This is an inofficial mirror of http://metamath.tirix.org for personal testing of a visualizer extension only.

Metamath Proof Explorer

Theorem cdleme17a

Description: Part of proof of Lemma E in Crawley p. 114, first part of 4th paragraph. F , G , and C represent f(s), f_s(p), and s_1 respectively. We show, in their notation, f_s(p)=(p \/ q) /\ (q \/ s_1). (Contributed by NM, 11-Oct-2012)

Ref Expression
Hypotheses cdleme17.l 
|- .<_ = ( le ` K )
cdleme17.j 
|- .\/ = ( join ` K )
cdleme17.m 
|- ./\ = ( meet ` K )
cdleme17.a 
|- A = ( Atoms ` K )
cdleme17.h 
|- H = ( LHyp ` K )
cdleme17.u 
|- U = ( ( P .\/ Q ) ./\ W )
cdleme17.f 
|- F = ( ( S .\/ U ) ./\ ( Q .\/ ( ( P .\/ S ) ./\ W ) ) )
cdleme17.g 
|- G = ( ( P .\/ Q ) ./\ ( F .\/ ( ( P .\/ S ) ./\ W ) ) )
cdleme17.c 
|- C = ( ( P .\/ S ) ./\ W )
Assertion cdleme17a 
|- ( ( ( K e. HL /\ W e. H ) /\ ( ( P e. A /\ -. P .<_ W ) /\ Q e. A /\ ( S e. A /\ -. S .<_ W ) ) /\ -. S .<_ ( P .\/ Q ) ) -> G = ( ( P .\/ Q ) ./\ ( Q .\/ C ) ) )

Proof

Step Hyp Ref Expression
1 cdleme17.l 
 |-  .<_ = ( le ` K )
2 cdleme17.j 
 |-  .\/ = ( join ` K )
3 cdleme17.m 
 |-  ./\ = ( meet ` K )
4 cdleme17.a 
 |-  A = ( Atoms ` K )
5 cdleme17.h 
 |-  H = ( LHyp ` K )
6 cdleme17.u 
 |-  U = ( ( P .\/ Q ) ./\ W )
7 cdleme17.f 
 |-  F = ( ( S .\/ U ) ./\ ( Q .\/ ( ( P .\/ S ) ./\ W ) ) )
8 cdleme17.g 
 |-  G = ( ( P .\/ Q ) ./\ ( F .\/ ( ( P .\/ S ) ./\ W ) ) )
9 cdleme17.c 
 |-  C = ( ( P .\/ S ) ./\ W )
10 1 2 3 4 5 6 7 8 9 cdleme7a 
 |-  G = ( ( P .\/ Q ) ./\ ( F .\/ C ) )
11 1 2 3 4 5 6 7 9 cdleme9 
 |-  ( ( ( K e. HL /\ W e. H ) /\ ( ( P e. A /\ -. P .<_ W ) /\ Q e. A /\ ( S e. A /\ -. S .<_ W ) ) /\ -. S .<_ ( P .\/ Q ) ) -> ( F .\/ C ) = ( Q .\/ C ) )
12 11 oveq2d 
 |-  ( ( ( K e. HL /\ W e. H ) /\ ( ( P e. A /\ -. P .<_ W ) /\ Q e. A /\ ( S e. A /\ -. S .<_ W ) ) /\ -. S .<_ ( P .\/ Q ) ) -> ( ( P .\/ Q ) ./\ ( F .\/ C ) ) = ( ( P .\/ Q ) ./\ ( Q .\/ C ) ) )
13 10 12 eqtrid 
 |-  ( ( ( K e. HL /\ W e. H ) /\ ( ( P e. A /\ -. P .<_ W ) /\ Q e. A /\ ( S e. A /\ -. S .<_ W ) ) /\ -. S .<_ ( P .\/ Q ) ) -> G = ( ( P .\/ Q ) ./\ ( Q .\/ C ) ) )