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Review We can use truth tables to evaluate the validity of arguments. Example: P1: B(C v A) P2: B v C C: A B(C v A) / B v C // A Like This B(C v A) / B v C / A T T T T T T T T T T T T T F T T T F T T F T T T T F T T F F F F T T F F F T T T T F T T T

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review
Review

We can use truth tables to evaluate the validity of arguments.

Example:

P1: B(C v A)

P2: B v C

C: A

B(C v A) / B v C // A

like this
Like This

B(C v A) / B v C / A

T T T T T T T T T

T T T T F T T T F

T T F T T T T F T

T F F F F T T F F

F T T T T F T T T

F T T T F F T T F

F T F T T F F F T

F T F F F F F F F

Row 2 proves that the argument is invalid (so does row 6)

indirect truth tables
Indirect Truth Tables

Consider an argument symbolized thusly:

P1. (B * D)~C

P2. D v B

C1. ~CB

If we were to use a full truth table to evaluate its validity, we would need 8 rows

shorter method
Shorter Method
  • There is a shorter way to test validity:

Start by assuming the argument is invalid

Represent this by putting T’s under the main logical operator of the premises and an F under the main logical operator of the conclusion

like this5
Like This

(B * D)~C / D v B // ~CB

T T F

Next, determine the truth values of the other statements, given the truth values of the premises and conclusion:

(B * D)~C / D v B // ~CB

F F T TT F T T F TF F F

notice
Notice
  • We have a consistent assignment of truth values that makes the premises true and the conclusion false
  • Therefore, the argument is invalid, as assumed
another example
Another Example

Consider this argument:

P1: C(BD)

P2: ~(BD)

C1: ~C

C(BD) / ~(BD) // ~C

T T F

a contradiction
A Contradiction!

C(BD) / ~(BD) // ~C

T T TF F T T F F FT

Notice that we cannot give a consistent assignment of truth values. The first premise, if we assume it to be true, is false. Because we get a contradiction, the argument is valid.

a complication
A Complication
  • Sometimes there is more than one way for the premises to be true and the conclusion false

Example:

A v B / BA / ~A v ~B // ~B * A

There are three ways the conclusion can be false!

slide10
See?

A v B / BA / ~A v ~B // ~B * A

T T T TF F F

T T T FT F T

T T T FT F F

To test the validity of the argument, we need to see if we can find a consistent set of truth value assignments on at least one of the three rows

start with the first row
Start with the First Row

A v B / BA / ~A v ~B // ~B * A

F T F TF F F

FT F T

FT F F

Notice that we have a contradiction in the first premise: we assumed it is true, but that assumption is inconsistent with the truth values of A and B on that row

can we stop
Can we Stop?
  • No! That there is a contradiction on the first row does not prove the argument is valid, for there may be a consistent assignment of truth values on the other two rows. Remember, we are looking for at least one consistent assignment of truth values to prove the argument invalid
go to second row
Go To Second Row

A v B / BA / ~A v ~B // ~B * A

F T F TF F F

FT T FT FT F T

FT F F

We have a contradiction in the third premise on the second row. So me must look at the third row

go to third row
Go to Third Row

A v B / BA / ~A v ~B // ~B * A

F T F TF F F

FT T FT FT F T

T T F FT F F

We have a contradiction in the second premise. Because there is a contradiction on every row, we know the argument is valid (there are no rows with a consistent assignment of truth values that makes the premises true and the conclusion false)

a good procedure to follow
A Good Procedure to Follow
  • Determine the number of ways the conclusion can be false and represent them with a truth table under the conclusion
  • Start with the first row and check if a contradiction is produced. If so, go to the second row. If not, stop; the argument is invalid
  • Repeat the second step until there are no more rows. If a contradiction is produced on every row, the argument is valid.
testing consistency with indirect truth tables
Testing Consistency with Indirect Truth Tables
  • We can determine whether a set of statements is consistent by first assuming that all the statements are true

Example:

~AB/ B(~CA)/ B~C/ ~A

T T T T

Next, compute the truth values of the other statements, based on this assumption

like this17
Like This

~AB/ B(~CA)/ B~C/ ~A

TF TT T T TFF F T T TF TF

Notice that there is a contradiction in the second statement.

Also notice that we need only one row, because there is only one way for the last statement to be true (and thus the assumption that all of them are true).

Because we derived a contradiction, the statements cannot all be true—thus, the statements are inconsistent.

a complication18
A Complication
  • Sometimes there is more than one way for a set of statements to come out true

Example:

~A v (B * C)/ ~C v ~A/ ~BA

T T T

In this case, there are three ways each of the statements can be true (to see this, recall the truth tables for disjunction and conditional)

what to do
What to do?

Select one of the statements and represent the three ways it can be true. Let’s select the last statement

~A v (B * C)/ ~C v ~A/ ~BA

T T FT T F

FT T T

TF T T

look at the first row
Look at the First Row

~A v (B * C)/ ~C v ~A/ ~BA

TF T T T TF FT T F

FT T T

TF T T

Notice that we do not have to assign a truth value to C in the first row, because the first two statements will come out true regardless. Since there is a consistent assignment of truth values on at least one row, the statements are consistent.

the procedure for testing consistency
The Procedure for Testing Consistency
  • Look at the statements and pick the one that has the fewest number of ways it can come out true.
  • Represent those ways with a truth table under the statement
  • Start with the first row and determine the truth values of the other statements.
  • If no contradiction is produced, stop. The statements are consistent. If not, go to the second row and repeat the previous step
  • If a contradiction is produced on every row, the statements are inconsistent.
common valid and invalid argument forms
Common Valid and Invalid Argument Forms
  • An argument form is valid just in case every substitution instance of that form is such that it is impossible for the premises to be true and the conclusion false
  • In the language of truth tables, there are no rows where the premises are true and the conclusion false
continued
Continued
  • How shall we define an invalid argument form?
  • Should we say that an argument form is invalid just in case every substitution instance of that form is invalid (e.g. there is at least one row on the truth table where the premises are true and the conclusion false?)
slide24
No
  • Here’s why: consider this argument form

pq

q

p

It is in an invalid form, but there are substitution instances of the form that are valid!

slide25
Huh?
  • See? Consider this substitution instance

(A v ~A)B

B

(A v ~A)

It is impossible for the premises to be true and the conclusion false, because the conclusion is a tautology (it is true on every row of the truth table)

slide26
So…
  • Let’s define an invalid argument form this way:

An argument form is invalid just in case there is at least one substitution instance of that form that is invalid.

In the language of truth tables, there is at least one substitution instance where at least one row has true premises and a false conclusion