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Subsection 6.1.2 Weaker Statements/Stronger Statements

The logical framework that we’ve just described gives us the tools we need to make statements that are exactly as strong as the facts warrant:

Strongest: [1] ∀x (Q(x) ∧ R(x))

[2] ∀x (Q(x))

[3] ∀x (P(x) ∨ Q(x)) [4] ∀x (P(x) → Q(x))

Weakest: [5] ∀x ((P(x) ∧ R(x)) → Q(x))

There’s also what may appear to be an even weaker claim:

[6] ∃x (Q(x))

But this one isn’t so straightforward. We’ll look at existential quantification on the next slide.

Meanwhile, in examples [1] – [5], we’ve said:

[1] Both Q and R are true of every x.

[2] Q is true of every x. We aren’t committing to anything about R.

[3], [4] Since [4] can be rewritten (using Conditional Disjunction) as ∀xP(x) ∨ Q(x)), we see that [3] and [4] are equally strong/weak. Which one we pick depends just on whether we’ve defined P in a positive or a negative way (and thus whether we wish to assert it or its complement).

[5] Q(x) can only be guaranteed to be true if both P(x) and R(x) are true.

Here’s an important way to view versions [4] and [5]: They contain “guards” that limit the circumstances in which their conclusions are guaranteed to be true. So, in [4], Q is only required to be true for those values of x for which P is true. And, in [5], both P and R must be true in order to guarantee the truth of Q for a given x.

Big Idea

Given an implication of the from P \(\rightarrow\) Q, we can think of P as a guard: it limits the circumstances in which we can guarantee that Q must be true.

Consider the following statements about people entering a big deal, special event. Define:

Ticket(x): True if x must have a ticket to enter.

ID(x): True if x must have a photo ID to enter.

Wrist(x): True if x must have a wristband to enter.

SignIn (x): True if x must sign in and be photographed at a special sign-in window.

Student(x): True if x has a special student ticket.

HasID(x): True if x is able to present a student ID.

Strongest: [1] \(\forall \)(Ticket(x) \(\wedge \) ID(x) \(\wedge \) SignIn(x))

[2] \(\forall \)x (Ticket(x) \(\wedge \) SignIn(x))

[3] \(\forall \)x ((Ticket(x) \(\vee \) Wrist(x)) \(\wedge \) SignIn(x)) [4] \(\forall \)x (Ticket(x) \(\wedge \) (Student(x) \(\rightarrow \) SignIn(x)))

Weakest: [5] \(\forall \)x (Ticket(x) \(\forall \) ((Student(x)  HasID(x))  SignIn(x)))

[1] says that everyone must have a ticket and an ID and must go to the sign-in window. [2] relaxes the ID requirement. [3] allows the option of a wristband or a ticket. [4] relaxes the requirements in a different way. It says that one only needs to sign in if one has a special student ticket. And [5] is even weaker. It says that one must sign in only if one both has a special student ticket and is not carrying a student ID.

Exercises Exercises

1.

1. Consider the following claim:

[1] ∀x (S(x) ∨ T(x))

Which of the following sets of premises is strong enough to enable us to prove [1]:

  1. x ((P(x) ∧ (Q(x) ∧ R(x))) → S(x))

x (P(x) ∨ R(x))

  1. x ((P(x) ∧ (Q(x) ∨ R(x))) → S(x))

x (P(x) ∨ Q(x))

  1. x ((P(x) ∨ (Q(x) ∨ R(x))) → S(x))

x (P(x) ∨ R(x))

  1. x (P(x) ∨ (Q(x) ∨ R(x)))

x (P(x) → T(x))

Answer.
Correct answer is C.

2.

2. Consider the following claim:

[2] ∀x (R(x) → T(x))

Which of the following sets of premises is too weak to enable us to prove [2]:

  1. x (P(x) ∧ T(x))

x (Q(x) → P(x))

  1. x (P(x) → ¬R(x))

x (P(x))

  1. x (P(x) → ¬R(x))

x (P(x) ∨ Q(x))

  1. x (Q(x) ∧ T(x))

x (P(x) → Q(x))

Answer.
Correct answer is C

3.

3. Consider the following claim:

[3] ∀xQ(x))

Which of the following sets of premises is so strong that [3] cannot be true:

x (P(x) ∨ Q(x))

x (R(x) → ¬P(x))

x (P(x) ∨ Q(x))

xP(x))

x (P(x) ∨ Q(x))

xP(x) ∨ ¬R(x))

x (P(x) ∨ Q(x))

x (P(x) → ¬Q(x))

Answer.
Correct answer is B.