Metamath Proof Explorer


Theorem clsk1indlem1

Description: The ansatz closure function ( r e. ~P 3o |-> if ( r = { (/) } , { (/) , 1o } , r ) ) does not have the K1 property of isotony. (Contributed by RP, 6-Jul-2021)

Ref Expression
Hypothesis clsk1indlem.k K = r 𝒫 3 𝑜 if r = 1 𝑜 r
Assertion clsk1indlem1 s 𝒫 3 𝑜 t 𝒫 3 𝑜 s t ¬ K s K t

Proof

Step Hyp Ref Expression
1 clsk1indlem.k K = r 𝒫 3 𝑜 if r = 1 𝑜 r
2 tpex 1 𝑜 2 𝑜 V
3 snsstp1 1 𝑜 2 𝑜
4 2 3 elpwi2 𝒫 1 𝑜 2 𝑜
5 df3o2 3 𝑜 = 1 𝑜 2 𝑜
6 5 pweqi 𝒫 3 𝑜 = 𝒫 1 𝑜 2 𝑜
7 4 6 eleqtrri 𝒫 3 𝑜
8 2 a1i 1 𝑜 2 𝑜 V
9 3 a1i 1 𝑜 2 𝑜
10 0ex V
11 10 snss 1 𝑜 2 𝑜 1 𝑜 2 𝑜
12 9 11 sylibr 1 𝑜 2 𝑜
13 snsstp3 2 𝑜 1 𝑜 2 𝑜
14 13 a1i 2 𝑜 1 𝑜 2 𝑜
15 2oex 2 𝑜 V
16 15 snss 2 𝑜 1 𝑜 2 𝑜 2 𝑜 1 𝑜 2 𝑜
17 14 16 sylibr 2 𝑜 1 𝑜 2 𝑜
18 12 17 prssd 2 𝑜 1 𝑜 2 𝑜
19 8 18 sselpwd 2 𝑜 𝒫 1 𝑜 2 𝑜
20 19 mptru 2 𝑜 𝒫 1 𝑜 2 𝑜
21 20 6 eleqtrri 2 𝑜 𝒫 3 𝑜
22 simpl 𝒫 3 𝑜 2 𝑜 𝒫 3 𝑜 𝒫 3 𝑜
23 sseq1 s = s t t
24 fveq2 s = K s = K
25 24 sseq1d s = K s K t K K t
26 25 notbid s = ¬ K s K t ¬ K K t
27 23 26 anbi12d s = s t ¬ K s K t t ¬ K K t
28 27 rexbidv s = t 𝒫 3 𝑜 s t ¬ K s K t t 𝒫 3 𝑜 t ¬ K K t
29 28 adantl 𝒫 3 𝑜 2 𝑜 𝒫 3 𝑜 s = t 𝒫 3 𝑜 s t ¬ K s K t t 𝒫 3 𝑜 t ¬ K K t
30 simpr 𝒫 3 𝑜 2 𝑜 𝒫 3 𝑜 2 𝑜 𝒫 3 𝑜
31 fveq2 t = 2 𝑜 K t = K 2 𝑜
32 31 sseq2d t = 2 𝑜 K K t K K 2 𝑜
33 32 notbid t = 2 𝑜 ¬ K K t ¬ K K 2 𝑜
34 33 cleq2lem t = 2 𝑜 t ¬ K K t 2 𝑜 ¬ K K 2 𝑜
35 34 adantl 𝒫 3 𝑜 2 𝑜 𝒫 3 𝑜 t = 2 𝑜 t ¬ K K t 2 𝑜 ¬ K K 2 𝑜
36 1oelpr 1 𝑜 1 𝑜
37 iftrue r = if r = 1 𝑜 r = 1 𝑜
38 prex 1 𝑜 V
39 37 1 38 fvmpt 𝒫 3 𝑜 K = 1 𝑜
40 39 adantr 𝒫 3 𝑜 2 𝑜 𝒫 3 𝑜 K = 1 𝑜
41 36 40 eleqtrrid 𝒫 3 𝑜 2 𝑜 𝒫 3 𝑜 1 𝑜 K
42 1n0 1 𝑜
43 42 neii ¬ 1 𝑜 =
44 eqcom 1 𝑜 = 2 𝑜 2 𝑜 = 1 𝑜
45 df-2o 2 𝑜 = suc 1 𝑜
46 df-1o 1 𝑜 = suc
47 45 46 eqeq12i 2 𝑜 = 1 𝑜 suc 1 𝑜 = suc
48 suc11reg suc 1 𝑜 = suc 1 𝑜 =
49 44 47 48 3bitri 1 𝑜 = 2 𝑜 1 𝑜 =
50 42 49 nemtbir ¬ 1 𝑜 = 2 𝑜
51 43 50 pm3.2ni ¬ 1 𝑜 = 1 𝑜 = 2 𝑜
52 elpri 1 𝑜 2 𝑜 1 𝑜 = 1 𝑜 = 2 𝑜
53 51 52 mto ¬ 1 𝑜 2 𝑜
54 53 a1i 𝒫 3 𝑜 2 𝑜 𝒫 3 𝑜 ¬ 1 𝑜 2 𝑜
55 eqeq1 r = 2 𝑜 r = 2 𝑜 =
56 id r = 2 𝑜 r = 2 𝑜
57 55 56 ifbieq2d r = 2 𝑜 if r = 1 𝑜 r = if 2 𝑜 = 1 𝑜 2 𝑜
58 15 prid2 2 𝑜 2 𝑜
59 2on0 2 𝑜
60 nelsn 2 𝑜 ¬ 2 𝑜
61 59 60 ax-mp ¬ 2 𝑜
62 nelneq2 2 𝑜 2 𝑜 ¬ 2 𝑜 ¬ 2 𝑜 =
63 58 61 62 mp2an ¬ 2 𝑜 =
64 63 iffalsei if 2 𝑜 = 1 𝑜 2 𝑜 = 2 𝑜
65 57 64 eqtrdi r = 2 𝑜 if r = 1 𝑜 r = 2 𝑜
66 prex 2 𝑜 V
67 65 1 66 fvmpt 2 𝑜 𝒫 3 𝑜 K 2 𝑜 = 2 𝑜
68 67 adantl 𝒫 3 𝑜 2 𝑜 𝒫 3 𝑜 K 2 𝑜 = 2 𝑜
69 54 68 neleqtrrd 𝒫 3 𝑜 2 𝑜 𝒫 3 𝑜 ¬ 1 𝑜 K 2 𝑜
70 nelss 1 𝑜 K ¬ 1 𝑜 K 2 𝑜 ¬ K K 2 𝑜
71 41 69 70 syl2anc 𝒫 3 𝑜 2 𝑜 𝒫 3 𝑜 ¬ K K 2 𝑜
72 snsspr1 2 𝑜
73 71 72 jctil 𝒫 3 𝑜 2 𝑜 𝒫 3 𝑜 2 𝑜 ¬ K K 2 𝑜
74 30 35 73 rspcedvd 𝒫 3 𝑜 2 𝑜 𝒫 3 𝑜 t 𝒫 3 𝑜 t ¬ K K t
75 22 29 74 rspcedvd 𝒫 3 𝑜 2 𝑜 𝒫 3 𝑜 s 𝒫 3 𝑜 t 𝒫 3 𝑜 s t ¬ K s K t
76 7 21 75 mp2an s 𝒫 3 𝑜 t 𝒫 3 𝑜 s t ¬ K s K t