1.5 Rules of Inference and Valid Arguments
In this section we look at rules of inference (also called logical implications) which form
the basis of logical arguments.
Example 1. Here is a typical example of something someone might say.
If someone works in the accounting department then they make more than 50K.
John works in the accounting department. Therefore John makes more than 50K.
Is this a valid argument? Notice the first part "If someone works in the accounting
department then they make more than 50K" involves the "for every" () operation. In
this section we want to work at a simpler level and only work with propositions. We will
postpone valid arguments involving predicates until the next section.
Example 2. With this in mind, let's rephrase Example 1 as
(1)
If John works in the accounting department then John makes more than 50K.
John works in the accounting department. Therefore John makes more than 50K.
Is this valid argument? What do we mean by a valid argument? We shall say it is an "ifthen" statement that is always true. To see what we mean by this, let
p = "John works in the accounting department"
q = "John makes more than 50K"
Then (1) has the form
(2)
[(p  q)p]  q
In words we are saying "if p implies q is true and p true then q is true. [(p  q)p]  q
has the form
(3)
PQ
where P and Q are the logical expressions P = (p  q)p and Q = q.
If P and Q are logical expressions then we say
P logically implies Q if P  Q evaluates to true for every possible combination
of logical values for the logical variables appearing in them.
We write P  Q if this holds and say P  Q is a rule of inference or logical
implication or valid argument.
1.5 - 1
The expression p(p  q)  q is called modus ponens which is sometimes translated as
the way that affirms by affirming. It is one of the most famous rules of inference. Modus
ponens is also often written as
p
pq
q
To show that modus ponens is in fact a valid argument we make a truth table.
p
0
0
1
1
q
0
1
0
1
pq
1
1
0
1
p(pq)
0
0
0
1
p(pq)  q
1
1
1
1
Note that there is 1 (true) in the last column for p(p  q)  q. So p(p  q)  q is in
fact a valid argument. In this case a statement of the form p(p  q)  q is always true
no matter whether p and q are individually true of false.
Note that in showing p(p  q)  q is always true, it was not really necessary to make the
last column of the truth table. It would have been sufficient to note that whenever there is
a 1 in a row in the column for p(p  q) there is also a 1 in that row for the column of q.
This implies that the column for p(p  q)  q has all 1's. This is a general rule.
P  Q if whenever there is a 1 in a row for the truth table of P there is a 1 in the
truth table of Q
One can also show P  Q by showing P  Q = 1 using Boolean algebra.
Example 2. Show p(p  q)  q = 1 using Boolean algebra.
Solution. We use p  q = p' + q. Then (p(p  q))  q = (p(p' + q))' + q =
(pp' + pq)' + q = (0 + pq)' + q = (pq)' + q = p' + q' + q = p' +1 = 1.
To summarize, there are three possible ways to show P  Q.
1. Show the truth table of P  Q has a 1 in every row.
2. Show each row of the truth table of P with a 1 also has a 1 in the truth table of Q.
3. Show P  Q = 1 using Boolean algebra
Some books state that something is a logical implication in terms of the notion of a
tautology.
An expression P is a tautology if it is true for all values of the logical
variables in the expression, i.e. P = 1.
1.5 - 2
Example 3. Show that p + p' is a tautology.
Solution. It is true if p is true or if p is false. Hence it is a tautology.
In fact it was noted in formula (11) of section 1.3 that p + p' = 1. For example the
statement "Either it is raining or it is not raining." is always true.
Note that
P  Q if and only if P  Q is a tautology
So, another way of stating that modus ponens is a valid argument is to say that
"(p(p  q))  q is a tautology".
Another famous valid argument is modus tollens which is (p  q)q'  p'. In words,
if p implies q and q is false then p is false.
Example 4. The following two statements have the form of modus tollens and hence are
always true.
If Fido is a bird implies Fido can fly and if Fido can not fly then Fido is not a bird.
If John works in the accounting department implies John makes more than $50K
and if John doesn't make more than $50K then John doesn't work in the accounting
department.
Example 5. Show modus tollens is a valid
argument.
