let S1, S2 be Function of NAT,ExtREAL; :: thesis: ( S1 . 0 = N . 0 & ( for n being Element of NAT
for y being R_eal st y = S1 . n holds
S1 . (n + 1) = y + (N . (n + 1)) ) & S2 . 0 = N . 0 & ( for n being Element of NAT
for y being R_eal st y = S2 . n holds
S2 . (n + 1) = y + (N . (n + 1)) ) implies S1 = S2 )
assume that
A1: S1 . 0 = N . 0 and
A2: for n being Element of NAT
for y being R_eal st y = S1 . n holds
S1 . (n + 1) = y + (N . (n + 1)) and
A3: S2 . 0 = N . 0 and
A4: for n being Element of NAT
for y being R_eal st y = S2 . n holds
S2 . (n + 1) = y + (N . (n + 1)) ; :: thesis: S1 = S2
defpred S₁[ set ] means S1 . $1 = S2 . $1;
for n being set st n in NAT holds
S₁[n]

proof

let n be set ; :: thesis: ( n in NAT implies S₁[n] )
assume A5: n in NAT ; :: thesis: S₁[n]
then reconsider n = n as Element of REAL ;
reconsider n = n as Element of NAT by A5;
A6: for k being Element of NAT st S₁[k] holds
S₁[k + 1]

proof

let k be Element of NAT ; :: thesis: ( S₁[k] implies S₁[k + 1] )
reconsider y2 = S2 . k as R_eal ;
assume S1 . k = S2 . k ; :: thesis: S₁[k + 1]
hence S1 . (k + 1) = y2 + (N . (k + 1)) by A2
.= S2 . (k + 1) by A4 ;
:: thesis: verum

end;

A7: S₁[ 0 ] by A1, A3;
for k being Element of NAT holds S₁[k] from NAT_1:sch 1(A7, A6);
then S1 . n = S2 . n ;
hence S₁[n] ; :: thesis: verum

end;

hence S1 = S2 by FUNCT_2:12; :: thesis: verum