let X be set ; for B being Subset-Family of X
for F being Dependency-set of X holds X deps_encl_by B is full_family
let B be Subset-Family of X; for F being Dependency-set of X holds X deps_encl_by B is full_family
let F be Dependency-set of X; X deps_encl_by B is full_family
set F = X deps_encl_by B;
per cases
( B is empty or not B is empty )
;
suppose
not
B is
empty
;
X deps_encl_by B is full_family then reconsider B =
B as non
empty Subset-Family of
X ;
thus
X deps_encl_by B is
(F1)
ARMSTRNG:def 14 ( X deps_encl_by B is (F2) & X deps_encl_by B is (F3) & X deps_encl_by B is (F4) )A5:
now for x, y being Subset of X
for c being Element of B st [x,y] in X deps_encl_by B & x c= c holds
y c= clet x,
y be
Subset of
X;
for c being Element of B st [x,y] in X deps_encl_by B & x c= c holds
y c= clet c be
Element of
B;
( [x,y] in X deps_encl_by B & x c= c implies y c= c )assume that A6:
[x,y] in X deps_encl_by B
and A7:
x c= c
;
y c= cconsider a,
b being
Subset of
X such that A8:
[x,y] = [a,b]
and A9:
for
c being
set st
c in B &
a c= c holds
b c= c
by A6;
A10:
y = b
by A8, XTUPLE_0:1;
x = a
by A8, XTUPLE_0:1;
hence
y c= c
by A7, A9, A10;
verum end; hence
X deps_encl_by B is
(F2)
by Th18;
( X deps_encl_by B is (F3) & X deps_encl_by B is (F4) )thus
X deps_encl_by B is
(F3)
X deps_encl_by B is (F4) proof
let a,
b,
a9,
b9 be
Subset of
X;
ARMSTRNG:def 12 ( [a,b] in X deps_encl_by B & [a,b] >= [a9,b9] implies [a9,b9] in X deps_encl_by B )
assume that A15:
[a,b] in X deps_encl_by B
and A16:
[a,b] >= [a9,b9]
;
[a9,b9] in X deps_encl_by B
A17:
b9 c= b
by A16;
A18:
a c= a9
by A16;
hence
[a9,b9] in X deps_encl_by B
;
verum
end; thus
X deps_encl_by B is
(F4)
verumproof
let a,
b,
a9,
b9 be
Subset of
X;
ARMSTRNG:def 13 ( [a,b] in X deps_encl_by B & [a9,b9] in X deps_encl_by B implies [(a \/ a9),(b \/ b9)] in X deps_encl_by B )
assume that A21:
[a,b] in X deps_encl_by B
and A22:
[a9,b9] in X deps_encl_by B
;
[(a \/ a9),(b \/ b9)] in X deps_encl_by B
hence
[(a \/ a9),(b \/ b9)] in X deps_encl_by B
;
verum
end; end; end;