Invalid Arguments

Invalid Arguments

Argument by the Converse

Argument by the Inverse

A Strategy for Invalidating an Argument

Invalid Arguments

Argument by the Converse

Argument by the Inverse

A Strategy for Invalidating an Argument

The Book of
Foundational Mathematics

Example 2.5.1

Example 2.5.2

Example 2.5.3

Example 2.5.4

Example 2.5.5

Counter Example

Example 2.5.6

The Book ofFoundational Mathematics

Example 2.5.1

Example 2.5.2

Example 2.5.3

Example 2.5.4

Example 2.5.5

Counter Example

Example 2.5.6

The Book of
Foundational Mathematics

Detailed Contents

Propositional Logic

Arguments and Proof

Verifying Mathematical Arguments in Maple

Set Theory

Set Theory in Maple

All of us at one point were probably presented with arguments that don’t seem to be correct. Just because someone can string together a group of premises and assert some conclusion does not mean that conclusion necessarily follows from the premises.

Remember that an argument is valid if the argument’s implication is a logical implication. This means that no matter what truth value the argument’s propositions have, the overall implication always evaluates to 1. This means that if we can come up with even one truth value assignment reducing to the form of

1 → 0,

then the argument is not valid. In other words, for any given argument

(p₁ ∧ p₂ ∧ p₃ ∧ … ∧ p_n) → c,

with

c = 0,

then the argument is invalid (because all premises are assumed to have truth value 1.)

Here, we’ll examine some of the most common fallacies made when constructing an argument, and how to detect if a given argument is invalid.

Consider the following argument using propositions p and q:

	p → q
	q
∴	p

We have two premises and a conclusion:

p₁:	p → q
p₂:	q
c:	p

Remember that for an argument, we assume that (p₁ ∧ p₂) = 1, meaning we need both p₁ = 1 and p₂ = 1.

Since p₂ is just q, that means we assume q = 1.

Now, we need p₁ = 1, meaning we need p → q = 1. Since we know that q = 1, this means that p = 0 or p = 1 since both values give us that p → q = 1.

However, notice that if we assume that p = 0, then our conclusion c has truth value 0 because c = p. Writing this out in a tabular form, we see that

p₁	=	p → q	=	0 → 1	=	1
p₂	=	q	=	1
c	=	p	=	0

Thus, since we can simultaneously make (p₁ ∧ p₂) = 1 and c = 0, this argument is invalid.

The argument

	p → q
	q
∴	p

is invalid.

Figure 2.5.1: The above invalid argument is sometimes referred to as an “argument by the converse.”

Before showing another invalid argument, we should take a moment to compare the invalid argument shown in Example 2.5.1 (and displayed to the left in Figure 2.5.1) to the closely related, but actually valid Modus Ponens argument.

	p → q
	p
∴	q

Modus Ponens

	p → q
	q
∴	p

Argument by the Converse

In the above Modus Ponens argument, p is a premise, while q is the conclusion. In the argument by converse above, q is a premise, while p is the conclusion. Even though these arguments look similar, it’s important not to get them mixed up.

Johnny can be a bit of a reckless driver. He tends to ignore speed limits, and often does not ensure all of his lights are functioning properly. As such, he is prone to getting pulled over by the police a lot more than anyone else. Consider the following propositions:

s:	Johnny is speeding.
r:	Johnny runs a red light.
t:	Johnny gets pulled over and is issued a ticket.

Now consider the following argument:

	s → t
	t
∴	s

Even though this argument asserts that Johnny was speeding, do we actually know that Johnny was speeding? All we know to be true is that Johnny got a ticket (this is a premise of the above argument.) He could have gotten a ticket because he ran a red light. He may have also gotten a ticket for some other reason, such as malfunctioning tail lights, or an expired registration.

Knowing that Johnny got pulled over and issued a ticket is not enough to determine whether or not he was speeding.

Charlotte is an aspiring entrepreneur. She’s working very hard to promote her robot engineering company and is producing a lot of advertising for her business. Consider the following propositions:

m:	Charlotte advertises in magazines.
s:	Charlotte advertises on social media.
b:	Charlotte advertises on billboards
c:	Charlotte’s business gains a new customer.

Now consider the following argument:

	s → c
	c
∴	s

Just like in Example 2.5.1, knowing that c is true does not mean that s must be true as well. If s is 0, then the implication s → c is trivially true, even when c = 1 (which it is since it’s a premise of the argument.) Again, maybe her business gained a new customer because of advertisements in a magazine, or possibly on a billboard.

