let m, n be Nat; ( 2 * (m + 1) <= n implies card ((Choose (n,(m + 1),1,0)) \ (Domin_0 (n,(m + 1)))) = card (Choose (n,m,1,0)) )
defpred S1[ object , object ] means for p being XFinSequence of NAT
for k being Nat st $1 = p & (2 * k) + 1 = min* { N where N is Nat : 2 * (Sum (p | N)) > N } holds
ex r1, r2 being XFinSequence of NAT st
( $2 = r1 ^ r2 & len r1 = (2 * k) + 1 & len r1 = len (p | ((2 * k) + 1)) & p = (p | ((2 * k) + 1)) ^ r2 & ( for o being Nat st o < (2 * k) + 1 holds
r1 . o = 1 - (p . o) ) );
assume A1:
2 * (m + 1) <= n
; card ((Choose (n,(m + 1),1,0)) \ (Domin_0 (n,(m + 1)))) = card (Choose (n,m,1,0))
A2:
m <= m + m
by XREAL_1:31;
m <= m + 1
by NAT_1:13;
then
2 * m <= 2 * (m + 1)
by XREAL_1:64;
then
2 * m <= n
by A1, XXREAL_0:2;
then
m <= n
by A2, XXREAL_0:2;
then
(card n) choose m > 0
by Th26;
then reconsider W = Choose (n,m,1,0) as non empty finite set by CARD_1:27, CARD_FIN:16;
set Z = Domin_0 (n,(m + 1));
set CH = Choose (n,(m + 1),1,0);
A3:
for x being object st x in (Choose (n,(m + 1),1,0)) \ (Domin_0 (n,(m + 1))) holds
ex y being object st
( y in W & S1[x,y] )
proof
let x be
object ;
( x in (Choose (n,(m + 1),1,0)) \ (Domin_0 (n,(m + 1))) implies ex y being object st
( y in W & S1[x,y] ) )
assume A4:
x in (Choose (n,(m + 1),1,0)) \ (Domin_0 (n,(m + 1)))
;
ex y being object st
( y in W & S1[x,y] )
x in Choose (
n,
(m + 1),1,
0)
by A4, XBOOLE_0:def 5;
then consider p being
XFinSequence of
NAT such that A5:
p = x
and A6:
dom p = n
and A7:
rng p c= {0,1}
and A8:
Sum p = m + 1
by CARD_FIN:40;
not
p in Domin_0 (
n,
(m + 1))
by A4, A5, XBOOLE_0:def 5;
then
not
p is
dominated_by_0
by A6, A8, Def2;
then consider o being
Nat such that A9:
(
(2 * o) + 1
= min* { N where N is Nat : 2 * (Sum (p | N)) > N } &
(2 * o) + 1
<= dom p &
o = Sum (p | (2 * o)) &
p . (2 * o) = 1 )
by A7, Th15;
set q =
p | ((2 * o) + 1);
consider r2 being
XFinSequence of
NAT such that A10:
p = (p | ((2 * o) + 1)) ^ r2
by Th1;
rng (p | ((2 * o) + 1)) c= {0,1}
by A7;
then consider r1 being
XFinSequence of
NAT such that A11:
len (p | ((2 * o) + 1)) = len r1
and A12:
rng r1 c= {0,1}
and A13:
for
i being
Nat st
i <= len (p | ((2 * o) + 1)) holds
(Sum ((p | ((2 * o) + 1)) | i)) + (Sum (r1 | i)) = 1
* i
and A14:
for
i being
Nat st
i in len (p | ((2 * o) + 1)) holds
r1 . i = 1
- ((p | ((2 * o) + 1)) . i)
by Th25;
take R =
r1 ^ r2;
( R in W & S1[x,R] )
len p = (len r1) + (len r2)
