Proving Set Identity: A ∪ (B - A) = ∅ Explanation

[Mr Davis 97]

Homework Statement

The problem is to prove that $A \cup (B - A) = \varnothing$

Homework Equations

The Attempt at a Solution

The solution in the textbook is that
$A \cup (B-A) = \{x~ |~ x \in A \land (x \in B \land x \not\in A) \} = \{x~ |~ x \in A \land x \not\in A \land x \in B \} = \{x~ | ~ F\} = \varnothing$. I am just confused as to why $\{x~ | ~ F\} = \varnothing$. Why is that logically a consequence?

 

[fresh_42]

Homework Statement

The problem is to prove that $A \cup (B - A) = \varnothing$

Homework Equations

The Attempt at a Solution

The solution in the textbook is that
$A \cup (B-A) = \{x~ |~ x \in A \land (x \in B \land x \not\in A) \} = \{x~ |~ x \in A \land x \not\in A \land x \in B \} = \{x~ | ~ F\} = \varnothing$. I am just confused as to why $\{x~ | ~ F\} = \varnothing$. Why is that logically a consequence?

Well, first of all, it should be an intersection $\cap$ (and), not a union $\cup$ (or).
Then $\{x\,\vert \,F\}$ is the set of all $x$, which satisfy $"false"$. But $"false"$ is never satisfied, thus $\emptyset$.

 

[Mark44]

$A \cup (B-A) = \{x~ |~ x \in A \land (x \in B \land x \not\in A) \}$

To expand on what fresh_42 said, the expression on the right is the definition of $A \cap (B - A)$; i.e., the intersection of A and B - A, not the union of these two sets.