Creating a single formula $phi$ in the pure language of equality.
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I am trying to build a formula $phi$ in the pure language of equality with variables $x,y,z$ such that 1. $x$ is free for substitution for $y$ but not for $z$, 2. $y$ is not free for substitution for $x$ or $z$, and 3. $z$ is free for substitution for $x$ but not for $y$.
Here is the formula I started building:
$phi = (forall x(xdot{=}y)) wedge (forall x (xdot{=}z)) wedge (forall z (ydot{=}z))$
Since $y$ is unbound everywhere in $phi$, $x$ is free for $y$. Since $z$ is bound in the third conjunct, $x$ is not free for $z$. Then $phi$ satisfies Condition 1.
Both $x$ and $z$ are bound in $phi$, and so $y$ is not free for either $x$ or $z$. Then $phi$ satisfies Condition 2.
However, I am stuck on satifying Condition 3. Nothing works, so I assume the formula I wrote is wrong. But it seems like one condition is never satisfied no matter what I choose.
Here are some definitions I can use, I have been trying to use (3):
logic predicate-logic
asked Nov 14 at 21:47
numericalorange
1,598311
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I am trying to build a formula $phi$ in the pure language of equality with variables $x,y,z$ such that 1. $x$ is free for substitution for $y$ but not for $z$, 2. $y$ is not free for substitution for $x$ or $z$, and 3. $z$ is free for substitution for $x$ but not for $y$.
Here is the formula I started building:
$phi = (forall x(xdot{=}y)) wedge (forall x (xdot{=}z)) wedge (forall z (ydot{=}z))$
Since $y$ is unbound everywhere in $phi$, $x$ is free for $y$. Since $z$ is bound in the third conjunct, $x$ is not free for $z$. Then $phi$ satisfies Condition 1.
Both $x$ and $z$ are bound in $phi$, and so $y$ is not free for either $x$ or $z$. Then $phi$ satisfies Condition 2.
However, I am stuck on satifying Condition 3. Nothing works, so I assume the formula I wrote is wrong. But it seems like one condition is never satisfied no matter what I choose.
Here are some definitions I can use, I have been trying to use (3):
logic predicate-logic
asked Nov 14 at 21:47
numericalorange
1,598311
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to build a formula $phi$ in the pure language of equality with variables $x,y,z$ such that 1. $x$ is free for substitution for $y$ but not for $z$, 2. $y$ is not free for substitution for $x$ or $z$, and 3. $z$ is free for substitution for $x$ but not for $y$.
Here is the formula I started building:
$phi = (forall x(xdot{=}y)) wedge (forall x (xdot{=}z)) wedge (forall z (ydot{=}z))$
Since $y$ is unbound everywhere in $phi$, $x$ is free for $y$. Since $z$ is bound in the third conjunct, $x$ is not free for $z$. Then $phi$ satisfies Condition 1.
Both $x$ and $z$ are bound in $phi$, and so $y$ is not free for either $x$ or $z$. Then $phi$ satisfies Condition 2.
However, I am stuck on satifying Condition 3. Nothing works, so I assume the formula I wrote is wrong. But it seems like one condition is never satisfied no matter what I choose.
Here are some definitions I can use, I have been trying to use (3):
logic predicate-logic
asked Nov 14 at 21:47
numericalorange
1,598311
I am trying to build a formula $phi$ in the pure language of equality with variables $x,y,z$ such that 1. $x$ is free for substitution for $y$ but not for $z$, 2. $y$ is not free for substitution for $x$ or $z$, and 3. $z$ is free for substitution for $x$ but not for $y$.
Here is the formula I started building:
$phi = (forall x(xdot{=}y)) wedge (forall x (xdot{=}z)) wedge (forall z (ydot{=}z))$
Since $y$ is unbound everywhere in $phi$, $x$ is free for $y$. Since $z$ is bound in the third conjunct, $x$ is not free for $z$. Then $phi$ satisfies Condition 1.
Both $x$ and $z$ are bound in $phi$, and so $y$ is not free for either $x$ or $z$. Then $phi$ satisfies Condition 2.
However, I am stuck on satifying Condition 3. Nothing works, so I assume the formula I wrote is wrong. But it seems like one condition is never satisfied no matter what I choose.
Here are some definitions I can use, I have been trying to use (3):
logic predicate-logic
logic predicate-logic
asked Nov 14 at 21:47
numericalorange
1,598311
asked Nov 14 at 21:47
numericalorange
1,598311
asked Nov 14 at 21:47
numericalorange
1,598311
asked Nov 14 at 21:47
numericalorange
1,598311
asked Nov 14 at 21:47
numericalorange
1,598311
1,598311
add a comment |
add a comment |
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