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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 http://www.philosophypages.com/lg/e11b.htm and http://www.philosophypages.com/lg/e11c.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

----

∴ q

Or

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.

An argument is valid if there is no line on which all the premises are true and the

conclusion false.

An argument is invalid if there is at least one line on which the premises are true and the

conclusion false.

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 (this is called Modus Ponens):

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 Q P P → Q Q (P(P → Q)) →Q

T T T T T T

T F T F F T

F T F T T T

F F F T F T

In the completed truth-table, the first two cells in each row give us the assignment of truth-

values, 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 fifth cell of the

row that the conclusion is also true under such an assignment. So this argument has been shown

to be valid. Please not in this case the statement (P(P → Q)) →Q is a tautology.

More examples

Show that the following argument is not valid:

P → Q

P

----------

∴ Q Solution:-

P Q P → Q P Q ((P → Q) P)→ Q

T T T F F T

T F F F T T

F T T T F F

F F T T T 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. Please not in this case the statement ((P → Q) P)→ Q is not a

tautology.

Exercise:- Determine whether or not the following argument is valid:

P → Q

Q

----------

∴ P

Answer:- This argument is valid and is called Modus Tollens (M.T.).

Another example:

P Q

(Q → P)

----------------

∴ Q P

Again we draw a truth-table for the premises and the conclusion:

P Q P Q→P PQ (Q→P) Q P (( P Q) ( (Q → P)))→(Q P)

T T F T T F F T

T F F T F F T T

F T T F T T T T

F F T T T F F T

As you can see, there is no assignment where the premises are true and the conclusion is false.

So it is valid. Please not in this case the statement (( P Q)( (Q → P)))→( Q P) is a

tautology.

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 (C.D.)

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 OR In logical operator notation with two premises

---------------

----------------

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The constructive dilemma is the disjunctive version of modus ponens.

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.

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.

Now let us show that the constructive

dilemma is a valid argument by construction

the truth table.

It is clear that the premises are true on lines 1, 3, 4, 9, and 13, and on each of these lines the conclusion

is also true. Thus, the argument is valid, and we can be sure that every argument that is a substitution-

instance of this argument form must be valid.

Destructive dilemma

In logic, a destructive dilemma is any logical argument of the following form:

--------------------

∴

The argument can be read in this way:

1. If P, then Q

2. If R, then S

3. Not Q or not S

4. Therefore, not P or not R

1st Premise 2nd Prem. Conc.

P Q R S P→Q R→S (P→Q)(R→S) P R Q S

T T T T T T T T T

T T T F T F F T T

T T F T T T T T T

T T F F T T T T T

T F T T F T F T T

T F T F F F F T F

T F F T F T F T T

T F F F F T F T F

F T T T T T T T T

F T T F T F F T T

F T F T T T T F T

F T F F T T T F T

F F T T T T T T T

F F T F T F F T F

F F F T T T T F T

F F F F T T T F F

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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. If it rains, we will stay inside.

2. If it is sunny, we will go for a walk.

3. Either we will not stay inside, or we will not go for a walk.

4. 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 P

P P P ----- ------------ --------------

∴Q ∴ (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 P Q P P P P (P P) → Any Proposition

T F T F T T

T F F F T T

F T T F T T

F T F F T T

Please note that ―(P P) → Any Proposition‖ is a tautology.

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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Another definition of a valid and invalid argument

Let us consider an argument p1, p2, p3, …. , pn / ∴ q. And look at the behavior of the statement

(p1p2p3 …. pn) → q:

1. If all those situations which make all the premises true also make the conclusion

true then (p1p2p3 …. pn) → q is a tautology.

2. If for all situations at hand at least one of the premises is false then again the

(p1p2p3 …. pn) → q is a tautology.

3. If there is at least one situation in which the premises are all true but the conclusion is false

then (p1p2p3 …. pn) → q is not a tautology.

Note under first two conditions p1, p2, p3, …. , pn / ∴ q will be a valid argument and under the

last condition p1, p2, p3, …. , pn / ∴ q will be an invalid argument.

So here is the standard mathematical definition of a valid argument:

The argument p1, p2, p3, …. , pn / ∴ q, is said to be valid if the statement

(p1p2p3 …. pn) → q

is a tautology. In other words, validity means that if all the premises are true, then the conclusion must

be true.

A sound argument is an argument with two features: (i) it is valid, and (ii) its

premises are all true (for at least one situation).

Rules of Inference

The Method of Proof

The construction of truth-tables provides a reliable method of evaluating the validity of

arguments in the propositional calculus. We can always tabulate the truth-values of premises and

conclusion, checking for a line on which the premises are true while the conclusion is false.

Although this method always works, however, it isn't always convenient, since the appropriate

truth-table must have 2n lines, where n is the number of simple propositions involved. Thus, an

argument with six different simple propositions would require the construction of a truth-table

with 64 lines.

Fortunately, there is another, shorter way to proceed, by constructing a formal proof of the

validity of an argument. The basic notion underlying this new method is that since a chain of

interrelated arguments is valid so long as each of its links is valid, we can demonstrate the

validity of an argument by starting with its premises, taking one tiny valid step at a time, and

finally reaching its conclusion. The only limitations we need to impose on this procedure are

that each of our tiny steps must be a substitution-instance of some valid argument form and that

we can discover a seamless path leading from premises to conclusion.

Although any valid pattern of inferences could be used in this proof procedure, we will make

things easier by relying on a very short list of valid argument forms. Each new step that we take

in constructing a proof must then be a substitution-instance of one of these rules of inference.

You've already seen three of them: Modus Ponens (M.P.), Modus Tollens (M.T.) and

Constructive Dilemma (C.D.).

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We'll add just six more, making a total of nine elementary valid argument forms to be used as rules of

inference.

Hypothetical Syllogism (H.S.)

A larger truth-table is required to demonstrate the validity

of the argument form called Hypothetical Syllogism (H.S.),

since it involves three statement variables instead of two,

and we must consider all eight of the possible combinations

of their truth-values:

p → q

q → r

_____

∴p → r

Example: "If Debbie is promoted, then Gene will be, too.

