What does it mean when x is 'free'?In the context of first-order logic and resolution (I'm trying to study Skolemization for a midterm tomorrow), I am seeing several references to x being free or not free. What does this mean?To pull all quantifiers in front of the formula and thus transform it into a prenex form, use the following equivalences, where x is not free in Q:Edit: ∀ x P ( x ) ∧ ∃ x Q ( x ) ≡ ∀ x ( P ( x ) ∧ ∃ x Q ( x ) ) ≡ ∀ x ( P ( x ) ∧ ∃ y Q ( y ) ) ≡ ∀ x ∃ y ( P ( x ) ∧ Q ( y ) ) .If the second x is bound by the existential quantifier - why can it be pulled into prenex form if the blurb above says that this can be done only if x is not free? Or am I misunderstanding something?Edit 2:OK, i'm clearly not all here at the moment. Ignore that question

Question

What does it mean when x is &#039;free&#039;?In the context of first-order logic and resolution (I&#039;m trying to study Skolemization for a midterm tomorrow), I am seeing several references to   x being free or not free. What does this mean?To pull all quantifiers in front of the formula and thus transform it into a prenex form, use the following equivalences, where   x is not free in Q:Edit:  ∀  x  P  (  x  )  ∧  ∃  x  Q  (  x  )  ≡  ∀  x  (  P  (  x  )  ∧  ∃  x  Q  (  x  )  )  ≡  ∀  x  (  P  (  x  )  ∧  ∃  y  Q  (  y  )  )  ≡  ∀  x  ∃  y  (  P  (  x  )  ∧  Q  (  y  )  )  .If the second x is bound by the existential quantifier - why can it be pulled into prenex form if the blurb above says that this can be done only if x is not free? Or am I misunderstanding something?Edit 2:OK, i&#039;m clearly not all here at the moment. Ignore that question

Bryson Carlson · Accepted Answer

A variable       x    1   in a formula   ϕ  (      x    1    ,      x    2    ,  ⋯  ,      x    n    ) is called bound if       x    1   is contained in some sub-formula of the form   ∀      x    1    (  ψ  (      x    1    ,  ⋯  )  ). If ia variable is not bound, it is free. That is, a variable is free if it is not being quantified over, so that it is &quot;free&quot; to assume different values. On the other hand, if a variable is being quantified over, it is almost like a placeholder -- it doesn&#039;t make sense to try and input different values for that variable.With regards to your second question, since   x is bound it is not free.

armlanna1sK · Accepted Answer

x is free in   Q if it is not quantified over.So in the       L          R      i      n      g       formula   x  =  1,   x is free.But in the       L          R      i      n      g       formula   ∀  x  (  x  =  1  ),   x is not free (in which case we call   x bound)