Question

If R is a regular expression, let ${\mathit{R}}^{\mathbf{\{}\mathbf{m}\mathbf{,}\mathbf{n}\mathbf{\}}}$ represent the expression

${\mathit{R}}^{\mathbf{m}}\mathbf{\cup }{\mathit{R}}^{\mathbf{m}\mathbf{+}\mathbf{1}}\mathbf{\cup }\mathbf{.}\mathbf{..}\mathbf{\cup }{\mathit{R}}^{\mathbf{n}}$

Show how to implement the ${\mathit{R}}^{\mathbf{\{}\mathbf{m}\mathbf{,}\mathbf{n}\mathbf{\}}}$ operator, using the ordinary exponentiation operator, but without “· · · ”.

Short answer

Their information processing record is a series of settings that all the program goes through it while evaluating the inputs.

Step-by-step solution

Regular Expression with ordinary operator

Consider the following Regular expression:

$R^n\cup R^{m+1}\cup \mathrm{...}\cup R^n$

Here, user needs to be proved the implementation of operator with the help of the ordinary operator but without using “…”.

Computation configuration history

This may have been demonstrated using calculation histories.

Its data processing history is indeed the succession of configurations that even the system does while analyzing the input.

M seems to be a Turing machine that has been considered. Its Turing machine somehow doesn't approve or disapprove a string if something doesn't come to a standstill on that input.

Calculation history for a certain input may be computed using predictable machines.

As a result, it may be claimed that almost all of the states in the Turing machine are essentially countable, as seems to be the organization of all the other symbols on the tape.

As a result, the Turing machine's combination of the two sets is said to have been countable.

As it has been established that all sequences are countable, the start of operations of the this series may be expressed as follows:

$R^{m,n}={\sum }_{m=0}^n\cup {\left(A\right)}^n$