let X be set ; :: thesis: for S being SigmaField of X
for N, F being sequence of S st F . 0 = N . 0 & ( for n being Nat holds
( F . (n + 1) = (N . (n + 1)) \ (N . n) & N . n c= N . (n + 1) ) ) holds
union (rng F) = union (rng N)

let S be SigmaField of X; :: thesis: for N, F being sequence of S st F . 0 = N . 0 & ( for n being Nat holds
( F . (n + 1) = (N . (n + 1)) \ (N . n) & N . n c= N . (n + 1) ) ) holds
union (rng F) = union (rng N)

let N, F be sequence of S; :: thesis: ( F . 0 = N . 0 & ( for n being Nat holds
( F . (n + 1) = (N . (n + 1)) \ (N . n) & N . n c= N . (n + 1) ) ) implies union (rng F) = union (rng N) )

assume that
A1: F . 0 = N . 0 and
A2: for n being Nat holds
( F . (n + 1) = (N . (n + 1)) \ (N . n) & N . n c= N . (n + 1) ) ; :: thesis: union (rng F) = union (rng N)
thus union (rng F) c= union (rng N) :: according to XBOOLE_0:def 10 :: thesis: union (rng N) c= union (rng F)
proof
let x be object ; :: according to TARSKI:def 3 :: thesis: ( not x in union (rng F) or x in union (rng N) )
assume x in union (rng F) ; :: thesis: x in union (rng N)
then consider Y being set such that
A3: x in Y and
A4: Y in rng F by TARSKI:def 4;
consider n being object such that
A5: n in dom F and
A6: Y = F . n by ;
reconsider n = n as Element of NAT by A5;
A7: ( ex k being Nat st n = k + 1 implies ex Z being set st
( x in Z & Z in rng N ) )
proof
given k being Nat such that A8: n = k + 1 ; :: thesis: ex Z being set st
( x in Z & Z in rng N )

reconsider k = k as Element of NAT by ORDINAL1:def 12;
Y = (N . (k + 1)) \ (N . k) by A2, A6, A8;
then x in N . (k + 1) by ;
hence ex Z being set st
( x in Z & Z in rng N ) by FUNCT_2:4; :: thesis: verum
end;
( n = 0 implies ex Z being set st
( x in Z & Z in rng N ) ) by ;
hence x in union (rng N) by ; :: thesis: verum
end;
let x be object ; :: according to TARSKI:def 3 :: thesis: ( not x in union (rng N) or x in union (rng F) )
assume x in union (rng N) ; :: thesis: x in union (rng F)
then consider Y being set such that
A9: x in Y and
A10: Y in rng N by TARSKI:def 4;
A11: ex n being object st
( n in dom N & Y = N . n ) by ;
ex Z being set st
( x in Z & Z in rng F )
proof
ex s being Element of NAT st x in F . s
proof
defpred S1[ Nat] means x in N . \$1;
A12: ex k being Nat st S1[k] by ;
ex k being Nat st
( S1[k] & ( for r being Nat st S1[r] holds
k <= r ) ) from then consider k being Nat such that
A13: x in N . k and
A14: for r being Nat st x in N . r holds
k <= r ;
A15: ( ex l being Nat st k = l + 1 implies ex s being Element of NAT st x in F . s )
proof
given l being Nat such that A16: k = l + 1 ; :: thesis: ex s being Element of NAT st x in F . s
take k ; :: thesis: ( k is Element of NAT & x in F . k )
reconsider l = l as Element of NAT by ORDINAL1:def 12;
A17: not x in N . l
proof
assume x in N . l ; :: thesis: contradiction
then l + 1 <= l by ;
hence contradiction by NAT_1:13; :: thesis: verum
end;
F . (l + 1) = (N . (l + 1)) \ (N . l) by A2;
hence ( k is Element of NAT & x in F . k ) by ; :: thesis: verum
end;
( k = 0 implies ex s being Element of NAT st x in F . s ) by ;
hence ex s being Element of NAT st x in F . s by ; :: thesis: verum
end;
hence ex Z being set st
( x in Z & Z in rng F ) by FUNCT_2:4; :: thesis: verum
end;
hence x in union (rng F) by TARSKI:def 4; :: thesis: verum