by A11, A10, AFINSQ_1:17;
then A15:
dom R = n
by A6, AFINSQ_1:def 3;
rng r2 c= rng p
by A10, AFINSQ_1:25;
then
rng r2 c= {0,1}
by A7;
then
(rng r1) \/ (rng r2) c= {0,1}
by A12, XBOOLE_1:8;
then A16:
rng R c= {0,1}
by AFINSQ_1:26;
(
(p | ((2 * o) + 1)) | (dom (p | ((2 * o) + 1))) = p | ((2 * o) + 1) &
r1 | (dom r1) = r1 )
;
then A17:
(Sum (p | ((2 * o) + 1))) + (Sum r1) = 1
* (len (p | ((2 * o) + 1)))
by A11, A13;
A18:
2
* o < (2 * o) + 1
by NAT_1:13;
then
Segm (2 * o) c= Segm ((2 * o) + 1)
by NAT_1:39;
then A19:
(p | ((2 * o) + 1)) | (2 * o) = p | (2 * o)
by RELAT_1:74;
A20:
Segm ((2 * o) + 1) c= Segm (len p)
by A9, NAT_1:39;
then A21:
dom (p | ((2 * o) + 1)) = (2 * o) + 1
by RELAT_1:62;
A22:
len (p | ((2 * o) + 1)) = (2 * o) + 1
by A20, RELAT_1:62;
then A23:
p | ((2 * o) + 1) = ((p | ((2 * o) + 1)) | (2 * o)) ^ <%((p | ((2 * o) + 1)) . (2 * o))%>
by AFINSQ_1:56;
2
* o in Segm ((2 * o) + 1)
by A18, NAT_1:44;
then
(p | ((2 * o) + 1)) . (2 * o) = p . (2 * o)
by A22, FUNCT_1:47;
then
Sum (p | ((2 * o) + 1)) = (Sum (p | (2 * o))) + (Sum <%(p . (2 * o))%>)
by A23, A19, AFINSQ_2:55;
then A24:
Sum (p | ((2 * o) + 1)) = o + 1
by A9, AFINSQ_2:53;
m + 1
= (Sum (p | ((2 * o) + 1))) + (Sum r2)
by A8, A10, AFINSQ_2:55;
then
(Sum r1) + (Sum r2) = ((m + 1) - ((2 * o) + 1)) + (2 * o)
by A17, A21, A24;
then
Sum (r1 ^ r2) = m
by AFINSQ_2:55;
hence
R in W
by A15, A16, CARD_FIN:40;
S1[x,R]
thus
S1[
x,
R]
verum
end;
consider F being Function of ((Choose (n,(m + 1),1,0)) \ (Domin_0 (n,(m + 1)))),W such that
A27:
for x being object st x in (Choose (n,(m + 1),1,0)) \ (Domin_0 (n,(m + 1))) holds
S1[x,F . x]
from FUNCT_2:sch 1(A3);
W c= rng F
proof
let x be
object ;
TARSKI:def 3 ( not x in W or x in rng F )
assume
x in W
;
x in rng F
then consider p being
XFinSequence of
NAT such that A28:
p = x
and A29:
dom p = n
and A30:
rng p c= {0,1}
and A31:
Sum p = m
by CARD_FIN:40;
set M =
{ N where N is Nat : 2 * (Sum (p | N)) < N } ;
m < m + 1
by NAT_1:13;
then
2
* m < 2
* (m + 1)
by XREAL_1:68;
then
2
* m < n
by A1, XXREAL_0:2;
then
( 2
* (Sum (p | n)) < n &
n in NAT )
by A29, A31, ORDINAL1:def 12;
then A32:
n in { N where N is Nat : 2 * (Sum (p | N)) < N }
;
{ N where N is Nat : 2 * (Sum (p | N)) < N } c= NAT
then reconsider M =
{ N where N is Nat : 2 * (Sum (p | N)) < N } as non
empty Subset of
NAT by A32;
ex
k being
Nat st
(
(2 * k) + 1
= min* M &
Sum (p | ((2 * k) + 1)) = k &
