set cn = the carrier of N_5;
let L be LATTICE; ( ex K being full Sublattice of L st N_5 ,K are_isomorphic iff ex a, b, c, d, e being Element of L st
( a,b,c,d,e are_mutually_distinct & a "/\" b = a & a "/\" c = a & c "/\" e = c & d "/\" e = d & b "/\" c = a & b "/\" d = b & c "/\" d = a & b "\/" c = e & c "\/" d = e ) )
A1:
the carrier of N_5 = {0,(3 \ 1),2,(3 \ 2),3}
by YELLOW_1:1;
thus
( ex K being full Sublattice of L st N_5 ,K are_isomorphic implies ex a, b, c, d, e being Element of L st
( a,b,c,d,e are_mutually_distinct & a "/\" b = a & a "/\" c = a & c "/\" e = c & d "/\" e = d & b "/\" c = a & b "/\" d = b & c "/\" d = a & b "\/" c = e & c "\/" d = e ) )
( ex a, b, c, d, e being Element of L st
( a,b,c,d,e are_mutually_distinct & a "/\" b = a & a "/\" c = a & c "/\" e = c & d "/\" e = d & b "/\" c = a & b "/\" d = b & c "/\" d = a & b "\/" c = e & c "\/" d = e ) implies ex K being full Sublattice of L st N_5 ,K are_isomorphic )proof
reconsider td = 3
\ 2 as
Element of
N_5 by A1, ENUMSET1:def 3;
reconsider dw = 2 as
Element of
N_5 by A1, ENUMSET1:def 3;
reconsider t = 3 as
Element of
N_5 by A1, ENUMSET1:def 3;
reconsider tj = 3
\ 1 as
Element of
N_5 by A1, ENUMSET1:def 3;
reconsider cl = the
carrier of
L as non
empty set ;
reconsider z =
0 as
Element of
N_5 by A1, ENUMSET1:def 3;
given K being
full Sublattice of
L such that A2:
N_5 ,
K are_isomorphic
;
ex a, b, c, d, e being Element of L st
( a,b,c,d,e are_mutually_distinct & a "/\" b = a & a "/\" c = a & c "/\" e = c & d "/\" e = d & b "/\" c = a & b "/\" d = b & c "/\" d = a & b "\/" c = e & c "\/" d = e )
consider f being
Function of
N_5,
K such that A3:
f is
isomorphic
by A2;
A4:
not
K is
empty
by A3, WAYBEL_0:def 38;
then A5:
(
f is
one-to-one &
f is
monotone )
by A3, WAYBEL_0:def 38;
reconsider K =
K as non
empty SubRelStr of
L by A3, WAYBEL_0:def 38;
reconsider ck = the
carrier of
K as non
empty set ;
A6:
ck = rng f
by A3, WAYBEL_0:66;
reconsider g =
f as
Function of the
carrier of
N_5,
ck ;
reconsider a =
g . z,
b =
g . td,
c =
g . dw,
d =
g . tj,
e =
g . t as
Element of
K ;
reconsider ck =
ck as non
empty Subset of
cl by YELLOW_0:def 13;
A7:
b in ck
;
A8:
c in ck
;
A9:
e in ck
;
A10:
d in ck
;
a in ck
;
then reconsider A =
a,
B =
b,
C =
c,
D =
d,
E =
e as
Element of
L by A7, A8, A10, A9;
take
A
;
ex b, c, d, e being Element of L st
( A,b,c,d,e are_mutually_distinct & A "/\" b = A & A "/\" c = A & c "/\" e = c & d "/\" e = d & b "/\" c = A & b "/\" d = b & c "/\" d = A & b "\/" c = e & c "\/" d = e )
take
B
;
ex c, d, e being Element of L st
( A,B,c,d,e are_mutually_distinct & A "/\" B = A & A "/\" c = A & c "/\" e = c & d "/\" e = d & B "/\" c = A & B "/\" d = B & c "/\" d = A & B "\/" c = e & c "\/" d = e )
take
C
;
ex d, e being Element of L st
( A,B,C,d,e are_mutually_distinct & A "/\" B = A & A "/\" C = A & C "/\" e = C & d "/\" e = d & B "/\" C = A & B "/\" d = B & C "/\" d = A & B "\/" C = e & C "\/" d = e )
take
D
;
ex e being Element of L st
( A,B,C,D,e are_mutually_distinct & A "/\" B = A & A "/\" C = A & C "/\" e = C & D "/\" e = D & B "/\" C = A & B "/\" D = B & C "/\" D = A & B "\/" C = e & C "\/" D = e )
