Prove that (A-B) - C ⊆ A - C

Question

I came up with a proof but I’m not sure if I’ve done it well. Specifically I’m not sure if number 6 is a valid argument.
Here it is.

proof:

Let x ∈ (A-B)-C

1. (A-B)-C = {x|x ∈ (A-B) ∧ x ∉ C} By definition of difference of sets

2. = {x|(x ∈ A ∧ x ∉ B) ∧ x ∉ C} By definition of difference of sets

3. ={x| x ∉ C ∧ (x ∈ A ∧ x ∉ B) } By commutative laws

4. = {x| (x ∉ C ∧ x ∈ A) ∧ (x ∉ B ∧ x ∉ B) } distrib. laws of L.E

5. = {x| (x ∉ C ∧ x ∈ A) ∧ ¬(x ∈ B v x ∈ B) }Demorgan’s first law of L.E

6. = {x| (x ∉ C ∧ x ∈ A) ∧ T } By 2. x ∉ B and x ∉ B

7. = {x| (x ∉ C ∧ x ∈ A)} By Identity laws of logical equivalence

8. = {x| ( x ∈ A ∧ x ∉ C )} By commutative laws

9. = {x|x ∈ (A-C)} By definition of difference of sets

10. = x ∈ (A-C) meaning of set builder notation

Since x ∈ (A-B)-C and x ∈ (A-C), then (A-B)-C is a subset of (A-C)

Q.E.D

Answer 1

You confused yourself trying to apply distribution in line 4.   There's no need to distribute $\wedge$ over $\wedge$.   Just use commutation and association.

You tried to force every step to be an equality; some need not be, since you are trying to prove the LHS is a subset of the RHS.

Here's a more compact summary of what you need.

$\begin{align}(A-B)-C~=~&\{x\mid (x\in A\wedge x\notin B)\wedge x\notin C\} &&\text{definition of set difference} \\=~& \{x\mid (x\in A\wedge x\notin C)\wedge x\notin B\} && \text{commutation and association over }\wedge\\~=~& (A-C)-B &&\text{definition of set difference}\\ ~\subseteq~& (A-C)&&\text{property of set difference}\end{align}$

Answer 2

Alternative:

Since $x\in A-B$ we have $x\in A\land x\not\in B$ so $x\in A$. Since $x\in (A-B)-C$ we have $x\in (A-B) \land x\not\in C$ so $x\not\in C$. By definition $x\in A-C$, so the subset relation is proved.

Answer 3

Hint (where $\overline X$ denotes the absolute complement of $X$): $$(A \setminus B) \setminus C = (A \cap \overline B)\cap \overline C = (A \cap \overline C)\cap \overline B \;\;\subseteq\;\; A \cap \overline C = A \setminus C$$