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It can be proved by showing that if $NP鈯�BPP$,then $SAT鈭�RP$.Consider a formula$蠒$ with n

Variables $x_1,x_2,鈥�x_n$be the input. $蠒$is satisfied$鈭�$truth assignment for$x_1,x_2,鈥�x_n$ so that$蠒(x_1,x_2,鈥�x_n)=1$

Assume A be an BPP algorithm with error probability at most$2^{-k}$ for SAT, where$k=\left|蠒\right|$ is the length of formula .$蠒$ Such$A$exist because of the assumption that.$SAT鈭�BPP$

First run$A$on .$蠒$ If A rejects, then reject, otherwise try to satisfy assignment for$蠒$ on variable at a time.

Initialize$x_i$ to 0 and then call $A$to determine that if the formula resulting is satisfiable: if$A$returns 鈥渁ccept鈥� then permanently set$x_i$ to 0; otherwise set $x_i$to 1. Then proceed with $x_2$similarly.

$蠒(x_1,x_2,鈥�x_n)=1$accept or otherwise rejects verify the assignment for $蠒$if managed to construct a satisfying assignment at end.

Here is the analysis. $蠒$Always reject either because $A$do not arrive at a satisfying assignment at the end, if it is unsatisfiable.

On the other hand, suppose is satisfiable. Proceed to show that it accepts with probability at least.$\frac{1}{2}$

Invoke A a total of$n+1$times. If$蠒$ is satisfiable and$A$returns 鈥渁ccept鈥� each time only for an assignment for $x_i$variable which is part of a satisfying assignment, then end up with a satisfying assignment.

Now show that the probability that at least one of theinvocations return 鈥渞eject鈥� for an assignment for variable which is part of a satisfying assignment is at most.

The probability that an invocation of$A$returns does so is at most $2-k$. So, probability that encounter it is at most $(n+1){.2}^{-k}$, which is at most because.$n+1鈮�k$

Since both thealgorithm$A$ andthe construction of satisfying assignment run in polynomial time, the whole procedure clearly runs in polynomial time. Hence,$\mathbf{SAT}鈭�\mathbf{RP}$ andtherefore.$\mathbf{NP}\mathbf{=}\mathbf{RP}$