let IT1, IT2 be Element of F; ( ( ex p being Element of S st
( not a _|_ & not x _|_ ) & ( for q being Element of S st not a _|_ & not x _|_ holds
IT1 = ((ProJ a,b,q) * (ProJ q,a,x)) * (ProJ x,q,y) ) & ( for q being Element of S st not a _|_ & not x _|_ holds
IT2 = ((ProJ a,b,q) * (ProJ q,a,x)) * (ProJ x,q,y) ) implies IT1 = IT2 ) & ( ( for p being Element of S holds
( a _|_ or x _|_ ) ) & IT1 = 0. F & IT2 = 0. F implies IT1 = IT2 ) )
thus
( ex p being Element of S st
( not a _|_ & not x _|_ ) & ( for q being Element of S st not a _|_ & not x _|_ holds
IT1 = ((ProJ a,b,q) * (ProJ q,a,x)) * (ProJ x,q,y) ) & ( for q being Element of S st not a _|_ & not x _|_ holds
IT2 = ((ProJ a,b,q) * (ProJ q,a,x)) * (ProJ x,q,y) ) implies IT1 = IT2 )
( ( for p being Element of S holds
( a _|_ or x _|_ ) ) & IT1 = 0. F & IT2 = 0. F implies IT1 = IT2 )proof
given p being
Element of
S such that A3:
( not
a _|_ & not
x _|_ )
;
( ex q being Element of S st
( not a _|_ & not x _|_ & not IT1 = ((ProJ a,b,q) * (ProJ q,a,x)) * (ProJ x,q,y) ) or ex q being Element of S st
( not a _|_ & not x _|_ & not IT2 = ((ProJ a,b,q) * (ProJ q,a,x)) * (ProJ x,q,y) ) or IT1 = IT2 )
consider r being
Element of
S such that A4:
( not
a _|_ & not
x _|_ )
by A3;
assume that A5:
for
q being
Element of
S st not
a _|_ & not
x _|_ holds
IT1 = ((ProJ a,b,q) * (ProJ q,a,x)) * (ProJ x,q,y)
and A6:
for
q being
Element of
S st not
a _|_ & not
x _|_ holds
IT2 = ((ProJ a,b,q) * (ProJ q,a,x)) * (ProJ x,q,y)
;
IT1 = IT2
IT1 = ((ProJ a,b,r) * (ProJ r,a,x)) * (ProJ x,r,y)
by A5, A4;
hence
IT1 = IT2
by A6, A4;
verum
end;
thus
( ( for p being Element of S holds
( a _|_ or x _|_ ) ) & IT1 = 0. F & IT2 = 0. F implies IT1 = IT2 )
; verum