set B = { x where x is Relation of X,X : x is Equivalence_Relation of X } ;
id X in { x where x is Relation of X,X : x is Equivalence_Relation of X } ;
then reconsider B = { x where x is Relation of X,X : x is Equivalence_Relation of X } as non empty set ;
defpred S₁[ Element of B, Element of B, Element of B] means ( $1 \/ $2 c= $3 & ( for x9 being Element of B st $1 \/ $2 c= x9 holds
$3 c= x9 ) );
A1: for x, y being Element of B ex z being Element of B st S₁[x,y,z]

let x, y be Element of B; :: thesis:
y in B ;
then A2: ex y99 being Relation of X,X st
( y = y99 & y99 is Equivalence_Relation of X ) ;
x in B ;
then ex x99 being Relation of X,X st
( x = x99 & x99 is Equivalence_Relation of X ) ;
then reconsider x1 = x, y1 = y as Equivalence_Relation of X by A2;
consider z being Equivalence_Relation of X such that
A3: x1 \/ y1 c= z and
A4: for x9 being Equivalence_Relation of X st x1 \/ y1 c= x9 holds
z c= x9 by EQREL_1:12;
z in B ;
then reconsider z9 = z as Element of B ;
take z9 ; :: thesis:
thus x \/ y c= z9 by A3; :: thesis:

hereby :: thesis:

let x9 be Element of B; :: thesis:
assume A5: x \/ y c= x9 ; :: thesis:
x9 in B ;
then ex x99 being Relation of X,X st
( x9 = x99 & x99 is Equivalence_Relation of X ) ;
then reconsider x19 = x9 as Equivalence_Relation of X ;
x \/ y c= x19 by A5;
hence z9 c= x9 by A4; :: thesis:

end;

consider B1 being BinOp of B such that
A6: for x, y being Element of B holds S₁[x,y,B1 . (x,y)] from BINOP_1:sch 3(A1);
defpred S₂[ Element of B, Element of B, Element of B] means $3 = $1 /\ $2;
A7: for x, y being Element of B ex z being Element of B st S₂[x,y,z]

proof

let x, y be Element of B; :: thesis:
y in B ;
then A8: ex y1 being Relation of X,X st
( y = y1 & y1 is Equivalence_Relation of X ) ;
x in B ;
then ex x1 being Relation of X,X st
( x = x1 & x1 is Equivalence_Relation of X ) ;
then reconsider x9 = x, y9 = y as Equivalence_Relation of X by A8;
set z = x9 /\ y9;
x9 /\ y9 in B ;
then reconsider z = x9 /\ y9 as Element of B ;
take z ; :: thesis:
thus S₂[x,y,z] ; :: thesis:

end;

consider B2 being BinOp of B such that
A9: for x, y being Element of B holds S₂[x,y,B2 . (x,y)] from BINOP_1:sch 3(A7);
reconsider L = LattStr(# B,B1,B2 #) as non empty LattStr ;

A10: now :: thesis:

let x, y be Equivalence_Relation of X; :: thesis:
A11: y in B ;
x in B ;
then reconsider x1 = x, y1 = y as Element of B by A11;
reconsider B19 = B1 . (x1,y1) as Element of B ;
B19 in B ;
then ex B199 being Relation of X,X st
( B19 = B199 & B199 is Equivalence_Relation of X ) ;
then reconsider B19 = B19 as Equivalence_Relation of X ;
A12: for x9 being Element of B st x1 \/ y1 c= x9 holds
B1 . (x1,y1) c= x9 by A6;

A13: now :: thesis:

let x9 be Equivalence_Relation of X; :: thesis:
assume A14: x \/ y c= x9 ; :: thesis:
x9 in B ;
then reconsider x19 = x9 as Element of B ;
x \/ y c= x19 by A14;
hence B1 . (x,y) c= x9 by A12; :: thesis:

end;

x1 \/ y1 c= B1 . (x1,y1) by A6;
then B19 = x "\/" y by A13, EQREL_1:def 2;
hence B1 . (x,y) = x "\/" y ; :: thesis:

end;

