Hints for direct proof that $(A \cap C = B \cap C$ and $A \cup C = B \cup C)$ implies $A = B$

Question

I am trying to use direct proof to prove that

$(A∩C=B∩C∧A∪C=B∪C)→A=B$ as Can you conclude that A = B if A, B, and C are sets such that...

But i am stuck with the followings,please help.

$1.∀((x∈A ∨ x∈C) ⇒ (x∈B ∨ x∈C))=premise$

$2.∀((x∈B ∨ x∈C) ⇒ (x∈A ∨ x∈C))=premise$

$3.∀((x∈A ∧ x∈C) ⇒ (x∈B ∧ x∈C))=premise$

$4.∀((x∈B ∧ x∈C) ⇒ (x∈A ∧ x∈C))=premise$

$5.x∈A ∧ x∈C=premise$

$6.x∈B∧x∈C$=(3,5 modus ponens)

$7.x∈C =(simpl.)$

$8.x∈B ∨ x∈C=(1,7modus ponens)$

$9.∀((x∈A∧x∈C)⇒(x∈B ∨ x∈C))$=(Universal generation)

we can get $A⋂C⊆B⋃C$

After the repeated process,i got these conclusion by rule of inference,does they helpful in deriving $A=B$.

Conclusion that derived by 1,2,3,4:

a)$A⋂C⊆B⋃C$

b)$A⋂C⊆A⋃C$

c)$B⋂C⊆B⋃C$

d)$B⋂C⊆A⋃C$

what should i do next?

Answer 1

Line 5 is where you started heading in the wrong direction. You should be trying to prove $\forall x(x \in A \Rightarrow x \in B)$ and also $\forall x(x \in B \Rightarrow x \in A)$. For the first of these, you should be assuming $x \in A$ (not $x \in A \wedge x \in C$ as you did) and then proving $x \in B$, after which you can generalize. Similarly for the second.

You might think that it is difficult to use the premises without knowing whether or not $x \in C$. (Perhaps that is why you were tempted to assume $x \in A \wedge x \in C$ in line 5.) One way to deal with that is to use reasoning by cases: assume $x \in C$ for case 1 and $x \notin C$ for case 2.

Answer 2

If $x \in A$, then either $x \in C$, so then $x \in A \cap C= B \cap C$ so $x \in B$, or $x \notin C$.

In the last case we do know $x \in A \cup C = B \cup C$ so $x \in C$ (which is not the case) or $x \in B$, which thus is the case.

In both cases $x \in B$ purely from $x \in A$ and the symmetric argument gives the other inclusion.

If you want the argument in some formal proof system (with inference rules etc.) then please specify the exact system you use, preferably from some online resource. But the basic idea of proving two inclusions based on the additional case of $x \in C$ or not will probably resurface there as well.

Alternatively note that first for any $D$: $$(D \cup C) \cap C^\complement = (D \cap C^\complement) \cup (C \cap C^\complement)= (D \cap C^\complement) \cup \emptyset= D \cap C^\complement\tag{1}$$

So $$A = A \cap (C \cup C^\complement) =(A \cap C) \cup (A \cap C^\complement) = (B \cap C) \cup ((A \cup C) \cap C^\complement)=\\= (B \cap C) \cup ((B \cup C) \cap C^\complement) = (B \cap C) \cup (B \cap C^\complement) = B \cap (C \cup C^\complement)= B$$

where we apply (1) with $D=A$ first, later with $D=B$ and use the given $A \cdot C= B \cdot C$ for $\cdot \in \{\cap,\cup\}$ as well. So that's a purely set-algebraic proof, which is quite direct and mirrors the first "verbal" argument as well.