But if Gene is promoted, then Kim will be angry. Therefore,

if Debbie is promoted, then Kim will be angry."

Despite its greater size, this truth-table establishes validity

in exactly the same way as its more compact predecessors:

both premises are true only on the first, fifth, seventh, and

eighth lines and the conclusion is also true on each of these

lines. It follows that all arguments sharing in this general

form must be valid.

Disjunctive

Syllogism

(D.S.) has the Following

argument form

p q

p

_____

∴q

The truth-table demonstration

of its validity should look

familiar by now. Whenever

the premises are true (on the

third line of the truth table),

so is the conclusion.

1st Premise 2

nd Premise Conclusion

p q p q p q

T T T F T

T F T F F

F T T T T

F F F T F

Example:- Either the meeting is in room 302, or it is in room 306. It is not in room 302.

Therefore, it is in room 306.

Absorption

(Abs.) has the

simpler form:

p → q

__________

∴p → (p q)

The truth-table at the right shows the validity of all

substitution-instances of this argument form.

Whenever its premise is true, the conclusion is true as

well. (In fact, you may notice that, in this unusual

instance, it is also true that the premise is true

whenever the conclusion is. The two statement forms

are logically equivalent to each other.)

Premise Conclusion

p q p → q p → (pq)

T T T T

T F F F

F T T T

F F T T

Simplification(Simp.) rule permits

us to infer the truth of a conjunct

from that of a conjunction.

pq pq

_____ OR _____

∴p ∴q

Its truth-table is at right. Notice

that Simp. warrants only an

inference to the first of the two

conjuncts, even though the truth

of the second conjunct could be

also be derived.

Premise Conclusion

p q p q p

T T T T

T F F T

F T F F

F F F F

Example:- Main floor seats cost R600 and balcony seats cost R450. So, main floor

seats cost R600. (One may also conclude that balcony seats cost R450.)

1

st Premise 2

nd Premise Conclusion

p q r p → q q → r p → r

T T T T T T

T T F T F F

T F T F T T

T F F F T F

F T T T T T

F T F T F T

F F T T T T

F F F T T T

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Conjunction (Conj.) permits the

derivation of a conjunction from

the truth of both of its conjuncts. p

q

_____

∴p q

As the truth-table at the right

illustrates, this is a natural

inference from our definition of

the connective.

1st

Premise 2nd

Premise Conclusion

p q p q T T T T F F F T F F F F

Example:- Main floor seats cost R600. Balcony seats cost R450. So, main floor seats

cost R600 and balcony seats cost R450.

Addition (Add.) is the

argument form:

p ______

∴ p q

This rule warrants the inference from any true statement

to its disjunction with anything whatsoever. This is an

amazingly powerful device, since it permits us to

introduce any new statement whatsoever into the context

of a proof. Our challenge in applying it will lie in

discovering an appropriate or helpful substitution for q in

each specific case.

Premise Conclusion

p q p p q T T T T T F T T

F T F T

F F F F

Example:- Los Angeles is on the Pacific Ocean. So, Either Los Angeles is on the Pacific Ocean

or Denver is on the Pacific Ocean.

Constructing a Proof

Now let's see how to use these nine rules of inference in order to demonstrate the validity of

arguments in the propositional calculus. Consider, for example, the argument:

A → (B C)

D → C

A

B

______________

∴ D

In order to construct a formal proof of the validity of this argument, we begin by numbering

each of its premises and indicating that we are assuming their truth as the premises of an

argument:

1. A → (B C) premise

2. D → C premise

3. A premise

4. B premise

Next, we notice that premise 1 has the form p → q and that premise 3 is the antecedent of that

conditional. That is, premises 1 and 3, taken together, are the premises of an argument that is a

substitution-instance of the valid argument form known as Modus Ponens. The conclusion of

that argument would be the consequent of the conditional, or B C. Thus, we can take the

tiny step of adding this conclusion to our list of established statements, indicating at the right a

simple justification that explains exactly where it came from, by listing the previous statements

used as premises of an argument that follows one of the rules of inference.

1. A → (B C) premise

2. D → C premise

3. A premise

4. B premise

5. B C 1, 3 M.P.

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In the same way, we can now use this new statement, together with statement 4, as the premises

of a substitution-instance of D.S., which justifies the further conclusion C.

1. A → (B C) premise

2. D → C premise

3. A premise

4. B premise

5. B C 1, 3 M.P.

6. C 5, 4 D.S.

Finally, this new statement and statement 2 are the premises of a substitution instance of M.T.

which justifies the conclusion ~D.

1. A → (B C) premise

2. D → C premise

3. A premise

4. B premise

5. B C 1, 3 M.P.

6. C 5, 4 D.S.

7. D 2, 6 M.T.

But this was the conclusion of the original argument, so by proceeding step by valid step, we

have shown that if the premises of that original argument (1-4) are true, then its conclusion (7)

must also be true. Since each step in our proof relies only upon a rule of inference and the

supposed truth of earlier statements, the entire chain of reasoning must be valid.

Another example: Establish the validity of the following arguments:

p ∨ q

r ∨ s

q ∧ r ________________

p ∧ s ___________________

Proof:- 1. p ∨ q premise

2. r ∨ s premise

3. q ∧ r premise

4. q 3 Simp.

5. r 3 Simp.

6. p 1, 4 D.S.

7. s 2, 5 D.S.

8. p ∧ s 6, 7 Conj. So done.

Example:- Classify the following arguments as valid or invalid. If the argument is valid you

have to prove it. In case of an invalid argument you need to provide a counter example:

a) p ˄ ¬q premise 1

q r premise 2

---------

r

b) p ˄ q premise 1

q r premise 2

---------

r

c) ¬p ˄

q premise 1

r p premise 2

¬r s premise 3

s t premise 4

---------

t

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Solution:-

a) This argument is clearly invalid. Here is the counter example:

Consider the circumstances in which p is true but q and r are both false. Then both

the premises are true but the conclusion is clearly false.

b) 1. p q premise 1

2. q r premise 2

3. q 1 Simplification

4. r 2, 3 Modus Ponens

which is the required conclusion. Hence the argument is valid.

c) 1. p q premise 1 .

