let seq be Real_Sequence; :: thesis: ( ( ( for n being Element of NAT holds seq . (n + 1) <= seq . n ) implies for n, k being Element of NAT holds seq . (n + k) <= seq . n ) & ( ( for n, k being Element of NAT holds seq . (n + k) <= seq . n ) implies for n, m being Element of NAT st n <= m holds
seq . m <= seq . n ) & ( ( for n, m being Element of NAT st n <= m holds
seq . m <= seq . n ) implies for n being Element of NAT holds seq . (n + 1) <= seq . n ) )
thus ( ( for n being Element of NAT holds seq . (n + 1) <= seq . n ) implies for n, k being Element of NAT holds seq . (n + k) <= seq . n ) :: thesis: ( ( ( for n, k being Element of NAT holds seq . (n + k) <= seq . n ) implies for n, m being Element of NAT st n <= m holds
seq . m <= seq . n ) & ( ( for n, m being Element of NAT st n <= m holds
seq . m <= seq . n ) implies for n being Element of NAT holds seq . (n + 1) <= seq . n ) )

proof

assume A1: for n being Element of NAT holds seq . (n + 1) <= seq . n ; :: thesis: for n, k being Element of NAT holds seq . (n + k) <= seq . n
let n be Element of NAT ; :: thesis: for k being Element of NAT holds seq . (n + k) <= seq . n
defpred S₁[ Element of NAT ] means seq . (n + $1) <= seq . n;
A2: S₁[ 0 ] ;

A3: now

let k be Element of NAT ; :: thesis: ( S₁[k] implies S₁[k + 1] )
assume A4: S₁[k] ; :: thesis: S₁[k + 1]
seq . ((n + k) + 1) <= seq . (n + k) by A1;
hence S₁[k + 1] by A4, XXREAL_0:2; :: thesis: verum

end;

thus for k being Element of NAT holds S₁[k] from NAT_1:sch 1(A2, A3); :: thesis: verum

end;

thus ( ( for n, k being Element of NAT holds seq . (n + k) <= seq . n ) implies for n, m being Element of NAT st n <= m holds
seq . m <= seq . n ) :: thesis: ( ( for n, m being Element of NAT st n <= m holds
seq . m <= seq . n ) implies for n being Element of NAT holds seq . (n + 1) <= seq . n )

proof

assume A5: for n, k being Element of NAT holds seq . (n + k) <= seq . n ; :: thesis: for n, m being Element of NAT st n <= m holds
seq . m <= seq . n
let n, m be Element of NAT ; :: thesis: ( n <= m implies seq . m <= seq . n )
assume n <= m ; :: thesis: seq . m <= seq . n
then consider k1 being Nat such that
A6: m = n + k1 by NAT_1:10;
k1 in NAT by ORDINAL1:def 13;
hence seq . m <= seq . n by A5, A6; :: thesis: verum

end;

assume A7: for n, m being Element of NAT st n <= m holds
seq . m <= seq . n ; :: thesis: for n being Element of NAT holds seq . (n + 1) <= seq . n
let n be Element of NAT ; :: thesis: seq . (n + 1) <= seq . n
n <= n + 1 by NAT_1:13;
hence seq . (n + 1) <= seq . n by A7; :: thesis: verum