Metamath Proof Explorer


Theorem pexmidlem1N

Description: Lemma for pexmidN . Holland's proof implicitly requires q =/= r , which we prove here. (Contributed by NM, 2-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 pexmidlem1N KHLXArXq˙Xqr

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 n0i rX˙X¬X˙X=
8 3 5 pnonsingN KHLXAX˙X=
9 8 adantr KHLXArXq˙XX˙X=
10 7 9 nsyl3 KHLXArXq˙X¬rX˙X
11 simprr KHLXArXq˙Xq˙X
12 eleq1w q=rq˙Xr˙X
13 11 12 syl5ibcom KHLXArXq˙Xq=rr˙X
14 simprl KHLXArXq˙XrX
15 13 14 jctild KHLXArXq˙Xq=rrXr˙X
16 elin rX˙XrXr˙X
17 15 16 syl6ibr KHLXArXq˙Xq=rrX˙X
18 17 necon3bd KHLXArXq˙X¬rX˙Xqr
19 10 18 mpd KHLXArXq˙Xqr