Metamath Proof Explorer


Theorem pexmidlem6N

Description: Lemma for pexmidN . (Contributed by NM, 3-Feb-2012) (New usage is discouraged.)

Ref Expression
Hypotheses pexmidlem.l ˙=K
pexmidlem.j ˙=joinK
pexmidlem.a A=AtomsK
pexmidlem.p +˙=+𝑃K
pexmidlem.o ˙=𝑃K
pexmidlem.m M=X+˙p
Assertion pexmidlem6N KHLXApA˙˙X=XX¬pX+˙˙XM=X

Proof

Step Hyp Ref Expression
1 pexmidlem.l ˙=K
2 pexmidlem.j ˙=joinK
3 pexmidlem.a A=AtomsK
4 pexmidlem.p +˙=+𝑃K
5 pexmidlem.o ˙=𝑃K
6 pexmidlem.m M=X+˙p
7 1 2 3 4 5 6 pexmidlem5N KHLXApAX¬pX+˙˙X˙XM=
8 7 3adantr1 KHLXApA˙˙X=XX¬pX+˙˙X˙XM=
9 8 fveq2d KHLXApA˙˙X=XX¬pX+˙˙X˙˙XM=˙
10 simpl1 KHLXApA˙˙X=XX¬pX+˙˙XKHL
11 3 5 pol0N KHL˙=A
12 10 11 syl KHLXApA˙˙X=XX¬pX+˙˙X˙=A
13 9 12 eqtrd KHLXApA˙˙X=XX¬pX+˙˙X˙˙XM=A
14 13 ineq1d KHLXApA˙˙X=XX¬pX+˙˙X˙˙XMM=AM
15 simpl2 KHLXApA˙˙X=XX¬pX+˙˙XXA
16 simpl3 KHLXApA˙˙X=XX¬pX+˙˙XpA
17 16 snssd KHLXApA˙˙X=XX¬pX+˙˙XpA
18 3 4 paddssat KHLXApAX+˙pA
19 10 15 17 18 syl3anc KHLXApA˙˙X=XX¬pX+˙˙XX+˙pA
20 6 19 eqsstrid KHLXApA˙˙X=XX¬pX+˙˙XMA
21 10 15 20 3jca KHLXApA˙˙X=XX¬pX+˙˙XKHLXAMA
22 3 4 sspadd1 KHLXApAXX+˙p
23 10 15 17 22 syl3anc KHLXApA˙˙X=XX¬pX+˙˙XXX+˙p
24 23 6 sseqtrrdi KHLXApA˙˙X=XX¬pX+˙˙XXM
25 simpr1 KHLXApA˙˙X=XX¬pX+˙˙X˙˙X=X
26 eqid PSubClK=PSubClK
27 3 5 26 ispsubclN KHLXPSubClKXA˙˙X=X
28 10 27 syl KHLXApA˙˙X=XX¬pX+˙˙XXPSubClKXA˙˙X=X
29 15 25 28 mpbir2and KHLXApA˙˙X=XX¬pX+˙˙XXPSubClK
30 3 4 26 paddatclN KHLXPSubClKpAX+˙pPSubClK
31 10 29 16 30 syl3anc KHLXApA˙˙X=XX¬pX+˙˙XX+˙pPSubClK
32 6 31 eqeltrid KHLXApA˙˙X=XX¬pX+˙˙XMPSubClK
33 5 26 psubcli2N KHLMPSubClK˙˙M=M
34 10 32 33 syl2anc KHLXApA˙˙X=XX¬pX+˙˙X˙˙M=M
35 24 34 jca KHLXApA˙˙X=XX¬pX+˙˙XXM˙˙M=M
36 3 5 poml4N KHLXAMAXM˙˙M=M˙˙XMM=˙˙X
37 21 35 36 sylc KHLXApA˙˙X=XX¬pX+˙˙X˙˙XMM=˙˙X
38 sseqin2 MAAM=M
39 20 38 sylib KHLXApA˙˙X=XX¬pX+˙˙XAM=M
40 14 37 39 3eqtr3rd KHLXApA˙˙X=XX¬pX+˙˙XM=˙˙X
41 40 25 eqtrd KHLXApA˙˙X=XX¬pX+˙˙XM=X