Solution. The truth table at the right does this
since every row that has a 1 for (p  q)q' also has
a 1 for p'.
p
0
0
1
1
q
0
1
0
1
pq
1
1
0
1
q'
1
0
1
0
(pq)q'
1
0
0
0
p'
1
1
0
0
An alternative way to see that modus tollens is use the following famous equivalence
p  q = q'  p'
This is called transposition. If we use transposition then (p  q)q' = q'(q'  p'). By
modus ponens one has q'(q'  p')  p'. So (p  q)q'  p'.
Rules of inference P  Q where P is the and of several other logical expression are
called valid arguments.
1.5 - 3
A valid argument is a rule of inference of the form (P1P2…Pn)  Q
where P1, P2, …, Pn and Q are logical expressions.
Remark. We shall usually use the term valid argument for the case where there are
several Pi. Also, P1P2…Pn is called the hypothesis and Q is called the conclusion and
P1, P2, …, Pn, are called premises. Often we write the valid argument P1P2…Pn  Q as
P1
P2
…
__
Pn
Q
Here is a list of basic rules of inference and equivalences that are used to verify that other
expressions are valid arguments.
Rules of Inference
(1)
p(p  q)  q
modus ponens (MP)
(2)
(p  q)q'  p'
modus tollens (MT)
(3)
(p  q)(q  r)  p  r
(4)
(p + q)p'  q
(5)
pp+q
(6)
pq  p
(7)
(p  q)(r  s)(p + r)  q + s
(8)
(p  q)(r  s)(q' + s')  p' + r'
hypothetical syllogism (HS)
disjunctive syllogism (DS)
addition or disjunction introduction (AD)
simplification (S)
constructive dilemma (CP)
destructive dilemma (DD)
Equivalences
(9)
p  q = (p  q)(q  p) = pq + p'q'
(10a)
p  q = p' + q = q'  p'
Transposition (T)
(10b)
pq1q2…qn  r = p' + q1' + q2' + … + qn' + r = r'q1q2…qn  p'
Transposition (T)
(11)
p  q  p  pq
(12)
(pq)  r = p  (q  r)
absorption (AB)
exportation (E)
That (1) – (8) are rules of inference and (9) – (12) are equivalences can be verified either
by means of truth tables or Boolean algebra. If one does this by Boolean algebra, then it
1.5 - 4
is useful to recall that (p  q) = p' + q and (p  q)' = pq'. At the end we shall show most
of the ones that have not been shown already.
The term syllogism has pretty much the same meaning at the term valid arugument.
In words, a hypothetical syllogism can be stated as "if p implies q and q implies r then p
implies r.
Example 11. The statement "If Tweetie is a bird implies Tweetie can fly and if Tweetie
can fly implies Tweetie has feathers then Tweetie is a bird implies Tweetie has have
feathers" is a true statement.
Example 12. Show a hypothetical
syllogism is a rule of inference.
In the truth table at the right every
row that has a 1 for (p  q)(q  r)
also has a 1 for p  r.
p
q
r
pq
qr
(pq)(qr)
pr
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
1
1
1
1
0
0
1
1
1
1
0
1
1
1
0
1
1
1
0
1
0
0
0
1
1
1
1
1
0
1
0
1
In addition to truth tables and
Boolean algebra there is another method to show expressions are valid arguments. This
method is based upon the following two properties of rules of inference.
(13)
If P  Q and Q  R then P  R
(14)
If P  Q and P  R then P  QR
transitivity
To see why (13) is true note that P  Q implies whenever there is a 1 in a row of the
truth table for P there is a 1 in the corresponding row of the truth table for Q. Similarly,
Q  R implies whenever there is a 1 in a row of the truth table for Q there is a 1 in the
corresponding row of the truth table for R. So if P  Q and Q  R are both true then if
there is a 1 in a row of the truth table for P then there is a 1 in the corresponding row of
the truth table of Q and hence there is a 1 in the corresponding row of the truth table of R,
so P  R is true.