2. r p premise 2

3. r s premise 3

4. s t premise 4

5. p 1 Simplification

6. r 2, 5 Modus Tollens

7. s 3, 6 Modus Ponens

8. t 4, 7 Modus Ponens

which is the required conclusion. Hence the argument is valid.

Rules of Replacement (also called Rules of Equivalence)

We complete our development of the proof procedure for the propositional calculus by making

use of another useful way of validly moving from step to step. Since two logically equivalent

statements have the same truth-value on every possible combination of truth-values for their

component parts, no change in the truth-value of any statement occurs when we replace one of

them with the other. Thus, when constructing proofs of validity, we can safely use a statement

containing either one of a pair of logical equivalents as the premise for a step whose conclusion

is exactly the same, except that it contains the other one.

Although this would work for any pair of logically equivalent statement forms, remembering all

of them would be cumbersome. Instead, we will once again rely upon a short list of ten rules of

replacement in our construction of proofs, and we have already examined few of them:

Here is the list of Laws of Logic (tautologous biconditionals to be used as rules of

replacement)

For any primitive statements p, q, r;

1) p p p Idempotent Laws or Tautology (Taut.)

p p p

2) p p Double Negation Law (D.N.)

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3) p q q p Commutative Laws (Comm.)

p q q p

4) (p q) r p (q r) Associative Laws (Assoc.)

(p q) r p (q r)

5) p (q r) (p q) (p r) Distributive Laws (Dist.)

p (q r) (p q) (p r)

6) (p q) p q De Morgan’s Laws (DeM.)

(p q) p q

Augustus

De Morgan

(1806-1871)

7) (p q) (q p) Contrapositive or Transitivity (Trans.)

8) p q (p q) Switcheroo or Implication (Impl.)

9) (p ↔ q) (p q) (q p) Meaning of Biconditional also called Equivalence (Equiv.)

(p ↔ q) (p q) (p q)

10) ((p q) r ) (p (q r)) Exportation (Exp.)

Sometimes we may also refer to the following replacement rules (here T0 is any tautology and F0

is any contradiction) which are also tautological equivalences:

11) p T0 p (Identity Laws)

p F0 p

12) p T0 T0 (Domination Laws)

p F0 F0

13) p (p q) p (Absorption Laws)

p (p q) p

14) pp T0 (Negation Laws / Inverse Laws)

pp F0 (Also called Trivial tautology/contradiction)

Replacement in Proofs

Using the rules of replacement in the construction of proofs is a fairly straightforward procedure. Since

the rules are biconditionals, the replacement can work in either direction—right side for left, or left side

for right. What is more, since the statement forms on either side are logically equivalent, they can be

used to replace each other wherever they occur, even as component parts of a line. (When applying the

nine rules of inference, on the other hand, we must always work with whole lines of a proof.)

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Consider the following argument:

A (B C)

A D

D C

____________

D

As before, we begin by numbering each of the premises:

1. A (B C) premise

2. A D premise

3. D C premise

Next, notice that we can use our rules of replacement and inference to derive some part of the information

conveyed by the first premise:

1. A (B C) premise

2. A D premise

3. D C premise

4. (A B) (A C) 1 Dist.

5. (A C) (A B) 4 Comm.

6. A C 5 Simp.

So long as each step is justified by reference to an earlier step (or steps) in the proof and to one of the nineteen

rules, it must be a valid derivation. Next, let's work with the third premise a bit:

1. A (B C) premise

2. A D premise

3. D C premise

4. (A B) (A C) 1 Dist.

5. (A C) (A B) 4 Comm.

6. A C 5 Simp.

7. C D 3 Trans.

8. C D 7 D.N.

Again, each step is justified by application of one of the rules of replacement to all or part of a preceding line in

the proof. Now conjoin the second premise with our eighth line, and we've set up a constructive dilemma:

1. A (B C) premise

2. A D premise

3. D C premise

4. (A B) (A C) 1 Dist.

5. (A C) (A B) 4 Comm.

6. A C 5 Simp.

7. C D 3 Trans.

8. C D 7 D.N.

9. (A D) (C D) 2, 8 Conj.

10. D D 9, 6 C.D.

All that remains is to apply Tautology in order to reach our intended conclusion, so the entire proof will look like

this:

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1. A (B C) premise

2. A D premise

3. D C premise

4. (A B) (A C) 1 Dist.

5. (A C) (A B) 4 Comm.

6. A C 5 Simp.

7. C D 3 Trans.

8. C D 7 D.N.

9. (A D) (C D) 2, 8 Conj.

10. D D 9, 6 C.D.

11. D 10 Taut.

Don't worry if the intermediate stages of this proof were a little puzzling as we went along; what matters for

right now is that you understand the use of the rules of replacement along with the rules of inference.

Proof-Procedures for Propositional Logic

Inventing Proofs

Taken together, the nine rules of inference and ten rules of replacement are a complete set in the sense

that using just these nineteen rules is sufficient to demonstrate the validity of every valid argument of

the propositional calculus. (In fact, there is a good deal of redundancy built into the system: if we

happened to forget about M.P. entirely, for example, we could always work around it by using Impl.,

D.N., D.S., and Impl. again.) What's more, showing the validity of an argument in this way is often more

convenient than building the huge truth-table required to cover every possible combination of truth-

values.

But constructing proofs isn't always easy, as you've discovered in your last couple of exercise sets. Even

if you know the rules inside-out, upside-down, and backwards-and-forwards (as you must) and even if

you are skilled at recognizing the intermediate arguments that are their substitution instances in the

context of a proof (as you should be), inventing a proof on your own may seem a daunting task. But

don't be discouraged. No matter how long and complex a proof becomes, each of its steps is always just

the application of one of our nineteen rules; we just have to figure out which ones lead us from premises

to conclusion. There's nothing magical about the process. No one simply "sees" the whole proof at the

outset, although some people may move a little faster than others. With practice, anyone can learn to

perform the necessary steps.

