let p be XFinSequence of ; :: thesis: for n being Element of NAT st rng p c= {0 ,n} holds
ex q being XFinSequence of st
( len p = len q & rng q c= {0 ,n} & ( for k being Element of NAT st k <= len p holds
(Sum (p | k)) + (Sum (q | k)) = n * k ) & ( for k being Element of NAT st k in len p holds
q . k = n - (p . k) ) )
let n be Element of NAT ; :: thesis: ( rng p c= {0 ,n} implies ex q being XFinSequence of st
( len p = len q & rng q c= {0 ,n} & ( for k being Element of NAT st k <= len p holds
(Sum (p | k)) + (Sum (q | k)) = n * k ) & ( for k being Element of NAT st k in len p holds
q . k = n - (p . k) ) ) )
assume A1:
rng p c= {0 ,n}
; :: thesis: ex q being XFinSequence of st
( len p = len q & rng q c= {0 ,n} & ( for k being Element of NAT st k <= len p holds
(Sum (p | k)) + (Sum (q | k)) = n * k ) & ( for k being Element of NAT st k in len p holds
q . k = n - (p . k) ) )
defpred S1[ set , set ] means for k being Element of NAT st k = $1 holds
$2 = n - (p . k);
A2:
for k being Element of NAT st k in len p holds
ex x being Element of {0 ,n} st S1[k,x]
consider q being XFinSequence of such that
A4:
( dom q = len p & ( for k being Element of NAT st k in len p holds
S1[k,q . k] ) )
from STIRL2_1:sch 6(A2);
( rng q c= {0 ,n} & {0 ,n} is Subset of NAT )
by RELAT_1:def 19;
then
rng q c= NAT
by XBOOLE_1:1;
then reconsider q = q as XFinSequence of by RELAT_1:def 19;
take
q
; :: thesis: ( len p = len q & rng q c= {0 ,n} & ( for k being Element of NAT st k <= len p holds
(Sum (p | k)) + (Sum (q | k)) = n * k ) & ( for k being Element of NAT st k in len p holds
q . k = n - (p . k) ) )
thus
len p = len q
by A4; :: thesis: ( rng q c= {0 ,n} & ( for k being Element of NAT st k <= len p holds
(Sum (p | k)) + (Sum (q | k)) = n * k ) & ( for k being Element of NAT st k in len p holds
q . k = n - (p . k) ) )
thus
rng q c= {0 ,n}
by RELAT_1:def 19; :: thesis: ( ( for k being Element of NAT st k <= len p holds
(Sum (p | k)) + (Sum (q | k)) = n * k ) & ( for k being Element of NAT st k in len p holds
q . k = n - (p . k) ) )
defpred S2[ Element of NAT ] means ( $1 <= len p implies (Sum (p | $1)) + (Sum (q | $1)) = n * $1 );
A5:
S2[ 0 ]
A6:
for k being Element of NAT st S2[k] holds
S2[k + 1]
proof
let k be
Element of
NAT ;
:: thesis: ( S2[k] implies S2[k + 1] )
assume A7:
S2[
k]
;
:: thesis: S2[k + 1]
set k1 =
k + 1;
assume A8:
k + 1
<= len p
;
:: thesis: (Sum (p | (k + 1))) + (Sum (q | (k + 1))) = n * (k + 1)
then
(
k + 1
c= len p &
len p = dom p )
by NAT_1:40;
then
(
dom (p | (k + 1)) = k + 1 &
dom (q | (k + 1)) = k + 1 &
k < k + 1 )
by A4, NAT_1:13, RELAT_1:91;
then
(
len (p | (k + 1)) = k + 1 &
len (q | (k + 1)) = k + 1 &
k in dom (p | (k + 1)) &
k in dom (q | (k + 1)) &
k c= k + 1 &
k < len p )
by A8, NAT_1:40, NAT_1:45, XXREAL_0:2;
then
(
p | (k + 1) = ((p | (k + 1)) | k) ^ <%((p | (k + 1)) . k)%> &
(p | (k + 1)) . k = p . k &
(p | (k + 1)) | k = p | k &
q | (k + 1) = ((q | (k + 1)) | k) ^ <%((q | (k + 1)) . k)%> &
(q | (k + 1)) . k = q . k &
(q | (k + 1)) | k = q | k &
k in len p )
by CARD_FIN:43, FUNCT_1:70, NAT_1:45, RELAT_1:103;
then
(
Sum (p | (k + 1)) = (Sum (p | k)) + (Sum <%(p . k)%>) &
q . k = n - (p . k) &
Sum (q | (k + 1)) = (Sum (q | k)) + (Sum <%(q . k)%>) )
by A4, Lm4;
then
(
Sum (p | (k + 1)) = (Sum (p | k)) + (p . k) &
Sum (q | (k + 1)) = (Sum (q | k)) + (n - (p . k)) )
by STIRL2_1:44;
hence
(Sum (p | (k + 1))) + (Sum (q | (k + 1))) = n * (k + 1)
by A7, A8, NAT_1:13;
:: thesis: verum
end;
for k being Element of NAT holds S2[k]
from NAT_1:sch 1(A5, A6);
hence
( ( for k being Element of NAT st k <= len p holds
(Sum (p | k)) + (Sum (q | k)) = n * k ) & ( for k being Element of NAT st k in len p holds
q . k = n - (p . k) ) )
by A4; :: thesis: verum