let X be set ; :: thesis: for S being SigmaField of X

for N being sequence of S ex F being sequence of S st

( F . 0 = N . 0 & ( for n being Nat holds F . (n + 1) = (N . (n + 1)) \/ (F . n) ) )

let S be SigmaField of X; :: thesis: for N being sequence of S ex F being sequence of S st

( F . 0 = N . 0 & ( for n being Nat holds F . (n + 1) = (N . (n + 1)) \/ (F . n) ) )

let N be sequence of S; :: thesis: ex F being sequence of S st

( F . 0 = N . 0 & ( for n being Nat holds F . (n + 1) = (N . (n + 1)) \/ (F . n) ) )

defpred S_{1}[ set , set , set ] means for A, B being Element of S

for c being Nat st c = $1 & A = $2 & B = $3 holds

B = (N . (c + 1)) \/ A;

A1: for c being Nat

for A being Element of S ex B being Element of S st S_{1}[c,A,B]

A2: ( F . 0 = N . 0 & ( for n being Nat holds S_{1}[n,F . n,F . (n + 1)] ) )
from RECDEF_1:sch 2(A1);

for n being Nat holds F . (n + 1) = (N . (n + 1)) \/ (F . n) by A2;

hence ex F being sequence of S st

( F . 0 = N . 0 & ( for n being Nat holds F . (n + 1) = (N . (n + 1)) \/ (F . n) ) ) by A2; :: thesis: verum

for N being sequence of S ex F being sequence of S st

( F . 0 = N . 0 & ( for n being Nat holds F . (n + 1) = (N . (n + 1)) \/ (F . n) ) )

let S be SigmaField of X; :: thesis: for N being sequence of S ex F being sequence of S st

( F . 0 = N . 0 & ( for n being Nat holds F . (n + 1) = (N . (n + 1)) \/ (F . n) ) )

let N be sequence of S; :: thesis: ex F being sequence of S st

( F . 0 = N . 0 & ( for n being Nat holds F . (n + 1) = (N . (n + 1)) \/ (F . n) ) )

defpred S

for c being Nat st c = $1 & A = $2 & B = $3 holds

B = (N . (c + 1)) \/ A;

A1: for c being Nat

for A being Element of S ex B being Element of S st S

proof

consider F being sequence of S such that
let c be Nat; :: thesis: for A being Element of S ex B being Element of S st S_{1}[c,A,B]

let A be Element of S; :: thesis: ex B being Element of S st S_{1}[c,A,B]

reconsider B = (N . (c + 1)) \/ A as Element of S ;

take B ; :: thesis: S_{1}[c,A,B]

thus S_{1}[c,A,B]
; :: thesis: verum

end;let A be Element of S; :: thesis: ex B being Element of S st S

reconsider B = (N . (c + 1)) \/ A as Element of S ;

take B ; :: thesis: S

thus S

A2: ( F . 0 = N . 0 & ( for n being Nat holds S

for n being Nat holds F . (n + 1) = (N . (n + 1)) \/ (F . n) by A2;

hence ex F being sequence of S st

( F . 0 = N . 0 & ( for n being Nat holds F . (n + 1) = (N . (n + 1)) \/ (F . n) ) ) by A2; :: thesis: verum