Workshop LOGIC and LANGUAGE. (Revised Oct 2013).
In this session, we shall investigate how normal language is governed by logic rules, which although may not be
apparent to us, nevertheless govern our language, and hence perhaps the way we think. You have seen some examples
of sentences which we shall not consider, due to bad grammar, ambiguity and so on which we will not consider. We
also throw out sentences of the type which are opinions or desires, such as
I think that music is great!
I would like to learn to program in Python.
so we are left with sentences like
If you come home early, you can watch TV.
Steve is cheerful and brave.
These sentences are declarative sentences and can be enclosed in special brackets like this
<Holmes and Watson were on the case.>
1. Logic Connectives
1.1 The AND Connective
Sentences refer to other sentences, and are therefore connected to other sentences. Think of the complex sentence;
<Holmes and Watson were on the case.>
This can be written as two simpler sentences
<Holmes was on the case> AND <Watson was on the case>
We can make life a little easier by writing a symbol for each sentence, like
P = <Holmes was on the case>
Q = <Watson was on the case>
and we can use Boolean algebra to represent the compound sentence as P.Q
Now let’s relate this to the truth tables we have been working with. Here’s the truth table for the AND gate. What goes
in the right hand column? (You know the answer from Digital Logic, but now let’s work it out using language).
Consider the first line. This means “Holmes was not on the case AND Watson was not on the case”. So the compound
sentence P.Q “Holmes and Watson were not on the case” is clearly false, 0. Now work out the entries of the remaining
rows.
P
Q
0
0
0
1
1
0
1
1
You should remember building the AND gate and you can apply this to the two sentences above.
1.2 A Simple Problem in Logic
Consider the sentence P = <Cows can fly>. Write down, in English, the meaning of the sentence 𝑷. 𝑷
Now work out the truth table for it, by thinking
P
0
1
and check it by building up this circuit
𝑷. 𝑷
1.3 The Conditional - IF
One of the most fundamental truth functors in our daily language is the “IF”. Just think how many times in a day we
say or think this “IF”. How on earth can we get this “IF” into digital electronics and formal logic? Well, let’s see.
The problems with the IF were recognized by Philo of Megara around 300 BC.
According to Sextus Empiricus, ca. 200 AD,
‘Philo says that a sound conditional is one that does not begin with a truth
and ends with a falsehood"
Let’s take for example, the sentence set “If Fred is at home, then he is asleep”. We may divide it into two atomic
sentences;
P = <Fred is at home >
Q = <Fred is asleep>
We shall represent the IF functor by the arrow; P  Q. The truth table corresponding to the IF functor is shown below
P
Q
PQ
0
0
1
0
1
1
1
0
0
1
1
1
Let’s take this line by line.
Line 4. No problem. Here Fred is at home and Fred is asleep. This is clearly true.
Line 3. OK but vital. Fred is at home and Fred is not asleep. This is clearly false.
Lines 1 and 2. In both, Fred is not at home. In one case he is asleep (perhaps on a park bench) and on the other day he
is not asleep. The point is, could P  Q be true here? Well yes, it could be true since we know that P is false, Fred is
not at home, so he could be either asleep or not asleep. So lines 1 and 2 have to be correct (together). Linguists agree
with this definition. It is false only in one case, and the definition leads to some interesting language paradoxes as you
will see if there is time.
Is it is possible to find a combination of AND, OR, NOT gates which will do the job? The answer is yes and the result
is shown in the circuit below
The Philonian Conditional; “If P then Q” can be represented by the truth functor
“Not P or Q”,
< P  Q> is equivalent to 𝑷 + 𝑸
Let’s test this out. We can easily build 𝑷 + 𝑸 Wire it up as shown below.
Now get its truth table and verify this is the same as the Philonian Conditional, as given in the table above.