take
E
;
( A,B,C,D,E are_mutually_distinct & A "/\" B = A & A "/\" C = A & C "/\" E = C & D "/\" E = D & B "/\" C = A & B "/\" D = B & C "/\" D = A & B "\/" C = E & C "\/" D = E )
thus
A <> B
by A5, Th4, WAYBEL_1:def 1;
ZFMISC_1:def 7 ( not A = C & not A = D & not A = E & not B = C & not B = D & not B = E & not C = D & not C = E & not D = E & A "/\" B = A & A "/\" C = A & C "/\" E = C & D "/\" E = D & B "/\" C = A & B "/\" D = B & C "/\" D = A & B "\/" C = E & C "\/" D = E )
thus
A <> C
by A5, WAYBEL_1:def 1;
( not A = D & not A = E & not B = C & not B = D & not B = E & not C = D & not C = E & not D = E & A "/\" B = A & A "/\" C = A & C "/\" E = C & D "/\" E = D & B "/\" C = A & B "/\" D = B & C "/\" D = A & B "\/" C = E & C "\/" D = E )
thus
A <> D
by A5, Th3, WAYBEL_1:def 1;
( not A = E & not B = C & not B = D & not B = E & not C = D & not C = E & not D = E & A "/\" B = A & A "/\" C = A & C "/\" E = C & D "/\" E = D & B "/\" C = A & B "/\" D = B & C "/\" D = A & B "\/" C = E & C "\/" D = E )
thus
A <> E
by A5, WAYBEL_1:def 1;
( not B = C & not B = D & not B = E & not C = D & not C = E & not D = E & A "/\" B = A & A "/\" C = A & C "/\" E = C & D "/\" E = D & B "/\" C = A & B "/\" D = B & C "/\" D = A & B "\/" C = E & C "\/" D = E )
hence
B <> C
;
( not B = D & not B = E & not C = D & not C = E & not D = E & A "/\" B = A & A "/\" C = A & C "/\" E = C & D "/\" E = D & B "/\" C = A & B "/\" D = B & C "/\" D = A & B "\/" C = E & C "\/" D = E )
hence
B <> D
;
( not B = E & not C = D & not C = E & not D = E & A "/\" B = A & A "/\" C = A & C "/\" E = C & D "/\" E = D & B "/\" C = A & B "/\" D = B & C "/\" D = A & B "\/" C = E & C "\/" D = E )
hence
B <> E
;
( not C = D & not C = E & not D = E & A "/\" B = A & A "/\" C = A & C "/\" E = C & D "/\" E = D & B "/\" C = A & B "/\" D = B & C "/\" D = A & B "\/" C = E & C "\/" D = E )
hence
C <> D
;
( not C = E & not D = E & A "/\" B = A & A "/\" C = A & C "/\" E = C & D "/\" E = D & B "/\" C = A & B "/\" D = B & C "/\" D = A & B "\/" C = E & C "\/" D = E )
thus
C <> E
by A5, WAYBEL_1:def 1;
( not D = E & A "/\" B = A & A "/\" C = A & C "/\" E = C & D "/\" E = D & B "/\" C = A & B "/\" D = B & C "/\" D = A & B "\/" C = E & C "\/" D = E )
hence
D <> E
;
( A "/\" B = A & A "/\" C = A & C "/\" E = C & D "/\" E = D & B "/\" C = A & B "/\" D = B & C "/\" D = A & B "\/" C = E & C "\/" D = E )
z c= td
;
then
z <= td
by YELLOW_1:3;
then
a <= b
by A3, WAYBEL_0:66;
then
A <= B
by YELLOW_0:59;
hence
A "/\" B = A
by YELLOW_0:25;
( A "/\" C = A & C "/\" E = C & D "/\" E = D & B "/\" C = A & B "/\" D = B & C "/\" D = A & B "\/" C = E & C "\/" D = E )
z c= dw
;
then
z <= dw
by YELLOW_1:3;
then
a <= c
by A3, WAYBEL_0:66;
then
A <= C
by YELLOW_0:59;
hence
A "/\" C = A
by YELLOW_0:25;
( C "/\" E = C & D "/\" E = D & B "/\" C = A & B "/\" D = B & C "/\" D = A & B "\/" C = E & C "\/" D = E )
Segm 2
c= Segm 3
by NAT_1:39;
then
dw <= t
by YELLOW_1:3;
then
c <= e
by A3, WAYBEL_0:66;
then
C <= E
by YELLOW_0:59;
hence
C "/\" E = C
by YELLOW_0:25;
( D "/\" E = D & B "/\" C = A & B "/\" D = B & C "/\" D = A & B "\/" C = E & C "\/" D = E )
tj <= t
by YELLOW_1:3;
then
d <= e
by A3, WAYBEL_0:66;
then
D <= E
by YELLOW_0:59;
hence
D "/\" E = D
by YELLOW_0:25;
( B "/\" C = A & B "/\" D = B & C "/\" D = A & B "\/" C = E & C "\/" D = E )
thus
B "/\" C = A
( B "/\" D = B & C "/\" D = A & B "\/" C = E & C "\/" D = E )
td <= tj
by Lm1, YELLOW_1:3;
then
b <= d
by A3, WAYBEL_0:66;
then
B <= D
by YELLOW_0:59;
hence
B "/\" D = B
by YELLOW_0:25;
( C "/\" D = A & B "\/" C = E & C "\/" D = E )
thus
C "/\" D = A
( B "\/" C = E & C "\/" D = E )
thus
B "\/" C = E
C "\/" D = E
thus
C "\/" D = E
verum
end;
thus
( ex a, b, c, d, e being Element of L st
( a,b,c,d,e are_mutually_distinct & a "/\" b = a & a "/\" c = a & c "/\" e = c & d "/\" e = d & b "/\" c = a & b "/\" d = b & c "/\" d = a & b "\/" c = e & c "\/" d = e ) implies ex K being full Sublattice of L st N_5 ,K are_isomorphic )
verumproof
given a,
b,
c,
d,
e being
Element of
L such that AAA:
a,
b,
c,
d,
e are_mutually_distinct
and A59:
a "/\" b = a
and A60:
a "/\" c = a
and A61:
c "/\" e = c
and A62:
d "/\" e = d
and A63:
b "/\" c = a
and A64:
b "/\" d = b
and A65:
c "/\" d = a
and A66:
b "\/" c = e
and A67:
c "\/" d = e
;
ex K being full Sublattice of L st N_5 ,K are_isomorphic
set ck =
{a,b,c,d,e};
reconsider ck =
{a,b,c,d,e} as
Subset of
L ;
set K =
subrelstr ck;
A68:
the
carrier of
(subrelstr ck) = ck
by YELLOW_0:def 15;
A69:
subrelstr ck is
meet-inheriting
proof
let x,
y be
Element of
L;
YELLOW_0:def 16 ( not x in the carrier of (subrelstr ck) or not y in the carrier of (subrelstr ck) or not ex_inf_of {x,y},L or "/\" ({x,y},L) in the carrier of (subrelstr ck) )
assume that A70:
x in the
carrier of
(subrelstr ck)
and A71:
y in the
carrier of
(subrelstr ck)
and
ex_inf_of {x,y},
L
;
"/\" ({x,y},L) in the carrier of (subrelstr ck)
per cases
( ( x = a & y = a ) or ( x = a & y = b ) or ( x = a & y = c ) or ( x = a & y = d ) or ( x = a & y = e ) or ( x = b & y = a ) or ( x = b & y = b ) or ( x = b & y = c ) or ( x = b & y = d ) or ( x = b & y = e ) or ( x = c & y = a ) or ( x = c & y = b ) or ( x = c & y = c ) or ( x = c & y = d ) or ( x = c & y = e ) or ( x = d & y = a ) or ( x = d & y = b ) or ( x = d & y = c ) or ( x = d & y = d ) or ( x = d & y = e ) or ( x = e & y = a ) or ( x = e & y = b ) or ( x = e & y = c ) or ( x = e & y = d ) or ( x = e & y = e ) )
by A68, A70, A71, ENUMSET1:def 3;
end;
end;
subrelstr ck is
join-inheriting
proof
let x,
y be
Element of
L;
YELLOW_0:def 17 ( not x in the carrier of (subrelstr ck) or not y in the carrier of (subrelstr ck) or not ex_sup_of {x,y},L or "\/" ({x,y},L) in the carrier of (subrelstr ck) )
assume that A84:
x in the
carrier of
(subrelstr ck)
and A85:
y in the
carrier of
(subrelstr ck)
and
ex_sup_of {x,y},
L
;
"\/" ({x,y},L) in the carrier of (subrelstr ck)
per cases
