Metamath Proof Explorer


Theorem pexmidlem7N

Description: Lemma for pexmidN . Contradict pexmidlem6N . (Contributed by NM, 3-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 pexmidlem7N K HL X A p A ˙ ˙ X = X X ¬ p X + ˙ ˙ X M X

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 simpl1 K HL X A p A ˙ ˙ X = X X ¬ p X + ˙ ˙ X K HL
8 simpl3 K HL X A p A ˙ ˙ X = X X ¬ p X + ˙ ˙ X p A
9 8 snssd K HL X A p A ˙ ˙ X = X X ¬ p X + ˙ ˙ X p A
10 simpl2 K HL X A p A ˙ ˙ X = X X ¬ p X + ˙ ˙ X X A
11 3 4 sspadd2 K HL p A X A p X + ˙ p
12 7 9 10 11 syl3anc K HL X A p A ˙ ˙ X = X X ¬ p X + ˙ ˙ X p X + ˙ p
13 vex p V
14 13 snss p X + ˙ p p X + ˙ p
15 12 14 sylibr K HL X A p A ˙ ˙ X = X X ¬ p X + ˙ ˙ X p X + ˙ p
16 15 6 eleqtrrdi K HL X A p A ˙ ˙ X = X X ¬ p X + ˙ ˙ X p M
17 3 5 polssatN K HL X A ˙ X A
18 7 10 17 syl2anc K HL X A p A ˙ ˙ X = X X ¬ p X + ˙ ˙ X ˙ X A
19 3 4 sspadd1 K HL X A ˙ X A X X + ˙ ˙ X
20 7 10 18 19 syl3anc K HL X A p A ˙ ˙ X = X X ¬ p X + ˙ ˙ X X X + ˙ ˙ X
21 simpr3 K HL X A p A ˙ ˙ X = X X ¬ p X + ˙ ˙ X ¬ p X + ˙ ˙ X
22 20 21 ssneldd K HL X A p A ˙ ˙ X = X X ¬ p X + ˙ ˙ X ¬ p X
23 nelne1 p M ¬ p X M X
24 16 22 23 syl2anc K HL X A p A ˙ ˙ X = X X ¬ p X + ˙ ˙ X M X