Is my proof by contradiction about the empty set correct? I am trying to learn about proofs and one of the exercice in my book (Maths ABC) is about proof by contradiction. I think I understand the concept but I would like to have a feedback on the following proof about whether or not it is correct.

Let $A$ be the following proposition: 
  $$[(S\cup P = P) \land (S\cap P = \emptyset)]\rightarrow(S = \emptyset)$$
  Prove by contradiction that $A$ is true.

My proof:
Let assume that $[(S\cup P = P) \land (S\cap P )= \emptyset)]\rightarrow\;S\neq \emptyset$ is true.
$$(S\cup P) = \{x\mid x\in S \lor x\in P\}\land P\{x\mid x\in \emptyset\lor x\in P\}$$
hence, if $(S\cup P=P)$ we can see by symmetry that $S$ has to be equal to $\emptyset$ which is absurd and contradict our assumption.
$[(S\cup P = P)\land(S\cap P = \emptyset)]$ cannot be true if one of the two statement is false thus it is enough for us to assert that $S=\emptyset$
QEA
 A: The structure of the proof is wrong.


*

*The negation of $[(S \cup P = P) \text{ and } (S \cap P= \emptyset)] \implies (S = \emptyset)$ to prove by contradiction is $[(S \cup P = P) \text{ and } (S \cap P = \emptyset)] \text{ and } (S \ne \emptyset)$, it is not $[(S \cup P = P) \text{ and } (S \cap P = \emptyset)] \implies (S \ne \emptyset)$

*The statement $(S \cup P = P)$ does not imply $S = \emptyset$, since if $S \subseteq P$ then $(S \cup P = P)$ is true.
Now
Proof. Suppose for sake of contradiction that $[(S \cup P = P) \text{ and } (S \cap P = \emptyset)] \text{ and } (S \ne \emptyset)$. Since $S \cup P = P$, we have $S \subseteq P$. Thus $S \cap P = S \ne \emptyset$, a contradiction to the hypothesis $S \cap P = \emptyset$.
Maybe you can try prove

Let $A, B$ be sets. Show that the three statements $A \subseteq B$, $A \cup B = B$, $A \cap B = A$ are logically equivalent (any one of them implies the other two).

A: Your proof is incorrect since $S \cup P = P$ does not imply that $S = \emptyset$.  For example, if $S = \{1\}$ and $P = \{1, 2\}$, then $S \cup P = \{1\} \cup \{1, 2\} = \{1, 2\} = P$.  
Assume $S \neq \emptyset$.  Then there exists $x \in S$.  Then $x \in S \cup P = P$, so $x \in P$.  Since $x \in S$ and $x \in P$, $x \in S \cap P = \emptyset$, which is a contradiction.  
A: I think your proof is wrong. The fact $S \cup P=P$ does not imply alone that $S$ is empty. It can be perfectly part of $P$. However together with the second condition $S\cap P=\emptyset$ they do imply that $S$ is empty.
