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Subsection 5.2.5 Quantifier Exchange Proof Problem: First Class

FirstClass

Assume that if anyone has a coach class ticket but there is no coach seat available then that passenger gets a first class ticket. Alice has a coach class ticket. But every coach seat is one that cannot be occupied by Alice. Thus, Alice gets a first class ticket.

Assign the following names to basic objects and statements:

A: Alice

C(x): True if x has a Coach class ticket.

O(x, y) True if Person x may Occupy seat y.

F(x) True if x gets a First class ticket.

Prove: ∀x ((C(x) ∧ ¬∃y (O(x, y))) → F(x)) If anyone has a coach class ticket but there is no coach seat then that passenger gets a first class ticket.

C(A) Alice has a coach class ticket.

∀y (¬O(A, y)) Every coach seat is one that cannot be occupied by Alice.

∴ F(A) Alice gets a first class ticket.

You should do this proof yourself.

You can also watch our video, which will outline our strategy for doing this.

Video cover image

Exercises Exercises

Exercise Group.

1.

Neighborhood Gangs : Assume the following premise:

[1] (∃x (MemberOfGangWeHate(x) ∧ InOurNeighborhood(x) ))→ HighAlert

(By the way, notice the predicate HighAlert. It looks like a Boolean proposition. It is. There’s no reason that we can’t, in predicate logic, use Boolean propositions. Think of them as predicates that don’t take any arguments.)

Suppose that we want to prove:

HighAlert

(Part I) Which of the following statements, would if combined with [1], be sufficient to enable us to prove HighAlert?

I. MemberOfGangWeHate(OneEye)

II. MemberOfGangWeHate(Bugsy)

III. InNeighborhood(Lefty)

IV. InNeighborhood(Bugsy)

V. ∀x (InNeighborhood(x))

  1. I and III together is the only sufficient combination.

  2. II and IV together is the only sufficient combination.

  3. I and V together is the only sufficient combination.

  4. I and V together is a sufficient combination but there are other sufficient combinations of these premises.

  5. No combination of these premises is sufficient.

Answer.
Correct answer is D.
Solution.
Explanation. All we need is to find one person who is both in the gang and in the neighborhood. We can do that with II and IV: the person is Bugsy. But we can also do it with V (which tells us that everyone is in the neighborhood) plus any claim about someone being in the gang. That could be either I or II.
2.

Let’s return to the Neighborhood Gangs problem. We’ll abbreviate MemberOfGangWeHate as Member and InOurNeighborhood as In. Prove HighAlert assuming the following premises:

[1] (∃x (Member(x) ∧ In(x) ))→ HighAlert Premise

[2] Member(Bugsy) Premise

[3] In(Bugsy) Premise

Answer.

questionId: Gangs1

problemType: gradeLogicProof

questionTitle: Generalizing from a Known Instance

questionDisplayText: Working with quantifiers when it’s not necessary to instantiate.

[1] (x (Member(x)  In(x) ))  HighAlert Premise

[2] Member(Bugsy) Premise

[3] In(Bugsy) Premise

[4] Member(Bugsy)  In(Bugsy) Conjunction [2], [3]

[5] x (Member(x)  In(x) ) Existential Generalization [4]

[6] High Alert Modus Ponens [1], [5]

3.

Back to the Neighborhood Gangs problem again. This time, prove HighAlert using the following premises:

[1] (∃x (Member(x) ∧ In(x) ))→ HighAlert Premise

[2] Member(OneEye) Premise

[3] ∀x (In(x)) Premise

Answer.

questionId: Gangs2

problemType: gradeLogicProof

questionTitle: Different Premises Lead to the Same Conclusion

questionDisplayText: Applying a general claim to a known instance, OneEye.

hints: Derive as much as you can about OneEye. Then see if you can describe something that must exist because OneEye does.

[1] (x (Member(x)  In(x) )) HighAlert Premise

[2] Member(OneEye) Premise

[3] x (In(x)) Premise

[4] In(OneEye) Universal Instantiation [3]

[5] Member(OneEye)  In(OneEye) Conjunction [2], [4]

[6] x (Member(x)  In(x) ) Existential Generalization [5]

[7] High Alert Modus Ponens [1], [6]