Question

Which of the following is/are tautology?A. a v b → b ^ cB. a ^ b → b v cC. a v b → (b → c)D. None of these​explain please

Answer 1

Step-by-step explanation:

Which of the following is/are tautology:

a)(a v b ) →(b ∧ c)

b)(a ∧ b) → (b v c)

c)(a v b) → (b → c)

d)(a → b) → (b → c)

Correct answer is option 'B'. Can you explain this answer?

Related Test: Test: Mathematical Logic- 1

GATE Question

(a) (a v b)→(b ∧ c)

= (a + b)'+ bc

= a' b' + bc

Therefore, ((a v b) → (b ∧ c)) is contingency and not tautology.

(b) (a ∧ b) →(b v c)

= ab → b + c

= (ab)' + b + c

= a' + b' + b + c

= a' + 1 + c

= 1

So ((a ∧ b) →(b v c)) is tautology,

(c) (a v b)→ (b → c)

= (a + b) → (b' + c)

= (a+ b)' + b' + c

= a' b' + b' + c

= b' + c

So ((a v b) → (b → c)) is contingency but not tautology.

(d) (a → b) → (b→ c)

= (a' + b) → (b' + c)

= (a' + b) + b' + c

= ab' + b' + c

= b' + c

Therefore; ((a → b) → (b → c)} is contingency but not tautology.

Answer 2

Answer:

Option (B) is correct answer.

Step-by-step explanation:

In mathematics, a tautology is a compound assertion that always has Truth value. No matter what each component's particular makeup is, a tautology's conclusion is always true.

Calculation for option (A):

$(a\vee b)\rightarrow(b \wedge c)=(a+b)'+bc\\(a\vee b)\rightarrow(b \wedge c)=a'b'+bc$

Therefore, $(a\vee b)\rightarrow(b \wedge c)$  is contingency and not tautology.

Calculation for option (B):

$(a\wedge b)\rightarrow(b \vee c)=ab\rightarrow b+c\\(a\wedge b)\rightarrow(b \vee c)=(ab)'+b+c\\(a\wedge b)\rightarrow(b \vee c)=a'+b'+b+c\\(a\wedge b)\rightarrow(b \vee c)=a'+1+c (\because b'+b=1)\\(a\wedge b)\rightarrow(b \vee c)=1$

Therefore, $(a\wedge b)\rightarrow(b \vee c)$ is tautology.

Calculation for option (C):

$(a \wedge b)\rightarrow (b\rightarrow c)=(a+b)\rightarrow (b'+c) \\(a \wedge b)\rightarrow (b\rightarrow c)=(a+b)'+b'+c\\(a \wedge b)\rightarrow (b\rightarrow c)=a'b'+b'+c\\(a \wedge b)\rightarrow (b\rightarrow c)=b'+c$

Therefore, $(a \wedge b)\rightarrow (b\rightarrow c)$  is contingency but not tautology.

Calculation for option (D):

$(a \rightarrow b) \rightarrow (b\rightarrow c)= (a'+ b) \rightarrow (b'+ c)\\(a \rightarrow b) \rightarrow (b\rightarrow c)= (a' + b) + b' + c\\(a \rightarrow b) \rightarrow (b\rightarrow c)= ab' + b' + c\\(a \rightarrow b) \rightarrow (b\rightarrow c)= b' + c$

Therefore, $(a \rightarrow b) \rightarrow (b\rightarrow c)$ is contingency but not tautology.

Thus, option (A) and option (B) is correct.

From the above calculation the required topology is $(a \wedge b) \rightarrow (b\vee c)$

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