. 44 and 45 from Exercise set 1.5
. 9, 18, 20, 22 from set 2.2

Assume that A, B are two unary predicates. For each of the
followings, either prove that it's true using logical equivalences
or define an interpretation (example) which the statement is false.

. \forall x (A(x)-->B(x)) ==> \exists x(A(x) /\ B(x))

Sol: False. Consider an example which makes A(x) false for all x, then
the left hand-side is always true; so there is no x which makes the
right-hand side true; example A(x): x^2 < -1

. \exists x (A(x)--> ~B(x)) <==> ~\forall x (A(x) /\ B(x))

Sol: true;  \exists x (A(x)--> ~B(x)) <==> \exists x(~A(x) \/ ~B(x))
                                      <==> \exists x ~(A(x) /\ B(x))
                                      <==> ~\forall x (A(x) /\ B(x))

. \exists x A(x) /\ \exists x ~A(x) <==> \exists x(A(x) /\ ~A(x))

Sol: False. Let A(x) be "x is even" and define the domain to be N (set of
naturals). Then the left-hand side says "there is a natural number x such that
it is even and there is a natural number (maybe different from x) that is odd".
Clearly this is true. But the right-hand side says "there is a natural number
x such that: x is even and odd", which clearly is not true.

. \exists x A(x) \/ \exists x ~A(x) <==> \exists x(A(x) \/ ~A(x))

sol: True.  We show both sides are true statements.

 \exists x A(x) \/ \exists x ~A(x) <==> ~\exists x A(x) --> \exists x ~A(x)
                                   <==> \forall x ~A(x) --> \exists x ~A(x)

 This is a true statement, because if for all values of x, A(x) is false then
 clearly there is at least one value which makes it false; no matter what 
 interpretation we have for A. So the LHS of original is always true.
 Also, no matter what interpretation we have for A, there is some value x of the
 domain for which either A(x) is true or false. So, the RHS is always true, too.