Prove that is is not possible to define the connective $\land$ in terms of $\lnot$ and ↔ 
[Exercise 12.2] Prove that is is not possible to define the connective $\land$ in terms of $\lnot$ and ↔.


Suggestion:  By  induction  on  the complexity  of  formulae  constructed  by  these  connectives, prove that every truth table has  an  even number of rows that assign the respective formula as true.

I think i have already solved this exercise, so i would like to know if my solution is correct:
All the possible WFFs (Well Formed Formulae) constructed using $\lnot$ and ↔ are: $\lnot\ (a↔  b)$, $\lnot\ (\lnot a↔  b)$, $\lnot\ ( a↔ \lnot b)$, $\lnot\ (\lnot a $↔$  \lnot b)$, $\ (\lnot a↔  b)$, $\ ( a↔  \lnot b)$ and $\ ( \lnot a↔  \lnot b)$.  Once all the truth-tables of the previous formulae have been verified, it is possible to determine that none of them match with the $a \land b$ operator truth-table , thus proving that the operator is not definable in terms of $\lnot$ and ↔.
In spite of the validity of my solution, i also would like to know how to actually solve this exercise using the concept of induction on the complexity of formulae, given the following definition:

Def. 2.4 (Complexity Degree of a WFF). For each formula on propositional logic we determine a natural number such that the following conditions hold:


1). Every atomic formula has complexity degree $0$.


2). If $A$ has a degree of complexity $n$, then$\ ( \lnot A )\ $ has a degree of complexity of $n+1$


3). If $A$ and $B$ have degrees of complexity $m$ and $n$, respectively, then$\ (A \land B )\ $, $\ (A \lor B )\ $, $\ (A \to B )\ $ and $\ (A ↔ B )\ $ have degrees of complexity $max$$\{m, n\}$$+1$, where $max$$\{m, n\}$ is the greatest value between $m$ and $n$.

 A: First thing you should notice is that you also get $\top$ and $\bot$ since $a\leftrightarrow a =\top$. Also notice that $(\neg a\leftrightarrow b)\equiv \neg(a\leftrightarrow b)$. We also can restrict ourselves to having only two atomic propositions $a$ and $b$. Consider the levels of complexity up to $2$. At complexity $0$ we have $a$ and $b$, for complexity $1$ $\top$ $\bot$ $\neg a$ $\neg b$ and $a\leftrightarrow b$ at $2$ we get $\neg a \leftrightarrow b$ which are all the sentences you can get up to logical equivalence by using $\leftrightarrow$ and $\neg$. If we prove that we are done since none of them are equivalent to $a \wedge b$. we show that applying $\leftrightarrow$ and $\neg$ to any sentence or sentences equivalent to the ones listed above we get another sentence equivalent to the one listed above. Then by induction on the complexity of the formula we are done.
We see that the sentences are closed under $\neg$ meaning if I apply $\neg$ to any sentence I get one that is logically equivalent to one of the ones listed above.
\begin{equation}
\begin{matrix} 
\top && a&& b && (a\leftrightarrow b)\\
\bot && \neg a && \neg b && \neg(a \leftrightarrow b)
\end{matrix}
\end{equation}
We also need to show they are closed under $\leftrightarrow$. This is a proof by case and there are $64$ pairs of sentences which would be silly to check all of them. The first trick is that $(\neg p\leftrightarrow q)\equiv \neg(p \leftrightarrow q)$ where $p$ and $q$ are not necessarily atomic formulas. So we just need to check that it's true for pairs in $\{\top,a,b,a\leftrightarrow b\}$ since we already establised the set is closed under negation. We are now left with $16$ pairs. Since $\leftrightarrow$ is symmetric and  for each $p$ we have $p\leftrightarrow p\equiv \top$ this leaves us with $6$ cases. For any sentence $p$ one has $\top \leftrightarrow p  \equiv p $ which covers all cases that have $\top$. For $a$ and $b$ it's immediate and finaly for $(a\leftrightarrow(a \leftrightarrow b))$ using truth tables you cans show that it's equivalent to $a\leftrightarrow b$. The same goes for $(b\leftrightarrow(b\leftrightarrow a))$
A: For the issues hinging on the expressive adequacy (or functional completeness) of propositional connectives, I'd like to point out, in a very restricted manner, an alternative to the method of exhausting possible cases, which is based on an ingenuous idea of Emil Post.
Let us call the connectives monotonic such that a truth-value change from F to T of one of the connected propositional variables, while the other connected propositional variable's truth-value is not changed from T to F, never makes the truth-value of the compound proposition change from T to F.
Consider $P\wedge Q$ as an example.

The truth-value of the compound proposition changes from T to F, only if truth-value of one of the propositional variables changes from T to F.
We shall call the connectives affine if a truth-value change of a connected propositional variable either results in a truth-value change of the compound proposition or never makes a difference. $\neg$ and $\leftrightarrow$ change truth-value whenever the truth-value of one of the propositional variable changes.
Can we emulate the behaviour of a monotonic connective by affine connectives? In order to keep the truth-value the same when a propositional variable changes, an even number times affine connectives have to be used. But then we cannot emulate the change from T to F of the compound proposition.
An excellent exposition of Post's approach is given in Odysseus Makridis' article The Sheffer Stroke. Also see Joel Hamkins' The Hierarchy of Logical Expressivity.
