CSC438F/2404F
Solutions to Problem Set 1
Fall, 2016
Due: Friday, September 30, beginning of tutorial
1. Do Exercise 7, page 12 of the Notes: Show that the contraction rules can be derived
from the cut rule (with weakenings and exchanges).
Solution:
To get contraction right, use weakenings and exchanges on A → A to get A, Γ → ∆, A.
Now, use a single cut on Γ → ∆, A, A and A, Γ → ∆, A to get Γ → ∆, A. Similarly for
contraction left.
2. Do Exercise 8, page 12 of the Notes: Give suitable PK rules for the connecitve ⊃.
Solution: The rules are
left
Γ → ∆, A B, Γ → ∆
(A ⊃ B), Γ → ∆
right
A, Γ → ∆, B
Γ → ∆, (A ⊃ B)
3. Do Excercies 11, page 16 of the Notes: Prove the equivalence of the three forms of
compactness.
Solution: As observed in the exercise, Form 3 is the contrapositive of Form 1, so
Form 1 and Form 3 are equivalent. So it suffices to show that Form 1 and Form 2 are
equivalent.
We use the fact that Φ |= A iff Φ ∪ {¬A} is unsatisfiable.
Form 1 → Form 2: Suppose Φ |= A. Then Φ ∪ {¬A} is unsatisfiable. Hence by Form
1 some finite subset Ψ of Φ ∪ {¬A} is unsatisfiable. Let Φ0 = Ψ − {¬A}. Then Φ0 is a
finite subset of Φ and Φ0 ∪ {¬A} is unsatisfiable. Hence Φ0 |= A. This proves Form 2.
Form 2 → Form 1: Suppose that Φ is an unsatisfiable set of formulas. Then Φ |=
(P ∧ ¬P ). By Form 2 there is a finite subset Φ0 of Φ such that Φ0 |= (P ∧ ¬P ). Thus
Φ0 ∪ {¬(P ∧ ¬P )} is unsatisfiable. But ¬(P ∧ ¬P ) is valid, so Φ0 is unsatisfiable. This
proves Form 1.
4. Do Excercise 1, page 19 of the Notes: Prove the Unique Readability Theorem for terms.
Solution:
First we assign weights to each term, as suggested on page 19, namely each n-ary
function symbol gets a weight of n − 1, and each variable gets weight -1. Notice that
a constant symbol c also gets weight -1 (because it is a 0-ary function symbol).
The analog of the Lemma at the top of page 3 is
Lemma: The weight of any term t is -1, but the weight of any proper initial segment
of t is ≥ 0.
Proof: Structural induction on terms. (See the definition of term on page 19.) For the
base case, the term is a variable. It gets weight -1 by definition, and the only proper
initial segment is the empty string, which gets weight 0.
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For the induction step, the term t has the form
f t1 ...tn
where f is an n-ary function symbol and t1 , ..., tn are terms. By the induction hypothesis, each term ti has weight -1, and by definition the weight of f is n − 1. Hence the
weight of t is n − 1 + (n · (−1)) = −1. On the other hand, any proper initial segment of
t is either empty (has weight 0) or includes f and at most n − 1 of the terms ti together
with possibly an initial segment of some tj (which has weight ≥ 0 by the induction
hypothesis). Therefore every proper initial segment of t has weight ≥ 0. Now we can prove the Unique Readability Theorem for terms from the Lemma. This
theorem states that every term t can be uniquely parsed. If t is a variable, this is
obvious. So suppose that t is f t1 ...tn , where f is an n-ary function symbol and t1 , ..., tn
are terms. Suppose that
f t1 ...tn =syn gu1 ...uk
where g is a k=ary function symbol and u1 , ..., uk are terms. Then clearly f =syn g.
Also either t1 is an initial segment of u1 or u1 is an initial segment of t1 . But by the
above Lemma, neither can be a proper initial segment, so we conclude t1 =syn u1 .
Now we proceed to t2 and u2 and conclude t2 = u2 by the same argument. In fact, by
induction on i we show in this way for i = 1, ..., n that k ≥ i and ti = ui . It follows that
k = n (since otherwise the f t1 ...tn is a proper initial segment of gu1 ...uk .) It follows
that k = n (since otherwise the f t1 ...tn is a proper initial segment of gu1 ...uk ). Thus
we have shown that there is only one way to parse a term t.
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Practice Exercises (Do not hand in):
• Exercise 6 page 12: Find PK proofs.
• Prove the Inversion Principle: For each PK rule except weakening, if the bottom
sequent is valid, then all top sequents are valid. (See page 13 of the Notes.)
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