( ( x = a & y = a ) or ( x = a & y = b ) or ( x = a & y = c ) or ( x = a & y = d ) or ( x = a & y = e ) or ( x = b & y = a ) or ( x = b & y = b ) or ( x = b & y = c ) or ( x = b & y = d ) or ( x = b & y = e ) or ( x = c & y = a ) or ( x = c & y = b ) or ( x = c & y = c ) or ( x = c & y = d ) or ( x = c & y = e ) or ( x = d & y = a ) or ( x = d & y = b ) or ( x = d & y = c ) or ( x = d & y = d ) or ( x = d & y = e ) or ( x = e & y = a ) or ( x = e & y = b ) or ( x = e & y = c ) or ( x = e & y = d ) or ( x = e & y = e ) )
by A68, A84, A85, ENUMSET1:def 3;
end;
end;
then reconsider K =
subrelstr ck as non
empty full Sublattice of
L by A69, YELLOW_0:def 15;
take
K
;
N_5 ,K are_isomorphic
thus
N_5 ,
K are_isomorphic
verumproof
reconsider t = 3 as
Element of
N_5 by A1, ENUMSET1:def 3;
reconsider td = 3
\ 2 as
Element of
N_5 by A1, ENUMSET1:def 3;
reconsider dw = 2 as
Element of
N_5 by A1, ENUMSET1:def 3;
A106:
now not dw = tdend;
reconsider tj = 3
\ 1 as
Element of
N_5 by A1, ENUMSET1:def 3;
reconsider z =
0 as
Element of
N_5 by A1, ENUMSET1:def 3;
defpred S1[
object ,
object ]
means for
X being
Element of
N_5 st
X = $1 holds
( (
X = z implies $2
= a ) & (
X = td implies $2
= b ) & (
X = dw implies $2
= c ) & (
X = tj implies $2
= d ) & (
X = t implies $2
= e ) );
A110:
now not tj = dwend;
A114:
now not tj = tdend;
A116:
for
x being
object st
x in the
carrier of
N_5 holds
ex
y being
object st
(
y in ck &
S1[
x,
y] )
proof
let x be
object ;
( x in the carrier of N_5 implies ex y being object st
( y in ck & S1[x,y] ) )
assume A117:
x in the
carrier of
N_5
;
ex y being object st
( y in ck & S1[x,y] )
per cases
( x = 0 or x = 3 \ 1 or x = 2 or x = 3 \ 2 or x = 3 )
by A1, A117, ENUMSET1:def 3;
suppose A118:
x = 0
;
ex y being object st
( y in ck & S1[x,y] )take y =
a;
( y in ck & S1[x,y] )thus
y in ck
by ENUMSET1:def 3;
S1[x,y]let X be
Element of
N_5;
( X = x implies ( ( X = z implies y = a ) & ( X = td implies y = b ) & ( X = dw implies y = c ) & ( X = tj implies y = d ) & ( X = t implies y = e ) ) )thus
(
X = x implies ( (
X = z implies
y = a ) & (
X = td implies
y = b ) & (
X = dw implies
y = c ) & (
X = tj implies
y = d ) & (
X = t implies
y = e ) ) )
by A118, Th3, Th4;
verum end; suppose A119:
x = 3
\ 1
;
ex y being object st
( y in ck & S1[x,y] )take y =
d;
( y in ck & S1[x,y] )thus
y in ck
by ENUMSET1:def 3;
S1[x,y]let X be
Element of
N_5;
( X = x implies ( ( X = z implies y = a ) & ( X = td implies y = b ) & ( X = dw implies y = c ) & ( X = tj implies y = d ) & ( X = t implies y = e ) ) )thus
(
X = x implies ( (
X = z implies
y = a ) & (
X = td implies
y = b ) & (
X = dw implies
y = c ) & (
X = tj implies
y = d ) & (
X = t implies
y = e ) ) )
by A110, A114, A112, A119, Th3;
verum end; suppose A120:
x = 2
;
ex y being object st
( y in ck & S1[x,y] )take y =
c;
( y in ck & S1[x,y] )thus
y in ck
by ENUMSET1:def 3;
S1[x,y]let X be
Element of
N_5;
( X = x implies ( ( X = z implies y = a ) & ( X = td implies y = b ) & ( X = dw implies y = c ) & ( X = tj implies y = d ) & ( X = t implies y = e ) ) )thus
(
X = x implies ( (
X = z implies
y = a ) & (
X = td implies
y = b ) & (
X = dw implies
y = c ) & (
X = tj implies
y = d ) & (
X = t implies
y = e ) ) )
by A110, A106, A120;
verum end; suppose A121:
x = 3
\ 2
;
ex y being object st
( y in ck & S1[x,y] )take y =
b;
( y in ck & S1[x,y] )thus
