Question

Consider the following argument. If I get a Christmas bonus, I'll buy a stereo. If I sell my motorcycle, I'll buy a stereo. ∴ If I get a Christmas bonus or I sell my motorcycle, then I'll buy a stereo. Let p = "if I get a Christmas bonus," q = "if I sell my motorcycle," and r = "I'll buy a stereo." Is the argument valid or invalid?
Select the answer that shows the symbolic form of the argument and justifies your conclusion. form:______.
p → r invalid, converse error
q → r
∴ p ∨ q → r
form:_____.
r → q valid, proof by division into cases
r → p
∴ r → p ∨ q
form:______.
r → q invalid, converse error
r → p
∴ r → p ∨ q
form:______.
p → r valid, proof by division into cases
q → r
∴ p ∨ q → r
form:_______.
r → q invalid, inverse error
r → p
∴ r → p ∨ q

Answer 1

Answer:

p → r valid, proof by division into cases

q → r

∴ p ∨ q → r

Step-by-step explanation:

Let

p = "if I get a Christmas bonus,"

q = "if I sell my motorcycle,"

and

r = "I'll buy a stereo."

This can be written as:

If I get a Christmas bonus, I'll buy a stereo

p → r

If I sell my motorcycle, I'll buy a stereo

q → r

∴ If I get a Christmas bonus or I sell my motorcycle, then I'll buy a stereo.

∴ p ∨ q → r

To prove this argument we partition the argument into a group of smaller statements that together cover all of the original argument and then we prove each of the smaller statements. If you see the conclusion ∴ p ∨ q -> r so if the conclusion contains a conditional argument of form "If A1  or A2 or... or An then C ”, then we prove "If A1 then C", "If A2 then C" and so on upto "If An then C" . This depicts that the conclusion  C is true no matter which if the Ai holds true. This method is called proof by division into cases. In the given example, this takes the form:

p → r

q → r

p ∨ q

∴ r

Since proof by division into cases is an inference rule thus given argument is valid. Lets make a truth table to show if this argument is valid

p   q   r   p ∨ q   p → r    q → r    p ∨ q → r

0   0   0     0        1           1             1

0   0   1      0        0          0            0

0   1    0     1         1           0            0

0   1    1      1         0          1               1

1    0   0     1         0          1             0    

1    0   1      1         1           0            1    

1    1    0     1         0          0            0    

1    1    1      1         1           1              1    

An argument is valid if all of the premises are true, then the conclusion is true. So the truth table shows that the conclusion is true i.e. 1 where all premises are true i.e. 1. So the argument is valid.

Hence

p → r valid, proof by division into cases

q → r

∴ p ∨ q → r