# Program Analysis

```Program Analysis - Exercise Sheet
5 March 2015
Let (L1 , v1 , t1 , u1 ) and (L2 , v2 , t2 , u2 ) be two complete lattices.
Exercise 1
Can you find an example where:
– α : L1 ⇒ L2 is completely join-preserving
– γ : L2 ⇒ L1 is completely meet-preserving
– γ ◦ α is extensive
yet (α, γ) does not form a Galois Connection ?
Solution
α ◦ γ is not reductive for y.
Exercise 2
Give example of a Galois connection which is not a Galois insertion.
1
Solution
γ is not injective.
Exercise 3
• Given two monotone functions f : L1 → L1 and g : L2 → L2 and a
Galois connection (α, γ) between L1 and L2 , prove that the following
statements are equivalent:
– ∀z ∈ L2 : α(f (γ(z))) v2 g(z)
– ∀z ∈ L2 : f (γ(z)) v1 γ(g(z))
• If (α, γ) is not a Galois connection, show that the statements above
are equivalent or give a counter-example.
Solution
• We prove the first statement implies the second one.
"=>" Let us apply monotone function γ to the first statement:
∀z ∈ L2 : γ(α(f (γ(z)))) v1 γ(g(z))
By γ ◦ α extensive:
∀z ∈ L2 : f (γ(z)) v1 γ(α(f (γ(z)))) v1 γ(g(z))
"<=" By applying monotone function α to the second statement:
∀z ∈ L2 : α(f (γ(z))) v2 α(γ(g(z)))
By α ◦ γ reductive:
∀z ∈ L2 : α(f (γ(z))) v2 α(γ(g(z))) v2 g(z)
2
• If (α, γ) is not a Galois connection, then the statements are not equivalent. It is sufficient to look at the example from Exercise 1, and define
f and g as identity functions on L1 , respectively L2 .
Exercise 4
Assume L2 = {⊥, >}, with ⊥v2 > and ∀e ∈ L1 : α(e) = >. Can you find γ
such that (α, γ) is a Galois connection?
Solution
In this case, α is not completely join-preserving, thus we cannot build a
Galois Connection using this α. We now show by contradiction why α is
not completely join-preserving:
We assume α is completely join-preserving.
This means: ∀Y ⊆ L1 : α(tY ) = t{α(y)|y ∈ Y }
Let us pick Y = ∅. By substitution, we obtain:
α(t∅) = t{α(y)|y ∈ ∅}
This is equivalent with:
α(⊥1 ) = t∅ (⊥1 is the least element in L1 )
≡ > =⊥ (⊥ is the least element in L2 )
We reached contradiction. This implies α is not completely join-preserving.
3
```