WS 2019/2020 20.11.2019 Exercises to the lecture

WS 2019/2020 20.11.2019 Exercises to the lecture

Semantics Sheet 2 Prof. Dr. Roland Meyer,

Sören van der Wall Delivery until — at —

In the lecture, we constructed a set of types from a given APTA A, namely θ ::= q | τ → θ

τ ::= ^

{(θ ₁ , m 1 ), . . . , (θ k , m k )},

where q ∈ Q is a state of A and 0 ≤ k, i.e. the conjunct may be empty.

Intuitivly, a type describes the (powerset-constructional) behaviour of A on any tree: A tree typed by a type q is accepted by A from state q. A tree t typed by a type

^ {(θ _1,1 , m 1,1 ), . . . , (θ _1,k

, m _1,k

)} → . . . → ^

{(θ _n,1 , m n,1 ), . . . , (θ _n,k

, m _n,k

)} → q is an uncomplete tree, that requires n trees s ₁ , . . . , s _n , each s _i having all types θ _i,1 , . . . , θ _i,k

, in order to be accepted by A from state q. Whenever s _i is used in t, the just mentioned run of A visits the root of s i in a state q i,j for some 1 ≤ j ≤ k i , where θ i,j = τ → . . . → q i,j , and this path of the run from the root of t up to the root of s _i sees maximal priority m _i,j .

Exercise 2.1 (Language of APTA)

Look at the APTA A = (Σ, Q, δ, q ₀ , Ω) with Σ = {a : σ → σ → σ, b : σ}, Q = {q ₀ , q ₁ }, Ω = {(q ₀ , 2), (q 1 , 1)} and transition relation

δ(q ₀ , a) = (1, q ₁ ) ∧ (1, q ₀ ) ∧ (2, q ₀ ) δ(q ₁ , a) = (1, q ₁ )

δ(q ₀ , b) = tt.

Describe its language.

Exercise 2.2 (Well-formed types of APTAs)

Which of the following types are well-formed for the APTA A from Ex. 1?

1. V {(q ₀ , 2), ((q ₁ , 1) → q ₀ , 1)} → q ₀ 2. V

{((q ₀ , 2) → q 1 , 2), ((q 1 , 1) → q 1 , 1)} → V

{(q ₀ , 2), (q 1 , 1)} → q 0

3. V

{(q ₀ , 2), (q 1 , 1), (q 1 , 2)} → q 0

4. V

{(q ₂ , 2)} → q 0

List the atomic types θ with θ :: _a σ → σ.

Exercise 2.3 (Type Judgements) Set the types

θ x,q

= ^

{(q ₀ , 2), (q 1 , 2)} → ^

{(q ₀ , 2), (q 1 , 2)} → q 0 , θ _x,q

= ^

{(q ₀ , 2), (q ₁ , 1)} → ^

{(q ₀ , 2), (q ₁ , 2)} → q ₁ , θ _H,q

= ^

{(θ _x,q

, 2), (θ _x,q

, 2)} → θ _x,q

, θ _H,q

= ^

{(θ _x,q

, 2), (θ _x,q

, 2)} → θ _x,q

, θ F = ^

{(θ _x,q

, 2), (θ x,q

, 2)} → ^

{(q ₀ , 2), (q 1 , 2)} → q 0 . Show a formal deduction for the type judgement

{H : (θ _H,q

, 2) ^f , H : (θ _H,q

, 2) ^f , F : (θ _F , 2) ^f } ` λxλy.a(xyy)(F (Hx)y) : θ _F .