Philosophy 112
What is This?
This is a supplement to section 2.1 of the Logic Text. You should read that before you read this.
Section 2.1 introduces three new two-place connectives: conjunction, disjunction, and the biconditional. It briefly discusses what they mean and how to parse sentences containing them.
New Connectives
To briefly review, the conjunction sign is ‘∧’, and is read as ‘and’, so ‘(P∧Q)’ is read as ‘P and Q’. The two parts of a conjunction (here, P and Q) are called its conjuncts. The disjunction sign is ‘∨’, read this as ‘or’, so ‘(P∨Q)’ is read as ‘P or Q’. The two parts of a disjunction (here, P and Q) are called its disjuncts. The biconditional sign is ‘↔’, read this as ‘if and only if’, so ‘(P↔Q)’ is read as ‘P if and only if Q’. The two parts of a biconditional are called its constituents.
Test Your Understanding
- What do we call the sentence P when it occurs in the conjunction, (P∧Q)?
We say that P is a conjunct. If we want to be more specific, we can say that it is the left conjunct.
- What do we call the sentence Q when it occurs in the disjunction, (P∨Q)?
We say that Q is a disjunct. If we want to be more specific, we can say that it is the right disjunct.
- How would you symbolize the sentence, ‘Penny sits and Quinn runs’?
You’d need to decide on a scheme of abbreviation. Let’s use P for ‘Penny sits’ and Q for ‘Quinn runs’. Then the sentence is symbolized as ‘(P∧Q)’.
- What do we call the sentence (S → R), when it occurs in the biconditional, ((S → R)↔Q)?
We call it a constituent, or, more precisely, the left constituent. It is also, of course, a conditional.
Syntax
These new connectives have the same syntax as the conditional. To create a sentence using one of them, you stick it between two sentences, and surround the whole thing in parentheses:
- if ◻ and ○ are both symbolic sentences, then so are (◻ → ○), (◻∧○), (◻∨○), and (◻↔○).
As before, we have conventions for dropping parentheses, and writing sentences in informal notation. The new conventions are summarized in the box on page 2 of chapter 2.
Test Your Understanding
For each of the following sentences, determine whether or not it is in official notation, informal notation, or not well-formed:
- (P∧Q
- A↔B
- (P∨Q)
- P∧R
- P → (Q∧R)
- P → Q∧R
- P∨Q∨R
- P∨Q∧R
- Not well-formed: note that it doesn’t have an even number of parentheses.
- Not well-formed: A and B are not sentence letters
- Official Notation
- Informal Notation: the outermost parentheses have been dropped. In official notation, (P∧R)
- Informal Notation: the outermost parentheses have been dropped. The main connective is the conditional, so in official notation, (P → (Q∧R)).
- Informal Notation: the outer most parentheses have been dropped. Also, the parentheses around the consequent, Q∧R, have been dropped, as allowed by the first of our new conventions. So, in official notation, (P → (Q∧R)).
- Informal Notation, following the second of our new conventions. In official notation, this sentence would be written as ((P∨Q)∨R).
- Not well-formed. We don’t have any convention that allows us to drop parentheses when we have a mixed series of conjunctions and disjunctions.
Characteristic Truth Tables
The meaning of a connective can be expressed with a truth table—that is, a table that summarizes the truth or falsehood of a complex sentence, given the truth or falsehood of its parts.
Chapter 2, section 1 presents truth tables for our three new connectives. Truth tables for the conditional and negation can be found on pages 4 and 5 of chapter 1. Before continuing, you should write down all five of those tables on a piece of paper and think a little bit about what they say.
Test Your Understanding
For each of the following sentences and the supplied schemes of abbreviation, first decide whether or not you think it is true, then look at the truth tables from 2.1, and calculate its truth value.
- P∧R (P: Obama is President; Q: Romney is President)
False. Here is the table for conjunction:
| ◻ | ○ | ◻∧○ |
|---|---|---|
| T | T | T |
| T | F | F |
| F | T | F |
| F | F | F |
Since P is true and R is false, the situation for P∧R is represented by the second row of this table. As that row indicates, when the left conjunct is true, and the right conjunct is false, the conjunction is false.