To see why (14) is true suppose P  Q and P  R . So whenever there is a 1 in a row of
the truth table for P there is a 1 in the corresponding rows of the truth tables for Q and R
and hence the corresponding row of QR. Thus P  QR.
Note that to show P1P2…Pn  Q it is sufficient to do the following.
1.5 - 5
(15)



Find Pn+1, Pn+1, …, Pn+m such that
P1P2…Pn  Pn+1
P1P2…Pn  Pn+2
...
P1P2…Pn  Pn+m
Pn+m = Q
Next note that to show P1P2…Pn  Pn+1 it is sufficient to do the following.
(16)
Find a subset Pj1, Pj2, …, Pjk of P1,P2, …, Pn such that

 Pj1Pj2…Pjk  Pn+1
The reason for this is that we can use simplification to conclude that
P1P2…Pn  Pj1Pj2…Pjk and then use transitivity to conclude P1P2…Pn  Pn+1.
To show P1P2…Pn  Pn+2 it is again sufficient to do the following.
(17)
Find a subset Pj1, Pj2, …, Pjk of P1,P2, …, Pn+1 such that

 Pj1Pj2…Pjk  Pn+2
The reason for this is that we can repeatedly use (14) to conclude that
P1P2…Pn  Pj1Pj2…Pjk and then use transitivity to conclude P1P2…Pn  Pn+2.
In general, showing
(18)
Find a subset Pj1, Pj2, …, Pjk of P1,P2, …, Pn+k-1 such that

 Pj1Pj2…Pjk  Pn+k
for k = 1, 2, …, m will show (15).
Often when people use (15) and (18) to verify that P1P2…Pn  Q is a valid argument
they display the logic something like the following.
Step
1.
2.
(19)
Reason
P1
P2
. . .
n
Pn
n +1
Pn+1
n +2
Pn+2
. . .
n +m
Pn+m = Q
Premise
Premise
Premise
Rule r1 using j1,1, j1,2, …, j1,k1
Rule r2 using j2,1, j2,2, …, j2,k2
Rule rm using jm,1, jm,2, …, jm,km
1.5 - 6
where r1, r2, …, rm are rules of inference. Here is an example.
Example 13. Show that the following is a valid argument.
If
and
and
and
then
John is not optimistic and John is busy
John will play the lottery only if he is optimistic
If John does not play the lottery then he will visit the casino
If John visits the casino then he will be broke
John is broke
p
q
r
s
t
John is optimistic
John is busy
John will play the lottery
John will visit the casino
John is broke
Let
=
=
=
=
=
Then the above is
If
and
and
and
then
p'q
rp
r'  s
st
t
or p'q(r  p)(r'  s)(s  t)  t. The following shows this is a valid argument.
Step
1.
2.
3.
4.
5.
6.
7.
Reason
Premise
Premise
Modus tollens using 1 and 2
Premise
Modus ponens using 3 and 4
Premise
Modus ponens using 5 and 6
p'
rp
r'
r'  s
s
st
t
When using the above method of establishing valid arguments, the following rule of
inference is useful.
(20)
pq  pq
Conjunction
This is clearly true. It is typically applied in the following type of situation. Suppose you
have an argument of the following form.
1.5 - 7
Step
1.
2.
Reason
P1
P2
. . .
n
Pn
n +1
Pn+1
n +2
Pn+2
. . .
n +m
Pn+m
Premise
Premise
Premise
Rule r1 using j1,1, j1,2, …, j1,k1
Rule r2 using j2,1, j2,2, …, j2,k2
Rule rm using jm,1, jm,2, …, jm,km
At this point you can add a line of the form
n +m+1
PjPk
Conjunction using j and k
In this context (20) is follows from (14).
Example 14. Show that the following is a valid argument
If
and
and
then
Tweetie is a bird
Tweetie is a bird implies Tweetie flies
Tweetie is a bird implies Tweetie has feathers
Tweetie flies and has feathers
If we let
p = Tweetie is a bird
q = Tweetie flies
r = Tweetie has feathers
then what we want to show has the form [p(p  q)(p  r)]  qr. We show this is a
valid argument as follows.
