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?
Best Answer
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.