What does it mean when x is 'free'?In the context of first-order logic and resolution...





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:
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

Answer & Explanation

Bryson Carlson

Bryson Carlson


2022-11-26Added 10 answers

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 "free" to assume different values. On the other hand, if a variable is being quantified over, it is almost like a placeholder -- it doesn'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.



2022-11-27Added 1 answers

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)

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