Exercise 1.
By induction on . is the statement that if then .
holds as and by an axiom of , for all .
Assuming , let so where . So by we have . Also and this is provable in by a theorem on a previous web page. And by the distributivity axiom, the commutativity axiom and the axiom on multiplication by we have . So puuting these together using first-order logic rules for we have , as required.
Exercise 2.
Induction on the complexity of sentences. It's important to
get the induction hypothesis right. Let be the statement that
If is a sentence with a total of
at most occurences of , , , or ,
then:
if is true ; and
if is false .
The rest is left to the reader.
Exercise 3.
To prove and , work, as usual, in a model of the appropriate theory and appeal to completeness.
If then so they are not equal, using axioms of .
If then since implies so and implies . So , a contradiction.
Exercise 4.
Work in an arbitrary model of . At the end we appeal to the completeness theorem to show the required statement is provable. Let with .
We first show . Suppose not, and suppose . Then . It follows by multiplying by and from the axioms that . Similarly and putting these together, . But this is impossible as and hence . Therefore and by symmetry so .
Now as it follows that and hence . But then for would mean and similiarly the other way round. Finally, as and it follows from the axioms that and we are done.
Exercise 5.
Again, work in an arbitrary model of and let be arbitrary.
We first claim that there is such that . Indeed, by , there is such that . If just let and note that . Otherwise so and there is such that . Then and the claim is done.
Next we find such that . Indeed so by an axiom of there is such that . By there is such that . So or . We show the latter is impossible. If let using again. So or . If then contradicting the lemma. Similarly if then contradicting the lemma.
Finally given we need to check for if so there is such that as required. But if then and so contradicting our choice of .
Exercise 6.
It is actually helpful for this to assume the result in the exercise above that shows is injective in .
The proof is by induction on the complexity of existential sentences. Again
it is important to get the induction hypothesis right. Let be
the statement that If is an existential formula with a total of
at most occurences of , , ,
then there is a formula in the form shown, and in the
same free variables, provably equivalent to .
BASE CASE: In the case the statement is either in the form given already, in which case there is nothing to prove, or else is of the form
in which case it is equivalent in to
as you can check.
STEP FOR : To transform the statement
First rename the variables so that they are diofferent from the as necessary, and then write it as
STEP FOR : To transform the statement
Write it as
using the pairing function of the exercise above, but note that this statement will have to be transformed to the required form using elementary algebraic means, all of which are available in .
STEP FOR : To transform the statement
write it as
and use previous transformations for and .