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 ˙ = join K
pexmidlem.a A = Atoms K
pexmidlem.p + ˙ = + 𝑃 K
pexmidlem.o ˙ = 𝑃 K
pexmidlem.m M = X + ˙ p
Assertion pexmidlem1N K HL X A r X q ˙ X q r

Proof

Step Hyp Ref Expression
1 pexmidlem.l ˙ = K
2 pexmidlem.j ˙ = join K
3 pexmidlem.a A = Atoms K
4 pexmidlem.p + ˙ = + 𝑃 K
5 pexmidlem.o ˙ = 𝑃 K
6 pexmidlem.m M = X + ˙ p
7 n0i r X ˙ X ¬ X ˙ X =
8 3 5 pnonsingN K HL X A X ˙ X =
9 8 adantr K HL X A r X q ˙ X X ˙ X =
10 7 9 nsyl3 K HL X A r X q ˙ X ¬ r X ˙ X
11 simprr K HL X A r X q ˙ X q ˙ X
12 eleq1w q = r q ˙ X r ˙ X
13 11 12 syl5ibcom K HL X A r X q ˙ X q = r r ˙ X
14 simprl K HL X A r X q ˙ X r X
15 13 14 jctild K HL X A r X q ˙ X q = r r X r ˙ X
16 elin r X ˙ X r X r ˙ X
17 15 16 syl6ibr K HL X A r X q ˙ X q = r r X ˙ X
18 17 necon3bd K HL X A r X q ˙ X ¬ r X ˙ X q r
19 10 18 mpd K HL X A r X q ˙ X q r