Let's look at one more instance together:

Establish the validity of the following arguments:

p ∨ q

(p ∧ q) → r

q ∧ r

_____________________

p

_____________________

Proof:- 1. p ∨ q premise

2. (p ∧ q) → r premise

3. q ∧ r premise

4. q 3 Simp.

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5. r 3 Simp.

6. (p ∧ q) 2, 5 M.T.

7. p ∨ q 6 DeM.

8. p 4, 7 D.S. ~ so the argument is valid.

A Detailed Example

Consider the following argument:

A (B C)

(B A) [D (E ↔ F)]

D (G E) ___________________________

(G F) The first step, as always is to number and label the premises of the argument as the starting-point for our proof:

1. A (B C) premise

2. (B A) [D (E ↔ F)] premise

3. D (G E) premise

Now, begin working from the top down. Notice that the first premise is a conjunction whose right conjunct is

itself another conjunction. We have several possibilities here, since this statement could be a substitution

instance of statement forms that occur in several of our rules: Simp. and Comm. can be applied to any

conjunction, and Assoc. will work on one of this form, too.

Which one should we use? Well, let's think ahead a bit. Premise 2 is a conditional statement, and we

know that if its antecedent were on a line by itself, then that line together with premise 2 would be a

substitution-instance of the premises required for an application of M.P. That is (leaving lots of room for

the work we may have to do in between) we would like to be able to build the proof like this:

1. A (B C) premise

2. (B A) [D (E ↔ F)] premise

3. D (G E) premise

B A ? ? ?

D (E ↔ F) 2, ? M.P.

But in order to make this work, we have to derive B A. Looking back at the first premise now, we notice that

an application of Assoc. would change the grouping in such a way as to get A and B together, from which we

could then simplify to have them on a line of their own:

1. A (B C) premise

2. (B A) [D (E ↔ F)] premise

3. D (G E) premise

4. (A B) C 1 Assoc.

5. A B 4 Simp.

B A ? ? ?

D (E ↔ F) 2, ? M.P.

But now it's pretty obvious that all we have to do is commute line 5 to get the missing premise for our M.P.:

1. A (B C) premise

2. (B A) [D (E ↔ F)] premise

3. D (G E) premise

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4. (A B) C 1 Assoc.

5. A B 4 Simp.

6. B A 5 Comm.

7. D (E ↔ F) 2, 6 M.P.

This new line is a disjunction, so it would serve as one of the premises of a disjunctive syllogism if we only had

D. But we can get that easily enough by simplifying our third premise:

1. A (B C) premise

2. (B A) [D (E ↔ F)] premise

3. D (G E) premise

4. (A B) C 1 Assoc.

5. A B 4 Simp.

6. B A 5 Comm.

7. D (E ↔ F) 2, 6 M.P.

8. D 3 Simp.

9. E ↔ F 7, 8 D.S.

Suppose that at this point, we've run out of ideas about what to do working from the top down. So let's work

from the bottom up instead—remember the conclusion we're trying to establish here:

(G F)

Since this conclusion is the negation of a disjunction, we naturally think of De Morgan's Theorems, which state

its equivalence to a conjunction of negations. And of course, we can form a conjunction by deriving both

conjuncts on their own and then using Conj. So we're now guessing that the end of our argument will look

something like this:

1. A (B C) premise

2. (B A) [D (E ↔ F)] premise

3. D (G E) premise

4. (A B) C 1 Assoc.

5. A B 4 Simp.

6. B A 5 Comm.

7. D (E ↔ F) 2, 6 M.P.

8. D 3 Simp.

9. E ↔ F 7, 8 D.S.

G ? ? ?

F ? ? ?

G F ?, ? Conj.

(G F) ? DeM.

All we need to do now is to close the gap between the top part of the proof and the ending that would provide

our desired conclusion. So, in effect, we're starting all over: we now have lines 1-9 available as premises, and we

need to invent two little proofs, one with the conclusion G and the other with the conclusion F.

For the first of these, we need to go back up to line 3, since that's the only place where G occurs. If we commute that line and then simplify, we'll have:

1. A (B C) premise

2. (B A) [D (E ↔ F)] premise

3. D (G E) premise

4. (A B) C 1 Assoc.

5. A B 4 Simp.

6. B A 5 Comm.

7. D (E ↔ F) 2, 6 M.P.

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8. D 3 Simp.

9. E ↔ F 7, 8 D.S.

10. (G E) D 3 Comm.

11. (G E) 10 Simp.

G ? ? ?

F ? ? ?

G F ?, ? Conj.

(G F) ? DeM.

Now applying De Morgan's Theorems and simplifying will complete the first of our little arguments:

1. A (B C) premise

2. (B A) [D (E ↔ F)] premise

3. D (G E) premise

4. (A B) C 1 Assoc.

5. A B 4 Simp.

6. B A 5 Comm.

7. D (E ↔ F) 2, 6 M.P.

8. D 3 Simp.

9. E ↔ F 7, 8 D.S.

10. (G E) D 3 Comm.

11. (G E) 10 Simp.

12. G E 11 DeM.

13. G 12 Simp.

F ? ? ?

G F ?, ? Conj.

(G F) ? DeM.

Now our only goal is to derive F, and that is likely to come from line 9, which is the simplest line in which F

occurs. So let's use Equiv. to unpack line 9 and simplify the result:

1. A (B C) premise

2. (B A) [D (E ↔ F)] premise

3. D (G E) premise

4. (A B) C 1 Assoc.

5. A B 4 Simp.

6. B A 5 Comm.

7. D (E ↔ F) 2, 6 M.P.

8. D 3 Simp.

9. E ↔ F 7, 8 D.S.

10. (G E) D 3 Comm.

11. (G E) 10 Simp.

12. G E 11 DeM.

13. G 12 Simp.

14. (E F) (F E) 9 Equiv.

