Question 1162984: Is the following argument valid? Provide a proof for your answer.
p =⇒ (q ∧ r)
∼q
− − − − − − − − − −
∼p
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
The argument is valid. We can prove it as such using a proof by contradiction.
The conclusion is ~p, read as "not p". This is the opposite of statement p. Assume that p is the case.
If p is true, then so is q ^ r. This is because of the first line reading p -> (q ^ r). Use the modus ponens rule here.
q ^ r means that both q and r are true at the same time. Therefore, q is the case. But the second line says that ~q is also the case as well.
We have q and ~q clash to form the contradiction.
In short, we assume the opposite of the conclusion (we assume p) and it leads to a contradiction (q and ~q). So making the assumption that p is the case will not work.
Therefore, the opposite of the original assumption must be true and the conclusion of ~p is valid.
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Let's look at a truth table. This is where we list out all the possible truth values of p,q,r and build logical compound statements from them to form the argument as a whole
The first three columns are p,q,r which alternate with true (T) or false (F) values. This lists all the possible ways to have three T's or F's placed together.
The fourth column has us combine q and r to get q ^ r. This column is only true if both q and r are true together; otherwise it is false.
Then we construct a conditional p -> (q ^ r). This is only false if p is true while (q ^ r) is false.
Next up, form the column for ~q, which is the opposite of q. If q is true, then ~q is false, and vice versa.
Lastly, form the column for ~p, which is the conclusion. This will be the opposite of the p column.
Here's what the truth table looks like
| p | q | r | q ^ r | p -> (q ^ r) | ~q | ~p | | T | T | T | T | T | F | F | | T | T | F | F | F | F | F | | T | F | T | F | F | T | F | | T | F | F | F | F | T | F | | F | T | T | T | T | F | T | | F | T | F | F | T | F | T | | F | F | T | F | T | T | T | | F | F | F | F | T | T | T |
and here's a truth table showing just the premises and the conclusion (I removed columns p, q, r, and q ^ r)
| p -> (q ^ r) | ~q | ~p | | T | F | F | | F | F | F | | F | T | F | | F | T | F | | T | F | T | | T | F | T | | T | T | T | | T | T | T |
This is to help give the table a more simplified look.
We can see that there are no cases where we have all true premises lead to a false conclusion. An invalid argument is one where we have all true premises but a false conclusion. A valid argument is the opposite of that.
So this is another way to see we have a valid argument.
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