Description: Lemma for kur14 . Since the set T is closed under closure and complement, it contains the minimal set S as a subset, so S also has at most 1 4 elements. (Indeed S = T , and it's not hard to prove this, but we don't need it for this proof.) (Contributed by Mario Carneiro, 11-Feb-2015)
Ref | Expression | ||
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Hypotheses | kur14lem.j | |
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kur14lem.x | |
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kur14lem.k | |
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kur14lem.i | |
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kur14lem.a | |
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kur14lem.b | |
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kur14lem.c | |
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kur14lem.d | |
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kur14lem.t | |
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kur14lem.s | |
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Assertion | kur14lem9 | |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | kur14lem.j | |
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2 | kur14lem.x | |
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3 | kur14lem.k | |
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4 | kur14lem.i | |
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5 | kur14lem.a | |
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6 | kur14lem.b | |
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7 | kur14lem.c | |
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8 | kur14lem.d | |
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9 | kur14lem.t | |
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10 | kur14lem.s | |
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11 | vex | |
|
12 | 11 | elintrab | |
13 | ssun1 | |
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14 | ssun1 | |
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15 | ssun1 | |
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16 | 15 9 | sseqtrri | |
17 | 14 16 | sstri | |
18 | 13 17 | sstri | |
19 | 2 | topopn | |
20 | 1 19 | ax-mp | |
21 | 20 | elexi | |
22 | 21 5 | ssexi | |
23 | 22 | tpid1 | |
24 | 18 23 | sselii | |
25 | 1 2 3 4 5 6 7 8 9 | kur14lem7 | |
26 | 25 | simprd | |
27 | 26 | rgen | |
28 | 25 | simpld | |
29 | 21 | elpw2 | |
30 | 28 29 | sylibr | |
31 | 30 | ssriv | |
32 | 21 | pwex | |
33 | 32 | elpw2 | |
34 | 31 33 | mpbir | |
35 | eleq2 | |
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36 | sseq2 | |
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37 | 36 | raleqbi1dv | |
38 | 35 37 | anbi12d | |
39 | eleq2 | |
|
40 | 38 39 | imbi12d | |
41 | 40 | rspccv | |
42 | 34 41 | mpi | |
43 | 24 27 42 | mp2ani | |
44 | 12 43 | sylbi | |
45 | 44 | ssriv | |
46 | 10 45 | eqsstri | |
47 | 1 2 3 4 5 6 7 8 9 | kur14lem8 | |
48 | 1nn0 | |
|
49 | 4nn0 | |
|
50 | 48 49 | deccl | |
51 | 46 47 50 | hashsslei | |