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

for M being sigma_Measure of S

for F being sequence of (COM (S,M))

for G being sequence of S ex H being sequence of (bool X) st

for n being Element of NAT holds H . n = (F . n) \ (G . n)

let S be SigmaField of X; :: thesis: for M being sigma_Measure of S

for F being sequence of (COM (S,M))

for G being sequence of S ex H being sequence of (bool X) st

for n being Element of NAT holds H . n = (F . n) \ (G . n)

let M be sigma_Measure of S; :: thesis: for F being sequence of (COM (S,M))

for G being sequence of S ex H being sequence of (bool X) st

for n being Element of NAT holds H . n = (F . n) \ (G . n)

let F be sequence of (COM (S,M)); :: thesis: for G being sequence of S ex H being sequence of (bool X) st

for n being Element of NAT holds H . n = (F . n) \ (G . n)

let G be sequence of S; :: thesis: ex H being sequence of (bool X) st

for n being Element of NAT holds H . n = (F . n) \ (G . n)

defpred S_{1}[ Element of NAT , set ] means for n being Element of NAT

for y being set st n = $1 & y = $2 holds

y = (F . n) \ (G . n);

A1: for t being Element of NAT ex A being Subset of X st S_{1}[t,A]

for t being Element of NAT holds S_{1}[t,H . t]
from FUNCT_2:sch 3(A1);

then consider H being sequence of (bool X) such that

A2: for t, n being Element of NAT

for y being set st n = t & y = H . t holds

y = (F . n) \ (G . n) ;

take H ; :: thesis: for n being Element of NAT holds H . n = (F . n) \ (G . n)

thus for n being Element of NAT holds H . n = (F . n) \ (G . n) by A2; :: thesis: verum

for M being sigma_Measure of S

for F being sequence of (COM (S,M))

for G being sequence of S ex H being sequence of (bool X) st

for n being Element of NAT holds H . n = (F . n) \ (G . n)

let S be SigmaField of X; :: thesis: for M being sigma_Measure of S

for F being sequence of (COM (S,M))

for G being sequence of S ex H being sequence of (bool X) st

for n being Element of NAT holds H . n = (F . n) \ (G . n)

let M be sigma_Measure of S; :: thesis: for F being sequence of (COM (S,M))

for G being sequence of S ex H being sequence of (bool X) st

for n being Element of NAT holds H . n = (F . n) \ (G . n)

let F be sequence of (COM (S,M)); :: thesis: for G being sequence of S ex H being sequence of (bool X) st

for n being Element of NAT holds H . n = (F . n) \ (G . n)

let G be sequence of S; :: thesis: ex H being sequence of (bool X) st

for n being Element of NAT holds H . n = (F . n) \ (G . n)

defpred S

for y being set st n = $1 & y = $2 holds

y = (F . n) \ (G . n);

A1: for t being Element of NAT ex A being Subset of X st S

proof

ex H being sequence of (bool X) st
let t be Element of NAT ; :: thesis: ex A being Subset of X st S_{1}[t,A]

F . t is Element of COM (S,M) ;

then reconsider A = (F . t) \ (G . t) as Subset of X by XBOOLE_1:1;

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

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

end;F . t is Element of COM (S,M) ;

then reconsider A = (F . t) \ (G . t) as Subset of X by XBOOLE_1:1;

take A ; :: thesis: S

thus S

for t being Element of NAT holds S

then consider H being sequence of (bool X) such that

A2: for t, n being Element of NAT

for y being set st n = t & y = H . t holds

y = (F . n) \ (G . n) ;

take H ; :: thesis: for n being Element of NAT holds H . n = (F . n) \ (G . n)

thus for n being Element of NAT holds H . n = (F . n) \ (G . n) by A2; :: thesis: verum