defpred S_{1}[ set , set , set ] means for x, y being Subset of X

for s being Nat st s = $1 & x = $2 & y = $3 holds

y = x \/ (A1 . (s + 1));

A1: for n being Nat

for x being Subset of X ex y being Subset of X st S_{1}[n,x,y]

A2: F . 0 = A1 . 0 and

A3: for n being Nat holds S_{1}[n,F . n,F . (n + 1)]
from RECDEF_1:sch 2(A1);

take F ; :: thesis: ( F . 0 = A1 . 0 & ( for n being Nat holds F . (n + 1) = (F . n) \/ (A1 . (n + 1)) ) )

thus F . 0 = A1 . 0 by A2; :: thesis: for n being Nat holds F . (n + 1) = (F . n) \/ (A1 . (n + 1))

let n be Nat; :: thesis: F . (n + 1) = (F . n) \/ (A1 . (n + 1))

reconsider n = n as Element of NAT by ORDINAL1:def 12;

S_{1}[n,F . n,F . (n + 1)]
by A3;

hence F . (n + 1) = (F . n) \/ (A1 . (n + 1)) ; :: thesis: verum

for s being Nat st s = $1 & x = $2 & y = $3 holds

y = x \/ (A1 . (s + 1));

A1: for n being Nat

for x being Subset of X ex y being Subset of X st S

proof

consider F being SetSequence of X such that
let n be Nat; :: thesis: for x being Subset of X ex y being Subset of X st S_{1}[n,x,y]

let x be Subset of X; :: thesis: ex y being Subset of X st S_{1}[n,x,y]

take x \/ (A1 . (n + 1)) ; :: thesis: S_{1}[n,x,x \/ (A1 . (n + 1))]

thus S_{1}[n,x,x \/ (A1 . (n + 1))]
; :: thesis: verum

end;let x be Subset of X; :: thesis: ex y being Subset of X st S

take x \/ (A1 . (n + 1)) ; :: thesis: S

thus S

A2: F . 0 = A1 . 0 and

A3: for n being Nat holds S

take F ; :: thesis: ( F . 0 = A1 . 0 & ( for n being Nat holds F . (n + 1) = (F . n) \/ (A1 . (n + 1)) ) )

thus F . 0 = A1 . 0 by A2; :: thesis: for n being Nat holds F . (n + 1) = (F . n) \/ (A1 . (n + 1))

let n be Nat; :: thesis: F . (n + 1) = (F . n) \/ (A1 . (n + 1))

reconsider n = n as Element of NAT by ORDINAL1:def 12;

S

hence F . (n + 1) = (F . n) \/ (A1 . (n + 1)) ; :: thesis: verum