let Y be non empty set ; :: thesis: for a being Element of Funcs (Y,BOOLEAN) holds B_SUP (a,(%I Y)) = a
let a be Element of Funcs (Y,BOOLEAN); :: thesis: B_SUP (a,(%I Y)) = a
consider k3 being Function such that
A1: B_SUP (a,(%I Y)) = k3 and
A2: dom k3 = Y and
rng k3 c= BOOLEAN by FUNCT_2:def 2;
consider k4 being Function such that
A3: a = k4 and
A4: dom k4 = Y and
rng k4 c= BOOLEAN by FUNCT_2:def 2;
for y being Element of Y holds (B_SUP (a,(%I Y))) . y = a . y

proof

let y be Element of Y; :: thesis: (B_SUP (a,(%I Y))) . y = a . y

A5: now

EqClass (y,(%I Y)) in %I Y ;
then EqClass (y,(%I Y)) in { B where B is Subset of Y : ex z being set st
( B = {z} & z in Y ) } by PARTIT1:35;
then ex B being Subset of Y st
( EqClass (y,(%I Y)) = B & ex z being set st
( B = {z} & z in Y ) ) ;
then consider z being set such that
A6: EqClass (y,(%I Y)) = {z} and
z in Y ;
A7: y in {z} by A6, EQREL_1:def 8;
assume that
A8: ex x being Element of Y st
( x in EqClass (y,(%I Y)) & a . x = TRUE ) and
A9: a . y <> TRUE ; :: thesis: contradiction
consider x1 being Element of Y such that
A10: x1 in EqClass (y,(%I Y)) and
A11: a . x1 = TRUE by A8;
x1 = z by A10, A6, TARSKI:def 1;
hence contradiction by A9, A11, A7, TARSKI:def 1; :: thesis: verum

end;

A12: now

assume that
A13: for x being Element of Y holds
( not x in EqClass (y,(%I Y)) or not a . x = TRUE ) and
A14: a . y <> TRUE ; :: thesis: (B_SUP (a,(%I Y))) . y = a . y
a . y = FALSE by A14, XBOOLEAN:def 3;
hence (B_SUP (a,(%I Y))) . y = a . y by A13, Def20; :: thesis: verum

end;

y in EqClass (y,(%I Y)) by EQREL_1:def 8;
hence (B_SUP (a,(%I Y))) . y = a . y by A5, A12, Def20; :: thesis: verum

end;

then for u being set st u in Y holds
k3 . u = k4 . u by A1, A3;
hence B_SUP (a,(%I Y)) = a by A1, A2, A3, A4, FUNCT_1:9; :: thesis: verum