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Let $a, b, c, d$ be propositions. Assume that the equivalence $a ⇔ ( b \vee \neg b)$ and $b ⇔c$ hold. Then the truth-value of the formula $(a ∧ b) → (a ∧ c) ∨ d$ is always

  1. True
  2. False
  3. Same as the truth-value of $b$
  4. Same as the truth-value of $d$
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Equivalence holds given. This means, that for the expression if we choose LHS to be true then RHS must be true and vice versa, or if we choose LHS to be false then RHS must be false and vice versa.

$a \leftrightarrow (b \lor \neg b)$  is equivalent to or can be written as $a \leftrightarrow True$.
From this we can infer that $a$ is always true. Because if $a$ is false then $a \leftrightarrow T$ will become false and thus our assumption that $a$ is equivanlent to $(b \lor \neg b)$ will become false.

Similarly, $b \leftrightarrow c$  tell us that the truth value of $b$ and $c$ will be same, i.e if $b$ is $True$ then $c$ is $True$ and if $b$ is false then $c$ is false.

Now, $(a \land b) \rightarrow (a \land c) \lor d$

          $\equiv (T \land b) \rightarrow (T \land c) \lor d$

          $\equiv b \rightarrow c \lor d$

Case 1: If we take $b=False$ , then the above expression will be true as false impiles anything will be true.

Case 2: If we take $b=True$ then $c$ has to be true because of the equivalence given. So we have,

            $T \rightarrow (T \lor d)$

           $\equiv T \rightarrow T$

           $\equiv T$

Hence the given expression is always true.
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Here it is given that 

a ↔(b v ¬b) ≡ a ↔ True

so value of a will be true

Now b ↔ c means when b is true c will be true true or when b is false c will be false

case 1: Let's take b as TRUE

so, c will be TRUE

so, for expression [(a∧b)→(a∧c)∨d ]

a is true and b  is true so a∧b true and  a∧c true so both LHS and RHS are true hence the expression will be true

when b is false c is also false  as LHS is false the expression will always be true 

so answer is (A) true

Answer:

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