# Space of Invariants

## Problem Statement

Let be a set (e.g. the set of all integers). Let and . Define

For each element we have and we think of as an invariant.

Prove each of the following statements or give a counterexample.

**a):** If , then .

**b):** The set is finite.

**c):** If then also .

**d):** If then also .

**e):** Let and let

in other words, by definition of ,

Prove that then . What do we obtain for ?

**f):** Let and let

in other words, by definition of ,

Prove that then . What do we obtain for ?

**g):** Is the set a lattice? If it is a lattice, is it complete? If it is not a lattice, can it be modified to become a
lattice?

## Solutions

### Solution a)

True.

First we prove the following lemme using induction on .

#### Lemma

**Base.** Assume .

We show .

This holds according to the definition of invariants.

**Inductive step.** Assume .
We show .

By the induction assumption we know .

We can easily show for two invariants and that . This proves the lemma.

From the lemma we can conclude .

The set is non-empty, so there exists such that . This shows that .

### Solution b)

False.

You can easily find counter-examples.

### Solution c,d)

Both true.

The assumption gives us the following.

From the set theory we know that if and then we can conclude both and .

We can use this fact on the on the corresponding components:

Computing the union gives us:

Due of the disjunctivity of we can conclude that is an invariant.

For the intersection case we replace the invariant of with a stronger one . Similarly for .

This can be done due to the monotonicity of .

The proof is similar to the union case.

### Solution e)

Consider the following equivalence.

We show that each conjunct is valid.

- . Holds since is included in every invariant.

Let be an arbitrary member of .

Nested quantifiers can be swapped, but the reverse does not necessarily hold.

- . Holds if is non-empty.

If and we claim .

We proved in (a) that . It remains to prove the other two conditions.

Which shows that is included in .

This is correct because

To prove that is the bottom element we have to prove the following.

Which was proved in a lemma in part (a).

### Solution f)

Consider the following equivalence.

We show that each conjunct is valid.

- . Holds since is included in every invariant.

Let be an arbitrary member of .

- . Holds if is non-empty.

If and we claim .

To prove that is the top element we have to prove the following.

We prove the following lemme using induction on .

**Base.** Assume .

We show .

This holds according to the definition of invariants.

**Inductive step.** Assume .
We show .

By the induction assumption we know .

We can also easily show for two invariants and that . Note that the condition is equivalent to .

This proves the lemma.

### Solution g)

The ordered set is a lattice. We proved in part (c ) and (d) that for any both and exist and are equal to and respectively.

We showed in part (e) and (f) that for every subset both lub and gub exist: and .

The lattice is complete. The top element is and the bottom element is .