Test: Mathematical Logic- 1 - Electrical Engineering (EE) MCQ

(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.

Now since ¬ p → q is given true, we reduce the truth table as follows:

In the reduced truth table we need to find the truth value of ¬ pv(p→q} = p' +(p → q)
≡ p' + p' + q ≡ p' + q
The truth value of p' + q in the reduced truth table is given below:

Since in the reduced truth table also, the given expression is sometimes true and sometimes false, therefore the truth value of proposition ¬p v (p → q) can not be determined.

a ⇔ (b v ¬ b ) 
a ⇔ 4 True
So a is true, i.e. a = 1
b ⇔ c holds. So b = c
Now the given expression is
(a ∧ b) → ((a ∧ c) v d) ≡ (a - b ) → ((a . c) + d)
Putting a = 1 in above expression we get 
1 • b → ((1 • c) + d) 
≡ b→ c + d
≡ b' + c+ d
Now putting b = c in above expression we get 
≡ c' + c+ d=1 + d=1
So the expression is always true.

(P → (Q v R)) → ((P ∧ Q) → R)
≡ (P→ Q+ R) → (PQ → R)
≡ [P' + Q + R] → [(PQ)' + R]
≡ [P' + Q + R] → [P' + Q' + R]
≡ (P' + Q + R)' + P' + Q' + R 
≡ PQ' R' + P' + Q' + R
≡ Q' + Q' PR' + P' + R
≡ Q' + P' + R (by absorption law)
Which is a contingency (i.e. satisfiable but not valid).

By using min terms we can define 



∴ Only, choice (d) = A ∧ B
[Note: This problem can also be done by constructing truth table for each choice and - comparing with truth table for A ∧ B]

In conjunction normal form, for any particular assignment of truth values, all except one clause will always evaluate to true. So, the proportion of clauses which evaluate to true to the total number of clauses is equal to 
Now putting 
All of these proportions are ≥ 1/2 and so choice (a) atleast half of the clauses evaluate to true, is. the correct answer.

The given table can be converted into boolean function by adding mrnterms corresponding to true rows.

Since there is only one false in the above truth table, we can represent the function P Q more efficiently, in conjunctive normal form.
Translates P Q = P + Q' (the max -term corresponding to the third row, where the function is false).
Now, we can easily translate the choices into boolean algebra as follows:

As we can clearly see only choice (b)  is equivalent to P+ Q.

Clearly, choices I and IV are equivalent.

Clearly II and III are not equivalent to each other or to I and IV.

In other words it says ``Some rainy days are not cold" 

Given statement is
¬∀d[R(d)→C(d)]

≡¬∀d[¬R(d)∨C(d)
≡∃d[R(d)∧¬C(d)]