1

I want to formally prove that $X\subseteq A\cap B\Leftrightarrow X\subseteq A\wedge X\subseteq B$. This means $\forall x(x\in X\Rightarrow x\in A\wedge x\in B)$. By the use of logic ($P\Rightarrow (Q\wedge R)\equiv (P\Rightarrow Q)\wedge(P\Rightarrow R)$) we get $$ \forall x((x\in X\Rightarrow x\in A)\wedge (x\in X\Rightarrow x\in B)). $$ How can we conclude that this implies $\forall x(x\in X\Rightarrow x\in A)\wedge \forall x(x\in X\Rightarrow x\in B)$ (which is $X\subseteq A\wedge X\subseteq B$)?

hmakholm left over Monica
  • 286,031
Ivan
  • 53
  • Use that the relation $\subset$ is transitive, i.e.: $A \subset B \subset C$ implies $A \subset C$. – sqtrat Mar 18 '16 at 12:54
  • 1
    You're not revealing which formal system you're conducting your proof in, but any self-respecting proof system for first-order logic ought to allow you to prove $\forall x(\varphi\land\psi) \to (\forall x.\varphi)\land(\forall x.\psi)$ fairly straightforwardly. The details depend on information you haven't told us, though. – hmakholm left over Monica Mar 18 '16 at 12:55
  • @HenningMakholm what about union (ie disjunction?) – Ivan Mar 18 '16 at 12:57
  • @Ivan: What about that? – hmakholm left over Monica Mar 18 '16 at 12:57
  • consider an element $x\in (A\cap B)$.

    by definition of the intersection, we have $(x\in A) \land (x\in B)$.

    now consider the collection of all $x$ which fit this criteria

     – homedoggieo Mar 18 '16 at 12:57
  • @HenningMakholm If $X\subseteq A\cup B$, it doesnt mean that $X\subseteq A\vee X\subseteq B$, but if I use similar argument, I would get same thing as above, right? – Ivan Mar 18 '16 at 12:59
  • @Ivan: A self-respecting proof system shouldn't allow you to prove $\forall x(\varphi\lor\psi)\to(\forall x.\varphi)\lor(\forall x.\psi)$, because that is not valid in general! – hmakholm left over Monica Mar 18 '16 at 13:04
  • @HenningMakholm Well, are there some resources about this? At the moment, at university, we just started talking about ZF set theory (not much about logic behind it though) – Ivan Mar 18 '16 at 13:06
  • 1
    @Ivan: Are you sure you're actually doing formal proofs? It doesn't really sound like it, based on your comments and your apparent inability to say which formal system your proofs are supposed to be done with. – hmakholm left over Monica Mar 18 '16 at 13:12
  • @HenningMakholm No, im not sure actually... – Ivan Mar 18 '16 at 13:14

1 Answers1

2

$$\forall x (F \wedge G) \vdash \dashv (\forall x (F))\wedge(\forall x(G))$$ So you already have what you need!

Just keep in mind that thiw is not the case with $\vee$ since $$\forall x (F \vee G) \not \vdash (\forall x (F))\vee(\forall x(G))$$

christina_g
  • 150
  • 1
  • 2
  • 13
  • 1
    @Ivan: If you're not familiar with formal systems, just use common sense first. The first statement in this answer states that conjunction under a universal quantifier is equivalent to conjunction of the statements separately universally quantified. Why? Because if every object satisfies two properties $P,Q$, then surely every object satisfies $P$ and every object satisfies $Q$, and vice versa. This is obviously not the case for disjunction. Simple counter-example: everything is either a rock or not a rock, but that does not imply that either everything is a rock or everything is not a rock! – user21820 Mar 19 '16 at 17:13
  • 1
    @Ivan: But you might benefit greatly from reading the books I recommend at http://math.stackexchange.com/a/1684208. – user21820 Mar 19 '16 at 17:15