[EM] Can someone prove/disprove a relationship for me? [g is rational (II)(n) on X implies g is rational (I)(n) on X]

[David Catchpole]

General conditions:

Let X be some set. Let g be some function which returns some set over the
domain of all subsets of X, including X. Let f be some function such that
for any of the subsets Y of X, including X, f(Y) is equal to the
convex set over Y and g(Y) [that is, the set of all elements of Y which
are also elements of g(Y)]. Let n be some natural number.
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Rational (I)(n):

g is rational (I)(n) on X if and only if;

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For any of the subsets Y of X, including X, for any of the subsets Z of X,
including X, for any of the subsets W of X, including W;

If;

f(Y) is any of the subsets of Z, including Z, AND W is any of the
subsets of Y, including Y, AND the cardinality of W [number of elements in
W] is greater than or equal to n, AND the cardinality of Z is greater
than or equal to n, AND;

f(V) is any of the subsets of W, including W, OR f(V) is any of the
subsets of f(Z), including f(Z), where V is equal to the concave set over
W and f(Z) [that is, the set of all elements of W or f(Z)];

Then it must be that;

f(V) is equal to f(Z), where V is equal to the concave set over W and
f(Z).
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Rational (II)(n):

g is rational (II)(n) on X if and only if;

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For any of the subsets Y of X, including X, for any of the subsets Z of X,
including X;

If;

f(Y) is any of the subsets of Z, including Z, AND the cardinality of Y 
[number of elements in Y] is greater than or equal to n, AND the 
cardinality of Z is greater than or equal to n, AND;

f(V) is any of the subsets of Y, including Y, OR f(V) is any of the
subsets of f(Z), including f(Z), where V is equal to the concave set over
Y and f(Z) [that is, the set of all elements of Y or f(Z)];

Then it must be that;

g*(V) is equal to g*(Z), where V is equal to the concave set over Y and
f(Z).
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THEOREM:

g is rational (II)(n) on X implies g is rational (I)(n) on X
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I'm having BIG problems proving/disproving this theorem... Could someone
give me a hand?

Thanks,

David