I am trying to prove this claim but I am not sure about the step $$x ∈ A ∧ x ∉ B = A$$ and the step $$A → A = A$$ in my proof. Could someone verify my proof?
Let $x ∈ A-B$.\begin{align*} &\mathrel{\phantom{=}}{} A-B⊆ A\\ &= ∀x \{x ∈ A ∧ x ∉ B → x ∈ A \} &&\text{[Definition of subset]}\\ &= ∀x \{x ∈ A → x ∈ A \} &&[A ∧ x ∉ B = A]\\ &= ∀x \{x ∈(A → A ) \} &&\text{[Distributivity]}\\ &= ∀x \{x ∈ A \} &&[A → A = A]\\ &= A. \end{align*}
I don't understand this notation.. Never seen it before. It makes no sense: an inclusion statement is not a set.
But $A - B \subseteq A$ is trivial the standard way: let $x \in A- B$. Then $x \in A$ and $x \notin B$, so in particular $x \in A$ (we can use one half of an "and"-clause). So $A -B \subseteq A$ has been shown.