What does x is free for substitution for y in $varphi$, but not for z means?
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I am studying about first order predicate logic, and I have some difficulties understanding substitution and free variables.
If x is free for substitution for y in $varphi$, but not for z, where $varphi$ is a single formula, what would this mean?
Also x,y and z are variables.
Can you give a formula $varphi$ that satisfies this condition?
logic predicate-logic
asked Nov 19 at 21:42
astro_chae
62
add a comment |
up vote
0
down vote
favorite
I am studying about first order predicate logic, and I have some difficulties understanding substitution and free variables.
If x is free for substitution for y in $varphi$, but not for z, where $varphi$ is a single formula, what would this mean?
Also x,y and z are variables.
Can you give a formula $varphi$ that satisfies this condition?
logic predicate-logic
asked Nov 19 at 21:42
astro_chae
62
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am studying about first order predicate logic, and I have some difficulties understanding substitution and free variables.
If x is free for substitution for y in $varphi$, but not for z, where $varphi$ is a single formula, what would this mean?
Also x,y and z are variables.
Can you give a formula $varphi$ that satisfies this condition?
logic predicate-logic
asked Nov 19 at 21:42
astro_chae
62
I am studying about first order predicate logic, and I have some difficulties understanding substitution and free variables.
If x is free for substitution for y in $varphi$, but not for z, where $varphi$ is a single formula, what would this mean?
Also x,y and z are variables.
Can you give a formula $varphi$ that satisfies this condition?
logic predicate-logic
logic predicate-logic
asked Nov 19 at 21:42
astro_chae
62
asked Nov 19 at 21:42
astro_chae
62
asked Nov 19 at 21:42
astro_chae
62
asked Nov 19 at 21:42
astro_chae
62
asked Nov 19 at 21:42
astro_chae
62
62
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Consider the open wff with two free variables
$Fy to forall x Gzx$
Then substituting $x$ for $y$ gives us
$Fx to forall x Gzx$
Again a wff with two free variables (check you understand this claim). The two expressions can in effect be thought of as available to express the same binary relation.
Substituting $x$ for $z$ in the first wff however gives us
$Fy to forall x Gxx$
a wff now with one free variable. What's happened is that, in the second case, what was a free variable gets captured by the quantifier, and so the quantificational structure of the wff is changed. We don't want merely changing the letter we use for a variable to have that sort of effect -- hence, as we say, '$x$' is here not free to be substituted for '$z$' (while preserving quantificational structure).
answered Nov 19 at 23:54
Peter Smith
40.3k339118
Thanks for the answer :) What I understood is if we substitute a variable (for instance y to x) and if it is not restricted by the quantifiers, then it is free. Additionally, I was wondering if there is any example of closed formula. Would there be an example?
– astro_chae
Nov 20 at 4:26
Think about $exists z(forall yFy to forall xGxz)$. Which of $y$ and $z$ could be systematically replaced by $x$, and why?
– Peter Smith
Nov 20 at 14:21
Thanks for the reply :) What I am understanding is, If I replace y by x, then y is free because it is not restricted by any quantifiers, but if we replace z by x then because of $exists$x every x in the bracket becomes restricted. I hope I understood correctly. Thank you :)
– astro_chae
Nov 20 at 22:44
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Consider the open wff with two free variables
$Fy to forall x Gzx$
Then substituting $x$ for $y$ gives us
$Fx to forall x Gzx$
Again a wff with two free variables (check you understand this claim). The two expressions can in effect be thought of as available to express the same binary relation.
Substituting $x$ for $z$ in the first wff however gives us
$Fy to forall x Gxx$
a wff now with one free variable. What's happened is that, in the second case, what was a free variable gets captured by the quantifier, and so the quantificational structure of the wff is changed. We don't want merely changing the letter we use for a variable to have that sort of effect -- hence, as we say, '$x$' is here not free to be substituted for '$z$' (while preserving quantificational structure).
answered Nov 19 at 23:54
Peter Smith
40.3k339118
Thanks for the answer :) What I understood is if we substitute a variable (for instance y to x) and if it is not restricted by the quantifiers, then it is free. Additionally, I was wondering if there is any example of closed formula. Would there be an example?
– astro_chae
Nov 20 at 4:26
Think about $exists z(forall yFy to forall xGxz)$. Which of $y$ and $z$ could be systematically replaced by $x$, and why?
– Peter Smith
Nov 20 at 14:21
Thanks for the reply :) What I am understanding is, If I replace y by x, then y is free because it is not restricted by any quantifiers, but if we replace z by x then because of $exists$x every x in the bracket becomes restricted. I hope I understood correctly. Thank you :)
– astro_chae
Nov 20 at 22:44
add a comment |
up vote
1
down vote
accepted
Consider the open wff with two free variables
$Fy to forall x Gzx$
Then substituting $x$ for $y$ gives us
$Fx to forall x Gzx$
Again a wff with two free variables (check you understand this claim). The two expressions can in effect be thought of as available to express the same binary relation.
Substituting $x$ for $z$ in the first wff however gives us
$Fy to forall x Gxx$
a wff now with one free variable. What's happened is that, in the second case, what was a free variable gets captured by the quantifier, and so the quantificational structure of the wff is changed. We don't want merely changing the letter we use for a variable to have that sort of effect -- hence, as we say, '$x$' is here not free to be substituted for '$z$' (while preserving quantificational structure).
answered Nov 19 at 23:54
Peter Smith
40.3k339118
Thanks for the answer :) What I understood is if we substitute a variable (for instance y to x) and if it is not restricted by the quantifiers, then it is free. Additionally, I was wondering if there is any example of closed formula. Would there be an example?
