Writing some ZF axioms with FOL symbols

[plum356]

Homework Statement

Write the following axioms in the language of FOL.

Relevant Equations

The symbols used in FOL.

The axioms:

My work:
Extension:$$\forall x\forall y,\,(x=y)\iff(\forall z,\,(z\in x\iff z\in y))$$
Empty Set:$$\exists x|\forall y,\,\neg(y\in x)$$
Pair Set:$$\forall c\forall d,\,\exists e|(c\in e)\wedge(d\in e)\wedge[\forall f,\,\neg((f=c)\vee(f=d))\implies\neg(f\in e)]$$
If you consider any two sets $c$ and $d$, there is a set (pair set) $e$ such that:

1. Both $c$ and $d$ are in $e$.
2. If you consider any set, then if it is not the case that it is $c$ or $d$, then it is not in $e$.

Union:$$\forall d,\,\exists e|\forall c,\,(c\in d)\implies(\forall f,\,(f\in c)\implies(f\in e))$$
$\forall c$ selects the sets in $d$, and $\forall f$ selects the sets in those $c$ and adds them to $e$ (the union).
Power Set:$$\forall d\,\exists e|\forall c,\,(c\in d)\implies(c\in d)$$
I am not very sure about this one. Specifically, I am not set on how I should express "subset" of $d$. However, since a subset is a set and $\forall c$ is selecting all of the sets, then it should also select the subsets in $d$.
Regularity:$$\forall d,\,\text{either }(\forall y\,\neg(y\in d)),\text{ or }\exists e|(e\in d)\wedge(\forall y,\,(y\in e)\implies\neg(y\in d))$$
I am a bit uneasy about this one. If I say that $e\in d$ is true, then how can it be the case that $(y\in e)\implies\neg(y\in d)$? If a set is in another, then should it be the case that its elements are also in its parent?
I used "either or" for $(x\wedge\neg y)\vee(\neg x\wedge y)$ (exclusive or).
Comprehension:$$\forall d,\,\exists e|\forall c,\,((c\in d)\wedge\varphi(c))\implies(c\in e)$$

If you could just give me hints or counterexamples instead of flat out correct my mistakes, it would be perfect. Thanks!

 

[Mark44]

The symbols used in FOL.

I guess this acronym expands to first order logic, but I'm not sure that every reader would be able to figure this out. You shouldn't assume that everyone understands somewhat obscure acronyms.

You posted a lot of fairly dense symbolism, not all of which I've taken the time to parse. Here are a couple of comments on two of your entries.

First, in my experience, sets are often written in capital letters, and elements of sets in lowercase letters. For example, we can denote an element x of a set named X as $x \in X$. Your use of lowercase letters for both sets and set elements makes the symbolism more difficult to parse, and your lack of distinction between sets and elements of sets makes your explanations harder to understand.
Second, use $\in$ for "element of" and $\subset$ or $\subseteq$ for set containment.

Union:$$\forall d,\,\exists e|\forall c,\,(c\in d)\implies(\forall f,\,(f\in c)\implies(f\in e))$$
$\forall c$ selects the sets in $d$, and $\forall f$ selects the sets in those $c$ and adds them to $e$ (the union).

I'll just tackle your explanation here.
$\forall c$ selects the sets elements of set D, and likewise for elements f of set C.
Given two sets A and B, if a is an element of A, and b is an element of B, then a is in element of C and b is an element of C. I.e., $a \in A \cup B \land b \in A \cup B$
In symbols: $\exists C | a \in A \land b \in B \implies a \in C \land b \in C$

Power Set:$$\forall d\,\exists e|\forall c,\,(c\in d)\implies(c\in d)$$
I am not very sure about this one. Specifically, I am not set on how I should express "subset" of $d$. However, since a subset is a set and $\forall c$ is selecting all of the sets, then it should also select the subsets in $d$.

The notation often used for the power set of a set A is P(A). So, given a set A, if $B \subseteq A$, then $B \subset P(A)$. Alternatively, you could talk about elements of a given subset of A (there's always a subset), and that element will also be an element of P(A).

 

[plum356]

I guess this acronym expands to first order logic, but I'm not sure that every reader would be able to figure this out. You shouldn't assume that everyone understands somewhat obscure acronyms.

You posted a lot of fairly dense symbolism, not all of which I've taken the time to parse. Here are a couple of comments on two of your entries.

First, in my experience, sets are often written in capital letters, and elements of sets in lowercase letters. For example, we can denote an element x of a set named X as $x \in X$. Your use of lowercase letters for both sets and set elements makes the symbolism more difficult to parse, and your lack of distinction between sets and elements of sets makes your explanations harder to understand.
Second, use $\in$ for "element of" and $\subset$ or $\subseteq$ for set containment.

I'll just tackle your explanation here.
$\forall c$ selects the sets elements of set D, and likewise for elements f of set C.
Given two sets A and B, if a is an element of A, and b is an element of B, then a is in element of C and b is an element of C. I.e., $a \in A \cup B \land b \in A \cup B$
In symbols: $\exists C | a \in A \land b \in B \implies a \in C \land b \in C$

The notation often used for the power set of a set A is P(A). So, given a set A, if $B \subseteq A$, then $B \subset P(A)$. Alternatively, you could talk about elements of a given subset of A (there's always a subset), and that element will also be an element of P(A).

Yes, it's first order logic. The "edit" button disappeared, though, so I can't even modify the post.
For the symbols $\subseteq$ and "proper subset", I can't utilise them yet, because the exercise wants me to write the axioms using only FOL's language + $\in$.

I think that you're right for the $f$s, but the axiom says that $d$ is a set of sets, so it seems like I got lucky with $c$. I get your point; $a\in b$ is short for "$a$ is an element of $b$", so my wording is incorrect.
Ok, I understand your formula with two sets, but I am supposed to iterate over all of the sets in $D$. I can surely do this in pairs, but isn't that a bit extra? I think that $\forall C,\,(C\in D)\implies(C\subseteq E)$ does capture the idea of the union of all $C$s in $D$, since I am putting all of their elements in $E$.
And now I think that it should have the form $\forall D,\,\varphi(D)\implies(\text{what I wrote before})$, where $\varphi(D)$ expresses the fact that $D$ is a "set of sets".

Right, I can correct this. For all sets $D$, there exists a set $P(D)$ such that for all sets $C$, if $C\subseteq D$, then $C\in P(D)$.
$$\forall D,\,\exists P(D)|\forall C,\,\left[\forall e,\,(e\in C)\implies(e\in D)\right]\implies(C\in P(D))$$

Thank you!