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Let

  • $\text{M}(x)$ denote the predicate “$x$ is a mobile”;
  • $\text{B}(x)$ denote the predicate “$x$ is black”;
  • $\text{C}(x)$ denote the predicate “$x$ has calculator”.

Suppose that the universe is set of all mobiles.

Which of the following pair of statements and their equivalent expressions in first order logic, is/are correct?

  1. Only mobiles are black $: \forall x (\text{M}(x) \rightarrow \text{B}(x) )$
  2. Only mobiles are black $: \forall x (\text{M}(x) \vee \text{B}(x) )$
  3. All and Only black mobiles have calculator $: \forall x ( (\text{C}(x) \wedge \text{M}(x) ) \leftrightarrow \text{B}(x) )$
  4. All mobiles have calculator $: \forall x ( \text{M}(x) \wedge \text{C}(x) )$
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Universe is set of all mobiles. Here, M(x) is true for all x.


Option A : Only mobiles are black ≡ If x is not mobile then x is not black ≡ If x is black then x is mobile ≡ True.

∀x ( M(x) → X(x) ) ≡ ∀x ( T → B(x) ) ≡ ∀x ( B(x) ).

Statement doesn’t match expression.Thus, not correct.


Option B : Only mobiles are black ≡ If x is not mobile then x is not black  ≡ If x is black then x is mobile ≡ True.

∀x ( M(x) v B(x) ) ≡ ∀x ( T v B(x) ) ≡ True.

Statement matches expression.Thus, correct pair.


Option C : All and Only black mobiles have calculator ≡ All blacks have calculator and only blacks have calculator ≡ If x is black then x have calculator and if x have calculator then x is black.

∀x ((C(x) ^ M(x)) ↔ B(x)) ≡ ∀x (( C(x) T) ↔ B(x)) ≡ ∀x (C(x) ↔ B(x))

≡ ∀x ( (B(x) → C(x)) ^ (C(x) → B(x) ).

Statement matches expression.Thus, correct pair.


Option D : All mobiles have calculator ≡ All x have calculator.

∀x ( M(x) ^ C(x) ) ≡ ∀x ( T ^ C(x) ) ≡ ∀x ( C(x) ).

Statement matches expression.Thus, correct pair.


Answer:- B, C, D.

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