WeBWorK Standalone Renderer

Order 9 of the following sentences so that they form a logical proof of the statement:

For any non-empty sets $A$ and $E$ with $E \subsetneq A$ , then

$A - E = A \cap E^c$

Choose from these sentences.
You can assume all sets are nonempty.

Hence, $x \in A$ and $x \in E^c$
Let $x \in A \cap E^c$
Assume the dog ate my homework.
Assume $\exists x \in A \cap E^c$ but $x \not \in A - E$
By definition, $x \in A$ and $x \not \in E$
$A - E = A \cap E^c$
$A - E \subseteq A \cap E^c$
$x \in A$ and $x \in E^c$ and either $x \not \in A$ or $x \in E$
$x \in A \cap E^c$
Assume $\exists y \in E$ but $y \not \in A$
$x \in A\cap A^c$ or $x \in E \cap E^c$
Let $x \in A - E$

Your Proof (showing containment left to right first):

Choose from these sentences.
You can assume all sets are nonempty.

Hence, $x \in A$ and $x \in E^c$
Let $x \in A \cap E^c$
Assume the dog ate my homework.
Assume $\exists x \in A \cap E^c$ but $x \not \in A - E$
By definition, $x \in A$ and $x \not \in E$
$A - E = A \cap E^c$
$A - E \subseteq A \cap E^c$
$x \in A$ and $x \in E^c$ and either $x \not \in A$ or $x \in E$
$x \in A \cap E^c$
Assume $\exists y \in E$ but $y \not \in A$
$x \in A\cap A^c$ or $x \in E \cap E^c$
Let $x \in A - E$

Your Proof (showing containment left to right first):