# Metamath Proof Explorer

## Theorem 4atexlem7

Description: Whenever there are at least 4 atoms under P .\/ Q (specifically, P , Q , r , and ( P .\/ Q ) ./\ W ), there are also at least 4 atoms under P .\/ S . This proves the statement in Lemma E of Crawley p. 114, last line, "...p \/ q/0 and hence p \/ s/0 contains at least four atoms..." Note that by cvlsupr2 , our ( P .\/ r ) = ( Q .\/ r ) is a shorter way to express r =/= P /\ r =/= Q /\ r .<_ ( P .\/ Q ) . With a longer proof, the condition -. S .<_ ( P .\/ Q ) could be eliminated (see 4atex ), although for some purposes this more restricted lemma may be adequate. (Contributed by NM, 25-Nov-2012)

Ref Expression
Hypotheses 4that.l
4that.j
4that.a ${⊢}{A}=\mathrm{Atoms}\left({K}\right)$
4that.h ${⊢}{H}=\mathrm{LHyp}\left({K}\right)$
Assertion 4atexlem7

### Proof

Step Hyp Ref Expression
1 4that.l
2 4that.j
3 4that.a ${⊢}{A}=\mathrm{Atoms}\left({K}\right)$
4 4that.h ${⊢}{H}=\mathrm{LHyp}\left({K}\right)$
5 simp11l
6 simp1r1
7 6 3ad2ant1
8 simp1r2
9 8 3ad2ant1
10 simp2
11 simp3l
12 10 11 jca
13 simp1r3
14 13 3ad2ant1
15 simp3r
16 simp12
17 simp13
18 eqid ${⊢}\mathrm{meet}\left({K}\right)=\mathrm{meet}\left({K}\right)$
19 1 2 18 3 4 4atexlemex6
20 5 7 9 12 14 15 16 17 19 syl323anc
21 20 rexlimdv3a
22 21 3exp
23 22 3impd
24 23 3impia