Prove $\neg B \wedge A$ cannot be expressed using only exclusive or (XOR) and negation of $A$ and $B$.
Exclusive or gives whether $A,B$ are different. So any statement's truth table is invariant to exchanging $A,B$ which is not true for $\neg B \wedge A$. This proves $\neg B \wedge A$ cannot be expressed as just exclusive or. But with negation, this is really not quite intuitive to me. I tried to prove by induction on number of xors. And I prove a stronger case where any XOR and negation alone cannot express any statement with truth table of three $T$ or three $F$. The base case is easy to check. The inductive case is to show "given that a statement with less than $n$ xors does not express aforementioned clause, show a statement with less than $n+1$ xors cannot either". Fix such a statement with $n$ xors, we may write it as $G(A,B) \oplus H(A,B)$. Suppose to contrary that $G \oplus H$ expresses some $3T$ or $3F$ statement, say $\neg B \wedge A$. Then it may be checked that one of $G,H$ can too. For example, to express $\neg B \wedge A$, draw truth table for both $G,H$; their $T,F$ entry must differ and all other 3 entries match. But this will force $3T,3F$ to appear in either $H$ or $G$'s truth table, contradicting inductive hypothesis.
I'm not a logician so this is the best I can think of myself.
Questions: what is the standard way of proving this? and are there more intuitive/easier way to understand this phenomenon. Thank you!