let p be XFinSequence of NAT ; ( rng p c= {0,1} & not p is dominated_by_0 implies ex k being Nat st
( (2 * k) + 1 = min* { N where N is Nat : 2 * (Sum (p | N)) > N } & (2 * k) + 1 <= dom p & k = Sum (p | (2 * k)) & p . (2 * k) = 1 ) )
assume that
A1:
rng p c= {0,1}
and
A2:
not p is dominated_by_0
; ex k being Nat st
( (2 * k) + 1 = min* { N where N is Nat : 2 * (Sum (p | N)) > N } & (2 * k) + 1 <= dom p & k = Sum (p | (2 * k)) & p . (2 * k) = 1 )
set M = { 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 Subset of NAT ;
consider k being Nat such that
A3:
k <= dom p
and
A4:
2 * (Sum (p | k)) > k
by A1, A2;
reconsider k = k as Nat ;
k in M
by A4;
then reconsider M = M as non empty Subset of NAT ;
min* M in M
by NAT_1:def 1;
then consider n being Nat such that
A5:
min* M = n
and
A6:
2 * (Sum (p | n)) > n
;
A7:
Sum (p | 0) = 0
;
Sum (p | n) > 0
by A6;
then
n > 0
by A7;
then reconsider n1 = n - 1 as Nat by NAT_1:20;
reconsider S = Sum (p | n1) as Nat ;
take
S
; ( (2 * S) + 1 = min* { N where N is Nat : 2 * (Sum (p | N)) > N } & (2 * S) + 1 <= dom p & S = Sum (p | (2 * S)) & p . (2 * S) = 1 )
k in M
by A4;
then A8:
k >= n
by A5, NAT_1:def 1;
then A9:
dom p >= n
by A3, XXREAL_0:2;
A10:
2 * (Sum (p | n1)) = n1
proof
A11:
n1 < n1 + 1
by NAT_1:13;
then
Segm n1 c= Segm (n1 + 1)
by NAT_1:39;
then A12:
(p | n) | n1 = p | n1
by RELAT_1:74;
( (
n = len p &
p | (dom p) = p ) or
n < len p )
by A9, XXREAL_0:1;
then A13:
len (p | n) = n1 + 1
by AFINSQ_1:11;
then
n1 in Segm (len (p | n))
by A11, NAT_1:44;
then A14:
(p | n) . n1 in rng (p | n)
by FUNCT_1:3;
p | n = ((p | n) | n1) ^ <%((p | n) . n1)%>
by A13, AFINSQ_1:56;
then
Sum (p | n) = (Sum (p | n1)) + (Sum <%((p | n) . n1)%>)
by A12, AFINSQ_2:55;
then A15:
(2 * (Sum (p | n1))) + (2 * (Sum <%((p | n) . n1)%>)) >= n + 1
by A6, NAT_1:13;
n1 < n1 + 1
by NAT_1:13;
then
not
n1 in M
by A5, NAT_1:def 1;
then A16:
2
* (Sum (p | n1)) <= n1
;
(p | n) . n1 in {0,1}
by A1, A14;
then A17:
(
(p | n) . n1 = 0 or
(p | n) . n1 = 1 )
by TARSKI:def 2;
assume
2
* (Sum (p | n1)) <> n1
;
contradiction
then
(
Sum <%((p | n) . n1)%> = (p | n) . n1 & 2
* (Sum (p | n1)) < n1 )
by A16, AFINSQ_2:53, XXREAL_0:1;
then
(2 * (Sum (p | n1))) + (2 * (Sum <%((p | n) . n1)%>)) < n1 + 2
by A17, XREAL_1:8;
hence
contradiction
by A15;
verum
end;
p . n1 = 1
proof
Segm n c= Segm (len p)
by A9, NAT_1:39;
then A18:
dom (p | n) = n1 + 1
by RELAT_1:62;
A19:
(
Sum <%0%> = 0 &
p | n = ((p | n) | n1) ^ <%((p | n) . n1)%> )
by A18, AFINSQ_1:56, AFINSQ_2:53;
assume A20:
p . n1 <> 1
;
contradiction
A21:
n1 < n1 + 1
by NAT_1:13;
then
n1 < len p
by A9, XXREAL_0:2;
then A22:
n1 in dom p
by AFINSQ_1:86;
Segm n1 c= Segm n
by A21, NAT_1:39;
then A23:
(p | n) | n1 = p | n1
by RELAT_1:74;
n1 in Segm n
by A21, NAT_1:44;
then
n1 in (dom p) /\ n
by A22, XBOOLE_0:def 4;
then A24:
(p | n) . n1 = p . n1
by FUNCT_1:48;
A25:
n1 < n1 + 1
by NAT_1:13;
p . n1 in rng p
by A22, FUNCT_1:3;
then
p . n1 = 0
by A1, A20, TARSKI:def 2;
then
Sum (p | n) = (Sum (p | n1)) + 0
by A19, A24, A23, AFINSQ_2:55;
hence
contradiction
by A6, A10, A25;
verum
end;
hence
( (2 * S) + 1 = min* { N where N is Nat : 2 * (Sum (p | N)) > N } & (2 * S) + 1 <= dom p & S = Sum (p | (2 * S)) & p . (2 * S) = 1 )
by A3, A5, A8, A10, XXREAL_0:2; verum