Definition of an argument (in this handout we used the material form)
http://terpconnect.umd.edu/~bschan/170/june3.pdf and http://www.philosophypages.com/lg/e11a.htm
An argument (in the context of logic) is defined as a set of premises and a conclusion where
the conclusion and premises are separated by some trigger word, phrase or mark known
as a turnstile.
For example:
1 I think; therefore I am.
There is only one premise in this argument, I think. The conclusion is I am and the turnstile is
therefore.
2 All men are mortal. Socrates was a man. So, Socrates was mortal.
In this example there are two premises and the turnstile is so. This may also be written as:
All men are mortal.
Socrates was a man.
------------------------∴ Socrates was mortal.
In general we will write an argument in either of the following two ways:
p1
p2
p3
⁞
pn
Or
---∴q
p1, p2, p3, …. , pn
/ ∴q
Valid and Invalid arguments
Of greater interest to the logician are valid arguments. A valid argument is an argument for which
there is no possible situation in which the premises are all true and the conclusion is false. In other
words an argument is valid if for all those situations which make all the premises true will also
make the conclusion true.
An invalid argument is an argument for which there is at least one situation in which the
premises are all true but the conclusion is false.
One way to check the validity of an argument is to use full truth-table method. In this method
we draw a truth-table where the top row contains all the different sentence letters in the
argument, followed by the premises, and then the conclusion. If an argument is valid, then every
assignment where the premises are all true is also an assignment where the conclusion is true.
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But, if the table contains at least one row where the premises are all true and the conclusion is
false, then the argument is invalid.
We now study how to make use of full truth-table method to check the validity of an argument.
Let us consider the following argument:
P
P→Q
----------∴Q
To prove that it is valid, we draw a table where the top row contains all the different sentence
letters in the argument, followed by the premises, and then the conclusion:
Then using the same method as in drawing complex truth-tables, we list all the possible
assignments of truth-values to the sentence letters on the left. In our particular example, since
there are only two sentence letters, there should be 4 assignments:
P
T
T
F
F
Q
T
F
T
F
P P→Q
T
T
T
F
F
T
F
T
Q
T
F
T
F
In the completed truth-table, the first two cells in each row give us the assignment of truthvalues, and the next three cells tell us the truth-values of the premises and the conclusion under
each of the assignment. If an argument is valid, then every assignment where the premises are
all true is also an assignment where the conclusion is true. It so happens that there is only one
assignment (the first row) where both premises are true. We can see from the last cell of the row
that the conclusion is also true under such an assignment. So this argument has been shown to
be valid.
More examples
Show that the following argument is not valid:
P→Q
P
---------∴ Q
Solution:-
P Q P→Q
T
T
F
F
T
F
T
F
T
F
T
T
P
F
F
T
T
Q
F
T
F
T
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We know that an argument is invalid if there is at least one assignment where the premises are
true and the conclusion is false. The third row of the table shows that this is indeed the case.
When “P” is false and “Q” is true, the two premises are true but the conclusion is false. So the
argument is not valid.
Exercise:- Determine whether or not the following argument is valid:
P→Q
P
---∴ Q
Another example:
PQ
(Q → P)
---------------∴ Q P
Again we draw a truth-table for the premises and the conclusion:
P Q
T
T
F
F
T
F
T
F
P Q→P
F
F
T
T
T
T
F
T
PQ
T
F
T
T
(Q→P) Q P
F
F
T
F
F
T
T
F
As you can see, there is no assignment where the premises are true and the conclusion is false.
So it is valid.
Reconsider validity:
Now, it is probably clear to you why both a valid argument and an invalid argument can have
(e.g.) true premises and a true conclusion. (Why? Hint: An actual case is only one row in the
truth table.)
Constructive dilemma
In logic, a constructive dilemma is a formal logical argument that takes the form:
1a) P → Q.
1b) R → S.
2) Either P or R is true.
Therefore, either Q or S is true.
In logical operator notation with three premises
In logical operator notation with two premises
-----------------------------.
In sum, if two conditionals are true and at least one of their antecedents is, then at least one of their
consequents must be too.
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An example:
If I win a million dollars, I will donate it to an orphanage.
If my friend wins a million dollars, he will donate it to a wildlife fund.
Either I win a million dollars, or my friend wins a million dollars.
Therefore, either an orphanage will get a million dollars, or a wildlife fund will get a million dollars.
Another example of constructive dilemma.
If the statements “if it snows, then the schools close” and “if it is sunny, then Sam will go
to the park” are true, and if it snows or is sunny, then it is also true that the schools close
or Sam will go to the park.
The constructive dilemma is the disjunctive version of modus ponens.
1st Premise
2nd Prem. Conc.