(2 * k) + 1
<= dom p )
proof
set mm =
min* M;
min* M in M
by NAT_1:def 1;
then A33:
ex
o being
Nat st
(
min* M = o & 2
* (Sum (p | o)) < o )
;
then reconsider m1 =
(min* M) - 1 as
Nat by NAT_1:20;
A34:
2
* (Sum (p | (min* M))) < m1 + 1
by A33;
A35:
m1 < m1 + 1
by NAT_1:13;
then
Segm m1 c= Segm (min* M)
by NAT_1:39;
then A36:
(p | (min* M)) | m1 = p | m1
by RELAT_1:74;
min* M <= dom p
by A29, A32, NAT_1:def 1;
then A37:
Segm (min* M) c= Segm (len p)
by NAT_1:39;
then
dom (p | (min* M)) = min* M
by RELAT_1:62;
then
m1 in Segm (dom (p | (min* M)))
by A35, NAT_1:44;
then A38:
(p | (min* M)) . m1 = p . m1
by FUNCT_1:47;
m1 < m1 + 1
by NAT_1:13;
then
not
m1 in M
by NAT_1:def 1;
then
2
* (Sum (p | m1)) >= m1
;
then A39:
(
Sum <%(p . m1)%> = p . m1 &
(2 * (Sum (p | m1))) + (2 * (p . m1)) >= m1 + 0 )
by AFINSQ_2:53, XREAL_1:7;
reconsider S =
Sum (p | (min* M)) as
Nat ;
take
S
;
( (2 * S) + 1 = min* M & Sum (p | ((2 * S) + 1)) = S & (2 * S) + 1 <= dom p )
A40:
min* M <= dom p
by A29, A32, NAT_1:def 1;
len (p | (min* M)) = m1 + 1
by A37, RELAT_1:62;
then
p | (min* M) = ((p | (min* M)) | m1) ^ <%((p | (min* M)) . m1)%>
by AFINSQ_1:56;
then
Sum (p | (min* M)) = (Sum (p | m1)) + (Sum <%(p . m1)%>)
by A38, A36, AFINSQ_2:55;
hence
(
(2 * S) + 1
= min* M &
Sum (p | ((2 * S) + 1)) = S &
(2 * S) + 1
<= dom p )
by A40, A39, A34, NAT_1:9;
verum
end;
then consider k being
Nat such that A41:
(2 * k) + 1
= min* M
and A42:
Sum (p | ((2 * k) + 1)) = k
and A43:
(2 * k) + 1
<= dom p
;
set k1 =
(2 * k) + 1;
consider q being
XFinSequence of
NAT such that A44:
p = (p | ((2 * k) + 1)) ^ q
by Th1;
rng (p | ((2 * k) + 1)) c= {0,1}
by A30;
then consider r being
XFinSequence of
NAT such that A45:
len r = len (p | ((2 * k) + 1))
and A46:
rng r c= {0,1}
and A47:
for
i being
Nat st
i <= len (p | ((2 * k) + 1)) holds
(Sum ((p | ((2 * k) + 1)) | i)) + (Sum (r | i)) = 1
* i
and A48:
for
i being
Nat st
i in len (p | ((2 * k) + 1)) holds
r . i = 1
- ((p | ((2 * k) + 1)) . i)
by Th25;
set rq =
r ^ q;
A49:
dom (r ^ q) = (len (p | ((2 * k) + 1))) + (len q)
by A45, AFINSQ_1:def 3;
A50:
m = k + (Sum q)
by A31, A42, A44, AFINSQ_2:55;
dom (r ^ q) = (len (p | ((2 * k) + 1))) + (len q)
by A45, AFINSQ_1:def 3;
then A51:
dom (r ^ q) = dom p
by A44, AFINSQ_1:def 3;
(
(p | ((2 * k) + 1)) | (dom (p | ((2 * k) + 1))) = p | ((2 * k) + 1) &
r | (dom r) = r )
;
then A52:
(Sum (p | ((2 * k) + 1))) + (Sum r) = 1
