Let - - - be such that p - q -p - q - r is a tautology- Then p - q - r is logically equivalent to-

Case I If ∇ is same as ∧, then (p ∧ q) ⇒((p Δ q) ∧ r) is equivalent to ∼(p ∧ q) ∨ ((p Δ q) ∧ r) is equivalent to (∼(p ∧ q) ∨(p Δ q)) ∧(∼(p ∧ q) ∨ r which canno ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Tautology

Let Δ, ∇∈∧, ∨...

Question

Let $Δ, \nabla \in {\land, \lor}$ be such that $p \nabla q \Rightarrow ((p Δ q) \nabla r)$ is a tautology. Then $(p \nabla q) Δ r$ is logically equivalent to:

(p \land r) Δ q

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(p \nabla r) \land q

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(p Δ r) \lor q

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(p Δ r) \land q

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $(p Δ r) \lor q$
Case-I

If $\nabla$ is same as $\land$

Then $(p \land q) \Rightarrow ((p Δ q) \land r)$ is equivalent to $\sim (p \land q) \lor$ $((p Δ q) \land r)$ is equivalent to $(\sim (p \land q) \lor (p Δ q)) \land (\sim (p \land$ q) $\lor r$

Which cannot be a tautology

For both $Δ$ (i.e. $\lor or \land$ )

Case-II

If $\nabla$ is same as $\lor$

Then $(p \lor q) \Rightarrow ((p Δ q) \lor r)$ is equivalent to $\sim (p \lor q) \lor (p Δ q) \lor r$ which can be a tautology if $Δ$ is also same as $\lor$ .

Hence both $Δ$ and $\nabla$ are same as $v$ . Now $(p \nabla q) Δ r$ is equivalent to $(p \lor q \lor r)$ .

Suggest Corrections

Similar questions

Q. Let

Δ \in {\land, \lor, \Rightarrow, \Leftrightarrow}

be such that

(p \land q) Δ ((p \lor q) \Rightarrow q)

is a tautology. Then

Δ

is equal to :

Q. Let

r \in {p, q, \sim p, \sim q}

be such that the logical statement

r \lor (\sim p) \Rightarrow (p \land q) \lor r

is a tautology. Then

r

is equal to :

Q. The number of choices for

Δ \in {\land, \lor, \Rightarrow, \Leftrightarrow}

, such that

(p Δ q) \Rightarrow ((p Δ \sim q) \lor ((\sim p) Δ q))

is a tautology, is

Q. The statement

[(p \Rightarrow q) \land (q \Rightarrow r)] \Rightarrow (p \Rightarrow r)

Q. Let

*, □ \in {\land, \lor}

be such that Boolean expression

(p * \sim q) \Rightarrow (p □ q)

is a tautology. Then:

Join BYJU'S Learning Program

Let Δ,∇∈{∧,∨} be such that p∇q⇒((pΔq)∇r) is a tautology. Then (p∇q)Δr is logically equivalent to:

Let $Δ, \nabla \in {\land, \lor}$ be such that $p \nabla q \Rightarrow ((p Δ q) \nabla r)$ is a tautology. Then $(p \nabla q) Δ r$ is logically equivalent to: