## Subsection7.2.4Logical Ambiguity

Sometimes ambiguity arises more directly from the logic of the claim that is being made. This kind of ambiguity often happens when the English sentence contains some sort of negative. Often, in such statements, the issue is the scope of the not.

Consider: I didn’t take both of my pills this morning.

This could mean either of these things:

[1] I may have taken one of my pills this morning, but not both of them.

[2] For each of my pills, it’s the case that I didn’t take it this morning.

Or, writing the claims in our logical language:

[1] (x (pill(x) → take(x)))

[2] x ((pill(x)  take(x)))

Consider: Mo doesn’t want Jo and Bo to come to the party.

This could mean either of these things:

[1] Mo doesn’t want either Jo or Bo to come to the party.

[2] Mo doesn’t want it to happen that both of them come (maybe that would make things too exciting.)

We can write (very simple versions of) these meanings in Boolean logic if we define:

J: Mo wants Jo to come to the party.

B: Mo wants Bo to come to the party.

[1] J  B

[2] (J  B)

### ExercisesExercises

#### 1.

Consider: Smoking is not permitted on all Korean Air flights.

We want to encode the meaning of this sentence. Define:

KAf(x) : True if x is a Korean Air flight.

SP(x): True if smoking is permitted in or on x.

Shown here five logical expressions. Mark the two that correspond to the two reasonable interpretations of this sentence. (Hint: first write out the two interpretations in unambiguous English. Then translate each of those into logic.)

a) ∃x (¬(KAf(x) ∨ SP(x))) b) ∀x (KAf(x) → ¬SP(x))

c) ∀xKAf(x) ∧ ¬SP(x)) d) ∃x (KAf(x) ∧ ¬SP(x))

e) ¬∃x (KAf(x) → ¬SP(x))

Correct answer is B and D.
Solution.
Explanation: The original sentence has two interpretations: In one (almost surely the intended one), on all Korean Air flights smoking is banned. In the other, it is not the case that, on all KA flights, smoking is permitted (In other words, there’s at least one KA flight on which it is banned, although it might be permitted on some flights).

#### 2.

Consider: Cruz forgot to invite the whole class.

We want to encode the meaning of this sentence. So that we can focus on the issue of its ambiguity, we’ll simplify the problem. So we’ll ignore the issue of why Cruz didn’t invite people. We’ll just describe who was invited. Define:

Inclass(x): True if x is in the class.

Invited(x, y): True if x invited y.

Shown here are five logical expressions. Mark the two that correspond to the two reasonable interpretations of this sentence. (Hint: First write out the two interpretations in unambiguous English.   Next, translate each of those into logic.  Then compare your results to the ones listed below.  Be careful: It is possible that one of these is not identical to what you have written but is logically equivalent to something you have written.  In that case, mark it as one of the reasonable ones.)

a) ∀x (Inclass(x) → ¬Invited(Cruz, x)) b) ¬∀x (Inclass(x) ∧ Invited(Cruz, x))

c) ∃x (Inclass(x) ∨ ¬Invited(Cruz, x)) d) ∃x (Inclass(x) ∧ ¬Invited(Cruz, x))

e) ∀xInclass(x) → ¬Invited(Cruz, x))

Correct answer is A and B.
Solution.
Explanation: The original sentence has two interpretations: In one, Cruz didn’t invite anyone. The forgetting applied to the whole class. In the other, it’s not the case that the entire class was invited. Someone was left out. But perhaps some members were invited.

#### 3.

Consider: We couldn’t do all of these shows without the staff.

We want to encode the meaning of this sentence. So that we can focus on the issue of its ambiguity, we’ll simplify a bit and define these predicates:

Staff: True if the staff exists.

Show(x): True if x is a show.

Possible(x): True if we can do x.

Shown here are five logical expressions. Mark the two that correspond to the two reasonable interpretations of this sentence. (Hint: first write out the two interpretations in unambiguous English. Then translate each of those into logic.)

a) ¬Staff → (∀x (Show(x) ∨ ¬Possible(x))) b) ¬Staff → (∀x (Show(x) → ¬Possible(x)))

c) ¬Staff → (∃x (Show(x) ∧ ¬Possible(x))) d) (∀x (Show(x) ∨ ¬Possible(x))) → Staff

e) (∀x (Show(x) → ¬Possible(x))) → ¬Staff