* (len (p | ((2 * k) + 1)))
by A45, A47;
rng q c= rng p
by A44, AFINSQ_1:25;
then
rng q c= {0,1}
by A30;
then
(rng r) \/ (rng q) c= {0,1}
by A46, XBOOLE_1:8;
then A53:
rng (r ^ q) c= {0,1}
by AFINSQ_1:26;
A54:
Segm ((2 * k) + 1) c= Segm (len p)
by A43, NAT_1:39;
then A55:
len (p | ((2 * k) + 1)) = (2 * k) + 1
by RELAT_1:62;
then A56:
(r ^ q) | ((2 * k) + 1) = r
by A45, AFINSQ_1:57;
A57:
((2 * k) + 1) + 1
> (2 * k) + 1
by NAT_1:13;
then A58:
(2 * k) + 1
< 2
* (Sum r)
by A42, A52, A55;
A59:
2
* (Sum r) > (2 * k) + 1
by A42, A52, A55, A57;
then consider j being
Nat such that A60:
(
(2 * j) + 1
= min* { N where N is Nat : 2 * (Sum ((r ^ q) | N)) > N } &
(2 * j) + 1
<= dom (r ^ q) &
j = Sum ((r ^ q) | (2 * j)) &
(r ^ q) . (2 * j) = 1 )
by A53, A56, Th2, Th15;
set j1 =
(2 * j) + 1;
A61:
len ((p | ((2 * k) + 1)) ^ q) = (len (p | ((2 * k) + 1))) + (len q)
by AFINSQ_1:def 3;
not
r ^ q is
dominated_by_0
by A59, A56, Th2;
then A62:
not
r ^ q in Domin_0 (
n,
(m + 1))
by Th20;
set rqj =
(r ^ q) | ((2 * j) + 1);
Sum (r ^ q) = (Sum r) + (Sum q)
by AFINSQ_2:55;
then
r ^ q in Choose (
n,
(m + 1),1,
0)
by A29, A42, A52, A55, A51, A53, A50, CARD_FIN:40;
then A63:
r ^ q in (Choose (n,(m + 1),1,0)) \ (Domin_0 (n,(m + 1)))
by A62, XBOOLE_0:def 5;
then consider r1,
r2 being
XFinSequence of
NAT such that A64:
F . (r ^ q) = r1 ^ r2
and A65:
len r1 = (2 * j) + 1
and A66:
(
len r1 = len ((r ^ q) | ((2 * j) + 1)) &
r ^ q = ((r ^ q) | ((2 * j) + 1)) ^ r2 )
and A67:
for
i being
Nat st
i < (2 * j) + 1 holds
r1 . i = 1
- ((r ^ q) . i)
by A27, A60;
A68:
dom (r ^ q) = (len r1) + (len r2)
by A66, AFINSQ_1:def 3;
then A69:
len (r1 ^ r2) = len ((p | ((2 * k) + 1)) ^ q)
by A49, A61, AFINSQ_1:17;
reconsider K =
{ N where N is Nat : 2 * (Sum ((r ^ q) | N)) > N } as non
empty Subset of
NAT by A60, NAT_1:def 1;
(r ^ q) | ((2 * k) + 1) = r
by A45, A55, AFINSQ_1:57;
then
(2 * k) + 1
in K
by A58;
then A70:
(2 * k) + 1
>= (2 * j) + 1
by A60, NAT_1:def 1;
then
Segm ((2 * j) + 1) c= Segm ((2 * k) + 1)
by NAT_1:39;
then A71:
(p | ((2 * k) + 1)) | ((2 * j) + 1) = p | ((2 * j) + 1)
by RELAT_1:74;
(2 * j) + 1
in K
by A60, NAT_1:def 1;
then A72:
ex
N being
Nat st
(
N = (2 * j) + 1 & 2
* (Sum ((r ^ q) | N)) > N )
;
(Sum ((p | ((2 * k) + 1)) | ((2 * j) + 1))) + (Sum (r | ((2 * j) + 1))) = ((2 * j) + 1) * 1
by A47, A55, A70;
then
2
* (Sum (r | ((2 * j) + 1))) = (2 * ((2 * j) + 1)) - (2 * (Sum (p | ((2 * j) + 1))))
