Fuzzy logic of Godel In Gödel logic, is conjunction definable from implication, negation , and disjunction?
We know that conjunction in that logic is not definable from negation and implication.
 A: I assume that by "definable" you mean "expressible by a term".  In that case the answer is "no". 
Let $t(x,y)$ be any term (involving the variables $x$, $y$, and connectives $\to$, $\lnot$, $\vee$ -- possibly not all of them).   


*

*If we interpret $t$ on the set $\Delta:= \{(x,y)\in (0,1)^2: x>y\}$ (under the Gödel interpretation, so in particular $p\to q$ has value $1$ for $p\le q$, and value $q$ otherwise), then $t$ must be constant with value $1$ or $0$, or equivalent to one of the terms $x$ or $y$.   

*Similarly for $\nabla:= \{(x,y)\in (0,1)^2: x< y\}$. 


We will show that there is no term which is equivalent to $y$ on $\Delta$, and equivalent to $x$ on $\nabla$. 
So let $t$ be such a term of minimal length. Obviously $t$ is not of the form $\lnot t'$, and not equal to $x$ or $y$. 
So $t$ must be an implication $t_1\to t_2$ (or a disjunction $t_3\vee t_4$; this case is left as an exercise). 
Now consider all the possible behaviours of $t_1$ and $t_2$ on $\Delta$.
$t_1\restriction \Delta$ can be $x$, $y$ or $1$, similarly $t_2\restriction \Delta$.
Considering the 9 possible cases, we see that we get the desired result $y$ ($= x\wedge y$ on $\Delta$) in only 2 cases, as $x\to y$ and as $1\to y$. In both cases we must have $t_2=y$ (on $\Delta$). 
Similarly we must have  $t_2=x$ on $\nabla$ to get $t_1\to t_2$ equivalent to $x$ on $\nabla$. But then $t$ was not minimal. 
If you think this proof is ugly, consider my second proof, which is worse, but has the advantage that it can be outsourced to a computer: Consider binary functions on the interval $[0,1]$.   Start with the projections $x$ and $y$, close off under disjunction, negation, Gödel implication. This stops after finitely many steps. (19, if I counted correctly; at least if you consider only the behavior on the open square.)  Check that you have not generated the conjunction.