y in ck
by ENUMSET1:def 3;
S1[x,y]let X be
Element of
N_5;
( X = x implies ( ( X = z implies y = a ) & ( X = td implies y = b ) & ( X = dw implies y = c ) & ( X = tj implies y = d ) & ( X = t implies y = e ) ) )thus
(
X = x implies ( (
X = z implies
y = a ) & (
X = td implies
y = b ) & (
X = dw implies
y = c ) & (
X = tj implies
y = d ) & (
X = t implies
y = e ) ) )
by A114, A106, A108, A121, Th4;
verum end; suppose A122:
x = 3
;
ex y being object st
( y in ck & S1[x,y] )take y =
e;
( y in ck & S1[x,y] )thus
y in ck
by ENUMSET1:def 3;
S1[x,y]let X be
Element of
N_5;
( X = x implies ( ( X = z implies y = a ) & ( X = td implies y = b ) & ( X = dw implies y = c ) & ( X = tj implies y = d ) & ( X = t implies y = e ) ) )thus
(
X = x implies ( (
X = z implies
y = a ) & (
X = td implies
y = b ) & (
X = dw implies
y = c ) & (
X = tj implies
y = d ) & (
X = t implies
y = e ) ) )
by A112, A108, A122;
verum end; end;
end;
consider f being
Function of the
carrier of
N_5,
ck such that A123:
for
x being
object st
x in the
carrier of
N_5 holds
S1[
x,
f . x]
from FUNCT_2:sch 1(A116);
reconsider f =
f as
Function of
N_5,
K by A68;
A124:
now for x, y being Element of N_5 st f . x = f . y holds
x = ylet x,
y be
Element of
N_5;
( f . x = f . y implies x = y )assume A125:
f . x = f . y
;
x = ythus
x = y
verumproof
per cases
( ( x = z & y = z ) or ( x = tj & y = tj ) or ( x = td & y = td ) or ( x = dw & y = dw ) or ( x = t & y = t ) or ( x = z & y = tj ) or ( x = z & y = dw ) or ( x = z & y = td ) or ( x = z & y = t ) or ( x = tj & y = z ) or ( x = tj & y = dw ) or ( x = tj & y = td ) or ( x = tj & y = t ) or ( x = dw & y = z ) or ( x = dw & y = tj ) or ( x = dw & y = td ) or ( x = dw & y = t ) or ( x = td & y = z ) or ( x = td & y = tj ) or ( x = td & y = dw ) or ( x = td & y = t ) or ( x = t & y = z ) or ( x = t & y = tj ) or ( x = t & y = dw ) or ( x = t & y = td ) )
by A1, ENUMSET1:def 3;
end;
end; end;
A146:
rng f c= ck
A149:
dom f = the
carrier of
N_5
by FUNCT_2:def 1;
ck c= rng f
then A156:
rng f = ck
by A146;
A157:
for
x,
y being
Element of
N_5 holds
(
x <= y iff
f . x <= f . y )
proof
let x,
y be
Element of
N_5;
( x <= y iff f . x <= f . y )
thus
(
x <= y implies
f . x <= f . y )
( f . x <= f . y implies x <= y )proof
assume A158:
x <= y
;
f . x <= f . y
per cases
( ( x = z & y = z ) or ( x = z & y = td ) or ( x = z & y = dw ) or ( x = z & y = tj ) or ( x = z & y = t ) or ( x = td & y = z ) or ( x = td & y = td ) or ( x = td & y = dw ) or ( x = td & y = z ) or ( x = td & y = tj ) or ( x = td & y = t ) or ( x = dw & y = z ) or ( x = dw & y = td ) or ( x = dw & y = dw ) or ( x = dw & y = tj ) or ( x = dw & y = t ) or ( x = tj & y = z ) or ( x = tj & y = td ) or ( x = tj & y = dw ) or ( x = tj & y = tj ) or ( x = tj & y = t ) or ( x = t & y = z ) or ( x = t & y = td ) or ( x = t & y = dw ) or ( x = t & y = tj ) or ( x = t & y = t ) )
by A1, ENUMSET1:def 3;
end;
end;
thus
(
f . x <= f . y implies
x <= y )
verum
end;
take
f
;
WAYBEL_1:def 8 f is isomorphic
f is
one-to-one
by A124;
hence
f is
isomorphic
by A68, A156, A157, WAYBEL_0:66;
verum
end;
end;