P
Q
0
0
0
1
1
0
1
1
PQ
Yep! The truth tables are identical. The equality does sound reasonable, the sentences "If I am a boy, then I am male"
is the same as "I am not a boy or else I am male"
2. CONSISTENCY
2.1
The method of Consistency Checks.
A set of declarative sentences is said to be “consistent” if one or more combinations can be found which is true. Think
about this set of two sentences;
a) Holmes and Watson are on the case
b) Either Holmes was not on the case, or else Watson was.
They could be consistent if a combination can be found which is true. Can you think it out? You must say who was on
the case, if you can.
It’s easy to see how to check consistency of two sentences, they must both be true, that is sentence one is true AND
sentence two is true. Note the AND connective.
So first we extract the following “atomic” sentences from the two compound sentences and we find
P = <Holmes was on the case >
Q = <Watson was on the case>
Second, we must transcribe the original sentences and in terms of the atomic sentences P and Q. In this case the two
sentences become.
a)
P.Q
b) 𝑷 + 𝑸
Third, to test for consistency, we AND these sentences, to look for any case where both a) and b) are true together. So
we form
(P.Q) . (𝑷 + 𝑸)
Fourth, we make the electronic wire-up. The circuit is given below, but it is built up as follows;
1.
build P.Q
2.
build 𝑷 + 𝑸
3.
AND the outputs of 1 and 2 and connect to a light
Finally, make up the truth table by trying the various P’s and Q’s (0 and 1) and look for any
true combination of P and Q.
P
Q
0
0
0
1
1
0
1
1
If your circuit was correct, there should be a single true combination, P = 1, Q = 1. What does this mean? The only
consistent combination of P and Q is
a) Holmes is on the case
b) Watson is on the case
So we have proved that the original set of sentences (a) and (b) is consistent and we know exactly who was on the
case.
2.2
Watson and Holmes in Confusion
Test the consistency of this set of sentences using the method above.
a) Holmes and Watson were on the case.
b) If Holmes was on the case, then Watson was not.
First find the atomic sentences P and Q;
P
=
Q
=
Now transcribe the set of sentences a) and b) in terms of P and Q using Boolean Algebra
a)
b)
Now AND these sentences using Boolean algebra
c)
Now build the circuit (generate the terms for a) and then for b) and then AND the outputs, and then get the truth table.
Draw your circuit and the truth table.
Did you expect this result? Look at the starting sentences and think out the answer.
2.3
Smiley's People.
Test the consistency of this set of sentences;
"Smiley is an English spy. Smiley is not both a Russian and an English spy. If Smiley is a Cad, then he
is a Russian spy".
P
=
Q
=
R
=
Transcribe each of the three sentences in to a boolean expression.
Now write down the boolean expression for all three sentences together
Build a circuit using the boolean expression and find the truth table.
P
Q
R
0
0
0
0
0
1
0
1
0
0
1
1
1
0
0
1
0
1
1
1
0
1
1
1
If all has gone well, you should find only one consistent solution. Write this out in English
2.4
Too many Detectives.
Test the consistency of these sentences;
"If Holmes. solved the crime, then Lestrade took the criminal. Of course, if the criminal escaped, then
Lestrade did not take him. On this particular day, Holmes solved the crime, but the criminal escaped"
P
Q
R
=
=
=
Transcribe each of the three sentences in to a boolean expression.
Now write down the boolean expression for all three sentences together
Build a circuit using the boolean expression and find the truth table.
Identify any consistent solution.
2.5
Gunmen and Hostages.
Again, we have a set of four sentences, and we wish to find if there is/are any consistent solution(s) to this set.
"If the gunmen are tired, then they are on edge. If they are armed and on edge, then the hostages are in
danger. The gunmen are armed but the hostages are not in danger."
P
Q
R
S
=
=
=
=
Transcribe each of the three sentences in to a boolean expression.
Now write down the boolean expression for all three sentences together
Build a circuit using the boolean expression and find the truth table.
Identify any consistent solution.