– astro_chae
Nov 20 at 4:26
Think about $exists z(forall yFy to forall xGxz)$. Which of $y$ and $z$ could be systematically replaced by $x$, and why?
– Peter Smith
Nov 20 at 14:21
Thanks for the reply :) What I am understanding is, If I replace y by x, then y is free because it is not restricted by any quantifiers, but if we replace z by x then because of $exists$x every x in the bracket becomes restricted. I hope I understood correctly. Thank you :)
– astro_chae
Nov 20 at 22:44
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Consider the open wff with two free variables
$Fy to forall x Gzx$
Then substituting $x$ for $y$ gives us
$Fx to forall x Gzx$
Again a wff with two free variables (check you understand this claim). The two expressions can in effect be thought of as available to express the same binary relation.
Substituting $x$ for $z$ in the first wff however gives us
$Fy to forall x Gxx$
a wff now with one free variable. What's happened is that, in the second case, what was a free variable gets captured by the quantifier, and so the quantificational structure of the wff is changed. We don't want merely changing the letter we use for a variable to have that sort of effect -- hence, as we say, '$x$' is here not free to be substituted for '$z$' (while preserving quantificational structure).
answered Nov 19 at 23:54
Peter Smith
40.3k339118
Consider the open wff with two free variables
$Fy to forall x Gzx$
Then substituting $x$ for $y$ gives us
$Fx to forall x Gzx$
Again a wff with two free variables (check you understand this claim). The two expressions can in effect be thought of as available to express the same binary relation.
Substituting $x$ for $z$ in the first wff however gives us
$Fy to forall x Gxx$
a wff now with one free variable. What's happened is that, in the second case, what was a free variable gets captured by the quantifier, and so the quantificational structure of the wff is changed. We don't want merely changing the letter we use for a variable to have that sort of effect -- hence, as we say, '$x$' is here not free to be substituted for '$z$' (while preserving quantificational structure).
answered Nov 19 at 23:54
Peter Smith
40.3k339118
edited Nov 20 at 0:22
answered Nov 19 at 23:54
Peter Smith
40.3k339118
answered Nov 19 at 23:54
Peter Smith
40.3k339118
answered Nov 19 at 23:54
Peter Smith
40.3k339118
40.3k339118
Thanks for the answer :) What I understood is if we substitute a variable (for instance y to x) and if it is not restricted by the quantifiers, then it is free. Additionally, I was wondering if there is any example of closed formula. Would there be an example?
– astro_chae
Nov 20 at 4:26
Think about $exists z(forall yFy to forall xGxz)$. Which of $y$ and $z$ could be systematically replaced by $x$, and why?
– Peter Smith
Nov 20 at 14:21
Thanks for the reply :) What I am understanding is, If I replace y by x, then y is free because it is not restricted by any quantifiers, but if we replace z by x then because of $exists$x every x in the bracket becomes restricted. I hope I understood correctly. Thank you :)
– astro_chae
Nov 20 at 22:44
add a comment |
Thanks for the answer :) What I understood is if we substitute a variable (for instance y to x) and if it is not restricted by the quantifiers, then it is free. Additionally, I was wondering if there is any example of closed formula. Would there be an example?
– astro_chae
Nov 20 at 4:26
Think about $exists z(forall yFy to forall xGxz)$. Which of $y$ and $z$ could be systematically replaced by $x$, and why?
– Peter Smith
Nov 20 at 14:21
Thanks for the reply :) What I am understanding is, If I replace y by x, then y is free because it is not restricted by any quantifiers, but if we replace z by x then because of $exists$x every x in the bracket becomes restricted. I hope I understood correctly. Thank you :)
– astro_chae
Nov 20 at 22:44
Thanks for the answer :) What I understood is if we substitute a variable (for instance y to x) and if it is not restricted by the quantifiers, then it is free. Additionally, I was wondering if there is any example of closed formula. Would there be an example?
– astro_chae
Nov 20 at 4:26
Thanks for the answer :) What I understood is if we substitute a variable (for instance y to x) and if it is not restricted by the quantifiers, then it is free. Additionally, I was wondering if there is any example of closed formula. Would there be an example?
– astro_chae
Nov 20 at 4:26
Think about $exists z(forall yFy to forall xGxz)$. Which of $y$ and $z$ could be systematically replaced by $x$, and why?
– Peter Smith
Nov 20 at 14:21
Think about $exists z(forall yFy to forall xGxz)$. Which of $y$ and $z$ could be systematically replaced by $x$, and why?
– Peter Smith
Nov 20 at 14:21
Thanks for the reply :) What I am understanding is, If I replace y by x, then y is free because it is not restricted by any quantifiers, but if we replace z by x then because of $exists$x every x in the bracket becomes restricted. I hope I understood correctly. Thank you :)
– astro_chae
Nov 20 at 22:44
Thanks for the reply :) What I am understanding is, If I replace y by x, then y is free because it is not restricted by any quantifiers, but if we replace z by x then because of $exists$x every x in the bracket becomes restricted. I hope I understood correctly. Thank you :)
– astro_chae
Nov 20 at 22:44
add a comment |
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