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:

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:

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.

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:

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:

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:

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.

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

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.