a) Using long or short truth tables (symbolize as necessary and show tables) determine and state whether the following argument is valid, briefly explaining your table; and b) if valid, demonstrate validity by deductions using Group I rules.

Arg

If Peter is perceptive, then Quincy is a quack. Quincy, however, is not a quack. So Peter's not perceptive. (Hint: this is a valid argument)

To determine the validity of the given argument, we can use a truth table. Let's symbolize the statements as follows:

P: Peter is perceptive
Q: Quincy is a quack

The argument can be represented as:

Premise 1: P → Q
Premise 2: ¬Q
Conclusion: ¬P

Now, we can construct a truth table to evaluate all the possible truth value combinations of the premises and the conclusion.

| P | Q | P → Q | ¬Q | ¬P |
|----|----|-------|-----|-----|
| T | T | T | F | F |
| T | F | F | T | F |
| F | T | T | F | T |
| F | F | T | T | T |

In this truth table, we consider all possible truth values for P and Q. Then, we evaluate the truth values of the premises (P → Q and ¬Q) and the conclusion (¬P).

To determine validity, we look for a row in the truth table where all the premises are true (T) and the conclusion is also true (T).

In this case, we can see that in the last row of the truth table, where both premises, P → Q and ¬Q, are true and the conclusion ¬P is also true.

Therefore, since there is at least one row in the truth table where all premises are true and the conclusion is true, we can conclude that the argument is valid.

Now, let's demonstrate the validity using Group I rules (also known as natural deduction). The argument uses the following rules:

1. Modus Tollens (MT): If we have a conditional statement P → Q and its negation ¬Q, then we can infer the negation of P, ¬P.

Using Group I rules, we can demonstrate the validity as follows:

1. Assume P is true (Assumption)
2. Apply Modus Tollens to Premise 1: P → Q and ¬Q
3. Since ¬Q is true, infer ¬P (Modus Tollens)
4. Discharge the assumption: since we reached the conclusion ¬P assuming P, we can conclude that if Peter is perceptive (P), then Quincy is a quack (Q), and if Quincy is not a quack (¬Q), then Peter is not perceptive (¬P).

In this way, we have shown that the argument is valid using Group I rules.