Read the definition of in Problem 7.46.
a. Show that $\mathit{M}\mathit{I}\mathit{N}\mathbf{-}\mathit{F}\mathit{O}\mathit{R}\mathit{M}\mathit{U}\mathit{L}\mathit{A}\mathbf{\in }\mathit{P}\mathit{S}\mathit{P}\mathit{A}\mathit{C}\mathit{E}$.

b. Explain why this argument fails to show that$\mathit{M}\mathit{I}\mathit{N}\mathbf{-}\mathit{F}\mathit{O}\mathit{R}\mathit{M}\mathit{U}\mathit{L}\mathit{A}\mathbf{\in }\mathit{c}\mathit{o}\mathit{N}\mathit{P}$.If $\mathit{\varphi }\mathbf{\in }\mathit{M}\mathit{I}\mathit{N}\mathbf{-}\mathit{F}\mathit{O}\mathit{R}\mathit{M}\mathit{U}\mathit{L}\mathit{A}$, then$\mathit{\varphi }$ has a smaller equivalent formula. An NTM can verify that$\mathit{\varphi }\mathbf{\in }\mathit{M}\mathit{I}\mathit{N}\mathbf{-}\mathit{F}\mathit{O}\mathit{R}\mathit{M}\mathit{U}\mathit{L}\mathit{A}$ by guessing that formula.

Two Boolean formulae are the same if they contain the same set of components and are valid for the same set of operations. When $P=NP$, the following condition is valid for :

$MIN-FORMULA\subset P$.

(a)

For the input $\varphi$, construct the following machine $M$.

$M=\text{'}Oninput\varphi$

1. Any of the formulas smaller than f will be used next..

2. We'll go over each of the $\varphi$variable tasks in the following order:

We'll keep going till the task with the following formula f differs.

If this is not the case, the task will be completed.

When they are identical, $\varphi$both $f$and will be rejected.

The machine will be approved at that point.

Both the labour and the formula will require storage. Because when the preceding technique is one dimensional in terms of proportions, only one formula and one task will be saved.

Thus, it shows that role="math" localid="1663241725448" $MIN-FORMULA\in PSPACE$.

(b)

The condition$\varphi \in MIN-FORMULA$

Here,$\varphi$has smaller equivalents.

• An NTM verifies the following by using the formula.
• $\varphi \in MIN-FORMULA$
  
• To test and validate if two formulations are same or not, exponential time is required. When $\varphi \in MIN-FORMULA$.
• Even though $\varphi \in MIN-FORMULA$is true, the formula There is no NP informat in the formula f.
• Due to the fact that f cannot be verified in multinomial time. This is the case. The argument fails to demonstrate that $MIN-FORMULA\in coNP$.

Therefore, it shows that $MIN-FORMULA\in PSPACE$. And that the argument fails to show that $MIN-FORMULA\in coNP$.