Lets take a look on the following sets of logical connectives: $$K_{1}=\{\rightarrow,\equiv\},\ K_{2} = \{\vee, \wedge\} $$
For each logical connective $*\ $ in $\ K_{1}\cup K_{2}$, answer the following question: is there a statement, built only with logical connectives of the other group (the group of which $*$ isn't in) that represents the truth-function of $\ *\ $?
My partial solution:
I have prooved the following: Let A be a logical statement build only from logical connectives in $\{\vee, \wedge\}$, and $\mathcal{AP}=\{p_1,p_2\}$. Let $M$ be the assigning where for every atomic proposition p, $\ M(p)=F$ then $$M\not\models{A}$$
and therefore there is no such logical statement that represents the truth-functions of $\equiv, \rightarrow$.
I was able to find a logical statement A that built only from logical conectives in ${K_1}$ and represents $\wedge$: $$A=({p_1}\rightarrow{p_2})\equiv{p_1}$$
For the last option of representing $\wedge$ i think i should prove the following: Let A be a logical statement build only from logical connectives in $K_{1}$, and $\mathcal{AP}=\{p_1,p_2\}$. Let $M_1, M_2$ be the assignings: $$M_1(p1)=T, M_2(p1)=F$$ $$M_1(p2)=F, M_2(p2)=T$$
Then $$M_1\not\models{A}\ or\ M_2\not\models{A}$$
An answer, or even better, a general way of dealing with this kind of questions will be very helpful.
Thanks.