set IR = the InternalRel of R;
set CR = the carrier of R;
per cases
( the carrier of R = {} or not the carrier of R is empty )
;
suppose A1:
the
carrier of
R = {}
;
ex b1 being Relation of (Class (EqRel R)) st
for A, B being object holds
( [A,B] in b1 iff ex a, b being Element of R st
( A = Class ((EqRel R),a) & B = Class ((EqRel R),b) & a <= b ) )reconsider IT =
{} as
Relation of
(Class (EqRel R)) by RELSET_1:12;
take
IT
;
for A, B being object holds
( [A,B] in IT iff ex a, b being Element of R st
( A = Class ((EqRel R),a) & B = Class ((EqRel R),b) & a <= b ) )let A,
B be
object ;
( [A,B] in IT iff ex a, b being Element of R st
( A = Class ((EqRel R),a) & B = Class ((EqRel R),b) & a <= b ) )thus
(
[A,B] in IT implies ex
a,
b being
Element of
R st
(
A = Class (
(EqRel R),
a) &
B = Class (
(EqRel R),
b) &
a <= b ) )
;
( ex a, b being Element of R st
( A = Class ((EqRel R),a) & B = Class ((EqRel R),b) & a <= b ) implies [A,B] in IT )given a,
b being
Element of
R such that
A = Class (
(EqRel R),
a)
and
B = Class (
(EqRel R),
b)
and A2:
a <= b
;
[A,B] in IT
the
InternalRel of
R = {}
by A1;
hence
[A,B] in IT
by A2;
verum end; suppose
not the
carrier of
R is
empty
;
ex b1 being Relation of (Class (EqRel R)) st
for A, B being object holds
( [A,B] in b1 iff ex a, b being Element of R st
( A = Class ((EqRel R),a) & B = Class ((EqRel R),b) & a <= b ) )then reconsider R9 =
R as non
empty RelStr ;
set IT =
{ [(Class ((EqRel R),a)),(Class ((EqRel R),b))] where a, b is Element of R9 : a <= b } ;
set Y =
Class (EqRel R);
{ [(Class ((EqRel R),a)),(Class ((EqRel R),b))] where a, b is Element of R9 : a <= b } c= [:(Class (EqRel R)),(Class (EqRel R)):]
proof
let x be
object ;
TARSKI:def 3 ( not x in { [(Class ((EqRel R),a)),(Class ((EqRel R),b))] where a, b is Element of R9 : a <= b } or x in [:(Class (EqRel R)),(Class (EqRel R)):] )
assume
x in { [(Class ((EqRel R),a)),(Class ((EqRel R),b))] where a, b is Element of R9 : a <= b }
;
x in [:(Class (EqRel R)),(Class (EqRel R)):]
then consider a,
b being
Element of
R9 such that A3:
x = [(Class ((EqRel R),a)),(Class ((EqRel R),b))]
and
a <= b
;
A4:
Class (
(EqRel R),
a)
in Class (EqRel R)
by EQREL_1:def 3;
Class (
(EqRel R),
b)
in Class (EqRel R)
by EQREL_1:def 3;
hence
x in [:(Class (EqRel R)),(Class (EqRel R)):]
by A3, A4, ZFMISC_1:def 2;
verum
end; then reconsider IT =
{ [(Class ((EqRel R),a)),(Class ((EqRel R),b))] where a, b is Element of R9 : a <= b } as
Relation of
(Class (EqRel R)) ;
take
IT
;
for A, B being object holds
( [A,B] in IT iff ex a, b being Element of R st
( A = Class ((EqRel R),a) & B = Class ((EqRel R),b) & a <= b ) )let A,
B be
object ;
( [A,B] in IT iff ex a, b being Element of R st
( A = Class ((EqRel R),a) & B = Class ((EqRel R),b) & a <= b ) )hereby ( ex a, b being Element of R st
( A = Class ((EqRel R),a) & B = Class ((EqRel R),b) & a <= b ) implies [A,B] in IT )
assume
[A,B] in IT
;
ex a9, b9 being Element of R st
( A = Class ((EqRel R),a9) & B = Class ((EqRel R),b9) & a9 <= b9 )then consider a,
b being
Element of
R such that A5:
[A,B] = [(Class ((EqRel R),a)),(Class ((EqRel R),b))]
and A6:
a <= b
;
reconsider a9 =
a,
b9 =
b as
Element of
R ;
take a9 =
a9;
ex b9 being Element of R st
( A = Class ((EqRel R),a9) & B = Class ((EqRel R),b9) & a9 <= b9 )take b9 =
b9;
( A = Class ((EqRel R),a9) & B = Class ((EqRel R),b9) & a9 <= b9 )thus
(
A = Class (
(EqRel R),
a9) &
B = Class (
(EqRel R),
b9) &
a9 <= b9 )
by A5, A6, XTUPLE_0:1;
verum
end; given a,
b being
Element of
R such that A7:
A = Class (
(EqRel R),
a)
and A8:
B = Class (
(EqRel R),
b)
and A9:
a <= b
;
[A,B] in ITthus
[A,B] in IT
by A7, A8, A9;
verum end; end;