15. E F 14 Simp.

F ? ? ?

G F ?, ? Conj.

(G F) ? DeM.

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But now we're almost done: a little fiddling with line 12 will get us E, the other premise we need for an M.P.

on line 15, and D.N. will fill the gap completely, so we need only supply the missing line numbers in order to

complete the entire proof:

1. A (B C) premise

2. (B A) [D (E ↔ F)] premise

3. D (G E) premise

4. (A B) C 1 Assoc.

5. A B 4 Simp.

6. B A 5 Comm.

7. D (E ↔ F) 2, 6 M.P.

8. D 3 Simp.

9. E ↔ F 7, 8 D.S.

10. (G E) D 3 Comm.

11. (G E) 10 Simp.

12. G E 11 DeM.

13. G 12 Simp.

14. (E F) (F E) 9 Equiv.

15. E F 14 Simp.

16. E G 12 Comm.

17. E 16 Simp.

18. F 15, 17 M.P.

19. F 18 D.N.

20. G F 13, 19 Conj.

21. (G F) 20 DeM.

Although this turned out to be a very long proof, it wasn't really difficult, since each of the shorter segments in

which we invented it involved only a few simple steps. We simply worked from the top down and from the

bottom up until we closed the gap step by step in order to demonstrate the validity of the original argument.

Let us visit the Constructive Dilemma again and consider the following question:

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?

Constructive Dilemma Proof:

1. ( P → Q ) premise

2. ( R→ S ) premise

3. ( P R ) premise

4. ( Q → P ) 1 , Contraposition (or Trans.)

5. ( P → R ) 3 , Implication (or Impl.)

6. ( Q → R ) 4 , 5 Chain Argument (or H.S.)

7. ( Q → S ) 6 , 2 Chain Argument (or H.S.)

8. ( Q S ) 7 Implication (or Impl.) , Q.E.D.

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Another Example:- Consider the following proof:

1. A → B

2. B → C

3. C

4. A ∨ D

5. ∴D (This is the conclusion to be derived)

6. A → C 1, 2 H.S.

7. A 3, 6 M.T.

8. D 4, 7 D.S.

In practice there are three types of proofs: (a) direct, (b) conditional, and (c) indirect.

• Direct Proof: as its name implies, it goes directly from the given premises to its conclusion (see the

example above –lines 1-8).

• Conditional Proof:

When to use: whenever you are trying to prove a conditional statement (i.e., P → Q)

How to use?

(i) assume the antecedent (P) of the conditional statement you are trying to prove (i.e., the argument’s

conclusion) and

(ii) prove the consequent (Q) using the antecedent and the available premises

The rationale behind this – If the truth value of P is false then the truth value of P → Q will be true

(i.e. the conclusion is valid irrespective of the truth values of the premises). If the truth value of P is true

and we can prove Q with the help of P and the available premises, then again P → Q will be true. So in

both cases the conclusion P → Q is valid (in first instance by default value (where P is false) and in

second instance with the help of true value of P and available premises.

Conditional Derivation:

Prove the argument: P → Q, Q → R / ∴ P → R. (This is the famous H.S. (Hypothetical Syllogism)

rule of inference already proved on page 7)

Proof:

1. P → Q premise

2. Q → R premise

3. P Assumption

4. Q 1, 3 M.P.

5. R 2, 4 M.P.

6. P → R 3 - 5 Conditional Proof

Prove that G → T, (T ˅ S) → K / G → K

Proof:-

1. G → T premise

2. (T ˅ S) → K premise

3. G Assumption

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4. T 1, 3 M.P.

5. T ˅ S 4 Add.

6. K 2, 5 M.P.

7. G → K 3-6 Conditional Proof

Prove that C → (A D), B → (A E) / (C ˅ B) ˅ A

Proof:-

1. C → (A D) premise

2. B → (A E) premise

3. C ˅ B Assumption

4. (A D) ˅ (A E) 1, 2, 3 C.D.

5. A (D ˅ E) 4 Dist.

6. A 5 Simp.

7. (C ˅ B) → A 3-6 Conditional Proof

8. (C ˅ B) ˅ A 7 Impl.

Indirect Proof

Basic idea: prove that is true, not by deriving directly, but by showing that must be false.

At some point in a proof, you decide you’d like to be able to derive on a line, but you can’t figure

out how. Add an assumption line consisting of , and then proceed using the rules. Keep deriving

lines until you derive an explicit contradiction. We know that contradictions are always false. But

we also know that our rules are truth preserving, and so if they are applied to only true statements

they will produce only true statements. But, we managed to produce a false statement, the explicit

contradiction. So, the set of statements we were applying the rules to must not all have been true.

But the only statement that is suspect is the one we added as an assumption: . So that one must

be the one that is false. And if is false, then must be true. So you are justified in writing a

new derived line consisting of .

In particular if is the conclusion of an argument then this procedure is indirect proof of the validity

of the argument provided you have arrived at a contradiction. (This proof is recommended when there

are many variables.) But on the other hand if you do not arrive at a contradiction, stop: the argument is

invalid.

Prove that A → B, B → B / A

Direct Proof Indirect Proof

1. A → B premise

2. B → B premise

3. B ˅ B 2 Impl.

4. B 3 Taut.

5. A 4, 1 M.T.

1. A → B premise

2. B → B premise

3. A AIP

4. B 1, 3 M.P.

5. B 4, 2 M.P.

6. B B 4, 5 Conj.

7. A 3 – 6 IP

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Rules for the use of IP

1. Start subproof (SP) by indenting and designating first line AIP

2. IP ends only when an explicit contradiction is derived

3. Mark off IP, closing SP and discharging assumption

4. The next line after the closed IP SP can only be the negation of the AIP

5. All SPs must be closed before the proof can end

6. Once a SP has been closed, no lines in it may be used or cited

Prove that ( M ˅ P) → (K L), K ˅ L / M ˅ K

1. ( M ˅ P) → (K L) premise

2. K ˅ L premise

3. M AIP

4. M ˅ P 3 Add.

5. K L 1, 4 M.P.

6. K 5 Simp.

7. L 5 Simp.

8. K 7, 2 D.S.

9. K K 6, 8 Conj.

10. M 3 – 9 IP

11. M ˅ K 10 Add.

Prove that (I ˅ S) → C, D → (P I) / D → C

1. (I ˅ S) → C premise

2. D → (P I) premise

3. (D → C) AIP

4. ( D ˅ C) 3 Impl.