Now let us show that the constructive P Q R S P→Q R→S (P→Q)(R→S) P  R Q  S
dilemma is a valid argument by T T T T T
T
T
T
T
construction the truth table.
T T T F
It is clear that the premises are true T T F T
on lines 1, 3, 4, 9, and 13, and on
T T F F
each of these lines the conclusion is
T
T
T
F
T
T
F
T
T
T
T
T
T
T
T
T F TT
F
T
F
T
T
T
T
T
F
F
F
F
F
F
F
T
T
T
T
T
F
F
T
T
F
F
F
T
F
T
F
T
F
F
F
F
T
T
T
T
F
T
T
T
F
T
T
F
F
F
T
F
T
T
T
T
T
T
T
F
F
F
T
F
T
T
T
T
F
F
F
F
F
F
F
F
T
T
F
F
T
F
T
F
T
T
T
T
T
F
T
T
T
F
T
T
T
T
F
F
T
F
T
F
also true. Thus, the argument is
valid, and we can be sure that every
argument that is a substitutioninstance of this argument form must
be valid.
Destructive dilemma
In logic, a destructive dilemma is any logical argument of the following form:
--------------------
∴
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The argument can be read in this way:
1.
2.
3.
4.
If P, then Q
If R, then S
Not Q or not S
Therefore, not P or not R
In sum, if two conditionals are true and at least one of their antecedents is false, then at least one
of their consequents must also be false.
Here is an example of the destructive dilemma in English:
1.
2.
3.
4.
If it rains, we will stay inside.
If it is sunny, we will go for a walk.
Either we will not stay inside, or we will not go for a walk.
Therefore, either it will not rain, or it will not be sunny.
The destructive dilemma is the disjunctive version of modus tollens.
Exercise:- prove the validity of the destructive dilemma argument.
Special case: Inconsistent premises will form a valid argument
The following three arguments are all valid:
P
P
----∴Q
P
P
-----------∴ ( P P)
P
P
-------------∴( P P)
All these arguments are valid because there is no assignment under which both P and P are
true and Q / (P P) / (P  P) are false. There is no such assignment simply because P and
P cannot be true simultaneously.
Here is the truth table to convince you the validity of above three arguments:
P
T
T
F
F
P
F
F
T
T
Q
T
F
T
F
P P
F
F
F
F
P P
T
T
T
T
Exercise:
Use the full truth-table method to determine the validity of these arguments:
Answer:
1. Invalid.
2. Valid (Notice that the first premise is true only under the 4th assignment. Once we have found
out that the first premise is true only under the 4th assignment, we only need to calculate the
truth-values of the second premise and the conclusion under the same assignment. There is no
need to draw the rest of the truth-table. This means we can determine validity a lot faster.)
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Constructive Dilemma
What is the proof of constructive dilemma in logic NOT with truth table if p is q
and if r is s and either p or r is true therefore either q or s is true?
In:Math [Edit categories]
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
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Answer:
Improve
Wha?????!
Please use the discussion area to explain this fascinating question
Constructive Dilemma Proof:
1. ( A -> B ) Premise
2. ( C -> D ) Premise
3. ( A or C ) Premise
4. ( - B -> - A ) 1 , Contraposition
5. ( - A -> C ) 3 , Implication
6. ( - B -> C ) 4 , 5 Chain Argument
7. ( - B -> D ) 6 , 2 Chain Argument
8. ( B or D ) 7 Implication , Q.E.D.
Read more:
http://wiki.answers.com/Q/What_is_the_proof_of_constructive_dilemma_in_logic_NOT_with_truth_table_if_p
_is_q_and_if_r_is_s_and_either_p_or_r_is_true_therefore_either_q_or_s_is_true#ixzz1lKSC6mpS
Constructive dilemma
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This article needs additional citations for verification. Please help improve this article by adding
citations to reliable sources. Unsourced material may be challenged and removed. (December 2009)
Rules of inference
Propositional calculus
Modus ponens
Modus tollens
Modus ponendo tollens
Conjunction introduction
Simplification
Disjunction introduction
Disjunction elimination
Disjunctive syllogism
Hypothetical syllogism
Constructive dilemma
Destructive dilemma
Biconditional introduction
Biconditional elimination
Predicate calculus
Universal generalization
Universal instantiation
Existential generalization
Existential instantiation

v

d

e
In logic, a constructive dilemma is a formal logical argument that takes the form:
1a) P → Q.
b) R → S.
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2) Either P or R is true.
Therefore, either Q or S is true.
In logical operator notation with three premises
.
In logical operator notation with two premises[1]
.
In sum, if two conditionals are true and at least one of their
antecedents is, then at least one of their consequents must be
too.
An example:
If I win a million dollars, I will donate it to an orphanage.
If my friend wins a million dollars, he will donate it to a wildlife fund.
Either I win a million dollars, or my friend wins a million dollars.
Therefore, either an orphanage will get a million dollars, or a wildlife fund will get a million dollars.
The dilemma derives its name because of the
transfer of disjunctive operants.
[edit] Proof
1
.
2
.
3
.
(addition
)
4
.
(simplific
ation)
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5
.
(distribu
tion)
6
.
(DeMorg
an's
Law)
7
.
(from
assumpti
on)
7
.
And to once again restate the argument, one can turn this argument into a conditional, where if the first
three premises, then not P or R:
Note: An argument has the following form:
(P1 ^ P2 ^ ... ^ Pn)  C
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