let X be set ; for S being with_empty_element cap-closed semi-diff-closed Subset-Family of X
for A, B, R being set st R = DisUnion S & A in R & B in R & A <> {} holds
B \ A in R
let S be with_empty_element cap-closed semi-diff-closed Subset-Family of X; for A, B, R being set st R = DisUnion S & A in R & B in R & A <> {} holds
B \ A in R
let A, B, R be set ; ( R = DisUnion S & A in R & B in R & A <> {} implies B \ A in R )
assume that
A1:
R = DisUnion S
and
A2:
( A in R & B in R & A <> {} )
; B \ A in R
consider A1 being Subset of X such that
A3:
( A = A1 & ex F being disjoint_valued FinSequence of S st A1 = Union F )
by A1, A2;
consider f1 being disjoint_valued FinSequence of S such that
A4:
A1 = Union f1
by A3;
consider B1 being Subset of X such that
A5:
( B = B1 & ex F being disjoint_valued FinSequence of S st B1 = Union F )
by A1, A2;
reconsider R1 = R as non empty set by A2;
defpred S1[ Nat, object ] means $2 = B1 \ (f1 . $1);
A7:
for k being Nat st k in Seg (len f1) holds
ex x being Element of R1 st S1[k,x]
consider F being FinSequence of R1 such that
A8:
( dom F = Seg (len f1) & ( for k being Nat st k in Seg (len f1) holds
S1[k,F . k] ) )
from FINSEQ_1:sch 5(A7);
then B7:
B \ A c= meet (rng F)
;
now for x being object st x in meet (rng F) holds
x in B \ Alet x be
object ;
( x in meet (rng F) implies x in B \ A )assume B8:
x in meet (rng F)
;
x in B \ Athen consider Y being
object such that B10:
Y in rng F
by SETFAM_1:1, XBOOLE_0:def 1;
consider k being
object such that B11:
(
k in dom F &
Y = F . k )
by B10, FUNCT_1:def 3;
reconsider k =
k as
Nat by B11;
x in F . k
by B8, B10, B11, SETFAM_1:def 1;
then
x in B1 \ (f1 . k)
by A8, B11;
then B12:
(
x in B1 & not
x in f1 . k )
by XBOOLE_0:def 5;
hence
x in B \ A
by A3, A4, A5, B12, XBOOLE_0:def 5;
verum end;
then
meet (rng F) c= B \ A
;
then B16:
B \ A = meet (rng F)
by B7;
defpred S2[ Nat] means meet (rng (F | $1)) in R;
( meet (rng (F | 0)) in S & S c= R )
by A1, lemma100, SETFAM_1:def 8, SETFAM_1:1;
then C1:
S2[ 0 ]
;
C2:
for k being Nat st S2[k] holds
S2[k + 1]
proof
let k be
Nat;
( S2[k] implies S2[k + 1] )
assume C3:
S2[
k]
;
S2[k + 1]
end;
for k being Nat holds S2[k]
from NAT_1:sch 2(C1, C2);
then
S2[ len F]
;
hence
B \ A in R
by B16, FINSEQ_1:58; verum