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Subsection 4.1.6 We Inherit the Boolean Operators

We can build more complex wffs by applying any of the Boolean operators to one or more simpler wffs.

Examples of wffs:

  1. \(\neg\) Bear(Smokey) A statement.

  2. Bear(Smokey)  Bear(Snowflake) A statement.

  3. Bear(x) \(\rightarrow \) HasTail(x) Not a statement because of unbound variable x.

As we use the Boolean operators to build larger wffs from smaller ones, we can use parentheses to group operators, just as we did in Boolean logic.

Here are two examples that use parentheses:

[4] (Bear(Smokey) \(\wedge \) Bear(Snowflake)) \(\vee \) Deer(Bambi)

[5] Bear(Smokey) \(\wedge \) (Bear(Snowflake) \(\vee \) Deer(Bambi))

Note that these two expressions are different. Because the two operators are different, parentheses matter. It’s possible for one of these statements to be true and the other to be false.

Now suppose that we want to write an expression that contains many terms, all of them connected with the same operator (either and or or). For example, we might write:

[6] ((Bear(Smokey) ∧ Bear(Snowflake)) ∧ Bear(Ling Ling)) ∧ Bear(Tai Shan)

We’ve been careful here (and in all our examples up until now) to use parentheses to indicate how the operators should be grouped, even though we know that Boolean and is associative. Grouping doesn’t matter. We’ve proved that claim in the case of two ands (and three operands). So we have that:

(p ∧ q) ∧ r is equivalent to p ∧ (q ∧ r )

We’ll see later that it is straightforward (using a proof technique called induction) to prove the extension of that result to any number of operators. So, in the case where all the operators are the same, parenthesization doesn’t matter. However, we’ll continue to indicate a particular parenthesization until we are more confident of what we are doing.

Exercises Exercises

Exercise Group.

1. For each of the following expressions, indicate whether or not it is a wff:

1.

Bear(Smokey) ∧ ¬¬Deer(Bambi)

  • True

  • False

Answer.
Correct answer is True.
2.

Prime(269) ∧ ∨ Prime(270)

  • True

  • False

Answer.
Correct answer is False.
3.

(Prime(269) ∨ Prime(270)) ∨ (Prime(271) ∨ Prime(272))

  • True

  • False

Answer.
Correct answer is True.
4.

Prime(age(Smokey))

  • True

  • False

Answer.
Correct answer is False.
Solution.
Explanation: The first one is a wff. Two nots in a row are allowed. The second one is not a wff. It is not allowed to have two binary operators in a row. The third one is a wff. The fourth one is a wff. It exploits a function that returns the value of Smokey’s age.

Exercise Group.

2. Let’s return to our example claims [4] and [5]:

  1. (Bear(Smokey) ∧ Bear(Snowflake)) ∨ Deer(Bambi)

  2. Bear(Smokey) ∧ (Bear(Snowflake) ∨ Deer(Bambi))

Recall that the difference between them is the way that parentheses are used to group subexpressions.

Consider the following situations:

I. Smokey, Snowflake and Bambi are all deer.

II. Smokey is a bear. Snowflake is a deer. Bambi is a bear.

III. Smokey is a deer. Snowflake is a deer. Bambi is a bear.

IV. Smokey is a deer. Snowflake is a bear. Bambi is a deer.

1.

(Part 1) Consider just claim [4]. For each of the four situations, indicate whether it makes claim [4] true or false:

  1. Situation I

  2. Situation II

  3. Situation III

  4. Situation IV

Answer.
a). True , b). False , c). False , d). True ,
Solution.
Explanation: [4] is true if Bambi is a deer (no matter what happens with Smokey and Snowflake). So it’s true in I and IV. If Bambi isn’t a deer, then [4] is only true if both Smokey and Snowflake are bears. That’s not true in either II or III, so they are false.
2.

(Part 2) Consider just claim [5]. For each of the four situations, indicate whether it makes claim [5] true or false:

  1. Situation I

  2. Situation II

  3. Situation III

  4. Situation IV

Answer.
a). False , b). False , c). False , d). False.
Solution.
Explanation: For [5] to be true, we need Smokey to be a bear and at least one of the two disjuncts also to be true. So I, III, and IV are false, no matter what happens with Snowflake and Bambi. That leaves II. For it to be true, we need at least one of Snowflake is a bear or Bambi is a deer to be true. But neither is. So II is also false.