Consider the following argument using propositions p and q:

	p → q
	¬p
∴	¬q

Again, we have two premises and a conclusion:

p₁:	p → q
p₂:	¬p
c:	¬q

Instead of just trying to work out truth values of propositions, we could just use a truth table.

Remember, we only care about truth value assignments of p and q that make both premises p₁ and p₂ true, so let’s highlight those rows in the truth table where they are both true.

Notice that there is a row where both premises are true, but the conclusion c is false. This happens when p = 0, and q = 1. Hence, the implication

[(p → q) ∧ (¬p)] → ¬q

is not a tautology. In other words, it is not a logical implication, so we write

[(p → q) ∧ (¬p)] ⇏ ¬q.

Thus the above argument is invalid.

The argument

	p → q
	¬p
∴	¬q

is invalid.

Figure 2.5.2: The above argument is sometimes referred to as an “argument by the inverse.”

Again, it’s worth taking the time to compare the invalid argument by the inverse (displayed to the left in Figure 2.5.2) to an actually valid argument that it sort of resembles.

	p → q
	¬q
∴	¬p

Modus Tollens

	p → q
	¬p
∴	¬q

Argument by the Inverse

Just like the other invalid argument, pay attention to the second premise presented in each argument directly above. In Modus Tollens, the second premise is ¬q, whereas in the argument by the inverse the second premise is ¬p.

While Modus Ponens and Modus Tollens can be safely used in determining if an argument is invalid, introducing an argument by converse or argument by inverse produces a fallacy in reasoning. Even if a given argument is valid, any justification for the argument being valid that uses one of the two above invalid rules will be incorrect.

Last time, Johnny was in the midst of getting pulled over and issued a ticket. Maybe it was for speeding, perhaps it was for running a red light, or perhaps some other violation of traffic law occurred. Regardless, now Johnny may be running late for work (another typical Johnny attribute.) Consider these propositions:

ℓ:	Johnny is late to work.
r:	Johnny is rude to his company’s clients.
f:	Johnny is fired from his job.

Now suppose Johnny makes the following argument:

	ℓ → f
	¬ℓ
∴	¬f

Here, Johnny is trying to argue that because he was not late to work, he was not fired. The problem is that even if it is true that Johnny was not late to work (perhaps he started speeding after getting pulled over to try and make up for lost time), he may still have been fired for being rude to his company’s clients.

As such, we can’t conclude that Johnny was not fired from his job.

We were not able to conclude how Charlotte’s robot engineering business gained a new customer from Example 2.5.2. Regardless of how it happened, her business got a new customer who wants to purchase a robot to fight in the Mech-Fighter Tournament. Every customer of her robot engineering business can choose a robot with one of three features. Consider the following propositions:

ℓ:	The robot can shoot laser beams from its eyes.
j:	The robot is equipped with a jet pack.
f:	The robot has flamethrowers built into its arms.
w:	The robot wins the Mech-Fighter Tournament.

Now consider the following argument:

	f → w
	¬f
∴	¬w

Can we conclude that this customer’s robot didn’t win the Mech-Fighter Tournament? We can’t because all we know is that the robot ordered did not have flamethrowers built into its arms. It may have still won the tournament, but it would have won by shooting laser beams, or by flying around the tournament’s arena.

It’s possible the customer’s robot won, just not using flamethrowers.

For propositions p, q, and r, consider the following argument:

	p → ¬q
	r
∴	p ∨ ¬r

Is this a valid argument? None of the rules of inference discussed here seem like they would do much good in helping us reach conclusion p ∨ ¬r, so perhaps this is not a valid argument.

To be sure, we construct a truth table, and highlight all the rows rows where all premises are equal to 1.

		p₂		p₁	p₁ ∧ p₂		c	p₁ ∧ p₂ → c
p	q	r	¬q	p → ¬q	(p → ¬q) ∧ r	¬r	p ∨ ¬r	[(p → ¬q) ∧ r] → [p ∨ ¬r]
0	0	0	1	1	0	1	1	1
0	0	1	1	1	1	0	0	0
0	1	0	0	1	0	1	1	1
0	1	1	0	1	1	0	0	0
1	0	0	1	1	0	1	1	1
1	0	1	1	1	1	0	1	1
1	1	0	0	0	0	1	1	1
1	1	1	0	0	0	0	1	1

Here, it looks like there are three rows where all premises are equal to 1. However, only one of those rows has a conclusion that is equal to 1, the other two have the conclusion equal to 0.

