@ValidandNotinUse,
Hey Valid!
Alright. I have 8 and 9 down with 2 steps. I'm not quite sure if you are required to do 3 steps, but I found solutions in 2. Problem number 6 is an issue in that I had to use a replacement rule to execute the proof, and it was a few more steps than what you are allowed to use. I'll try to hash it out, but I'll go over problem 8 and 9 and how to attack them.
So here is your first problem, complete with step-by step inferences to reach the conclusion with the strategy for attacking proof in general incorporated into it..
Of course, in box number 1 you have the whole problem put into the proof. Now the question now is "what do I do next?" This is usually how I approach problems like this. First, look for a single variable, which in this case is P in line 3. The reason you want to look at the single variables first is because the chances of finding a modus ponens or a modus tollens are drastically higher than finding a constructive dilemma or hypothetical syllogism. But also keep in mind that with single variables, you can add both both together with ADD or combine two variables together with the Conjunction rule. SO LOOK FOR THE SINGLE VARIABLE FIRST. Then look for any conditionals [i.e. P --> ~(Q&T)] or disjunctions with which you can utilize the single variable. In the case of our problem, the single variable P in line 3 matches the antecedent of the conditional in line 1. We know that we can infer ~(Q&T) from line 3 because the rule allows us that. We can now infer ~(Q&T) in line 4. SO LOOK FOR A CONDITIONAL OR DISJUNCTION THAT MATCHES YOUR SINGLE VARIABLE.
Now the best thing to do in any proof, is to completely break down every possible derivation possible at the start. You are not penalized for making dead end inferences if you do not need them. They just make your proof longer. But the main point is to build up your inference bank so that if your intuition for the strategy for your proof fails, you have a complete bank which you can use regardless if you succeed in your proof the first time round. SO LOOK FOR EVERY POSSIBLE DERIVATION AND PUT IT DOWN. So now you have every possible derivation, which in the case of this proof is ~(Q&T) in line 4. Now at this point, go back to the beginning of your strategy. Look for a conditional or a disjunction with which you could possibly use your new inferred line. We have one in our case, namely line 2 [i.e. S --> (Q&T)] Use the Modus Tollens rule to get your conclusion.
Because this is a short proof, there may not have been a big issue with seeing the conclusion clear in your proof strategy. But suppose there are more lines. What I always try to do if there are many possibilities is to deconstruct the proof from the conclusion backwards. By the time you get to predicate logic, this is the best way to travel. So in the case of our proof, we know we need to get ~S. Where are the possible places we can get ~S in our proof. The only place we could get it off hand is in line 2. Now think to yourself? what would it take to get ~S out of line 2. This is where you have to visualize the rules. You could use Modus tollens, but you need a negated (Q&T) to do it. I just so happens we have it in line 1. In order to get that by itself, we need a single P. Wouldn't you know it, we have a P. As soon as you see the obvious connection, you can attack the proof from the back forwards.
So follow these rules (which were not really in order for the proof we did, but how you should approach it in any proof you do)
STEP 1 - Look for the single variable
STEP 2 - Look for a conditional or a disjunction which are most likely to help start off
STEP 3 - Look for every derivation possible
STEP 4 - When in doubt, work backwards.
STEP 5 - Go back to step 1
You can apply everything again to problem number 9.
Let's approach this from the perspective of the 4 step process.
Step 1 - Look for the single variable. We do not have it.
Step 2 - Look for a conditional or a disjunction which are most likely to help us start off. We DO have that. We have two conditionals, but we also have a disjunction. It is usually the case that a proof is constructed by the teacher to set up a constructive dilemma. It is evident in this proof. A constructive dilemma needs two conditionals, a disjunction, and they need to fit the form where the antecedents of both conditionals match both disjuncts of the disjunction. We have this in lines 2, 3, and 4 and we can derive P v T. Go to step 3.
Step 3 - Look for every derivation possible. We need to reevaluate what we have done so far. Now at this point you can work from the back forwards if you are feeling stuck, or go onto step 4, which will lead you back to step 1. Re-apply step 1? do we have a single variable this time around? No. Re-apply step 2? do we have a useful conditional or disjunction? Yes. The new inference in line 5 will help us derive S (which we could have seen by working backwards as well). We can use Modus Ponens to derive S in line 6.
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