Proof of $(x\subseteq y)\leftrightarrow(x\cap y =x)\leftrightarrow(x\cup y=y)$

Question

Exercise 1.2.1(vii) from Page 5 of Keith Devlin's "The Joy of Sets":

Prove the following assertion directly from the definitions. The drawing of "Venn diagrams" is forbidden; this is an exercise in the manipulation of logical formalisms. $$(x\subseteq y)\leftrightarrow(x\cap y =x)\leftrightarrow(x\cup y=y)$$

Attempt at solution:

This is not a difficult claim to understand or prove in terms of sets, but I can't figure out the pure logical formalism. I guess I should break this into seven individual implications?

1. $\forall w(w\in x\rightarrow w\in y)\rightarrow \forall z(z\in x\rightarrow (z\in x\enspace\wedge\enspace z\in y)) $
2. $\forall w(w\in x\rightarrow w\in y)\rightarrow \forall z((z\in x\enspace\wedge\enspace z\in y)\rightarrow z\in x))$
3. $\forall z(z\in x\rightarrow(z\in x \enspace\wedge\enspace z\in y))\rightarrow \forall w(w\in x\rightarrow w\in y)$
4. $\forall z(z\in x\rightarrow(z\in x \enspace\wedge\enspace z\in y))\rightarrow \forall w((w \in x \enspace\lor\enspace w\in y)\rightarrow w\in y)$
5. $\forall z(z\in x\rightarrow(z\in x \enspace\wedge\enspace z\in y))\rightarrow \forall w(w\in y\rightarrow(w \in x \enspace\lor\enspace w\in y))$
6. $\forall w((w\in x\enspace\lor\enspace w\in y)\rightarrow w\in y)\rightarrow\forall z((z\in x \enspace\wedge\enspace z\in y)\rightarrow z\in x)$
7. $\forall w((w\in x\enspace\lor\enspace w\in y)\rightarrow w\in y)\rightarrow\forall z(z\in x \rightarrow (z\in x \enspace\wedge\enspace z\in y))$

In addition to wondering if I'm doing this correctly, I also feel like I gained nothing from writing out these implications.

Answer 1

Exercise 1.2.1(vii) from page 5 of Keith Devlin's "The Joy of Sets":

Prove the following assertion directly from the definitions. this is an exercise in the manipulation of logical formalisms. $$x\subseteq y\;\leftrightarrow\;x\cap y =x\;\leftrightarrow\;x\cup y=y.\tag{*}$$

Even though this exercise is requesting "logical formalism" and sentence $(*)$ does not technically mean that those three atomic statements are equivalent to one another, the author quite certainly is wanting you to prove just that.

I understand from your linked post that it is not necessary to prove all seven implications; proving $(A→B),\; (B→C)$ and $(C→A)$ would be sufficient.

Yes. So, we just need

1. $∀p\;(p\in X→p\in Y)→∀q\;(q\in X\cap Y ↔ q\in X)$
2. $∀q\;(q\in X\cap Y ↔ q\in X)→∀r\;(r\in X\cup Y ↔ r\in Y)$
3. $∀r\;(r\in X\cup Y ↔ r\in Y)→∀p\;(p\in X→p\in Y)$

or, alternatively,

1. $∀p\;(p\in X→p\in Y)→∀q\;(q\in X\cap Y ↔ q\in X)$
2. $∀p\;(p\in X→p\in Y)→∀r\;(r\in X\cup Y ↔ r\in Y)$
3. $∀q\;(q\in X\cap Y ↔ q\in X)→∀p\;(p\in X→p\in Y)$
4. $∀r\;(r\in X\cup Y ↔ r\in Y)→∀p\;(p\in X→p\in Y)$

(the definition of subset has been applied).

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Addendum 1

I believe my #1 and #2 from the OP are equivalent to your #1. If you look at my #1 for example, I really don't know what more I can do to "prove" this.

Okay, let's reformat your #1, using $Ab$ to denote $b\in A:$ $$∀p\:(Xp → Yp) → ∀q\:(Xq → (Xq ∧ Yq)).\tag 4$$ This is logically equivalent to $$∀q\:∃p\:\Big((Xp → Yp) → (Xq → (Xq ∧ Yq))\Big),\tag 3$$ which is logically equivalent to $$∀q\:∃p\:\Big((Xp → Yp) → (Xq → Yq)\Big),\tag 2$$ which is a logical consequence (choose $p:=q$) of $$∀q\:\Big((Xq → Yq) → (Xq → Yq)\Big),\tag 1$$ which is clearly true. Notice that this proof is purely logical, utilising no result from Set Theory.

On the other hand, after the relevant results have been introduced after page 5, then a proof of the first part of $(*),$ $$X\subseteq Y\implies X\cap Y =X,$$ can simply (with the relevant justifications) be: $$X⊆Y \implies X∩Y⊆X\\ \text{and }X⊆Y \implies X⊆X∩Y.$$

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Addendum 2

Thanks. I don't understand why $(4)$ and $(3)$ are logically equivalent.

Paring them down (here, $Mn$ denotes that object $n$ holds property $M):$ \begin{gather}∀p\:Fp → ∀q\:Gq\tag 4 \\∀q\:∃p\:\big(Fp → Gq\big)\tag 3\end{gather}

Sentence $(3)$ is the Prenex form of sentence $(4).$ They are logically equivalent because (here, $\psi$ is a sentence) \begin{gather}\psi → ∀q\:Gq\quad\equiv\quad ∀q\;(\psi → Gq),\tag{L1}\\ ∀p\:Fp → \psi \quad\equiv\quad ∃q\;(Fp → \psi).\tag{L2} \end{gather}

$(\mathrm L1)$ and $(\mathrm L2)$ contains four logical entailments; their formal proofs can be found in exercises 3 & 6 on p. 114 & 116 of this book. I'll illustrate the least obvious one $$∀p\:Fp → \psi \quad\models\quad ∃q\;(Fp → \psi)$$ with an example:

• premise: “Dark matter exists if every perfumier is French.”
• conclusion: “For some perfumier, dark matter exists if they're French.”

If every perfumier is French, then the conclusion clearly follows from the premise. On the other hand, if some perfumier isn't French, then for that perfumier, “dark matter exists if they're French” is indeed (vacuously) true, and the conclusion is again true.

Answer 2

You could try doing these proofs using Proof Designer:

https://djvelleman.people.amherst.edu/pd.html

I think the result would be the kind of proof that Devlin has in mind.