Explain the methods of proving the theorems by direct, indirect, contradiction and by cases.

1. Direct Method:

  The implication p→q can be proved by showing that if p is true then q must also be true.

Eg: “If n is odd then  is odd”

   Proof:

       Assume that n is odd. Then n=2K+1 for some integer K. Then,

            = = +4K+1= (2( +2K)+1)= 2M+1 where M is an integer, Which is similar to (2K+1). Hence  is odd and proves the theorem, “If n is odd then  is odd”.

2.   Indirect Method:

To prove a theorem by indirect method, assume the conclusion is false and using true facts show the hypothesis is false. i.e. showing, ¬q→¬p , we can prove p→q.

Eg: “If 3n+2 is odd then n is odd”.

Proof:

Assuming n is even, n=2K, then,

3n+2= 3(2K)+2 =6K+2 =2(3K+1) =2M, where M is an integer,

Thus, 3n+2 is even which proves that, ” If 3n+2 is odd then n is odd” by indirect method.

3.  Contradiction Method:

To prove the implication of the form p→q, assume p ˄ ¬q is true. Then try to show that the assumption is false then its negation is true which leads to ¬p∨q to be true which is logically equivalent to p→q. Hence p→q is true.

Eg: “If  is even then a is an even number”

Proof: Let us suppose,

     p:  is even number.

     q: a is an even number.

Let us assume p ˄ ¬q is true i.e.  is even and a is odd. Then,

a= 2K+1 so, =  =2( +2K)+1 =2M+1 which is always odd, hence is odd, which contradicts our assumption. Hence, “If  is even then a is an even number”.

4.   By Cases:

To prove the implication of the form (     …   ) →q. The tautology [(     …   ) →q ↔ ( →q) ˄ ( →q) ˄ ……˄( →q)] as a value of inference. This shows that the original hypothesis made with disjunction of proposition   …,  can be proved by showing each implication ( →q).

-Show that the proposition p∨(q ˄ r) and (r∨q) ˄ (p∨r) are logically equivalent.

We can show p∨(q ˄ r) and (r∨q) ˄ (p∨r) are logically equivalent by showing

p∨(q ˄ r)↔(p∨q) ˄ (p∨r) is a tautology.

p	q	r	q ˄ r	p∨q	p∨r	p∨(q ˄ r)	(p∨q) ˄ (p∨r)	p∨(q ˄ r)↔(p∨q) ˄ (p∨r)
F	F	F	F	F	F	F	F	T
F	F	T	F	F	T	F	F	T
F	T	F	F	T	F	F	F	T
F	T	T	T	T	T	T	T	T
T	F	F	F	T	T	T	T	T
T	F	T	F	T	T	T	T	T
T	T	F	F	T	T	T	T	T
T	T	T	T	T	T	T	T	T