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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

Q1) Translate “There is a student who has asked every professor a question”
Q2) Translate “There is a professor who has asked every student a question”
Q3) Translate “There is a professor who has been asked a question by every student”
Q4) Translate “There is a student who has been asked a question by every professor”

in Mathematical Logic edited by
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It is my self-doubt.

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3 Answers

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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) $\exists x \forall y \left( S(x) \land ( P(y) \implies A(y,x)) \right)$
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  1. ∃x∀y : S(x) ^ P(y) ^ A(x,y)
  2. ∃x∀y : P(x) ^ S(y) ^ A(x,y)
  3. ∀x∃y : S(x) ^ P(y) ^ A(x,y)
  4. ∀x∃y : P(x) ^ S(y) ^ A(x,y)

 

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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”

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

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