5. D C 4 DeM.

6. D 5 Simp.

7. C 5 Simp.

8. P I 6, 2 M.P.

9. I 8 Simp.

10. I ˅ S 9 Add.

11. C 1, 10 M.P.

12. C C 7, 11 Conj.

13. D → C 3 – 12 IP

Prove that P → Q, Q → R, P ↔ R / P

1. P → Q premise

2. Q → R premise

3. P ↔ R premise

4. P AIP

5. Q 1, 4 M.P.

6. R 2, 5 M.P.

7. (P → R) (R → P) 3 Equiv.

8. P → R 7 Simp.

9. R 4, 8 M.P.

10. R R 9, 6 Conj.

11. P 4 – 10 IP

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Let us revisit Conditional Proof (CP)

Basic idea: prove the conditional → ψ is true, by assuming and deriving ψ.

At some point in a proof, you decide you’d like to be able to derive → ψ on a line, but you can’t figure

out how. Add an assumption line consisting of , then proceed using the rules.

Keep deriving lines until you derive ψ. At this point, we don’t know whether is actually true, since we

just assumed it, but we have shown that if were true, then ψ would be true.

But this fact that the subproof demonstrated, that if is true, then ψ is true, just is what the conditional

→ ψ means. So the subproof shows that the conditional can be validly inferred.

Rules for the use of CP

1. Start subproof (SP) by indenting and designating first line ACP

2. CP ends any time you want

3. Mark off CP, closing SP and discharging assumption

4. The next line after the closed CP SP can only be a conditional whose antecedent is the ACP and

whose consequent is the last line of the CP SP

5. All SPs must be closed before the proof can end

6. Once a SP has been closed, no lines in it may be used or cited

Nested Subproofs So long as the rules for subproofs are followed, a single proof can have more than one subproof, and can

even have subproofs within subproofs.

Prove that A → (B C), D → C / D → A

01. A → (B C) premise

02. D → C premise

03. D ACP

04. A AIP

05. C 2, 3 M.P.

06. B C 1, 4 M.P.

07. C 6 Simp.

08. C C 5, 7 Conj.

09. A 4 – 8 IP

10. D → A 3 – 9 CP

Prove that C → (A D), B → (A E) / (C ˅ B) ˅ A

01. C → (A D) premise

02. B → (A E) premise

03. C ˅ B ACP

04. A AIP

05. (A D) ˅ (A E) 1,2,3 C.D.

06. A (D ˅ E) 5 Dist.

07. A 6 Simp.

08. A A 4, 7 Conj.

09. A 4 – 8 IP

10. (C ˅ B) → A 3 – 9 CP

11. (C ˅ B) ˅ A 10 Impl.

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Prove that ( M ˅ P) → (K L), K ˅ L / M ˅ K

01. ( M ˅ P) → (K L) premise

02. K ˅ L premise

03. M ACP

04. K AIP

05. M ˅ P 3 Add.

06. K L 5, 1 M.P.

07. K 6 Simp.

08. K K 4, 7 Conj.

09. K 4 – 8 IP

10. M → K 3 – 9 CP

11. M ˅ K 10 Impl.

Following is another example proof using both Conditional Proof and Indirect Proof:

(A ˅ B) → C, D → ( F G) / (A ˅ D) → (C F)

1. (A ˅ B) → C premise

2. D → ( F G) premise

3. C → (A ˅ B) 1 Trans. (and D.N. C C)

4. C F ACP

5. D AIP

6. F G 2, 5 M.P.

7. F 4 Simp.

8. F 6 Simp.

9. F F 7, 8 Conj.

10. D 5 – 9 IP

11. C 4 Simp.

12. (A ˅ B) 3, 11 M.P.

13. A B 12 DeM.

14. A 13 Simp.

15. A D 10, 14 Conj.

16. C F → A D 4 – 15 CP

17. ( A D) → (C F) 16 Trans.

18. ( A ˅ D) → (C F) 17 DeM.

19. (A ˅ D) → (C F) D.N.

Another simple example of CP:

N → O, N → P / N → (O P)

1. N → O premise

2. N → P premise

3. N ACP

4. O 1, 3 M.P.

5. P 2, 3 M.P.

6. O P 4, 5 Conj.

7. N → (O P) 3 – 6 CP

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Inference Rules for Quantifiers

• x p(x) Universal Instantiation / Universal Specification ________

p(o) (substitute any element o of u.d.)

• p(g) (for a general element g of u.d.) ________

x p(x) Universal Generalization

• x p(x) Existential Instantiation ________

p(c) (substitute some element c of u.d.)

• p(o) (for some element o of u.d.) ________

x p(x) Existential Generalization

Rule of Universal Specification

This is a fairly obvious rule, but one that is important:

If an open statement is true for all possible replacements in the designated universe, then that open

statement is true for each specific individual member in that universe.

Symbolically speaking, we have:

If x p(x) is true, then we know that p(o) is true, for each o in the universe for x.

Here is a simple example using this rule. Consider the following premises:

1) Each actor/actress on the TV show Friends is a millionaire.

2) Jennifer Aniston is an actress on the TV show Friends.

Therefore, Jennifer Aniston is a millionaire.

Symbolically, consider setting up these three statements:

p(x): x is an actor/actress on Friends.

q(x): x is a millionaire.

Now, the given information is x [p(x) → q(x)]

If we wish to determine the financial status of Jennifer Aniston, we add into our premise the

statement p(Jennifer Aniston) as being true.