by A71;
then
((2 * j) + 1) + (((2 * j) + 1) - (2 * (Sum (p | ((2 * j) + 1))))) > (2 * (Sum (p | ((2 * j) + 1)))) + (((2 * j) + 1) - (2 * (Sum (p | ((2 * j) + 1)))))
by A45, A55, A70, A72, AFINSQ_1:58;
then
(2 * j) + 1
> 2
* (Sum (p | ((2 * j) + 1)))
by XREAL_1:6;
then
(2 * j) + 1
in M
;
then
(2 * j) + 1
>= (2 * k) + 1
by A41, NAT_1:def 1;
then A73:
(2 * j) + 1
= (2 * k) + 1
by A70, XXREAL_0:1;
A74:
len ((p | ((2 * k) + 1)) ^ q) = len (r ^ q)
by A49, AFINSQ_1:def 3;
now for i being Nat st i < len (r1 ^ r2) holds
(r1 ^ r2) . i = ((p | ((2 * k) + 1)) ^ q) . ilet i be
Nat;
( i < len (r1 ^ r2) implies (r1 ^ r2) . i = ((p | ((2 * k) + 1)) ^ q) . i )assume A75:
i < len (r1 ^ r2)
;
(r1 ^ r2) . i = ((p | ((2 * k) + 1)) ^ q) . inow (r1 ^ r2) . i = ((p | ((2 * k) + 1)) ^ q) . iper cases
( i < len r1 or i >= len r1 )
;
suppose A76:
i < len r1
;
(r1 ^ r2) . i = ((p | ((2 * k) + 1)) ^ q) . ithen A77:
(
i in dom r1 &
r1 . i = 1
- ((r ^ q) . i) )
by A65, A67, AFINSQ_1:86;
A78:
i in len r1
by A76, AFINSQ_1:86;
A79:
(
len r1 = len (p | ((2 * k) + 1)) &
i in NAT )
by A54, A65, A73, ORDINAL1:def 12, RELAT_1:62;
then A80:
r . i = 1
- ((p | ((2 * k) + 1)) . i)
by A48, A78;
(
((p | ((2 * k) + 1)) ^ q) . i = (p | ((2 * k) + 1)) . i &
(r ^ q) . i = r . i )
by A45, A78, A79, AFINSQ_1:def 3;
hence
(r1 ^ r2) . i = ((p | ((2 * k) + 1)) ^ q) . i
by A80, A77, AFINSQ_1:def 3;
verum end; suppose A81:
i >= len r1
;
(r1 ^ r2) . i = ((p | ((2 * k) + 1)) ^ q) . ithen A82:
((p | ((2 * k) + 1)) ^ q) . i = q . (i - (len r))
by A45, A55, A65, A73, A69, A75, AFINSQ_1:19;
(
(r1 ^ r2) . i = r2 . (i - (len r1)) &
(r ^ q) . i = q . (i - (len r)) )
by A45, A55, A65, A73, A69, A74, A75, A81, AFINSQ_1:19;
hence
(r1 ^ r2) . i = ((p | ((2 * k) + 1)) ^ q) . i
by A66, A69, A74, A75, A81, A82, AFINSQ_1:19;
verum end; end; end; hence
(r1 ^ r2) . i = ((p | ((2 * k) + 1)) ^ q) . i
;
verum end;
then A83:
r1 ^ r2 = (p | ((2 * k) + 1)) ^ q
by A68, A49, A61, AFINSQ_1:9, AFINSQ_1:17;
r ^ q in dom F
by A63, FUNCT_2:def 1;
hence
x in rng F
by A28, A44, A64, A83, FUNCT_1:3;
verum
end;
then A84:
rng F = W
;
A85:
F is one-to-one
proof
let x,
y be
object ;
FUNCT_1:def 4 ( not x in dom F or not y in dom F or not F . x = F . y or x = y )
assume that A86:
x in dom F
and A87:
y in dom F
and A88:
F . x = F . y
;
x = y
x in Choose (
n,
(m + 1),1,
0)
by A86, XBOOLE_0:def 5;
then consider p being
XFinSequence of
