Metamath Proof Explorer


Theorem 4atlem0a

Description: Lemma for 4at . (Contributed by NM, 10-Jul-2012)

Ref Expression
Hypotheses 4at.l ˙ = K
4at.j ˙ = join K
4at.a A = Atoms K
Assertion 4atlem0a K HL P A Q A R A S A ¬ R ˙ P ˙ Q ¬ S ˙ P ˙ Q ˙ R ¬ R ˙ P ˙ Q ˙ S

Proof

Step Hyp Ref Expression
1 4at.l ˙ = K
2 4at.j ˙ = join K
3 4at.a A = Atoms K
4 simprr K HL P A Q A R A S A ¬ R ˙ P ˙ Q ¬ S ˙ P ˙ Q ˙ R ¬ S ˙ P ˙ Q ˙ R
5 simpl1 K HL P A Q A R A S A ¬ R ˙ P ˙ Q ¬ S ˙ P ˙ Q ˙ R K HL
6 simpl3l K HL P A Q A R A S A ¬ R ˙ P ˙ Q ¬ S ˙ P ˙ Q ˙ R R A
7 simpl3r K HL P A Q A R A S A ¬ R ˙ P ˙ Q ¬ S ˙ P ˙ Q ˙ R S A
8 simpl2l K HL P A Q A R A S A ¬ R ˙ P ˙ Q ¬ S ˙ P ˙ Q ˙ R P A
9 simpl2r K HL P A Q A R A S A ¬ R ˙ P ˙ Q ¬ S ˙ P ˙ Q ˙ R Q A
10 eqid Base K = Base K
11 10 2 3 hlatjcl K HL P A Q A P ˙ Q Base K
12 5 8 9 11 syl3anc K HL P A Q A R A S A ¬ R ˙ P ˙ Q ¬ S ˙ P ˙ Q ˙ R P ˙ Q Base K
13 simprl K HL P A Q A R A S A ¬ R ˙ P ˙ Q ¬ S ˙ P ˙ Q ˙ R ¬ R ˙ P ˙ Q
14 10 1 2 3 hlexch1 K HL R A S A P ˙ Q Base K ¬ R ˙ P ˙ Q R ˙ P ˙ Q ˙ S S ˙ P ˙ Q ˙ R
15 5 6 7 12 13 14 syl131anc K HL P A Q A R A S A ¬ R ˙ P ˙ Q ¬ S ˙ P ˙ Q ˙ R R ˙ P ˙ Q ˙ S S ˙ P ˙ Q ˙ R
16 4 15 mtod K HL P A Q A R A S A ¬ R ˙ P ˙ Q ¬ S ˙ P ˙ Q ˙ R ¬ R ˙ P ˙ Q ˙ S