Here is another way to summarize what the table tells us: the conjunction is true only if both of its conjuncts are true. So, since R is false, P∧R is false.
- P∨Q (P: Obama is President; Q: Romney is President)
True. Here is the table for disjunction:
| ◻ | ○ | ◻∨○ |
|---|---|---|
| T | T | T |
| T | F | T |
| F | T | T |
| F | F | F |
Since P is true and R is false, the situation for P∨R is represented by the second row of this table. As that row indicates, when the left disjunct is true, and the right disjunct is false, the disjunction is false.
Here is another way to summarize what the table tells us: the disjunction is true if either of its disjuncts is true. So, since P is true, P∨R is true.
- P → Q (P: Obama is President; Q: Romney is President)
False! Here is the table for the conditional:
| ◻ | ○ | ◻ → ○ |
|---|---|---|
| T | T | T |
| T | F | F |
| F | T | T |
| F | F | T |
Since P is true and R is false, the situation for P → R is represented by the second row of this table. As that row indicates, when the antecedent is true, and the consequent is false, the conditional is false.
- P↔Q (P: Obama is President; Q: Romney is President)
False! Here is the table for the biconditional:
| ◻ | ○ | ◻↔○ |
|---|---|---|
| T | T | T |
| T | F | F |
| F | T | F |
| F | F | T |
Since P is true and R is false, the situation for P↔R is represented by the second row of this table. As that row indicates, when the left consitutent is true, and the right constituent is false, the biconditional is false.
Here is another way to summarize what the table tells us: the biconditional is true only if both of its constituents agree in “truth value”: that is, if both are true or both are false. But P is true and R is false, so P↔R is false.
- P∧Q (P: Penguins are birds; Q: Quail are birds)
- P∨Q (P: Penguins are birds; Q: Quail are birds)
- P → Q (P: Penguins are birds; Q: Quail are birds)
- P↔Q (P: Penguins are birds; Q: Quail are birds)
Since P and Q are both true, we are looking at the first row of each truth table.
True! A conjunction is true if both of its conjuncts are true, as is the case here.
True! A disjunction is true if both of its disjuncts are true, as is the case here.
You might be inclined to think that the English sentence is false—that ‘or’ is exclusive—one or the other, but not both. But our symbol, ‘∨’ represents inclusive disjunction. The table for exclusive disjunction would look like this:
◻ ○ ◻ exclusive-or ○ T T F T F T F T T F F F We are not going to introduce a connective for exclusive-or. Those who do sometimes use the disjunction sign, ∨, with a line beneath it: ⊻.
True! A conditional with both true antecedent and consequent is true, as you can see on the first row of the truth table.
You might balk at this. Shouldn’t the truth of the conditional require something more—a non-accidental connection between the truth of the antecedent and the consequent?
In English, no doubt this is often the case. But our conditional is what logicians call the material conditional. The only way it can be false is when the antecedent is true and the consequent false (see the second row of its truth table). Otherwise, it is true.
True! A biconditional is true if both of its constituents agree in truth value, as they do here.
You might be inclined to think that the English sentence is false—that ‘if and only if’ requires a stronger relationship between the two consituents; that it cannot just be an accident that they happen to both have the same truth value. But our biconditional ‘↔’ is the material biconditional, and expresses nothing more than agreement of truth value.
- R∧Q (R: Rabbits are birds; Q: Quail are reptiles)
- R∨Q (R: Rabbits are birds; Q: Quail are reptiles)
- R → Q (R: Rabbits are birds; Q: Quail are reptiles)
- R↔Q (R: Rabbits are birds; Q: Quail are reptiles)
Since R and Q are both false, we are looking at the bottom row of each truth table.
- False! A conjunction is true only if both of its conjuncts are true.
- False! A disjunction is true only if at least one of its disjuncts is true. But both of the disjuncts here are false.
- True! A conditional is true whenever its antecedent is false!
