let rseq1, rseq2 be Real_Sequence; :: thesis: ( rseq1 . 0 = 1 & ( for n being Element of NAT holds rseq1 . (n + 1) = (rseq1 . n) * (n + 1) ) & rseq2 . 0 = 1 & ( for n being Element of NAT holds rseq2 . (n + 1) = (rseq2 . n) * (n + 1) ) implies rseq1 = rseq2 )
assume that
A2: rseq1 . 0 = 1 and
A3: for n being Element of NAT holds rseq1 . (n + 1) = (rseq1 . n) * (n + 1) and
A4: rseq2 . 0 = 1 and
A5: for n being Element of NAT holds rseq2 . (n + 1) = (rseq2 . n) * (n + 1) ; :: thesis: rseq1 = rseq2
defpred S₁[ Element of NAT ] means rseq1 . $1 = rseq2 . $1;
A6: S₁[ 0 ] by A2, A4;
A7: 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 rseq1 . k = rseq2 . k ; :: thesis: S₁[k + 1]
hence rseq1 . (k + 1) = (rseq2 . k) * (k + 1) by A3
.= rseq2 . (k + 1) by A5 ;
:: thesis: verum

end;

for n being Element of NAT holds S₁[n] from NAT_1:sch 1(A6, A7);
hence rseq1 = rseq2 by FUNCT_2:63; :: thesis: verum