Below you will find help with selected exercises from the book.

9-2,5
9-4,7
9-6,5
9-7,10
9-10, 7
9-14, 4

9-2, 5. If Parsons objects then if Quincy turns the radio down Rachel will close her door.

P → (Q →R). Note the "then if" phrase. It tells you (1) that P will be set apart from everything that follows, and (2) that a conditional claim will come inside the parentheses.

9-4, 7. (P & R) → Q

~Q

~P

Invalid. Let's do this with the short truth-table method. Very quickly, a false conclusion and a true second premise give us P's truth and Q's falsity:

P     Q     R

T     F

Looking up to the first premise, we see that its antecedent must be true to make the whole premise true (since Q is false). P is already true; R must be true as well. We get:

P     Q     R

T     F     T

No contradictions here; so the argument is invalid.

9-6, 5. (Q → T) → S

~S v ~P

R → P

~(Q → T) v ~R

Disjunctive dilemma. Note the negations of both consequents. Ignore the complexity of (Q → T). (In such cases, it helps to think of parentheses as a nutshell: You treat the nut as a single object, no matter how many bits may be rattling around inside.)

9-7, 10.

1. (T v M) → ~Q
2. (P → Q) & (R→ S)
3. T /..~P

Start by thinking strategically. The letter P will be in your conclusion, and premise 2 is the only line with a P in it, so you'll have to get the P from there. Premise 2 also contains an R and an S that you don't need; so use simplification to drop that whole part of the line. Now you have the line P → Q. How can you get ~P from that? Modus tollens will do it for you, if you can find a ~Q. But where can you get that? Your only choice is premise 1, which will give you a ~Q if you can only find a T v M. Premise 3 supplies the T. Now you can see what to do:

4. T v M 3, ADD
5. ~Q 1, 4, MP
6. P → Q 2, SIM
7. ~P 5, 6, MT

9-10, 7.

1. (M v R) & P
2. ~S → ~P
3. S → ~M /... R

Two tricks here. First, looking at the conclusion, you realize that premise 1 will be vital to your deductive efforts. Second, you should be struck by the oddity of premise 2: both S and P negated, though they're not negated elsewhere. Use this as a clue that premise 2 will be more helpful with its negations gone—which means, if you take the contrapositive of it to produce P → S. Then, if you can get S out of that conditional, premise 3 gives you ~M.

One more thing. You now see that both conjuncts in premise 1 will be necessary, the first because it contains the vital R that's in the conclusion, and the second because you've produced a conditional containing P. There's nothing wrong with going back to premise 1 twice and simplifying it both times, once to use P on your new conditional and the other time to use M v R to get you the desired conclusion. So:

4. P → S 2, CONTR
5. P 1, SIM
6. S 4, 5, MP
7. ~M 3, 6, MP

(Now we simplify premise 1 again:)

8. M v R 1, SIM
9. R 7, 8, DA

9-14, 4.

1. P → (Q v R)
2. T → (S & ~R) /... (P & T) → Q

It's obvious what your CP premise will be, namely P & T. But how will it lead to Q? The main strategy is to simplify that P & T once you've assumed it, and use the P and the T separately to get Q. Premise 1 will then do the trick; but premise 1 leaves you with a disjunction, which can't be simplified. That R needs to be eliminated with a ~R. Now you notice that premise 2's consequent contains a ~R that can be simplified out of the compound consequent. Now we begin:

3. P & T CP Premise
4. P 3, SIM
5. T 3, SIM
6. Q v R 1, 4, MP

(The remaining work consists in also getting a ~R from the premises, which you'll then use in a disjunctive argument on step 6:)

7. S & ~R 2, 5, MP
8. ~R 7, SIM
9. Q 6, 8, DA
10. (P & T) → Q 3–9, CP