Step
1.
2.
3.
4.
5.
6.
Reason
Premise
Premise
Modus ponens using 1 and 2
Premise
Modus ponens using 1 and 4
Conjunction using 3 and 5
p
pq
q
pr
r
qr
The list of basic rules of inference and equivalences (1) – (12) is probably not long
enough to always easily show that something is a valid argument using the method (19).
In that case it might be easier to use a truth table.
1.5 - 8
Example 15. Show that the following is a valid argument.
If
Fido is a dog implies Fido is a mammal
then Fido is a brown dog implies Fido is a brown mammal
If we let
p = Fido is a dog
q = Fido is a mammal
r = Fido is brown
then what we want to show has the form (p  q)  (pr  qr). It looks like this might be
hard to show using (1) – (12). In this case we can use a truth table. As you can see from
the truth table at the right,
p
q
r
pr
qr
pq
pr  qr
everywhere there is a 1 in the
0
0
0
1
0
0
1
column for p  q there is a 1 in
0
0
1
1
0
0
1
the column for pr  qr. So
0
1
0
1
0
0
1
0
1
1
1
0
1
1
(p  q)  (pr  qr).
1
1
1
1
0
0
1
1
0
1
0
1
0
0
1
1
0
1
0
1
0
0
0
1
1
0
1
1
It turns out that with a little bit of manipulation it is possible to deduce
(p  q)  (pr  qr) by an argument of the form (19). This is done as follows. To show
(p  q)  (pr  qr) we need to show [(p  q)  (pr  qr)] = 1. However, by (12)
(p  q)  (pr  qr) is equivalent to [(p  q)pr]  qr. So we need to show
[[(p  q)pr]  qr] = 1 which is the same as [(p  q)pr]  qr. This can be shown to be
a valid argument as follows.
Step
1.
2.
3.
4.
5.
Reason
Premise
Premise
Modus ponens using 1 and 2
Premise
Conjunction using 3 and 4
pq
p
q
r
qr
Example 16 (Problem #6, §1.6, Rosen, 7th edition). Show that the following is a valid
argument.
If
it does not rain or it is not foggy then the sailing race will be held
and the lifesaving demonstration will go on
and if the sailing race is held then the trophy will be awarded
and the trophy was not awarded
then it rained
Let
1.5 - 9
p
q
r
s
t
=
=
=
=
=
it rains
it is foggy
the sailing race is held
the lifesaving demonstration goes on
the trophy is awarded
Then the above is ((p' + q') rs)(r  t)t'  p The following shows this is a valid
argument.
Step
1.
2.
3.
4.
5.
6.
7.
8.
9.
Reason
Premise
Premise
MT using 1 and 2
Addition using 3
DeMorgan using 4
Premise
MT using 5 and 6
DeMorgan using 7
Simplification using 8
rt
t'
r'
r' + s'
(rs)'
(p' + q') rs
(p' + q')'
pq
p
Example 1. Show that pq  p and pq  q are rules of inference. (called simplification.)
p
q
pq
(pq)p
Solution. In pq  p one has P = pq and Q = p. At the right is
0
0
0
1
a truth table. Note that every row of the truth table of
0
1
0
1
1
0
0
1
(pq)  p has a 1, so pq  p by method 1. Alternatively,
1
1
1
1
whenever the truth table of pq has a 1 (only the last row) the
truth table of p also has a 1, so pq  p by method 2. If we use
method 3, then we need to show (pq)  p = 1. Recall that p  q = p' + q, so
(pq)  p = (pq)' + q. Using de Morgan and other basic equivalences of Boolean algebra,
one has (pq)' + q = p' + q' + q = p' + 1 = 1. We could use either of these three methods to
show pq  q. However, it is probably conceptually simpler to use the following
principle.
Suppose P  Q and suppose P' and Q' are obtained from P and Q by replacing
parts of P and Q by equivalent expressions. Then P'  Q'.
To apply this principle to show pq  q, first note that if we interchange the roles of p and
q then we conclude qp  q. Since qp = pq, we conclude by the principle that pq  q.
