let i, j, n, k1, k2 be Element of NAT ; ( (Decomp (n,2)) . i = <*k1,(n -' k1)*> & (Decomp (n,2)) . j = <*k2,(n -' k2)*> implies ( i < j iff k1 < k2 ) )
assume that
A1:
(Decomp (n,2)) . i = <*k1,(n -' k1)*>
and
A2:
(Decomp (n,2)) . j = <*k2,(n -' k2)*>
; ( i < j iff k1 < k2 )
A3:
j in dom (Decomp (n,2))
by A2, FUNCT_1:def 4;
then A4:
(Decomp (n,2)) . j = (Decomp (n,2)) /. j
by PARTFUN1:def 8;
consider A being finite Subset of (2 -tuples_on NAT) such that
A5:
Decomp (n,2) = SgmX ((TuplesOrder 2),A)
and
for p being Element of 2 -tuples_on NAT holds
( p in A iff Sum p = n )
by Def4;
field (TuplesOrder 2) = 2 -tuples_on NAT
by ORDERS_1:100;
then A6:
TuplesOrder 2 linearly_orders A
by ORDERS_1:133, ORDERS_1:134;
A7:
i in dom (Decomp (n,2))
by A1, FUNCT_1:def 4;
then A8:
(Decomp (n,2)) . i = (Decomp (n,2)) /. i
by PARTFUN1:def 8;
thus
( i < j implies k1 < k2 )
( k1 < k2 implies i < j )proof
assume A9:
i < j
;
k1 < k2
then
[<*k1,(n -' k1)*>,<*k2,(n -' k2)*>] in TuplesOrder 2
by A5, A6, A1, A2, A7, A3, A8, A4, PRE_POLY:def 2;
then A10:
<*k1,(n -' k1)*> <= <*k2,(n -' k2)*>
by Def3;
<*k1,(n -' k1)*> <> <*k2,(n -' k2)*>
by A5, A6, A1, A2, A7, A3, A8, A4, A9, PRE_POLY:def 2;
then
<*k1,(n -' k1)*> < <*k2,(n -' k2)*>
by A10, Def2;
then consider t being
Element of
NAT such that A11:
t in Seg 2
and A12:
<*k1,(n -' k1)*> . t < <*k2,(n -' k2)*> . t
and A13:
for
k being
Element of
NAT st 1
<= k &
k < t holds
<*k1,(n -' k1)*> . k = <*k2,(n -' k2)*> . k
by Def1;
end;
assume A15:
k1 < k2
; i < j
A16:
for k being Element of NAT st 1 <= k & k < 1 holds
<*k1,(n -' k1)*> . k = <*k2,(n -' k2)*> . k
;
A17:
<*k1,(n -' k1)*> . 1 = k1
by FINSEQ_1:61;
( 1 in Seg 2 & <*k2,(n -' k2)*> . 1 = k2 )
by FINSEQ_1:3, FINSEQ_1:61;
then A18:
<*k1,(n -' k1)*> < <*k2,(n -' k2)*>
by A15, A17, A16, Def1;
assume A19:
i >= j
; contradiction
per cases
( i = j or j < i )
by A19, XXREAL_0:1;
suppose A20:
j < i
;
contradictionthen
[<*k2,(n -' k2)*>,<*k1,(n -' k1)*>] in TuplesOrder 2
by A5, A6, A1, A2, A7, A3, A8, A4, PRE_POLY:def 2;
then A21:
<*k2,(n -' k2)*> <= <*k1,(n -' k1)*>
by Def3;
<*k2,(n -' k2)*> <> <*k1,(n -' k1)*>
by A5, A6, A1, A2, A7, A3, A8, A4, A20, PRE_POLY:def 2;
hence
contradiction
by A18, A21, Def2;
verum end; end;