6.6 Exercises

A. Exercises: Conditional Proofs. Prove the following arguments are valid. These will require conditional proofs.

1. Premises: P, Q. Conclusion: (P→Q).
   
2. Premises: (P→Q), (Q→R). Conclusion: (P→R)
   
3. Premise: (P→Q), (S→R). Conclusion: ((~Q & ~R) → (~P & ~S).
   
4. Premise: (P→Q). Conclusion: ((P & R) →Q)
   
5. Premise: ((R&Q) → S), (~P→(R&Q)). Conclusion: (~S→P)
   
6. Premise: (P→~Q). Conclusion: (Q→~P)
   
7. Premises: (P→Q), (P→R). Conclusion: (P→(Q&R))
   
8. Premises: (P→(Q→R)), Q. Conclusion: (P→R)
   

B. Exercises: Theorems. Prove the following theorems.

1. (P→P)
   
2. ((P→Q)→((R→P)→(R→Q)))
   
3. ((P→(Q→R))→((P→Q)→(P→R))
   
4. ((~P→Q) → (~Q→P))
   
5. (((P→Q) & (P→R)) → (P→(Q&R)))
   

C. Exercises: Tautologies and Theorems. Make a truth table for each of the following complex sentences. Identify which are tautologies. Prove the tautologies.

1. ((P→Q)→Q)
   

   P	Q	((P	→	Q)	→	Q)
   T	T	T	T	T	T	T
   T	F	T	F	F	T	F
   F	T	F	T	T	T	T
   F	F	F	T	F	F	F

2. (P→(P→Q))
   

   P	Q	((P	→	Q)	→	P)
   T	T	T	T	T	T	T
   T	F	T	F	F	T	T
   F	T	F	T	T	F	F
   F	F	F	T	F	F	F

3. (P→(Q→P))
   

   P	Q	((P	→	(Q	→	P)
   T	T	T	T	T	T	T
   T	F	T	T	F	T	T
   F	T	F	T	T	T	F
   F	F	F	T	F	F	F

   Look at the main connective, conditional: on every row it is true, so it is a tautology. ((P → (Q → P) is, therefore, also a theorem, and provable from no premises.

   

4. (P→~P)
   

   P	(P	→	~	P)
   T	T	F	F	T
   F	F	T	T	F

5. (P→~~P)
   

   P	(P	→	~	~	P)
   T	T	T	T	F	T
   F	F	T	F	T	F

   Look at the main connective, conditional: on every row it is true, so it is a tautology. (P → ~~P) is, therefore, also a theorem, and provable from no premises.