NAT such that A89:
p = x
and A90:
dom p = n
and A91:
rng p c= {0,1}
and A92:
Sum p = m + 1
by CARD_FIN:40;
not
p in Domin_0 (
n,
(m + 1))
by A86, A89, XBOOLE_0:def 5;
then
not
p is
dominated_by_0
by A90, A92, Def2;
then consider k1 being
Nat such that A93:
(
(2 * k1) + 1
= min* { N where N is Nat : 2 * (Sum (p | N)) > N } &
(2 * k1) + 1
<= dom p &
k1 = Sum (p | (2 * k1)) &
p . (2 * k1) = 1 )
by A91, Th15;
y in Choose (
n,
(m + 1),1,
0)
by A87, XBOOLE_0:def 5;
then consider q being
XFinSequence of
NAT such that A94:
q = y
and A95:
dom q = n
and A96:
rng q c= {0,1}
and A97:
Sum q = m + 1
by CARD_FIN:40;
not
q in Domin_0 (
n,
(m + 1))
by A87, A94, XBOOLE_0:def 5;
then
not
q is
dominated_by_0
by A95, A97, Def2;
then consider k2 being
Nat such that A98:
(
(2 * k2) + 1
= min* { N where N is Nat : 2 * (Sum (q | N)) > N } &
(2 * k2) + 1
<= dom q &
k2 = Sum (q | (2 * k2)) &
q . (2 * k2) = 1 )
by A96, Th15;
A99:
len q = n
by A95;
reconsider M =
{ N where N is Nat : 2 * (Sum (q | N)) > N } as non
empty Subset of
NAT by A98, NAT_1:def 1;
set qk =
q | ((2 * k2) + 1);
consider s1,
s2 being
XFinSequence of
NAT such that A100:
F . y = s1 ^ s2
and A101:
len s1 = (2 * k2) + 1
and A102:
len s1 = len (q | ((2 * k2) + 1))
and A103:
q = (q | ((2 * k2) + 1)) ^ s2
and A104:
for
i being
Nat st
i < (2 * k2) + 1 holds
s1 . i = 1
- (q . i)
by A27, A87, A94, A98;
A105:
len q = (len (q | ((2 * k2) + 1))) + (len s2)
by A103, AFINSQ_1:17;
then A106:
len q = len (s1 ^ s2)
by A102, AFINSQ_1:17;
reconsider K =
{ N where N is Nat : 2 * (Sum (p | N)) > N } as non
empty Subset of
NAT by A93, NAT_1:def 1;
set pk =
p | ((2 * k1) + 1);
consider r1,
r2 being
XFinSequence of
NAT such that A107:
F . x = r1 ^ r2
and A108:
len r1 = (2 * k1) + 1
and A109:
len r1 = len (p | ((2 * k1) + 1))
and A110:
p = (p | ((2 * k1) + 1)) ^ r2
and A111:
for
i being
Nat st
i < (2 * k1) + 1 holds
r1 . i = 1
- (p . i)
by A27, A86, A89, A93;
assume
x <> y
;
contradiction
then consider i being
Nat such that A112:
i < len p
and A113:
p . i <> q . i
by A89, A90, A94, A95, AFINSQ_1:9;
A114:
len p = (len (p | ((2 * k1) + 1))) + (len r2)
by A110, AFINSQ_1:17;
then A115:
len p = len (r1 ^ r2)
by A109, AFINSQ_1:17;
now contradictionper cases
( k1 = k2 or k1 > k2 or k1 < k2 )
by XXREAL_0:1;
suppose A116:
k1 = k2
;
contradictionnow contradictionper cases
( i < len (p | ((2 * k1) + 1)) or i >= len (p | ((2 * k1) + 1)) )
;
