**1. a.** [*p*(0, 0) $\wedge $
*p*(0, 1)] $\vee $
[*p*(1, 0) $\wedge $
*p*(1, 1)] .

**2. a.** $\forall $*x q*(*x*), where *x $\u220a$
* {0, 1}.

**c.** $\forall $
*y p*(*x*, *y*), where *y $\u220a$
* {0, 1}.

**e.** $\exists x$
* p*(*x*), where *x* is an odd natural number.

**3. a.** *x* is a term. Therefore, *p*(*x*) is a wff, and it follows that $\exists x$ *p*(*x*) and $\forall x$* p*(*x*) are wffs. Thus, $\exists x$* p*(*x*) $\to $
$\forall $* p*(*x*) is a wff.

**4.** It is illegal to have an atom, *p*(*x*) in this case, as an argument to a predicate.

**5. a.** The three occurrences of *x*, left to right, are free, bound, and bound. The four occurrences of *y*, left to right, are free, bound, bound, and free.

**c.** The three occurrences of *x*, left to right, are free, bound, and bound. Both occurrences of *y* are free.

**6.** $\forall x$* p*(*x*, *y*, *z*) $\to $
$\exists z$
* q*(*z*). ...

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