In the book How to prove it, page 102, the following problem is given:
Suppose A, B, and C are sets, A \ B ⊆ C, and x is anything at all.
If x ∈ A \ C then x ∈ B.
I can prove it the following manner:
x ∈ A \ C $\rightarrow$ (x ∈ A $\wedge$ x not ∈ C)
It is given that
A \ B ⊆ C
x not ∈ C => x not ∈ A \ B
x not ∈ A \ B can be expressed as:
$\neg$ (x ∈ A $\wedge$ x not ∈ B) = (x not ∈ A $\vee$ x ∈ B)
by conditional law:
(x not ∈ A $\vee$ x ∈ B) = (x ∈ A $\rightarrow$ x ∈ B).
It is given that x ∈ A hence x ∈ B.
But the proof is provided by contradiction as follows:
Proof. Suppose x ∈ A \ C. This means that x ∈ A and x not ∈ C.
Suppose x not ∈ B.
Then x ∈ A \ B, so since A \ B ⊆ C, x ∈ C. But this contradicts the fact that
x not ∈ C. Therefore x ∈ B. Thus, if x ∈ A \ C then x ∈ B.
Why is proof by contradiction necessary preferred here? Is not the first proof more intuitive and enough?
Your proof is correct. Which one is more intuitive is up to personal taste.
I like the proof by contradiction, as it allows to dircectly use the assumption $A\setminus B \subset C$.
In comparison, your direct proof assumes $x\in A\setminus C$. So in order to use the assumption, you have to negate it first, i.e. you look at the complements $C^c\subset (A \setminus B)^c$.
As a remark to your proof: Even though you wrote in the beginning that the goal is to prove
x ∈ A \ C → (x ∈ A ∧ x not ∈ C), during the proof you should make clear that you consider an $x \in A\setminus C$.