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

proof

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

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

proof

assume A5: for n, k being Element of NAT holds seq . n < seq . ((n + 1) + k) ; :: thesis: for n, m being Element of NAT st n < m holds
seq . n < seq . m
let n, m be Element of NAT ; :: thesis: ( n < m implies seq . n < seq . m )
assume n < m ; :: thesis: seq . n < seq . m
then ex k being Element of NAT st m = (n + 1) + k by Lm1;
hence seq . n < seq . m by A5; :: thesis: verum

end;

assume A6: for n, m being Element of NAT st n < m holds
seq . n < seq . m ; :: thesis: for n being Element of NAT holds seq . n < seq . (n + 1)
let n be Element of NAT ; :: thesis: seq . n < seq . (n + 1)
n < n + 1 by NAT_1:13;
hence seq . n < seq . (n + 1) by A6; :: thesis: verum