Metamath Proof Explorer


Theorem axcontlem9

Description: Lemma for axcont . Given the separation assumption, all values of F over A are less than or equal to all values of F over B . (Contributed by Scott Fenton, 20-Jun-2013)

Ref Expression
Hypotheses axcontlem9.1 D = p 𝔼 N | U Btwn Z p p Btwn Z U
axcontlem9.2 F = x t | x D t 0 +∞ i 1 N x i = 1 t Z i + t U i
Assertion axcontlem9 N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U n F A m F B n m

Proof

Step Hyp Ref Expression
1 axcontlem9.1 D = p 𝔼 N | U Btwn Z p p Btwn Z U
2 axcontlem9.2 F = x t | x D t 0 +∞ i 1 N x i = 1 t Z i + t U i
3 simpll N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U N
4 simprl1 N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U Z 𝔼 N
5 simplr1 N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U A 𝔼 N
6 simprl2 N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U U A
7 5 6 sseldd N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U U 𝔼 N
8 simprr N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U Z U
9 1 2 axcontlem2 N Z 𝔼 N U 𝔼 N Z U F : D 1-1 onto 0 +∞
10 3 4 7 8 9 syl31anc N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U F : D 1-1 onto 0 +∞
11 f1ofun F : D 1-1 onto 0 +∞ Fun F
12 fvelima Fun F n F A a A F a = n
13 12 ex Fun F n F A a A F a = n
14 10 11 13 3syl N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U n F A a A F a = n
15 fvelima Fun F m F B b B F b = m
16 15 ex Fun F m F B b B F b = m
17 10 11 16 3syl N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U m F B b B F b = m
18 reeanv a A b B F a = n F b = m a A F a = n b B F b = m
19 simplr3 N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U x A y B x Btwn Z y
20 breq1 x = a x Btwn Z y a Btwn Z y
21 opeq2 y = b Z y = Z b
22 21 breq2d y = b a Btwn Z y a Btwn Z b
23 20 22 rspc2v a A b B x A y B x Btwn Z y a Btwn Z b
24 19 23 mpan9 N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U a A b B a Btwn Z b
25 simplll N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U a A b B N
26 4 adantr N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U a A b B Z 𝔼 N
27 7 adantr N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U a A b B U 𝔼 N
28 25 26 27 3jca N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U a A b B N Z 𝔼 N U 𝔼 N
29 simplrr N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U a A b B Z U
30 1 axcontlem4 N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U A D
31 30 sseld N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U a A a D
32 simpl N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U N A 𝔼 N B 𝔼 N x A y B x Btwn Z y
33 1 axcontlem3 N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A Z U B D
34 32 4 6 8 33 syl13anc N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U B D
35 34 sseld N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U b B b D
36 31 35 anim12d N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U a A b B a D b D
37 36 imp N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U a A b B a D b D
38 1 2 axcontlem7 N Z 𝔼 N U 𝔼 N Z U a D b D a Btwn Z b F a F b
39 28 29 37 38 syl21anc N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U a A b B a Btwn Z b F a F b
40 24 39 mpbid N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U a A b B F a F b
41 breq12 F a = n F b = m F a F b n m
42 40 41 syl5ibcom N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U a A b B F a = n F b = m n m
43 42 rexlimdvva N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U a A b B F a = n F b = m n m
44 18 43 syl5bir N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U a A F a = n b B F b = m n m
45 14 17 44 syl2and N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U n F A m F B n m
46 45 ralrimivv N A 𝔼 N B 𝔼 N x A y B x Btwn Z y Z 𝔼 N U A B Z U n F A m F B n m