thus
( ex p being Element of S st
( not a _|_ & not x _|_ ) implies ex IT being Element of F st
for q being Element of S st not a _|_ & not x _|_ holds
IT = ((ProJ (a,b,q)) * (ProJ (q,a,x))) * (ProJ (x,q,y)) )
( ( for p being Element of S holds
( a _|_ or x _|_ ) ) implies ex b1 being Element of F st b1 = 0. F )proof
given p being
Element of
S such that A2:
( not
a _|_ & not
x _|_ )
;
ex IT being Element of F st
for q being Element of S st not a _|_ & not x _|_ holds
IT = ((ProJ (a,b,q)) * (ProJ (q,a,x))) * (ProJ (x,q,y))
take
((ProJ (a,b,p)) * (ProJ (p,a,x))) * (ProJ (x,p,y))
;
for q being Element of S st not a _|_ & not x _|_ holds
((ProJ (a,b,p)) * (ProJ (p,a,x))) * (ProJ (x,p,y)) = ((ProJ (a,b,q)) * (ProJ (q,a,x))) * (ProJ (x,q,y))
let q be
Element of
S;
( not a _|_ & not x _|_ implies ((ProJ (a,b,p)) * (ProJ (p,a,x))) * (ProJ (x,p,y)) = ((ProJ (a,b,q)) * (ProJ (q,a,x))) * (ProJ (x,q,y)) )
assume
( not
a _|_ & not
x _|_ )
;
((ProJ (a,b,p)) * (ProJ (p,a,x))) * (ProJ (x,p,y)) = ((ProJ (a,b,q)) * (ProJ (q,a,x))) * (ProJ (x,q,y))
hence
((ProJ (a,b,p)) * (ProJ (p,a,x))) * (ProJ (x,p,y)) = ((ProJ (a,b,q)) * (ProJ (q,a,x))) * (ProJ (x,q,y))
by A1, A2, Th26;
verum
end;
thus
( ( for p being Element of S holds
( a _|_ or x _|_ ) ) implies ex b1 being Element of F st b1 = 0. F )
; verum