let s1, s2 be ExtREAL_sequence; :: thesis: ( s1 . 0 = s . 0 & ( for n being Nat holds s1 . (n + 1) = (s1 . n) + (s . (n + 1)) ) & s2 . 0 = s . 0 & ( for n being Nat holds s2 . (n + 1) = (s2 . n) + (s . (n + 1)) ) implies s1 = s2 )
assume that
A2: s1 . 0 = s . 0 and
A3: for n being Nat holds s1 . (n + 1) = (s1 . n) + (s . (n + 1)) and
A4: s2 . 0 = s . 0 and
A5: for n being Nat holds s2 . (n + 1) = (s2 . n) + (s . (n + 1)) ; :: thesis: s1 = s2
defpred S₁[ Element of NAT ] means s1 . $1 = s2 . $1;
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] )
assume s1 . k = s2 . k ; :: thesis: S₁[k + 1]
hence s1 . (k + 1) = (s2 . k) + (s . (k + 1)) by A3
.= s2 . (k + 1) by A5 ;
:: thesis: verum

end;

A7: S₁[ 0 ] by A2, A4;
for n being Element of NAT holds S₁[n] from NAT_1:sch 1(A7, A6);
hence s1 = s2 by FUNCT_2:113; :: thesis: verum