A15: now :: thesis:

let a, b, c be Element of B; :: thesis:
b in B ;
then A16: ex x2 being Relation of X,X st
( b = x2 & x2 is Equivalence_Relation of X ) ;
c in B ;
then A17: ex x3 being Relation of X,X st
( c = x3 & x3 is Equivalence_Relation of X ) ;
a in B ;
then ex x1 being Relation of X,X st
( a = x1 & x1 is Equivalence_Relation of X ) ;
then reconsider U1 = a, U2 = b, U3 = c as Equivalence_Relation of X by A16, A17;
thus B1 . (a,(B1 . (b,c))) = B1 . (a,(U2 "\/" U3)) by A10
.= U1 "\/" (U2 "\/" U3) by A10
.= (U1 "\/" U2) "\/" U3 by EQREL_1:13
.= B1 . ((U1 "\/" U2),c) by A10
.= B1 . ((B1 . (a,b)),c) by A10 ; :: thesis:

end;

A18: for a, b, c being Element of L holds a "\/" (b "\/" c) = (a "\/" b) "\/" c

proof

let a, b, c be Element of L; :: thesis:
reconsider x = a, y = b, z = c as Element of B ;
A19: the L_join of L . (a,b) = a "\/" b by LATTICES:def 1;
thus a "\/" (b "\/" c) = the L_join of L . (a,(b "\/" c)) by LATTICES:def 1
.= B1 . (x,(B1 . (y,z))) by LATTICES:def 1
.= the L_join of L . (( the L_join of L . (a,b)),c) by A15
.= (a "\/" b) "\/" c by A19, LATTICES:def 1 ; :: thesis:

end;

A20: now :: thesis:

let a, b be Element of B; :: thesis:
b in B ;
then A21: ex x2 being Relation of X,X st
( b = x2 & x2 is Equivalence_Relation of X ) ;
a in B ;
then ex x1 being Relation of X,X st
( a = x1 & x1 is Equivalence_Relation of X ) ;
then reconsider U1 = a, U2 = b as Equivalence_Relation of X by A21;
thus B1 . (a,b) = U2 "\/" U1 by A10
.= B1 . (b,a) by A10 ; :: thesis:

end;

A22: for a, b being Element of L holds a "\/" b = b "\/" a

proof

let a, b be Element of L; :: thesis:
reconsider x = a, y = b as Element of B ;
thus a "\/" b = B1 . (x,y) by LATTICES:def 1
.= the L_join of L . (b,a) by A20
.= b "\/" a by LATTICES:def 1 ; :: thesis:

end;

A23: now :: thesis:

let x, y be Equivalence_Relation of X; :: thesis:
A24: y in B ;
x in B ;
then reconsider x9 = x, y9 = y as Element of B by A24;
B2 . (x9,y9) = x9 /\ y9 by A9;
hence B2 . (x,y) = x /\ y ; :: thesis:

end;

A25: now :: thesis:

let a, b, c be Element of B; :: thesis:
b in B ;
then A26: ex x2 being Relation of X,X st
( b = x2 & x2 is Equivalence_Relation of X ) ;
c in B ;
then A27: ex x3 being Relation of X,X st
( c = x3 & x3 is Equivalence_Relation of X ) ;
a in B ;
then ex x1 being Relation of X,X st
( a = x1 & x1 is Equivalence_Relation of X ) ;
then reconsider U1 = a, U2 = b, U3 = c as Equivalence_Relation of X by A26, A27;
thus B2 . (a,(B2 . (b,c))) = B2 . (a,(U2 /\ U3)) by A23
.= U1 /\ (U2 /\ U3) by A23
.= (U1 /\ U2) /\ U3 by XBOOLE_1:16
.= B2 . ((U1 /\ U2),c) by A23
.= B2 . ((B2 . (a,b)),c) by A23 ; :: thesis:

end;

