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24 votes

Which of the following is/are a tautology?

  1. $a \vee b \to b \wedge c$
  2. $a \wedge b \to b \vee c$
  3. $a \vee b \to \left(b \to c \right)$
  4. $a \to b \to \left(b \to c \right)$
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4 Answers

Best answer
29 votes
29 votes

Answer: (B)

$\left(a \wedge b \right) \to b \vee c$
$\implies \neg \left(a \wedge b \right) \vee b \vee c$
$\implies \neg a \vee \neg b \vee b \vee c$
$\implies T$

Option (A) is not TRUE when $c$ is FALSE.
Option (C) is not TRUE when $b$ is TRUE and  $c$ is FALSE.
Option (D) is not TRUE when $a$ and $b$ are TRUE and $c$ is FALSE.

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12 votes
12 votes

Option b is tautology.

0 votes
0 votes

Answer: (B).

Lets go over each of the options with case method where $b = True$.

A)     $a \vee b → b \wedge  c$

$=> (a \vee True) → (True \wedge  c)$

$=> (True) → (c)$

$=> c$

Option A will not be Tautology if $c=False$.

 

B)     $a \wedge  b → b \vee c$

$=> (a \wedge True) → (True \vee c)$

$=> (a) → True$

$=> True$ (Anything implies True is always True)

Option B is Tautology.

 

C)     $a \vee b → (b → c)$

$=> a \vee True → (True → c)$

$=> (True) → (c) $

$=> c $

Option C will not be Tautology if $c=False$.

 

C)     $a →  b → (b → c)$ 
$=> a →  (True → (True → c))$

$=> a →  (True → (c))$

$=> a →  c$

Option D will not be Tautology if $a=True$ and $c=False$

Answer:

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