Show that a language is decidable if some enumerator enumerates the language in the standard string order.

A language $\mathrm{L}$ is recursively enumerable if $\mathrm{L}$ is the set of strings accepted by some TM.

If L is a recursive enumerable language, then:

• A TM stops in its final state if role="math" localid="1658301672430" $\mathbf{W}\mathbf{\in }\mathbf{L}$ .
• A TM stops in a non-final state or loops indefinitely if role="math" localid="1658301682670" $\mathbf{W}\mathbf{\notin }\mathbf{L}$.

If the language is recursive, then:

• A TM stops in its ultimate state if$\mathbf{W}\mathbf{\in }\mathbf{L}$ .
• The TM halts in a non-final state if$\mathbf{W}\mathbf{\notin }\mathbf{L}$ .

The enumerator works by creating the strings in basic word sequence, checking each one for status in $A$ using $M$, and outputting the string if it is in $A$. if $A$ is decidable from some TM $M$ . First investigate two scenarios if $A$ is enumerated by some enumerator $E$ in lexicographic order.

Because all finite languages are decidable, if $A$ is finite, it is decidable (just hardwire each of the strings into the TM).

If A is infinite, the behaviour of a TM $\mathit{M}$ that determines $A$ is as follows. $E$ is run to alphabets enumerate every string in $A$ up until a string that appears after w as soon as $M$ receives the input w. Since $A$ is infinite, this must eventually take place. If $W$ is already on the list, accept it; otherwise, reject it.

It is vital to take the situation where $A$ is finite into consideration individually since the enumerator may keep looping when enumerating a finite language. Consequently, to show the language's decidability without needing to use the enumerator of the language.

Hence, the language is decidable.