Metamath Proof Explorer


Theorem pexmidlem8N

Description: Lemma for pexmidN . The contradiction of pexmidlem6N and pexmidlem7N shows that there can be no atom p that is not in X .+ ( ._|_X ) , which is therefore the whole atom space. (Contributed by NM, 3-Feb-2012) (New usage is discouraged.)

Ref Expression
Hypotheses pexmidALT.a A = Atoms K
pexmidALT.p + ˙ = + 𝑃 K
pexmidALT.o ˙ = 𝑃 K
Assertion pexmidlem8N K HL X A ˙ ˙ X = X X X + ˙ ˙ X = A

Proof

Step Hyp Ref Expression
1 pexmidALT.a A = Atoms K
2 pexmidALT.p + ˙ = + 𝑃 K
3 pexmidALT.o ˙ = 𝑃 K
4 nonconne ¬ X = X X X
5 simpll K HL X A ˙ ˙ X = X X K HL
6 simplr K HL X A ˙ ˙ X = X X X A
7 1 3 polssatN K HL X A ˙ X A
8 7 adantr K HL X A ˙ ˙ X = X X ˙ X A
9 1 2 paddssat K HL X A ˙ X A X + ˙ ˙ X A
10 5 6 8 9 syl3anc K HL X A ˙ ˙ X = X X X + ˙ ˙ X A
11 df-pss X + ˙ ˙ X A X + ˙ ˙ X A X + ˙ ˙ X A
12 pssnel X + ˙ ˙ X A p p A ¬ p X + ˙ ˙ X
13 11 12 sylbir X + ˙ ˙ X A X + ˙ ˙ X A p p A ¬ p X + ˙ ˙ X
14 df-rex p A ¬ p X + ˙ ˙ X p p A ¬ p X + ˙ ˙ X
15 13 14 sylibr X + ˙ ˙ X A X + ˙ ˙ X A p A ¬ p X + ˙ ˙ X
16 simplll K HL X A ˙ ˙ X = X X p A ¬ p X + ˙ ˙ X K HL
17 simpllr K HL X A ˙ ˙ X = X X p A ¬ p X + ˙ ˙ X X A
18 simprl K HL X A ˙ ˙ X = X X p A ¬ p X + ˙ ˙ X p A
19 simplrl K HL X A ˙ ˙ X = X X p A ¬ p X + ˙ ˙ X ˙ ˙ X = X
20 simplrr K HL X A ˙ ˙ X = X X p A ¬ p X + ˙ ˙ X X
21 simprr K HL X A ˙ ˙ X = X X p A ¬ p X + ˙ ˙ X ¬ p X + ˙ ˙ X
22 eqid K = K
23 eqid join K = join K
24 eqid X + ˙ p = X + ˙ p
25 22 23 1 2 3 24 pexmidlem6N K HL X A p A ˙ ˙ X = X X ¬ p X + ˙ ˙ X X + ˙ p = X
26 22 23 1 2 3 24 pexmidlem7N K HL X A p A ˙ ˙ X = X X ¬ p X + ˙ ˙ X X + ˙ p X
27 25 26 jca K HL X A p A ˙ ˙ X = X X ¬ p X + ˙ ˙ X X + ˙ p = X X + ˙ p X
28 16 17 18 19 20 21 27 syl33anc K HL X A ˙ ˙ X = X X p A ¬ p X + ˙ ˙ X X + ˙ p = X X + ˙ p X
29 nonconne ¬ X + ˙ p = X X + ˙ p X
30 29 4 2false X + ˙ p = X X + ˙ p X X = X X X
31 28 30 sylib K HL X A ˙ ˙ X = X X p A ¬ p X + ˙ ˙ X X = X X X
32 31 rexlimdvaa K HL X A ˙ ˙ X = X X p A ¬ p X + ˙ ˙ X X = X X X
33 15 32 syl5 K HL X A ˙ ˙ X = X X X + ˙ ˙ X A X + ˙ ˙ X A X = X X X
34 10 33 mpand K HL X A ˙ ˙ X = X X X + ˙ ˙ X A X = X X X
35 34 necon1bd K HL X A ˙ ˙ X = X X ¬ X = X X X X + ˙ ˙ X = A
36 4 35 mpi K HL X A ˙ ˙ X = X X X + ˙ ˙ X = A