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