suppose A117:
i < len (p | ((2 * k1) + 1))
;
contradictionthen
i in len (p | ((2 * k1) + 1))
by AFINSQ_1:86;
then A118:
(
r1 . i = (r1 ^ r2) . i &
s1 . i = (s1 ^ s2) . i )
by A108, A109, A101, A116, AFINSQ_1:def 3;
(
r1 . i = 1
- (p . i) &
s1 . i = 1
- (q . i) )
by A108, A109, A111, A104, A116, A117;
hence
contradiction
by A88, A107, A100, A113, A118;
verum end; suppose A119:
i >= len (p | ((2 * k1) + 1))
;
contradictionthen A120:
(s1 ^ s2) . i = s2 . (i - (len (p | ((2 * k1) + 1))))
by A90, A108, A109, A95, A101, A106, A112, A116, AFINSQ_1:19;
(
p . i = r2 . (i - (len (p | ((2 * k1) + 1)))) &
q . i = s2 . (i - (len (p | ((2 * k1) + 1)))) )
by A90, A108, A109, A110, A101, A102, A103, A99, A112, A116, A119, AFINSQ_1:19;
hence
contradiction
by A88, A107, A109, A100, A115, A112, A113, A119, A120, AFINSQ_1:19;
verum end; end; end; hence
contradiction
;
verum end; suppose A121:
k1 > k2
;
contradiction
len s1 <= len p
by A90, A95, A102, A105, NAT_1:11;
then A122:
Segm (len s1) c= Segm (len p)
by NAT_1:39;
2
* k2 < 2
* k1
by A121, XREAL_1:68;
then A123:
len s1 < len r1
by A108, A101, XREAL_1:6;
then
(s1 ^ s2) | (len s1) = r1 | (len s1)
by A88, A107, A100, AFINSQ_1:58;
then A124:
s1 = r1 | (len s1)
by AFINSQ_1:57;
A125:
for
k being
Nat st
k < len (q | ((2 * k2) + 1)) holds
(q | ((2 * k2) + 1)) . k = (p | (len (q | ((2 * k2) + 1)))) . k
proof
let k be
Nat;
( k < len (q | ((2 * k2) + 1)) implies (q | ((2 * k2) + 1)) . k = (p | (len (q | ((2 * k2) + 1)))) . k )
assume A126:
k < len (q | ((2 * k2) + 1))
;
(q | ((2 * k2) + 1)) . k = (p | (len (q | ((2 * k2) + 1)))) . k
A127:
k in len s1
by A102, A126, AFINSQ_1:86;
then A128:
k in (dom q) /\ (len s1)
by A90, A95, A122, XBOOLE_0:def 4;
k in (dom p) /\ (len s1)
by A122, A127, XBOOLE_0:def 4;
then A129:
p . k = (p | (len (q | ((2 * k2) + 1)))) . k
by A102, FUNCT_1:48;
A130:
k < len r1
by A102, A123, A126, XXREAL_0:2;
then A131:
r1 . k = 1
- (p . k)
by A108, A111;
k in dom r1
by A130, AFINSQ_1:86;
then
k in (dom r1) /\ (len s1)
by A127, XBOOLE_0:def 4;
then A132:
r1 . k = (r1 | (len s1)) . k
by FUNCT_1:48;
s1 . k = 1
- (q . k)
by A101, A102, A104, A126;
hence
(q | ((2 * k2) + 1)) . k = (p | (len (q | ((2 * k2) + 1)))) . k
by A101, A124, A131, A132, A128, A129, FUNCT_1:48;
verum
end;
(2 * k2) + 1
in M
by A98, NAT_1:def 1;
then A133:
ex
N being
Nat st
(
(2 * k2) + 1
= N & 2
* (Sum (q | N)) > N )
;
