Disc Math Logic statements (Homework check)

  • Thread starter Miike012
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  • #1
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My solution
d. [itex]\forallx[/itex][itex]\existsy[/itex](F(x)^S(y) → [itex]\negA(y,x)[/itex])
e. [itex]\existsx[/itex][itex]\forally[/itex](F(x)^S(y) → [itex]\negA(y,x)[/itex])
f.[itex]\existsx[/itex][itex]\forally[/itex](S(x)^F(y) → A(x,y))
 

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Answers and Replies

  • #2
pasmith
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My solution
d. [itex]\forall x[/itex] [itex]\exists y[/itex](F(x)^S(y) → [itex]\neg A(y,x)[/itex])
e. [itex]\exists x[/itex] [itex]\forall y[/itex](F(x)^S(y) → [itex]\neg A(y,x)[/itex])
f.[itex]\exists x[/itex] [itex]\forall y[/itex](S(x)^F(y) → A(x,y))

For d: You have "for every faculty member, there is a student who has not asked a question of that faculty member". That's not equivalent to "some student has not asked a question of any faculty member", because in the first it might not be the same student in each case. You need to swap the quantifiers.

The others appear to be correct.
 

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