Using the Rule of Universal Specification, and Rule of Detachment (Modus Ponens), we can

conclude that q(Jennifer Aniston) is true; that is Jennifer Aniston is a millionaire.

Let’s go ahead and look at another example in greater detail.

Consider each of these open statements for the next example:

p(x): x is a show on prime-time TV

q(x): x is a show on late-night TV

r(x): x is a soap opera

Now, consider the following argument:

No soap opera is on prime-time TV or late-night TV

All My Children is a soap opera.

Therefore, All My Children is not on prime-time TV.

Here is the proof of the argument:

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(Note: Let A stand for ―All My Children‖)

1. x [p(x) q(x) → r(x)] premise

2. r(A) premise

3. p(A) q(A) → r(A) 1 Rule of Universal Specification

4. r(A) → (p(A) q(A)) 3 Trans.

5. r(A) → (p(A) q(A)) 4 D.N.

6. r(A) → (p(A) q(A)) 5 DeM.

7. p(A) q(A) 2, 6 M.P.

8. p(A) 7 Simp.

The Rule of Universal Generalization

If an open statement p(x) is proved to be true when x is replaced by any arbitrarily chosen element c

from our universe, then the universally qualified statement x p(x) is true. (This rule also extends

beyond one variable.)

We can use this to formally show that IF

x [p(x) → q(x)] AND

x [q(x) → r(x)] THEN

x [p(x) → r(x)].

Here is the proof of the argument:

1. x [p(x) → q(x)] premise

2. p(c) → q(c) 1 Rule of Universal Specification

3. x [q(x) → r(x)] premise

4. q(c) → r(c) 3 Rule of Universal Specification

5. p(c) → r(c) 2, 4 H.S.

6. x [p(x) → r(x)] 5 Rule of Universal Generalization

Now, consider the following assumptions

x p(x) AND

x [p(x) q(x) → r(x)]

And use those to prove:

x [r(x) → q(x)].

Here is the proof of the argument:

1. x p(x) premise

2. p(c) 1 Rule of Universal Specification

3. x [p(x) q(x) → r(x)] premise

4. p(c) q(c) → r(c) 3 Rule of Universal Specification

5. q(c) ACP (using here idea of Conditional Proof)

6. p(c) q(c) 2, 5 Conj.

7. r(c) 4, 6 M.P.

8. q(c) → r(c) 5 – 7 CP

9. r(c) → q(c) 8 Trans.

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10. x [r(x) → q(x)] 9 Rule of Universal Generalization

To make this example more concrete, consider the following open statements for p(x), q(x) and r(x):

Let the universe x be of all 4 sided polygons.

p(x): x is a quadrilateral.

q(x): x has four equal angles.

r(x): x is a rectangle.

Using basic geometry definitions, we find that

x p(x) (All 4 sided polygons are quadrilaterals.)

x [p(x) q(x) → r(x)] (All 4 sided polygons that are quadrilaterals

and have four equal angles are rectangles.)

Therefore, all 4 sided polygons that are not rectangles do not have four equal angles, OR

x [r(x) → q(x)]

Examples illustrating proof techniques

We will use these definitions in the following problems:

An integer n is even if and only if there exists another integer r such that n = 2*r.

An integer n is odd if and only if there exists another integer r such that n = (2*r) + 1.

If y | x, which is read as ―x is divisible by y‖, or ―y divides evenly into x‖, then x = yc, for some

integer c. Remember in this definition, y must be non-zero.

In the book there are proofs showing that the sum of two odd numbers is even and that the product

of two odd numbers is odd. We will establish here a proof of slightly more interesting result.

The square of an even number k is divisible by 4.

Since we know k is even, we have k=2*r for some integer r.

Now, we can compute k2:

k2 = (2*r)2 = 4*r2, which is divisible by 4 since r2 is an integer.

Also, the square of an odd number k leaves a remainder of 1 when divided by 4.

Since we know that k is odd, we have k=2*r+1 for some integer r. Now, we can compute k2:

k2 = (2*r+1)2 = 4*r2+4*r+1 = 4(r2+r) + 1, which leaves a remainder of 1 when divided by 4 since 4

divides evenly into 4(r2+r). (Because r2+r must be an integer...)

In both of these examples, we used the Universal Rule of Generalization because we proved the

result for arbitrary odd and even integers, and that implies that the general statement is true for all of

them.

Prove that if n is an integer then n2+n is an even integer

Proof #1: direct proof by method of exhaustion using cases.

All integers are even or odd. (I have not formally proved this, but you can use this fact throughout

the class..) Thus, we have two different possibilities for n:

n is even: Then there is an integer r such that n=2r, then we have

n2+n = (2r)2 + 2r

= 4r2 + 2r

= 2(2r2 + r),

since this value is divisible by 2, it is even.

n is odd: Then there is an integer r such that n=2r+1, then we have

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n2+n = (2r+1)2 + (2r+1)

= 4r2 + 4r + 1 + 2r + 1

= 4r2 + 6r + 2

= 2(2r2 + 3r + 1),

and this value is also divisible by 2, thus it is even.

One way to analyze how and why this proof is sufficient is to break it down into a logical argument.

Let p(n), q(n), and r(n) be the following open statements:

p(n): n is an even number.

q(n): n is an odd number.

r(n): n2+n is even.

Since all integers are either even or odd, our goal is to prove

(p(n) q(n)) r(n).

Logically, we can show this is equivalent to:

(p(n) r(n)) (q(n) r(n))

The first part of our proof showed that (p(n) r(n)) is true, while the second part of our proof

showed that (q(n) r(n)) is true as well. Thus, together, these parts prove the original claim.

Proof #2: Contradiction

For contradiction’s sake, assume that n2+n is odd. We have the following: n2+n = n(n+1).

We know that for the product of two integers to be odd, both of them have to be. (Part of this is

shown in the book...)

However, it is impossible for both n and n+1 to be odd. (This is something you would have to prove

also.)

Thus, the way contradiction works is that you assume that the result is false. Then use algebra and

other rules to show that the premise must be false as well, or that something impossible occurs if the

incorrect result is assumed.