len (q | ((2 * k2) + 1)) = len (p | (len (q | ((2 * k2) + 1))))
by A102, A122, RELAT_1:62;
then
q | ((2 * k2) + 1) = p | (len (q | ((2 * k2) + 1)))
by A125, AFINSQ_1:9;
then
len (q | ((2 * k2) + 1)) in K
by A101, A102, A133;
hence
contradiction
by A93, A108, A102, A123, NAT_1:def 1;
verum end; suppose A134:
k1 < k2
;
contradiction
len r1 <= len q
by A90, A109, A95, A114, NAT_1:11;
then A135:
Segm (len r1) c= Segm (len q)
by NAT_1:39;
2
* k1 < 2
* k2
by A134, XREAL_1:68;
then A136:
len r1 < len s1
by A108, A101, XREAL_1:6;
then
(r1 ^ r2) | (len r1) = s1 | (len r1)
by A88, A107, A100, AFINSQ_1:58;
then A137:
r1 = s1 | (len r1)
by AFINSQ_1:57;
A138:
for
k being
Nat st
k < len (p | ((2 * k1) + 1)) holds
(p | ((2 * k1) + 1)) . k = (q | (len (p | ((2 * k1) + 1)))) . k
proof
let k be
Nat;
( k < len (p | ((2 * k1) + 1)) implies (p | ((2 * k1) + 1)) . k = (q | (len (p | ((2 * k1) + 1)))) . k )
assume A139:
k < len (p | ((2 * k1) + 1))
;
(p | ((2 * k1) + 1)) . k = (q | (len (p | ((2 * k1) + 1)))) . k
A140:
k in len r1
by A109, A139, AFINSQ_1:86;
then A141:
k in (dom p) /\ (len r1)
by A90, A95, A135, XBOOLE_0:def 4;
k in (dom q) /\ (len r1)
by A135, A140, XBOOLE_0:def 4;
then A142:
q . k = (q | (len (p | ((2 * k1) + 1)))) . k
by A109, FUNCT_1:48;
A143:
k < len s1
by A109, A136, A139, XXREAL_0:2;
then A144:
s1 . k = 1
- (q . k)
by A101, A104;
k in dom s1
by A143, AFINSQ_1:86;
then
k in (dom s1) /\ (len r1)
by A140, XBOOLE_0:def 4;
then A145:
s1 . k = (s1 | (len r1)) . k
by FUNCT_1:48;
r1 . k = 1
- (p . k)
by A108, A109, A111, A139;
hence
(p | ((2 * k1) + 1)) . k = (q | (len (p | ((2 * k1) + 1)))) . k
by A108, A137, A144, A145, A141, A142, FUNCT_1:48;
verum
end;
(2 * k1) + 1
in K
by A93, NAT_1:def 1;
then A146:
ex
N being
Nat st
(
(2 * k1) + 1
= N & 2
* (Sum (p | N)) > N )
;
len (p | ((2 * k1) + 1)) = len (q | (len (p | ((2 * k1) + 1))))
by A109, A135, RELAT_1:62;
then
p | ((2 * k1) + 1) = q | (len (p | ((2 * k1) + 1)))
by A138, AFINSQ_1:9;
then
len (p | ((2 * k1) + 1)) in M
by A108, A109, A146;
hence
contradiction
by A109, A98, A101, A136, NAT_1:def 1;
verum end; end; end;
hence
contradiction
;
verum
end;
dom F = (Choose (n,(m + 1),1,0)) \ (Domin_0 (n,(m + 1)))
by FUNCT_2:def 1;
then
(Choose (n,(m + 1),1,0)) \ (Domin_0 (n,(m + 1))),W are_equipotent
by A85, A84, WELLORD2:def 4;
hence
card ((Choose (n,(m + 1),1,0)) \ (Domin_0 (n,(m + 1)))) = card (Choose (n,m,1,0))
by CARD_1:5; verum