How to solve this class of problems? I was presented with the following problem:

Ricardo, Rogério and Renato are brothers. One of the is a medic, the other one is a teacher and the other one is a musician. It is known that:
  
  
*
  
*Ricardo is a medic or Renato is a medic.
  
*Ricardo is a teacher or Rogério is a musician.
  
*Renato is a musician  or Rogério is a musician.
  
*Rogério is a teacher or Renato is a teacher.
  
  
  Then, their professions are respectivelly:
  
  
*
  
*Teacher, Medic, Musician
  
*Medic, Teacher, Musician
  
*Teacher, Musician, Medic
  
*Musician, Medic, Teacher
  
*Medic, Musician, Teacher
  

And I know there may be some way to interpret the propositions and solve the problem in a fashion similar to the way we multiply numbers. But it's not very clear to me how to use this way, I've read about Propositional Calculus and the Method of Analytic Tableux but I'm still lost. I've also tried to think in a way to use Venn diagrams but I still don't have much clue on what to do. I'm looking for a general way of solving this class of problems. I've also tried some random explorations on Mathematica but couldn't achieve anything. Can you guide me?

I guess I have managed to make a solution for this (following the answers given by Vadim123 and Peter Smith, I've made a table with the propositions:
$$\begin{vmatrix}
{Ricardo Medic}&{Renato Medic}\\ 
{Ricardo Teacher}&{Rogério Musician}\\ 
{Renato Musician}&{Rogério Musician}\\ 
{Rogério Teacher}&{Renato Teacher}
\end{vmatrix}$$
And then another one to follow the consequences, then If Ricardo is a medic, I checkmark the corresponding option and mark the other professions that he could have with an X mark. If Ricardo is a medic, then it's impossible for him to be a teacher nor Renato be a musician due to the nature of the XOR operation.
$$\begin{vmatrix}
{\checkmark }&{X}\\ 
{X }&{ - }\\ 
{ -}&{ -}\\ 
{- }&{- }
\end{vmatrix}$$
Now I make a bet in the profession of the other brothers, I'll make a bet that Renato is a musician, which implies that he can't be a teacher and also that Rogerio can't be a musician. The only job left for Rogerio is the job of musician. 
$$\begin{vmatrix}
{\checkmark }&{X}\\ 
{X }&{ X }\\ 
{ \checkmark }&{X}\\ 
{X }&{\checkmark }
\end{vmatrix}$$
Now I have two problems:


*

*How to determine the number of solutions of the given exercise? How to determine that there is only one option that answers it? I know that I gave a list of answers, but ignore that it exists, how to know that Ricardo is a medic, Renato is a musician and Rogerio is a teacher is the only answer? I can only think of brute force and perhaps the truth table, but for the latter, I'm still not sure it works. I'm worried because using brute-force, I guess I've been able to find 3 solutions until now:


$$\begin{matrix}
{S_1}&{RicardoTeacher}&{RenatoMedic}&{RogerioMusician}\\ 
{S_2}&{RicardoMedic}&{RenatoMusician}&{RogerioTeacher}\\ 
{S_3}&{RicardoMedic}&{RenatoTeacher}&{RogerioMusician}\\ 
\end{matrix}$$


*

*Are there ways to algebrize this problem? I've tried to make a truth table for it but I'm not sure it works. I'm interested in having a general approach for this kind of problem, in this case, there are $4$ propositions with the $XOR$ conector, what if the connector could be any conector? What if one of the propositions has a $XOR$, the other one has a $NOR$ and the next one has an $AND$?

 A: Here is the information you've been given.  
$$\begin{array} ~& \text{Ricardo} & \text{Renato} & \text{Rogerio}\\
\text{Medic}& 1& 1& \\
\text{Teacher} &2 &4 &4\\
\text{Musician} &&3& 23\\
\end{array}$$
You know that one of the 1's is correct, one of the 2's is correct, etc.
Make a choice, then trace through the consequences.  Suppose that Ricardo is a Medic (1).  Then Ricardo is not a Teacher, so by 2 Rogerio must be a Musician.  If Rogerio is a Musician then by 4 Renato must be a Teacher.
A: It is evident from 4, since only one of the brothers is a teacher, that Ricardo isn't a teacher, and 2 then tells you

Rogerió is musician.

So he isn't a teacher so 4 tells you

Renato is a teacher.

So he isn't a medic, so 1 tells you

Ricardo is a medic.

Job done!

