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It presents two brand-new concepts. For starters, the inductively defined set can be built in more than two methods (in this case, there are three).

create an expression). Second, it introduces the concept of a proof (or a demonstration).

It's possible that the proving phase will have to be broken into two or more cases. Induction is one of the methods.

It will be necessary to divide the steps. Because of the nature of the situation, there is a schism. Distribute function definition It's worth noting that it only applies to a few forms.

Expressions is a collection of expressions.

Take a look at the definition of expressions and several equations that describe

them.

Expressions, two functions (value and distribute)

These are the ten facts you'll be able to use.

1) If n is an integer, then (Const $n$) is an Expression.

2) It x and y are Expression, then (Plus x y ) is an Expression.

3) It x and y are Expression, then (Times x y) is an Expression.

4) Nothing else is an Expression.

An example Expression: $\left(Times\left(Const3\right)\left(Plus\left(Const1\right)\left(Const6\right)\right)\right)$

5) value (Const n) = n

6) value (Plus x y) =$\left(valuex\right)+\left(valuey\right)$

7) value (Times x y) =$\left(valuex\right)*\left(valuey\right)$

8) distribute$\left(Timesx\left(Plusyz\right)\right)=Plus\left(Timesxy\right)\left(Timesxz\right)$

9) distribute x = x -- for every other shape of its argument

10)$\left(x*a\right)+\left(x*b\right)=x*\left(a+b\right)$

A distributive law of addition over multiplication Here is what you need to prove. Prove by induction that value$\left(distributex\right)=valuex.$

Here this is prove by induction that value$\left(distributex\right)=valuex.$

The following steps are applied in the solution.

鈥� Write down what you are to prove.

鈥� Make a formula (or function) parameterized by the induction variable.

鈥� Here, need to prove several cases. One for each way an Expression can be constructed. Express each case using your formula from above.

鈥� Then work through the steps for each case.

鈥� In one of the induction cases you will need to do a case analysis over one of the variables. Clearly state when you do so.

Let , $A=\left\{<R>|RisaregularexpressiondescribingalanguageL\right\}.$

L contains at least one string that has 111 as a sub string

$(i.e.w鈭�Landw=x111y,forsomexandy).$

Here prove that A is decidable.

Use the facts that regular languages are closed under complement and intersection, and that certain properties of DFA鈥檚 are decidable.

$INFINIT{E}_{PDA}=\left\{\left(M\right)|MisaPDAandL\left(M\right)isaninfinitelanguage\right\}.$

On input $<M>:$

1. Repeat until there are transitions:

a. Mark current state and delete the transition that moved you to the state

b. If an already marked state is reached, accept

2. If no other transition exists, reject

Here, $INFINIT{E}_{PDA}=\left\{\left(M\right)|MisaPDAandL\left(M\right)isaninfinitelanguage\right\}.$

INFINITE_PDA is decidable is proved.