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The answers to assignment 4 of the cs6100 course in spring 2004. The assignments cover various topics in theory of computation, including proofs for decidability of pcp over a unary alphabet and the relationship between relations r1 and r2. The document also includes models for given sentences.
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Proof. Let the given tiles A be {A 1 ,... An}, where an Ai = (1ℓ, 1 r). We will be interested only in the quantities ℓ and r. Here is the decision procedure:
Accept iff there exists Ap, Aq ∈ A such that Ap = (1i, 1 i+u) and Aq = (1ℓ+v, 1 ℓ)
This is a correct decision procedure:
c 1 i + c 2 (ℓ + v) = c 1 (i + u) + c 2 ℓ
and this is indeed always solvable, by letting c 1 = v and c 2 = u.
Proof. Let J = {(x, 0) | x ∈ A} ∪ {(x, 1) | x ∈ B}. Suppose J is decidable. Then we may decide if x ∈ A by asking if (x, 0) ∈ J, and similarly, x ∈ B is decided by asking if (x, 1) ∈ J.
(∀x.R 1 (x, x)) ∧ (∀xy.R 1 (x, y) ⇔ R 1 (y, x)) ∧ (∀xyz.R 1 (x, y) ∧ R 1 (y, z) ⇒ R 1 (x, z))
One model is (N , =). For another model, let the domain be expressions of the lambda calculus and R 1 be the relation of α-convertibility.
(∀x.R 1 (x, x)) ∧ (∀xy.R 1 (x, y) ⇔ R 1 (y, x)) ∧ (∀xyz.R 1 (x, y) ∧ R 1 (y, z) ⇒ R 1 (x, z)) ∧ (∀x.R 1 (x, x) ⇒ ¬R 2 (x, x)) ∧ (∀xy.¬R 1 (x, y) ⇒ R 2 (x, y) ⊕ R 2 (y, x)) ∧ (∀xyz.R 2 (x, y) ∧ R 2 (y, z) ⇒ R 2 (x, z)) ∧ (∀x.∃y. R 2 (x, y))
One model is (N , R 1 , R 2 , ⊕), where R 1 is equality, R 2 is < and ⊕ is ∨.
Proof. By Theorem 6.10, we have a decision procedure for Th(N , +), where + is a relation. We now use the fact that
x < y iff ∃u. (x + u = y) ∧ ¬(x + u = x)
To decide a formula F ∈ Th(N , <), simply replace each occurrence of < in F by ∃u. (x + u = y) ∧ ¬(x + u = x) and then invoke the decision procedure for Th(N , +).