The reason is that you came to an incorrect conclusion. You must have arrived at it by making an

incorrect step. The ONLY possible incorrect step that could have been taken was the assumption

made in the beginning. Hence, that is wrong, which implies the truth of what you are trying to

prove.

In this class, you will often be given statements that may or may not be true. Your job will be to

determine which is the case, and give proof of your answer. Typically, disproving an assertion is

easier than proving one. Here are a couple examples illustrating how to disprove an assertion:

1) For all prime numbers greater than 100, the sum of their digits is greater than 4.

This is not true. We can verify that 101 is prime by showing that 2, 3, 5, and 7 do not divide into it

evenly. Furthermore, we can see the sum of the digits in this number is 2, which is less than or equal

to 4. Thus, by finding one prime number greater than 100 that has digits that sum to 2, we have

disproved the claim.

Thus, to disprove a ―for all‖ statement, all we had to do was find one value for which the statement

did not hold.

Here is another example of method of exhaustion:

All integers in between 10 and 20 are either prime or have 2 or 3 as a factor.

Proof:

10 = 2*5

11 is prime

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12 = 3*4

13 is prime

14 = 2*7

15 = 3*5

16 = 2*8

17 is prime

18 = 2*9

19 is prime

20 = 2*10

Here is an example where we prove a statement by proving its contrapositive:

If x + y is odd, then exactly one of x and y is odd.

The contrapositive of this statement is:

If either neither or both of x and y is odd, then x+y is even.

If we prove this statement, we have proven the original.

Thus, we can split the problem up into two cases:

Case 1:

Neither of x and y is odd, thus both are even.

Let x = 2a and y = 2b for some integers a and b.

Then,

x + y = 2a + 2b

= 2(a+b).

At this point we can conclude that x+y is even since it is divisible by 2. (We know this because a+b

must be an integer.)

Case 2:

Both of x and y are odd.

Let x = 2a+1, y=2b+1, for some integers a and b.

Then,

x+y = 2a+1+2b+1

= 2a+2b+2

= 2(a+b+1)

As before, we can conclude that x+y is even since it is a multiple of 2.

More Practice of a proof based on contradiction:

Theorem: (For all integers n)

If 3n+2 is odd, then n is odd.

Proof: Suppose that the conclusion is false, i.e., that n is even. Then n=2k for some integer k.

Then 3n+2 = 3(2k)+2 = 6k+2 = 2(3k+1). Thus 3n+2 is even, because it equals 2j for integer j =

3k+1. So 3n+2 is not odd. We have shown that ¬(n is odd)→¬(3n+2 is odd), thus its contra-

positive (3n+2 is odd) → (n is odd) is also true.

Difference between contradiction and contra-positive

Let us recall from page 26, the following statement about the contradiction:

Thus, the way contradiction works is that you assume that the result is false. Then use

algebra and other rules to show that the premise must be false as well, or that something

impossible occurs if the incorrect result is assumed.

Here you either end up with something impossible or arrived at the conclusion that premises must be

false.

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If you end up with second possibility (premise must be false), then you have in fact used contra-

positive. So contra-positive is a special kind of contradiction.

[Another nice way to put it]How Is This Different From Proof by Contradiction?

The difference between the Contrapositive method and the Contradiction method is subtle. Let's

examine how the two methods work when trying to prove "If P, Then Q".

Method of Contradiction: Assume P and Not Q and prove some sort of contradiction.

Method of Contrapositive: Assume Not Q and prove Not P.

The method of Contrapositive has the advantage that your goal is clear: Prove Not P. In the method

of Contradiction, your goal is to prove a contradiction, but it is not always clear what the

contradiction is going to be at the start.

Example If 2 | 5n then n is even.

Proof: Assume, BWOC, that 2 | 5n and n is odd. Since 2 | 5n, we have 5n = 2d for some integer d.

Since n is odd, we have n = 2k + 1 for some integer k. Then we have

2d = 5n = 5(2k + 1) = 10k + 5.

So 2d = 10k + 5. Solve for 5 to get

5 = 2d – 10k = 2(d – 5k).

But this says that 5 is an even number, a contradiction. So the statement is true. QED.

PROPOSITION 1 For any integer n, if n is odd, then 3n is odd.

PROOF Let n be an odd integer. Then n = 2k + 1 for some integer k. Therefore 3n = 3(2k + 1) =

6k + 3 = (6k + 2) + 1 = 2(3k + 1) + 1 = 2j + 1. Where j = 3k + 1 which is an integer as it is

the sum and product of integers. Thus 3n equals 2 times the integer j plus 1, and so 3n is

odd.

The converse of Proposition 1 is also true.

PROPOSITION 2 For any integer n, if 3n is odd, then n is odd.

PROOF A direct proof of Proposition 2 would assume that 3n = 2k + 1 for some integer k, and then

use an argument like that in the proof of Proposition 1 to show that n has a similar form.

This method does not work here, primarily because to change 3n into n, we must divide both

sides of an equation by 3, unlike the proof of Proposition 1, where we multiplied both sides

of an equation by 3. (Dividing by 3 is not as algebraically simple as multiplying by 3. For

instance, if we divide an integer by 3, the quotient may not be an integer.) Instead, we can

prove Proposition 2 by verifying its contrapositive.

PROOF We prove the contrapositive: For any integer n, if n is not odd, then 3n is not odd. Because

every integer is either even or odd, but not both, we can restate the contrapositive as: For

any integer n, if n is even, then 3n is even.

Let n be an even integer, so that n = 2k for some integer k. Then 3n = 3(2k) = 2(3k).

Because 3k is a product of integers, it is an integer. Thus 3n equals twice the integer 3k, and

so 3n is even.

Exercises

Prove each of the following by the contrapositive method.

1. If x and y are two integers whose product is even, then at least one of the two must be even.

2. If x and y are two integers whose product is odd, then both must be odd.

3. If a and b real numbers such that the product a b is an irrational number, then either a or b

must be an irrational number.