OK: but this is an ad hoc solution to the particular problem. Can we mechanise the reasoning? In particular, the OP is interested whether we can use the propositional calculus. Well, yes ....
Use e, i, o for Renator, Ricardo, and Rogerió. 
Use D, T, M for Doctor, Teacher, Musician.
So we have nine atoms we can write $De, Di, Do; Te, Ti, To; Me, Mi, Mo$.
The basic structure of the situation is that we have there is exactly one doctor

$(De \land \neg Di \land \neg Do) \lor (\neg De \land Di \land \neg Do) \lor (\neg De \land \neg Di \land Do)$

similarly for teacher and musicians,
and there there is exactly one profession for Renato

$(De \land \neg Te \land \neg Me) \lor (\neg De \land  Te \land \neg Me) \lor (\neg De \land \neg Te \land Me)$

and similarly for the other two guys.
So that's six general "axioms": and then there are four specific axioms

$Di \lor Do$

and so on.
We've got nine atoms (so it's a 512 line truth-table to do!) and ten axioms. And we can then do a truth-table to test any axiom to see if it follows. Be my guest .... The principle of the thing is mildly interesting, but of course implementing it in practice would be stunningly boring!
A: You can complete the following table that checks all possibilities. $E$, $I$, $R$ stands for Renato, Ricardo and Rogerio, $t$, $m$, $u$ for teacher, medic and musician. $m(I)$ means Ricardo is musician. $s1$ is sentence 1: Ricardo is a medic or Renato is a medic. So $s1=m(I) \lor m(E)$.  $s$ is $s1 \land s2 \land s3 \land s4$ , the final result. 
$$
\begin{array}{l|l|l} 
E&I&O& m(I)&m(E)& t(I)&u(O)&u(E)&u(O)&t(O)&t(E)&s1&s2&s3&s4& s \\
\hline{} \\
t&m&u&1&0&0&1&0&1&0&1&1&1&1&1&1\\
t&u&m&0&0&0&0&0&0&0&1&0&0&0&1&0 \\
m&u&t &\cdots\\
m&t&u \\
u&t&m \\
u&m&t \\
\end{array}
$$
A: Here is a 'logical' approach.
Write $\;Ri = me\;$ for "Ricardo is the medic", $\;Re = te\;$ for "Renato is the teacher", etc.  Using this formalization, you are given that
\begin{align}
(1) & Ri = me \lor Re = me \\
(2) & Ri = te \lor Ro = mu \\
(3) & Re = mu \lor Ro = mu \\
(4) & Ro = te \lor Re = te \\
\end{align}
where $\;Ri, Ro, Re\;$ are distinct and also $\;me, te, mu\;$ are distinct.
Now we note that $(1),(3),(4)$ are all of the same shape, viz. $\;x = p \lor y = p\;$ for distinct $\;x,y\;$.  And from this we can conclude, for any other distinct $\;z\;$, that $\;z \not= p\;$:
\begin{align}
& x = p \lor y = p \\
\Rightarrow & \;\;\;\;\;\text{"use $\;x \not= z\;$ on left hand side; use $\;y \not= z\;$ on right hand side"} \\
& z \not= p \lor z \not= p \\
\equiv & \;\;\;\;\;\text{"simplify"} \\
& z \not= p \\
\end{align}
Therefore, since $\;Ri, Ro, Re\;$ are distinct, we can conclude from $(1),(3),(4)$ that
\begin{align}
(1') & Ro \not= me \\
(3') & Ri \not= mu \\
(4') & Ri \not= te \\
\end{align}
Now there are many ways to find the solution; below is what I believe to be the most direct one.  I discovered this just by looking at the shape of the formulae.
First, substituting $(4')$ in $(2)$ results in $\;Ro = mu\;$.  Then substituting $\;mu\;$ for $\;Ro\;$ in $(4)$, using $\;mu \not= te\;$, results in $\;Re = te\;$.  Finally substituting $\;te\;$ for $\;Re\;$ in $(1)$, using $\;te \not= me\;$, results in $\;Ri = me\;$.
In words: Rogério is the musician, Renato is the teacher, and Ricardo is the medic.
A: There's some puzzle books which just have problems like this.  Those books use grids.  You can represent such a grid as follows:
            Ricardo   Rogerio   Renato
Medic
Teacher
Musician

Suppose we put an X if the person does not work in that particular occupation, and an O if that person does work in that particular occupation.
Since Ricardo is a medic or Renato is a medic, it is not the case that Rogerio is a medic.  Thus we can put an X where Rogerio and Medic intersect yielding:
           Ricardo   Rogerio   Renato
Medic                   X
Teacher
Musician

Similarly, the third clue tells us that Ricardo is not the musician.  I'll denote the above result by X-1, and what the third clue tells us by X-3.  The fourth clue tells us that Ricardo is not the teacher.  So, we have:
           Ricardo   Rogerio   Renato
Medic        O         X-1       X
Teacher      X-4
Musician     X-3

Now Ricardo is a teacher or Rogerio is a musician.  But, Ricardo is not a teacher, so Rogerio is the musician.  Consequently, it follows that Renato is the teacher.
"How to determine the number of solutions of the given exercise?"
For an n by n box, we have n possibilities for the first row, (n-1) for the second row, ..., 1 possibility for the last row.  So, we have n! possibilities a priori given that we can't tell anything about what the clues entails.
"How to determine that there is only one option that answers it?"
If there does not exist one option, then there exist two distinct options, or any proposed option entails a contradiction.  The converse holds also, so you can show that a problem does not entail a unique solution.  Showing that there only exists one option requires that you can eliminate all ambiguity of language in the clues in such problems.  I'm not quite sure you can do that here, because you can't really prove that such language necessarily entails that such propositions have one of two truth values necessarily.  That said, if you can find more than one answer here, then the grid method fails for these problems.  I've never seen it fail unless the problem already had ambiguities in the clues to begin with.
