Truth Tables and Validity P1
p
q
r
S
pvq
T
T
T
T
T
T
T
T
T
T
r&s
~(p v q)
~p
T
F
F
T
F
T
T
F
T
F
T
~q
C
~p & ~q
~(p v q) ->
(r & s)
(~p & ~q)
-> (r & s)
F
F
T
T
F
F
F
T
T
T
F
F
F
T
T
F
T
F
F
F
T
T
T
T
T
F
F
F
T
T
F
T
F
T
F
F
F
T
T
T
F
F
T
T
F
F
F
T
T
T
F
F
F
T
F
F
F
T
T
F
T
T
T
T
F
T
F
F
T
T
F
T
T
F
T
F
T
F
F
T
T
F
T
F
T
T
F
T
F
F
T
T
F
T
F
T
T
F
T
F
F
T
T
F
F
T
T
F
T
T
T
T
T
T
T
F
F
T
F
F
F
T
T
F
F
F
F
T
F
F
T
T
F
F
F
F
F
F
F
T
T
F
Kareem Khalifa
Department of Philosophy
Middlebury College
Overview
• Why this matters
• How truth tables work
• How to work truth tables
– The basic method
– The shortcut
• Sample Exercises
Reminder: The Official
Definition of Deductive Validity
 A deductive argument is valid when, if all of its
premises are true, its conclusion must be true.
 This is the SINGLE MOST IMPORTANT
CONCEPT IN THE CLASS!!!!!!!!!
 Failure to define validity properly is an automatic
5 point penalty on anything you do! You’ll also
be very confused if you don’t get this concept.
Why this matters
• Thus far, testing for validity has involved
devising counterexamples.
– This method has limitations
• We may not be creative enough to devise a
counterexample for an invalid argument.
• So we will not always be reliable in testing for
validity.
• Truth tables provide an algorithm for
testing for validity, in which no creativity is
required, however…
• Truth tables can also enhance our
creativity in devising counterexamples.
How truth tables work
• Recall: An argument is valid if, when all of its
premises are true, its conclusion must also be
true.
• Equivalently, there is no way in which all of the
premises are true and the conclusion is false.
• Truth tables specify all the different ways in
which premises and conclusions can be true and
false.
– “Where there’s a row, there’s a way.”
• Thus, an argument is valid if there is no row on a
truth table in which all of its premises are true
and its conclusion is false.
Where there’s a row, there’s a
way
• Recall that the truth/falsity of compound
statements is a function of the truth/falsity
of simple statements
• Truth tables specify all of the different
combinations of truth and falsity of simple
statements
• So truth tables tell us all the different ways
compound statements can be true or false.
Where there’s a row, there’s a
way: Illustration with ‘&’
p
q
p&q
T
T
T
T
F
F
F
T
F
F
F
F
More on how truth tables work
• Note that validity says nothing
about cases in which:
–Some of the premises are false
–All of the premises are false
• Lesson: Focus on the rows in
which all of the premises are true.
How to work truth tables in
5 easy steps
1. Set up guide columns.
2. Set up an additional column for every premise and
the conclusion. Clearly mark these columns with
“P’s” for each premise, and “C” for the conclusion.
3. Add more columns to help interpret complex
propositions.
4. Using your knowledge of the truth-tables for
conjunction ‘&’ , disjunction ‘v’, negation ‘~’,
material implication ‘→’, and material equivalence
‘↔’, fill out the table.
5. Look at all of the columns with P’s and C’s above
them. If there are one or more rows in which all of
the P-columns have T’s and the C-column has an F,
then you have an invalid argument; otherwise the
argument is valid.
Step 1: Setting up guide columns
• First, count the number of simple
propositions in the argument or proposition
under consideration. Call this number N.
• Ex A. Suppose you are asked to discern
the validity of the following argument: ~q |[(p → q) → p] → p. There are two simple
propositions—p and q—thus N=2.
More on guide columns
• Next, set up a table with
2N + 1 rows. Place the
simple propositions in the
top row in the leftmost
columns. These columns
are called guide columns.
• Continuing with
~q |- [(p → q) → p] → p,
there would 2N +1 rows=
22 +1 = 4+1 = 5 rows, this
leads to the following:
p
q
Filling out guide columns
• For the leftmost guide
column, the top 2N-1 rows
will be True, the bottom
2N-1 rows will be false. In
other words, the top half
of the leftmost column will
be true, the bottom half
will be false.
• In our example, the top
2N-1 rows will be 22-1= 21 =
the top 2 rows. Thus we
fill in the top half of the
leftmost column with T’s,
and the bottom with F’s.
p
T
T
F
F
q
More on filling out guide columns
•
•
•
For the next column, the top 2N-2
rows will be T, the next
2N-2 rows will be F, the next
2N-2 rows will be T, etc. In this
case, the top quarter will be T, the
next quarter will be F, the third
quarter will be T, the last quarter
will be F.
Continue raising 2 to a lower
power until the rows alternate
between T and F, i.e., when the
(20th or) 1st row is T, the next (20 =
1) row is F, and so on.
