Metamath Proof Explorer


Theorem cdlemg3a

Description: Part of proof of Lemma G in Crawley p. 116, line 19. Show p \/ q = p \/ u. TODO: reformat cdleme0cp to match this, then replace with cdleme0cp . (Contributed by NM, 19-Apr-2013)

Ref Expression
Hypotheses cdlemg3.l
|- .<_ = ( le ` K )
cdlemg3.j
|- .\/ = ( join ` K )
cdlemg3.m
|- ./\ = ( meet ` K )
cdlemg3.a
|- A = ( Atoms ` K )
cdlemg3.h
|- H = ( LHyp ` K )
cdlemg3.u
|- U = ( ( P .\/ Q ) ./\ W )
Assertion cdlemg3a
|- ( ( ( K e. HL /\ W e. H ) /\ ( P e. A /\ -. P .<_ W ) /\ Q e. A ) -> ( P .\/ Q ) = ( P .\/ U ) )

Proof

Step Hyp Ref Expression
1 cdlemg3.l
 |-  .<_ = ( le ` K )
2 cdlemg3.j
 |-  .\/ = ( join ` K )
3 cdlemg3.m
 |-  ./\ = ( meet ` K )
4 cdlemg3.a
 |-  A = ( Atoms ` K )
5 cdlemg3.h
 |-  H = ( LHyp ` K )
6 cdlemg3.u
 |-  U = ( ( P .\/ Q ) ./\ W )
7 1 2 3 4 5 6 cdleme8
 |-  ( ( ( K e. HL /\ W e. H ) /\ ( P e. A /\ -. P .<_ W ) /\ Q e. A ) -> ( P .\/ U ) = ( P .\/ Q ) )
8 7 eqcomd
 |-  ( ( ( K e. HL /\ W e. H ) /\ ( P e. A /\ -. P .<_ W ) /\ Q e. A ) -> ( P .\/ Q ) = ( P .\/ U ) )