Proving functional completeness of $F(x,y)=\lnot x \lor y$

Question

I've seen the following question in one of the past exams of my college, Is the following set of operations $F(x,y)=\lnot x \lor y$ from a functional complete set of operations?

The student that did this test answered this question: "No, because it is impossible to express the $\lnot$ operation." And what surprising me, That answer marked as 100% correct.

I've tried to answer this question and got completely the opposite, It is possible to express the operation $\lnot$ using the set $F(x,y)=\lnot x \lor y$.

How?

Given the set: $F(x,y)=\lnot x \lor y$, We can express the $\lnot$ operation that way: $$F(x, false)=\lnot x \lor false = \lnot x$$

And, Once we have got the operator $\lnot$, We can express the $\lor$ operation:

$$F(\lnot x,y)=x\lor y$$

Once we have gor the operators $\lnot$, $\lor$ We can easily express the $\land$ operation:

$$F(\lnot x \lor \lnot y, false)=(x\land y)\lor false = x\land y$$

I'm sure not, But it seems like it is illegal to use the constants $0,1,true,false$ as parameters (Instead of variables $x,y$) in order to prove that a set of operations is functional complete set of operations, That is the only reason why the student that answered this test got 100% for his answer.

That's correct? Please tell me if i did something wrong in my process. And if it is not correct please explain why.

Thanks!!

Answer 1

In general, $false$ can represent a function of any arity $f:\mathbb B^n\to\mathbb B$ such that

$$f(x_1,\dots,x_n)=0$$ for all $x_1,\dots,x_n\in\mathbb B$.

So, in order for a set of boolean functions to be functionally complete, they also have to be able to express the constant functions of any arity.

Thus, it would need to be possible to define a function $false:\mathbb B\to\mathbb B$ such that $false(x)=0$ only using $F$ but this is not possible.

Note, on the contrary however, that $F(x,x)=1$ for all $x\in\mathbb B$ and thus $F$ can define the constant function $true:\mathbb B\to\mathbb B$, $x\mapsto 1$. This already gives you the hint that negation can not be definable by $F$ as otherwise $false$ would be a well.