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**S(x):****x is a Student**

*P(x):***x is a Professor**

**A(x, y):****x has asked a question to y
Domain not given, so we have to think about default domain**

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(A) $\exists x \left( S(x) \land \forall y (P(y) \implies A(x,y) )\right)$

$\equiv \exists x \forall y \left( S(x) \land ( P(y) \implies A(x,y)) \right)$

Similarly,

(B) $\exists y \forall x \left( P(y) \land ( S(x) \implies A(y,x)) \right)$

(C) $\exists y \forall x \left( P(y) \land ( S(x) \implies A(x,y)) \right)$

(D) $\exists x \forall y \left( S(x) \land ( P(y) \implies A(y,x)) \right)$

$\equiv \exists x \forall y \left( S(x) \land ( P(y) \implies A(x,y)) \right)$

Similarly,

(B) $\exists y \forall x \left( P(y) \land ( S(x) \implies A(y,x)) \right)$

(C) $\exists y \forall x \left( P(y) \land ( S(x) \implies A(x,y)) \right)$

(D) $\exists x \forall y \left( S(x) \land ( P(y) \implies A(y,x)) \right)$

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$S(x): x$ is a Student

$P(x): x$ is a Professor

$A(x, y): x$ has asked a question to $y$

Q1. “There is a student who has asked every professor a question”

$=\exists x(S(x) \land \forall y(P(y) \rightarrow A(x,y)))$

Q2. “There is a professor who has asked every student a question”

$=\exists x(P(x) \land \forall y(S(y) \rightarrow A(x,y)))$

Q3. “There is a professor who has been asked a question by every student”

$=\exists x(P(x) \land \forall y(S(y) \rightarrow A(y,x)))$

Q4. “There is a student who has been asked a question by every professor”

$=\exists x(S(x) \land \forall y(P(y) \rightarrow A(y,x)))$

$P(x): x$ is a Professor

$A(x, y): x$ has asked a question to $y$

Q1. “There is a student who has asked every professor a question”

$=\exists x(S(x) \land \forall y(P(y) \rightarrow A(x,y)))$

Q2. “There is a professor who has asked every student a question”

$=\exists x(P(x) \land \forall y(S(y) \rightarrow A(x,y)))$

Q3. “There is a professor who has been asked a question by every student”

$=\exists x(P(x) \land \forall y(S(y) \rightarrow A(y,x)))$

Q4. “There is a student who has been asked a question by every professor”

$=\exists x(S(x) \land \forall y(P(y) \rightarrow A(y,x)))$