Propositional Logic Exercise(logical statement, statement simplifying)

 

 

 

Chapter : Propositional Logic

Topic : Logical statement, statement simplifying

 

 

 

Solution : b∨~d

1. 주어진 statement를 simplifying하는 문제이다.

2. (~a∧(a∨b))∨~d∨(b∧d)

≡ [(~a∧a)∨(~a∧b)]∨[(~d∨b)∧(~d∨d)] --- 분배법칙

≡ (~a∧b)∨(~d∨b) --- (~a∧a)=c / (~d∨d)=t

≡ ((~a∧b)∨b)∨~d --- 결합법칙으로 b랑 ~d의 자리를 바꿔준 것

≡ ((~a∧b)∨(b∧t))∨~d --- 여기가 좀 까다로운데 흡수법칙을 사용했다. (흡수법칙 : p∨(p∧q)≡p)

≡ (b∧(~a∨t))∨~d --- 분배법칙

≡ (b∧t)∨~d --- (~a∨t)=t

≡ b∨~d

 

 

 

1. Solution : ~a∨~b

1. 주어진 statement를 simplifying하는 문제이다.

2. (aXORb)∨~(a∨(b∧d))

≡ ((a∨b)∧~(a∧b))∨~((a∨b)∧(a∧d)) --- XOR의 개념 그대로 : aXORb = (a∨b)∧~(a∧b)

≡ ((a∨b)∧(~a∨~b))∨((~a∧~b)∨(~a∧~d)) --- 분배법칙

좌측 (a∨b)∧(~a∨~b) 부분만 먼저 풀면

≡ ((a∨b)∧~a)∨((a∨b)∧~b) --- 분배법칙

≡ (~a∧a)∨(~a∧b)∨(a∧~b)∨(b∧~b) --- 분배법칙

≡ (~a∧b)∨(a∧~b) --- (~a∧a)=c / (b∧~b)=c

다시 전체 식에 대입해주면

≡ (~a∧b)∨(a∧~b)∨(~a∧~b)∨(~a∧~d)

≡ (~a∧(b∨~b))∨(a∧~b)∨(~a∧~d) --- (~a∧b)∨(~a∧~b)를 묶어주었다

≡ ~a∨(a∧~b)∨(~a∧~d)  --- (b∨~b)=t

≡ ~a∨(a∧~b) --- 흡수법칙 ~a∨(~a∧~d)≡~a

≡ (~a∨a)∧(~a∨~b) --- 분배법칙

≡ ~a∨~b --- (~a∨a)=t

 

 

 

1. Solution : 

1. 주어진 statement를 simplifying하는 문제이다.

2. (~a∧~b∧~d)∨(~a∧b)∨(a∧b)∨(a∧~b∧~d)

≡ (~a∧~b∧~d)∨(a∧~b∧~d)∨(b∧(~a∨a)) --- 중간의 (~a∧b)∨(a∧b)에 분배법칙 적용

≡ (~a∧~b∧~d)∨(a∧~b∧~d)∨b --- (~a∨a)=t

≡ ((~b∧~d)∧(~a∨a))∨b --- 분배법칙

≡ (~b∧~d)∨b --- (~a∨a)=t

≡ (b∨~b)∧(b∨~d) --- 분배법칙

≡ b∨~d