# Set Theory Question

## Homework Statement

##C \subseteq A \cap B \implies A \cap B \cap C = C##

## Homework Equations

How do I get rid of the "belongs to" term on the right hand side? I know I need to prove either the left hand or the right hand side of the "or" term is correct, I'm just not sure how to get there.

## The Attempt at a Solution

~##(C \subseteq A \cap B) \cup (A \cap B \cap C = C)##

right hand side (right of the "or"):
##C \subseteq A \cap B \cap C## (Trivial)
##A \cap B \cap C \subseteq C## (This is the one we want to prove)

So all together:

~##(C \subseteq A \cap B) \cup (A \cap B \cap C \subseteq C)##
##\exists x \in C \therefore x \in A \cap B)##
##(\sim a \cup \sim b) \cup (a \cap b \cap c \subseteq C)##

Mentor
2021 Award

## Homework Statement

##C \subseteq A \cap B \implies A \cap B \cap C = C##

## Homework Equations

How do I get rid of the "belongs to" term on the right hand side? I know I need to prove either the left hand or the right hand side of the "or" term is correct, I'm just not sure how to get there.

## The Attempt at a Solution

~##(C \subseteq A \cap B) \cup (A \cap B \cap C = C)##

right hand side (right of the "or"):
##C \subseteq A \cap B \cap C## (Trivial)
##A \cap B \cap C \subseteq C## (This is the one we want to prove)

So all together:

~##(C \subseteq A \cap B) \cup (A \cap B \cap C \subseteq C)##
##\exists x \in C \therefore x \in A \cap B)##
##(\sim a \cup \sim b) \cup (a \cap b \cap c \subseteq C)##
I don't really understand your complexity here. Can't you simply use ##(X \subseteq Y) \wedge (Y \subseteq X) \Longrightarrow X = Y ##?