Metamath Proof Explorer


Theorem pmltpclem1

Description: Lemma for pmltpc . (Contributed by Mario Carneiro, 1-Jul-2014)

Ref Expression
Hypotheses pmltpclem1.1 φ A S
pmltpclem1.2 φ B S
pmltpclem1.3 φ C S
pmltpclem1.4 φ A < B
pmltpclem1.5 φ B < C
pmltpclem1.6 φ F A < F B F C < F B F B < F A F B < F C
Assertion pmltpclem1 φ a S b S c S a < b b < c F a < F b F c < F b F b < F a F b < F c

Proof

Step Hyp Ref Expression
1 pmltpclem1.1 φ A S
2 pmltpclem1.2 φ B S
3 pmltpclem1.3 φ C S
4 pmltpclem1.4 φ A < B
5 pmltpclem1.5 φ B < C
6 pmltpclem1.6 φ F A < F B F C < F B F B < F A F B < F C
7 breq1 a = A a < b A < b
8 fveq2 a = A F a = F A
9 8 breq1d a = A F a < F b F A < F b
10 9 anbi1d a = A F a < F b F c < F b F A < F b F c < F b
11 8 breq2d a = A F b < F a F b < F A
12 11 anbi1d a = A F b < F a F b < F c F b < F A F b < F c
13 10 12 orbi12d a = A F a < F b F c < F b F b < F a F b < F c F A < F b F c < F b F b < F A F b < F c
14 7 13 3anbi13d a = A a < b b < c F a < F b F c < F b F b < F a F b < F c A < b b < c F A < F b F c < F b F b < F A F b < F c
15 breq2 b = B A < b A < B
16 breq1 b = B b < c B < c
17 fveq2 b = B F b = F B
18 17 breq2d b = B F A < F b F A < F B
19 17 breq2d b = B F c < F b F c < F B
20 18 19 anbi12d b = B F A < F b F c < F b F A < F B F c < F B
21 17 breq1d b = B F b < F A F B < F A
22 17 breq1d b = B F b < F c F B < F c
23 21 22 anbi12d b = B F b < F A F b < F c F B < F A F B < F c
24 20 23 orbi12d b = B F A < F b F c < F b F b < F A F b < F c F A < F B F c < F B F B < F A F B < F c
25 15 16 24 3anbi123d b = B A < b b < c F A < F b F c < F b F b < F A F b < F c A < B B < c F A < F B F c < F B F B < F A F B < F c
26 breq2 c = C B < c B < C
27 fveq2 c = C F c = F C
28 27 breq1d c = C F c < F B F C < F B
29 28 anbi2d c = C F A < F B F c < F B F A < F B F C < F B
30 27 breq2d c = C F B < F c F B < F C
31 30 anbi2d c = C F B < F A F B < F c F B < F A F B < F C
32 29 31 orbi12d c = C F A < F B F c < F B F B < F A F B < F c F A < F B F C < F B F B < F A F B < F C
33 26 32 3anbi23d c = C A < B B < c F A < F B F c < F B F B < F A F B < F c A < B B < C F A < F B F C < F B F B < F A F B < F C
34 14 25 33 rspc3ev A S B S C S A < B B < C F A < F B F C < F B F B < F A F B < F C a S b S c S a < b b < c F a < F b F c < F b F b < F a F b < F c
35 1 2 3 4 5 6 34 syl33anc φ a S b S c S a < b b < c F a < F b F c < F b F b < F a F b < F c