3. TESTING THE VALIDITY OF ARGUMENTS
Here is an example of an ARGUMENT:
"Worcester Uni. students who are take Computing may use Room CH1007"
"I am in a Worc. Uni. Student and I take Computing"
"Therefore I may use the Room CH1007"
The first two lines are called the PREMISES of the argument, and the final line is the CONCLUSION.
An argument is said to be VALID when, if there is no situation where all the premises are true
and the conclusion is false.
How to proceed? Well, via a somewhat vicious route. We take the premises as given, but then we take the conclusion
and negate (invert) it. Then we do a consistency check (as above). If we find any single consistent solution (with true
premises and false conclusion) then clearly the argument is incorrect! Only if there is no consistent solution to this
set of sentences (true premises and negated conclusion) do we conclude that the argument is correct. The method's
called REDUCTIO AD ABSURDUM.
3.1
A Vicarious Example
Look at this argument;
"Either the vicar or the butler shot the "Earl. If the butler shot the Earl, the butler was not drunk. Unless the
vicar is a liar, the butler was drunk. Therefore, either the vicar is a liar, or he shot the Earl.”
a) The first thing to do is as usual find the atomic sentences
P
Q
R
S
b)
=
=
=
=
<The vicar shot the Earl>
<The butler shot the Earl>
<The butler was drunk>
<The vicar is a liar>
Now transcribe the Premises:
1st sentence
P+Q
2nd sentence
𝑸+𝑹
3rd sentence S + R
The second line may confuse you. Look at the second sentence. If this had been “If the butler shot the earl
then the butler was drunk” would have been written as the Boolean expression 𝑸 + 𝑹
c)
Now negate the conclusion (3rd sentence)
𝑺+𝑹
d)
Now AND all the sentences
(𝑷 + 𝑸). ( 𝑸 + 𝑹).( 𝑺 + 𝑹)
e)
Construct the equivalent digital logic circuit and if all has gone well, you should have found the final column
to be all O’s. This means there is no consistent solution. There is no case where the argument could be false,
so it MUST be a VALID argument.
3.2
A Question of Dialect.
Test the following argument for validity;
"If Higgins born in Bristol, then Higgins is not a Cockney. Higgins is either a Cockney or an
impersonator. Higgins is not an impersonator. Therefore Higgins was born in Bristol”.
P=
Q=
R=
Now transcribe, not forgetting to negate the conclusion and build up the circuit. You should find here that there is One
Consistent solution. Write this out in English; This is a counterexample to the argument. Since it exists, then the
argument was false.
3.3
Problem for a Medic.
Check the Validity of this argument;
"If the boy has spots in his mouth, then he has measles. If the boy has a rash on his back, then he has
heatspots. The boy has a rash on his back. Therefore he hasn't got measles."
P
=
Q
=
R
=
S
=
Transcription (plus inverted conclusion), Circuit If you find any counter-examples, then write these down in English:
4. ZAP THEM WITH LOGIC!
Here followeth, dear student, some classic thorny bushes of logic. They are in the form of arguments, which you
should test using the validity method used in the previous section. Actually the first two bushes are not thorny at all.
4.1 Modus Ponens
Prove the validity of this argument: "If there is a fault, it will blow up. There is a fault to therefore it will blow up.
4.2
Modus Tollens
Prove the validity of this argument; "If there is a fault it will "blow up. It will not blow up, therefore there is no fault".
4.3
Fallacy of Denying the Antecedent.
Think about (and investigate) the validity of the following argument; "If there is a fault, it will blow up. There is no
fault, therefore it will not blow up”.
4.4
Fallacy of Affirming the Consequent.
Discuss the validity of this argument; "If there is a fault, it will blow up. It will blow up, therefore there is a fault"
4.5
Fallacy of Conversion.
Check out the truth of the following argument; "If there is a fault then it will blow up, therefore if it blows up, then
there is a fault!