1. 2. 1. G > T 2. (T v S) > K / G > K 3. 1. (G v H) > (S. T) 2. (T v U) > (C. D) / G > C 4. 1. A > ~(A v E) / A > F 5. 1. H > (I > N) 2. (H > ~I) > (M v N) 3. ~N / M Walk through these using CP. First: for each one, are you clear at a glance what the ACP line is? For #2, it’s G; For #3, it’s G; For #4, it’s A; #5 doesn’t look like a good candidate for CP!

2. 2. 1. G > T 2. (T v S) > K / G > K So, it is clear the indented ACP line is G; and that the work will be finished when K shows up. 3. G ACP Usually, as soon as that ACP is introduced, something can be done with a Rule of Inference 4. T 3,4 MP That possibility had probably not even occurred to you beforehand. Now, since K, the object, is a consequent of 2, we need MP for 2: we need T v S; this is now available since T showed up. It takes some getting used to, but AD lets you add whatever you want, using a wedge. 5.T v S AD 4 6.K MP 2,5 There’s K, so the work is done. Time to clean up: 7. G > K CP 3-6

3. 3. 1. (G v H) > (S. T) 2. (T v U) > (C. D) / G > C A glance at this will tell us to assume G, and work to C. Next: ask yourself where C is going to come from. From SM done to what is now the consequent of 2. So the antecedent of 2 will be needed. Where will T v U come from? One possibility is AD; if you get T by itself. T would come out, by SM, after doing MP to 1 The needed line is G v H. How to get it? G is already and waiting, thanks to the ACP. 3. G ACP 4. G v H AD 3 5. S. T MP 1,4 6. T CM, SM 5 7. T v U AD 6 8. C. D MP 2,7 9. C SM 8 And now it’s done (have to remember where we’re going, so we stop! 10. G > C CP 3-9

4. 4. 1. A > ~(A v E) / A > F A will be the ACP line. But where is F going to come from? It’s not even in the premises! That means it has to be ADDed. 2. A ACP Notice that as soon as the assumption is written, it creates an option to do something, so do it! 3. ~ (A v E) Whenever you see a negative on a parenthesis, see if DM will work, and do it! 4. ~A. ~E Since line 4 ( SM’d) and line 2 contradict each other, you can use either of them to write a DS and make any ADDed statement come out on a line alone. Like F. 5.A v F AD 2 6.~A SM 4 7. F DS 5,6 That’s what was needed, so it’s done. 8. A > F CP 2-7

5. 5. 1. H > (I > N) 2. (H > ~I) > (M v N) 3. ~N / M Not a candidate for CP? Well,CP can be used to get a conditional statement –it need not be the conclusion every time. Think from the M. Where will it come from?DS The missing piece for that is already on line 3; the challenge is how to get the consequent of 2 out on a line. MP of course; and notice that the antecedent of 2, the needed piece, is a conditional statement. CP can be used to get it. What is the assumption?H When will the CP be done?When ~I is reached Get to it! 4. H ACP 5. I > N MP 1,4 6. ~I MT 3,5 Do you see it? The CP is already completed. 7. H > ~I CP 4-6 8. M v N MP 2,7 9. N v M CM 8 10. M DS 3, 9