1i φ n Z m n X B n Z | m n X B
21 10 20 mpd φ n Z | m n X B
22 infssuzcl n Z | m n X B M n Z | m n X B sup n Z | m n X B < n Z | m n X B
23 8 21 22 syl2anc φ sup n Z | m n X B < n Z | m n X B
24 6 23 eqeltrd φ N n Z | m n X B
25 5 24 sselid φ N Z
26 nfcv _ n
27 nfcv _ n <
28 11 26 27 nfinf _ n sup n Z | m n X B <
29 4 28 nfcxfr _ n N
30 nfcv _ n Z
31 nfcv _ n
32 31 29 nffv _ n N
33 nfv n X B
34 32 33 nfralw n m N X B
35 nfcv _ m n
36 nfcv _ m
37 nfra1 m m n X B
38 nfcv _ m Z
39 37 38 nfrabw _ m n Z | m n X B
40 nfcv _ m
41 nfcv _ m <
42 39 40 41 nfinf _ m sup n Z | m n X B <
43 4 42 nfcxfr _ m N
44 36 43 nffv _ m N
45 fveq2 n = N n = N
46 35 44 45 raleqd n = N m n X B m N X B
47 29 30 34 46 elrabf N n Z | m n X B N Z m N X B
48 47 biimpi N n Z | m n X B N Z m N X B
49 48 simprd N n Z | m n X B m N X B
50 24 49 syl φ m N X B
51 25 50 jca φ N Z m N X B