Recall, in our discussion of the Church–Turing thesis, that we introduced the language is a polynomial in several variables having an integral root}. We stated, but didn’t prove, thatis undecidable. In this problem, you are to prove a different property of—namely, that$\mathit{D}$is -hard. A problem is -hard if all problems in are polynomial time reducible to it, even though it may not be in$\mathit{N}\mathit{P}$itself. So you must show that all problems in $NP$are polynomial time reducible to $\mathit{D}$ .

Consider the way to proof the $NP$-completeness property of the given language. The language$D$can be shown as $NP$- complete by using a reduction from $3SAT$.

An expression$L$in can be converted into$\$L^\{\backslash prime\}\$$by using the following facts:$\$\backslash sima=(1-a)\$$

(Where the sign$\$(\backslash sim)\$$ is defined as a$NOT$ logic operator)

$\$a\$$and$\$b=(a b)\$$

$\$a\$$or $\$b=(a+b-ab)\$$

Now, for simplicity, a function $\$f(a,b,c)\$ \$f$could be defined for each of the eight possibilities for all the clauses in $3SAT$. Here, eight possibilities will be taken because each variable$a,bandc$ will acquire a value$\$1/0\$$.

An algorithm can be used which replace every clause with one of the replacements and also used for putting the symbol multiplication between them. In other word, the description of above algorithm can be given as:

$\$\$$

$f(a\backslash text \{or\}b\backslash text\{or\}1\backslash simc\}=\left[\right(a+b-ab)+(1-c)-(a+b-ab\left)\right(1-c\left)\right]$

$\$\$$

The running time of the above given algorithm is linear. Therefore, the language$\$D\{\backslash text \{$ can be said as role="math" localid="1663228652087" $\left\}\right\}\$NP$-complete.

It is already known to be un-decidable, thus it will not exist in$NP$ . Therefore from the above explanation, it can be said that the language$D$ is $NP$-hard.