In our example, this means just
one more step—22-2 = 20 = 1 so
alternate every 1 row between T
and F. Thus, you’re done
constructing guide columns for
~q |- [(p → q) → p] → p
p
q
T
T
T
F
F
T
F
F
What you’ve done so far…
• You’ve presented all of the possible ways
in which to interpret the argument you’re
out to analyze.
Step 2: Columns for premises and
conclusions P
C
• Set up an additional
column for every premise
and the conclusion. Mark
“P” above the columns
corresponding to
premises and “C” above
the column corresponding
to the conclusion.
• In our example, we have
one premise, ~q, and one
conclusion,
[(p → q) → p] → p. Thus,
we add two columns to
our truth-table.
p
q
T
T
T
F
F
T
F
F
~q [(p → q) → p]
→p
Step 3: Add more columns to
interpret complex propositions
•
•
First, for each complex premise/conclusion,
count the number of logical operators. Call this
number M. Add M-1 columns to the truth table.
There are 3 logical operators in this
conclusion, so add 3-1 = 2 additional columns
P
p
T
T
F
F
q
T
F
T
F
~q
C
→ q) →
→ p] → p
[(p →
More on additional columns
• Next, analyze the compound statement, such
that each of the additional columns corresponds
to the scope of the operator in question.
P
p
q
T
T
T
F
F
T
F
F
p→q
(p → q)→p ~q
C
→ p]
→ q) →
[(p →
→p
Step 4: Filling out the table
• Using your knowledge of the truth-tables for
conjunction ‘&’ , disjunction ‘v’, negation ‘~’,
material implication ‘→’, and material
equivalence ‘↔’, fill out the table.
P
p
q
p→q
T
T
F
F
T
F
T
F
C
[(p → q) → p]
→p
T
(p → q) → ~q
p
F
T
F
T
T
T
T
F
F
T
F
T
T
T
T
Step 5: Reading the truth table
•
•
•
Look at all of the columns with P’s and C’s above them. If there are one or more
rows in which all of the P-columns have T’s and the C-column has an F, then you
have an invalid argument; otherwise the argument is valid.
IMPORTANT: This does NOT mean that if you have ONE row with F’s in the Pcolumns or T’s in the C-column that you have a valid argument—you need to
prove that there are NO rows in the ENTIRE truth table that have all T’s in the Pcolumns and an F in the C-column in order to have a valid argument.
In this example, there is NO row in which P = T and C = F. Therefore the argument
is valid.
P
C
p
q
p→q
(p → q) →
p
~q
[(p → q) → p]
→p
T
T
F
F
T
F
T
F
T
F
T
T
T
T
F
F
F
T
F
T
T
T
T
T
Shortcut in Step 4
• First, determine which of the premises and
conclusion is the easiest in terms of figuring out
the truth values. Fill this column first.
P
p
q
T
T
F
F
T
F
T
F
p→q
(p → q) →
p
~q
F
T
F
T
C
[(p → q) → p]
→p
Shortcut, Continued
• Since an invalid argument can only occur when
the premises are true and the conclusion is
false, any rows in which either a premise is false
or the conclusion is true can be ignored, since it
will be irrelevant to determining the invalidity of
the argument.
P
p
q
T
T
F
F
T
F
T
F
p→q
(p → q) →
p
IGNORE!
F
T
IGNORE!
T
F
~q
F
T
F
T
C
[(p → q) → p]
→p
IGNORE!
T
IGNORE!
T
Exercise B2
VALID!
• (C v D)  (C & D), C & D├ C v D
C
P2
P1
C v D C & D (C v D)  (C & D)
C
D
T
T
T
T
F
T
F
T
F
F
IGNORE!
T
F
F
IGNORE!
Exercise B10
INVALID!
• U  (V v W), (V&W)  ~U ├ ~U
U V W ~U V v W V &
C
W
U
(V v W)
(V&W)  ~U
P1
P2
T T
T
F
T
T
T
F
T T
F
F
T
F
T
T
T F
T
F
T
F
T
T
T F
F
F
F
F T
T
T
F T
F
T
F F
T
T
F F
F
T
IGNORE!
F
IGNORE!
IGNORE!
VALID!
Exercise C7
• If terrorists’ demands are met, then lawlessness will be rewarded. If
terrorists’ demands are not met, then innocent hostages will be
murdered. So either lawlessness will be rewarded or innocent
hostages will be murdered.
P1
P2
C
• TL, ~TI |- L v I
T L I ~T T  L ~T  I L v I
T T T
T
F
T
F
T
IGNORE
F T F
F F T
F F F
IGNORE
T
T
T
F
T
F
F
T
T
T
F
F
T
T
T
T
T
F
F
Exercise B8
VALID!
• (O v P)  Q, Q  (O & P)├ (O v P)  (O & P)
P1
P2
C
O P Q O v P O & P (OvP)  Q 
(OvP) 
(O&P) (O&P)
Q
T T T
T
T
T T F
T
T
T F T
T
F
T F F
T
F
F T T
T
F
F T F
T
F
F F T
F
F
F F F
F
F
IGNORE!
IGNORE!
F
IGNORE!
F
T
T
F
F
T
F
F
F
T
F
T
IGNORE!
T