A28: for a, b, c being Element of L holds a "/\" (b "/\" c) = (a "/\" b) "/\" c

proof

let a, b, c be Element of L; :: thesis:
reconsider x = a, y = b, z = c as Element of B ;
A29: the L_meet of L . (a,b) = a "/\" b by LATTICES:def 2;
thus a "/\" (b "/\" c) = the L_meet of L . (a,(b "/\" c)) by LATTICES:def 2
.= B2 . (x,(B2 . (y,z))) by LATTICES:def 2
.= the L_meet of L . (( the L_meet of L . (a,b)),c) by A25
.= (a "/\" b) "/\" c by A29, LATTICES:def 2 ; :: thesis:

end;

A30: for a, b being Element of L holds a "/\" (a "\/" b) = a

proof

let a, b be Element of L; :: thesis:
reconsider x = a, y = b as Element of B ;
A31: B2 . (x,(B1 . (x,y))) = x

proof

y in B ;
then A32: ex x2 being Relation of X,X st
( y = x2 & x2 is Equivalence_Relation of X ) ;
x in B ;
then ex x1 being Relation of X,X st
( x = x1 & x1 is Equivalence_Relation of X ) ;
then reconsider U1 = x, U2 = y as Equivalence_Relation of X by A32;
B1 . (x,y) = U1 "\/" U2 by A10;
hence B2 . (x,(B1 . (x,y))) = U1 /\ (U1 "\/" U2) by A23
.= x by EQREL_1:16 ;
:: thesis:

end;

thus a "/\" (a "\/" b) = the L_meet of L . (a,(a "\/" b)) by LATTICES:def 2
.= a by A31, LATTICES:def 1 ; :: thesis:

end;

A33: now :: thesis:

let a, b be Element of B; :: thesis:
b in B ;
then A34: ex x2 being Relation of X,X st
( b = x2 & x2 is Equivalence_Relation of X ) ;
a in B ;
then ex x1 being Relation of X,X st
( a = x1 & x1 is Equivalence_Relation of X ) ;
then reconsider U1 = a, U2 = b as Equivalence_Relation of X by A34;
thus B2 . (a,b) = U2 /\ U1 by A23
.= B2 . (b,a) by A23 ; :: thesis:

end;

A35: for a, b being Element of L holds a "/\" b = b "/\" a

proof

let a, b be Element of L; :: thesis:
reconsider x = a, y = b as Element of B ;
thus a "/\" b = B2 . (x,y) by LATTICES:def 2
.= the L_meet of L . (b,a) by A33
.= b "/\" a by LATTICES:def 2 ; :: thesis:

end;

for a, b being Element of L holds (a "/\" b) "\/" b = b

proof

let a, b be Element of L; :: thesis:
reconsider x = a, y = b as Element of B ;
A36: B1 . ((B2 . (x,y)),y) = y

proof

y in B ;
then A37: ex x2 being Relation of X,X st
( y = x2 & x2 is Equivalence_Relation of X ) ;
x in B ;
then ex x1 being Relation of X,X st
( x = x1 & x1 is Equivalence_Relation of X ) ;
then reconsider U1 = x, U2 = y as Equivalence_Relation of X by A37;
B2 . (x,y) = U1 /\ U2 by A23;
hence B1 . ((B2 . (x,y)),y) = U2 "\/" (U1 /\ U2) by A10
.= y by EQREL_1:17 ;
:: thesis:

end;

thus (a "/\" b) "\/" b = the L_join of L . ((a "/\" b),b) by LATTICES:def 1
.= b by A36, LATTICES:def 2 ; :: thesis:

end;

then ( L is join-commutative & L is join-associative & L is meet-absorbing & L is meet-commutative & L is meet-associative & L is join-absorbing ) by A22, A18, A35, A28, A30, LATTICES:def 4, LATTICES:def 5, LATTICES:def 6, LATTICES:def 7, LATTICES:def 8, LATTICES:def 9;
then reconsider L = L as strict Lattice ;
take L ; :: thesis:
thus the carrier of L = { x where x is Relation of X,X : x is Equivalence_Relation of X } ; :: thesis:
thus for x, y being Equivalence_Relation of X holds
( the L_meet of L . (x,y) = x /\ y & the L_join of L . (x,y) = x "\/" y ) by A10, A23; :: thesis: