let Y, X be non empty set ; for F, G being BinOp of Y st F is idempotent & F is commutative & F is associative & G is_distributive_wrt F holds
for B being Element of Fin X st B <> {} holds
for f being Function of X,Y
for a being Element of Y holds G . a,(F $$ B,f) = F $$ B,(G [;] a,f)
let F, G be BinOp of Y; ( F is idempotent & F is commutative & F is associative & G is_distributive_wrt F implies for B being Element of Fin X st B <> {} holds
for f being Function of X,Y
for a being Element of Y holds G . a,(F $$ B,f) = F $$ B,(G [;] a,f) )
assume that
A1:
F is idempotent
and
A2:
( F is commutative & F is associative )
and
A3:
G is_distributive_wrt F
; for B being Element of Fin X st B <> {} holds
for f being Function of X,Y
for a being Element of Y holds G . a,(F $$ B,f) = F $$ B,(G [;] a,f)
let B be Element of Fin X; ( B <> {} implies for f being Function of X,Y
for a being Element of Y holds G . a,(F $$ B,f) = F $$ B,(G [;] a,f) )
assume A4:
B <> {}
; for f being Function of X,Y
for a being Element of Y holds G . a,(F $$ B,f) = F $$ B,(G [;] a,f)
let f be Function of X,Y; for a being Element of Y holds G . a,(F $$ B,f) = F $$ B,(G [;] a,f)
let a be Element of Y; G . a,(F $$ B,f) = F $$ B,(G [;] a,f)
defpred S1[ Element of Fin X] means G . a,(F $$ $1,f) = F $$ $1,(G [;] a,f);
A5:
for B1, B2 being non empty Element of Fin X st S1[B1] & S1[B2] holds
S1[B1 \/ B2]
proof
let B1,
B2 be non
empty Element of
Fin X;
( S1[B1] & S1[B2] implies S1[B1 \/ B2] )
assume A6:
(
G . a,
(F $$ B1,f) = F $$ B1,
(G [;] a,f) &
G . a,
(F $$ B2,f) = F $$ B2,
(G [;] a,f) )
;
S1[B1 \/ B2]
thus G . a,
(F $$ (B1 \/ B2),f) =
G . a,
(F . (F $$ B1,f),(F $$ B2,f))
by A1, A2, Th30
.=
F . (F $$ B1,(G [;] a,f)),
(F $$ B2,(G [;] a,f))
by A3, A6, BINOP_1:23
.=
F $$ (B1 \/ B2),
(G [;] a,f)
by A1, A2, Th30
;
verum
end;
A7:
for x being Element of X holds S1[{.x.}]
for B being non empty Element of Fin X holds S1[B]
from SETWISEO:sch 3(A7, A5);
hence
G . a,(F $$ B,f) = F $$ B,(G [;] a,f)
by A4; verum