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 1oex 1 𝑜 V
37 36 prid2 1 𝑜 1 𝑜
38 iftrue r = if r = 1 𝑜 r = 1 𝑜
39 prex 1 𝑜 V
40 38 1 39 fvmpt 𝒫 3 𝑜 K = 1 𝑜
41 40 adantr 𝒫 3 𝑜 2 𝑜 𝒫 3 𝑜 K = 1 𝑜
42 37 41 eleqtrrid 𝒫 3 𝑜 2 𝑜 𝒫 3 𝑜 1 𝑜 K
43 1n0 1 𝑜
44 43 neii ¬ 1 𝑜 =
45 eqcom 1 𝑜 = 2 𝑜 2 𝑜 = 1 𝑜
46 df-2o 2 𝑜 = suc 1 𝑜
47 df-1o 1 𝑜 = suc
48 46 47 eqeq12i 2 𝑜 = 1 𝑜 suc 1 𝑜 = suc
49 suc11reg suc 1 𝑜 = suc 1 𝑜 =
50 45 48 49 3bitri 1 𝑜 = 2 𝑜 1 𝑜 =
51 43 50 nemtbir ¬ 1 𝑜 = 2 𝑜
52 44 51 pm3.2ni ¬ 1 𝑜 = 1 𝑜 = 2 𝑜
53 elpri 1 𝑜 2 𝑜 1 𝑜 = 1 𝑜 = 2 𝑜
54 52 53 mto ¬ 1 𝑜 2 𝑜
55 54 a1i 𝒫 3 𝑜 2 𝑜 𝒫 3 𝑜 ¬ 1 𝑜 2 𝑜
56 eqeq1 r = 2 𝑜 r = 2 𝑜 =
57 id r = 2 𝑜 r = 2 𝑜
58 56 57 ifbieq2d r = 2 𝑜 if r = 1 𝑜 r = if 2 𝑜 = 1 𝑜 2 𝑜
59 15 prid2 2 𝑜 2 𝑜
60 2on0 2 𝑜
61 nelsn 2 𝑜 ¬ 2 𝑜
62 60 61 ax-mp ¬ 2 𝑜
63 nelneq2 2 𝑜 2 𝑜 ¬ 2 𝑜 ¬ 2 𝑜 =
64 59 62 63 mp2an ¬ 2 𝑜 =
65 64 iffalsei if 2 𝑜 = 1 𝑜 2 𝑜 = 2 𝑜
66 58 65 eqtrdi r = 2 𝑜 if r = 1 𝑜 r = 2 𝑜
67 prex 2 𝑜 V
68 66 1 67 fvmpt 2 𝑜 𝒫 3 𝑜 K 2 𝑜 = 2 𝑜
69 68 adantl 𝒫 3 𝑜 2 𝑜 𝒫 3 𝑜 K 2 𝑜 = 2 𝑜
70 55 69 neleqtrrd 𝒫 3 𝑜 2 𝑜 𝒫 3 𝑜 ¬ 1 𝑜 K 2 𝑜
71 nelss 1 𝑜 K ¬ 1 𝑜 K 2 𝑜 ¬ K K 2 𝑜
72 42 70 71 syl2anc 𝒫 3 𝑜 2 𝑜 𝒫 3 𝑜 ¬ K K 2 𝑜
73 snsspr1 2 𝑜
74 72 73 jctil 𝒫 3 𝑜 2 𝑜 𝒫 3 𝑜 2 𝑜 ¬ K K 2 𝑜
75 30 35 74 rspcedvd 𝒫 3 𝑜 2 𝑜 𝒫 3 𝑜 t 𝒫 3 𝑜 t ¬ K K t
76 22 29 75 rspcedvd 𝒫 3 𝑜 2 𝑜 𝒫 3 𝑜 s 𝒫 3 𝑜 t 𝒫 3 𝑜 s t ¬ K s K t
77 7 21 76 mp2an s 𝒫 3 𝑜 t 𝒫 3 𝑜 s t ¬ K s K t