- True! A biconditional is true if both of its constituents agree in truth value. Since they are both false here, they both have the same truth value, so the biconditional is true.
Sometimes people get confused about the difference here between ‘and’ and ‘if and only if’. Sometimes people think that ‘P and Q’ should be true if P and Q are both false, since at least they have the same truth value. But ‘and’ requires not just agreement between the conjuncts, but that they both be true. If you want to express the weaker relationship of agreement, use the biconditional instead.
And sometimes people think that ‘P if and only if Q’ should be false if P and Q are both false, confusing the weak requirement of agreement with the stronger requirement that both be true. If you want to express that stronger relationship, use the conjunction instead.
Calculating the Truth Value of a Complex Sentence
Once you parse a complex sentence, you can use the truth tables to calculate its truth value, given an assignment of truth values to the sentence letters it contains.
You can find some examples that illustrate this on pages 3 and 4 of chapter 2. I will not repeat them here.
Test Your Understanding
Suppose P and Q are true. What is the truth value of ∼(P∧Q)?
Maybe you can calculate this one in your head, or maybe you cannot. Either way, get out a piece of paper and try to calculate it by hand, following the model presented in the text, of first parsing the sentence into a tree, and then calculating the truth value of the whole by working your way up from the truth values of the atomic sentences.
The solution is presented below in stages.
Here is what the sentence looks like when parsed:
| ∼ | ( | P | ∧ | Q | ) |
| ⌊ | |||||
| ( | P | ∧ | Q | ) | |
| ⋀ | |||||
| P | Q |
Is that what you had? If so, move on to the next step, assigning truth values to the atomic sentences.
Here is the parsed sentence, with truth values assigned to the atomic sentences:
| ∼ | ( | P | ∧ | Q | ) |
| ⌊ | |||||
| ( | P | ∧ | Q | ) | |
| ⋀ | |||||
| P | Q | ||||
| T | T |
The next step is to calculate the truth value of the conjunction…
Since both P and Q are true, P∧Q is true too:
| ∼ | ( | P | ∧ | Q | ) |
| ⌊ | |||||
| ( | P | ∧ | Q | ) | |
| T | |||||
| ⋀ | |||||
| P | Q | ||||
| T | T |
What next?
Since the conjunction is true, its negation is false:
| ∼ | ( | P | ∧ | Q | ) |
| F | |||||
| ⌊ | |||||
| ( | P | ∧ | Q | ) | |
| T | |||||
| ⋀ | |||||
| P | Q | ||||
| T | T |
And so we conclude that, when P and Q are both true, ∼(P∧Q) is false.
- Suppose P and S are true, but Q and R are false. What is the truth value of (P∧Q)∨(R → S)? Or, more succinctly,
| P | Q | R | S | (P∧Q)∨(R → S) |
|---|---|---|---|---|
| T | F | F | T | ? |
Again, the solution is presented in stages:
Here is the parsed sentence, with appropriate truth values assigned to the atomic sentences:
| ( | P | ∧ | Q | ) | ∨ | ( | R | → | S | ) |
| ⋀ | ||||||||||
| ( | P | ∧ | Q | ) | ( | R | → | S | ) | |
| ⋀ | ⋀ | |||||||||
| P | Q | R | S | |||||||
| T | F | F | T |
Now we calculate up the tree:
| ( | P | ∧ | Q | ) | ∨ | ( | R | → | S | ) |
| ⋀ | ||||||||||
| ( | P | ∧ | Q | ) | ( | R | → | S | ) | |
| F | T | |||||||||
| ⋀ | ⋀ | |||||||||
| P | Q | R | S | |||||||
| T | F | F | T |
And, again,
| ( | P | ∧ | Q | ) | ∨ | ( | R | → | S | ) |
| T | ||||||||||
| ⋀ | ||||||||||
| ( | P | ∧ | Q | ) | ( | R | → | S | ) | |
| F | T | |||||||||
| ⋀ | ⋀ | |||||||||
| P | Q | R | S | |||||||
| T | F | F | T |
And so we see that, in this situation, the complex sentence is true.