let Omega be non empty set ; for Sigma being SigmaField of Omega
for ASeq being SetSequence of Sigma st ASeq is disjoint_valued holds
for n, m being Nat st n < m holds
(Partial_Union ASeq) . n misses ASeq . m
let Sigma be SigmaField of Omega; for ASeq being SetSequence of Sigma st ASeq is disjoint_valued holds
for n, m being Nat st n < m holds
(Partial_Union ASeq) . n misses ASeq . m
let ASeq be SetSequence of Sigma; ( ASeq is disjoint_valued implies for n, m being Nat st n < m holds
(Partial_Union ASeq) . n misses ASeq . m )
assume A1:
ASeq is disjoint_valued
; for n, m being Nat st n < m holds
(Partial_Union ASeq) . n misses ASeq . m
let n, m be Nat; ( n < m implies (Partial_Union ASeq) . n misses ASeq . m )
assume A2:
n < m
; (Partial_Union ASeq) . n misses ASeq . m
assume
(Partial_Union ASeq) . n meets ASeq . m
; contradiction
then consider x being object such that
A3:
x in (Partial_Union ASeq) . n
and
A4:
x in ASeq . m
by XBOOLE_0:3;
reconsider ASeq = ASeq as SetSequence of Omega ;
consider k being Nat such that
A5:
k <= n
and
A6:
x in ASeq . k
by A3, Th13;
ASeq . k misses ASeq . m
by A1, A2, A5;
then
(ASeq . k) /\ (ASeq . m) = {}
;
hence
contradiction
by A4, A6, XBOOLE_0:def 4; verum