So this argument is not valid. One way to show this is by setting

p = 0
q = 0
r = 1,

which corresponds to the first highlighted row. By doing this, we see that

p₁ = p → ¬q = (0) → ¬(0) = 0 → 1 = 1

p₂ = r = 1

c = p ∨ ¬r = (0) ∨ ¬(1) = 0 ∨ 0 = 0.

These truth value assignments make both premises true and the conclusion false.

The second highlighted row also provides truth value assignments that yield a false implication.

Hence, the implication is not a logical implication, meaning the argument is not valid.

In Example 2.5.5, we noted that none of the rules of inference we knew about looked like they would help us reach the desired conclusion. As such, we suspected that the argument was not valid.

By constructing a truth table, we were able to see truth value assignments that lead to a false implication.

Consider a general argument

[p₁ ∧ p₂ ∧ … ∧ p_n] → c

where the premises and conclusion involve combinations of propositions s₁, s₂, …, and s_m.

A truth value assignment for each of the propositions s₁, s₂, …, and s_m that makes

[p₁ ∧ p₂ ∧ … ∧ p_n] = 1

and

c = 0

is called a counter example to the argument.

We invalidate an argument by providing counter examples to that argument. Note that providing an assignment of truth values that make the argument’s implication true does not prove that the argument is valid. To show that an argument is valid, we have to make sure that every truth value assignment that makes all premises true also makes the conclusion true.

This is not the case for invalidating an argument. One counter example is all that’s needed to invalidate an argument.

However, depending on the number of underlying propositions used for the premises and conclusion, we may want to avoid constructing truth tables. We can still work out truth value assignments by just assuring that all premises evaluate to 1 when the conclusion evaluates to 0.

For propositions a, b, c, d, and e, consider the following argument:

	e
	e ∨ d
	d → (b → c)
	a → b
∴	¬c → ¬a

Is this argument valid? We could try to invalidate it by assigning truth values to a, b, c, d, and e to make the premises true and the conclusion false.

Let’s start with the conclusion ¬c → ¬a. We need

¬c → ¬a = 0.

This means that we need

¬c = 1
¬a = 0

meaning we have that

c = 0
a = 1.

Let’s record that information down for now:

a = 1
b = ?
c = 0
d = ?
e = ?

Now let’s look at the premises starting with a → b. Since it’s a premise, we need a → b = 1, and since a = 1, that means we also need b = 1.

a = 1
b = 1
c = 0
d = ?
e = ?

Looking at the third premise, we also need d → (b → c) = 1. We currently know that b = 1 and c = 0, so b → c = 0. This means that we need d = 0 in order to make the entire implication true:

d → (b → c) = 0 → (1 → 0) = 0 → (0) = 1.

Let’s record that information down.

a = 1
b = 1
c = 0
d = 1
e = ?

So now we just need to figure out e. Notice that e is a premise by itself, which means we need e = 1. So now we have truth values for all propositions>

a = 1
b = 1
c = 0
d = 1
e = 1.

However, we still need to check that all premises still evaluate to 1. The only premise we have left is e ∨ d, so let’s check that one:

e ∨ d = (1) ∨ (1) = 1.

So the last premise also evaluates to 1. This means that we’ve found a combination of truth value assignments that make all premises true and the conclusion false:

a = 1
b = 1
c = 0
d = 1
e = 1.

This means we found a counter example to the given argument, thus invalidating it.

Notice what we did in Example 2.5.6. We started off by choosing values for the propositions that would make the conclusion false. We then used those truth value assignments to try and see if we can choose values for the other propositions to make the premises all true.

Sometimes, we may have choices regarding truth values we could assign to propositions. If that’s the case, there’s nothing wrong with experimenting and trying to see if all premises can be made true while maintaining falsity of the conclusion.

If we are unable to make all premises true while holding the conclusion false, then the argument may actually be valid. After all, if no counter example exists, then the argument would have to be valid. At that point, it may be worth trying to use the rules of inference to validate the argument.

Similarly, if assuming values for propositions that make the conclusion false yield a contradiction, then perhaps the argument can be shown to be valid by way of a proof by contradiction.

As discussed in the previous section, validity of an argument implies the validity of an equivalent argument. By that token, an argument being invalid also implies that any logically equivalent argument is also invalid. So invalidating an equivalent argument is another way to invalidate a given argument.

p₁ ∧ p₂

(p₁ ∧ p₂) → c

(p → q) ∧ (¬p)

[(p → q) ∧ (¬p)] → ¬q