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

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