Proving uniqueness of solution in Boolean algebra

Question

System: $a+x=1 \land a\cdot x=0$ has unique solution for x, for all values of $a \in B$. It is obvious that $x=a'$ is one solution , but how to prove the it is only one? I have tried assuming that there is another solution $b\neq a'$, but I keep getting nowhere with using that and cycling axioms.

Answer 1

Try making the truth table for this problem. Assume $0$ and $1$ as False and True respectively. Also:

$$P:a+x=1$$

$$Q:a.x=0$$

$$R:P\wedge Q$$ For instance when $a=0,x=0$ then then the statement $P:a+x=1$ is false so the value of $P=0$. So now let us consider every case. $$$$\begin{array}{|c|c|c|} \hline a & x & P:a+x=1&Q:a.x=0&R:(a+x=1)\wedge (a.x=0) \\ \hline 0 & 0 & 0&1&0 \\ \hline 0&1&1&1&1\\ \hline 1&0&1&1&1\\ \hline 1&1&1&0&0\\ \hline \end{array}$$$$ Now you can see there are only two cases for the statement $R$ to be True. The cases are $a=0,x=1$ and $a=1,x=0$. And you want to find the solution of $x$ in terms of $a$ so it is obvious that: $$x=\bar a$$

The solution is unique because we have considered every value of $a$ and $x$ in the truth table.Satisfying both the statements $P$ and $Q$ you find there is only one possibility.

Hope this helps...

Answer 2

There are a couple of ambiguities in the question. First, "Boolean Algebra" can mean a general complemented distributive lattice, or the special case of this with just the two elements $\ 0\ $ and $\ 1\ $. Evidently, SNEHIL SANYAL's answer has (perfectly reasonably, I think) has adopted the latter interpretation.

Second, while the symbol "+" is occasionally used to represent the join operation of a Boolean algebra, it is more commonly used—in my experience, at least—to represent the symmetric difference operation—that is, $\ a+b = a.b' \vee a'.b\ $, where "$\vee$" is here used to denote the join. In the following, I assume it was intended to denote the join, and I will replace it with "$\vee$" .

The equations to be solved are $$ a\vee x = 1,\ a.x = 0 $$ Taking the join of both sides of the second equation with $\ a'.x\ $, we get: $$ a'.x = a'.x \vee a.x = (a' \vee a).x = 1.x = x\ ,$$ by using the distributivity of meet over join. Now taking the meet of both sides of the first equation with $\ a'\ $, we get: $$ a' = a'.(a\vee x) = a'.a \vee a'.x = 0 \vee a'.x = a'.x\ ,$$ again using the distributivity of meet over join. Combining these two results, we get $\ a' = a'.x = x\ $.