- #1

plum356

- 10

- 5

- 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:

- Both ##c## and ##d## are in ##e##.
- 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!

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