Example 2. Show that the proposition “If John likes to watch the Tigers and John likes
to eat hot dogs then John likes to watch the Tigers” is true no matter whether John likes
to watch the Tigers or eat hot dogs.
Solution. If we let p = “John likes to watch the Tigers” and q = “John likes to eat hot
dogs” then the statement “If John likes to watch the Tigers and John likes to eat hot dogs
1.5 - 10
then John likes to watch the Tigers” has the form (pq)  p, i.e it is a simplification. As
we saw in Example 1, it is true no matter whether each of the statements p and q are true
or false.
Remark. We often just say "if P then Q" for P  Q. This is confusing since we also say
"if P then Q" for P  Q.
Example 3. Show that p  p + q is a rule of inference. (This rule of inference is called
addition.)
Solution. In this example, P = p and Q = p + q. To show
p  p + q note that in the truth table at the right whenever there
is a 1 in the p column, there is a 1 in the p + q column. So
p  p + q is a rule of inference. Alternatively, we could use
Boolean algebra: p  (p + q) = p' + p + q = 1 + q = 1.
p
0
0
1
1
q
0
1
0
1
p+q
0
1
1
1
Example 4. Show that the proposition “If John likes to watch the Tigers then John likes
to eat hot dogs or John likes to eat hot dogs” is true no matter whether John likes to watch
the Tigers or eat hot dogs.
Solution. If we let p = “John likes to watch the Tigers” and q = “John likes to eat hot
dogs” then the statement “If John likes to watch the Tigers then John likes to eat hot dogs
or John likes to eat hot dogs” has the form p  (p + q), i.e it is an addition. As we saw in
Example 3, it is true no matter whether each of the statements p and q is true or false.
Example 17. Show disjunctive syllogism (p + q)p'  q is a rule of inference.
Solution. (p + q)p' = pp' + qp' = 0 + qp' = qp'. Then qp'  q by simplification.
Example 18. Show p  q = p  pq is an equivalence.
Solution. p  pq = p' + pq = (p' + p)(p' + q) = (1)(p  q) = p  q.
In order to show constructive dilemma is a rule of inference, it will be helpful to use
another general method of proving valid arguments called Proof by Contradiction. It
says
PQ1Q2…Qn  R if and only if R'Q1Q2…Qn  P'
The reason is that PQ1Q2…Qn  R if and only if PQ1Q2…Qn  R = 1 and
R'Q1Q2…Qn  P' if and only if R'Q1Q2…Qn  P' = 1. However by (10b) the two
logical expressions PQ1Q2…Qn  R and R'Q1Q2…Qn  P' are equivalent.
Example 19. Show constructive dilemma (p  q)(r  s)(p + r)  q + s is a rule of
inference.
1.5 - 11
Solution. We prove this by contradiction. We must show
(q + s)'(p  q)(r  s)  (p + r)'. Using DeMorgan this is equivalent to
q's'(p  q)(r  s)  p'r'. By MT, q'(p  q)  p' and s'(r  s)  r'. Combining this
with simplification, (14) and transitivity proves CD is a rule of inference.
Destructive dilemma can be proved from constructive dilemma in the same way modus
tollens is proved from modus ponens.
Example 16 (Page 73, Rosen, 7th edition). Show that the following is a valid argument.
If
and
and
and
then
it is not sunny this afternoon and it is colder than yesterday
we will go swimming only if it is sunny
if we do not go swimming then we will take a canoe trip
if we take a canoe trip then we will be home by sunset
we will be home by sunset
p
q
r
s
t
it is sunny
it is colder than yesterday
we go swimming
we take a canoe trip
we will be home by sunset
Let
=
=
=
=
=
Then the above is p'q(r  p)(r'  s)(s  t)  t The following shows this is a valid
argument.
Step
1.
2.
3.
4.
5.
6.
7.
Reason
Premise
Premise
MT using 1 and 2
Premise
MP using 3 and 4
Premise
MP using 5 and 6
p'
rp
r'
r'  s
s
st
t
1.5 - 12