I need help with some 2 questions , thank for help There will be L ⊆ Σ* , Prove or disprove: 1. If each RL equivalence class is a regular language, then RL contains an infinite equivalence class 2. If L is a regular language, then RL contains an infinite equivalence class
RL: x RL y ( for each z in Σ* , xz in L if and only if yz L)
- If each RL equivalence class is a regular language, then RL contains an infinite equivalence class
Consider the language 0^(2^n) for n >= 0: 0, 00, 0000, …. Every single string in this language is distinguishable: every equivalence class consists of a single string. Finite languages are always regular. Therefore, we have each RL equivalence class as a regular language, but none of the RL equivalence classes is infinite. The claim is therefore false, having been disproven by counterexample.
- If L is a regular language, then RL contains an infinite equivalence class
Because L is regular, we know that there are finitely many equivalence classes under RL - indeed, we know there is exactly one for each state in a minimal deterministic finite automaton for the language. Because all of the infinitely many strings over the alphabet must belong to one of these equivalence classes, it follows that at least one of the classes must contain infinitely many of those strings. Suppose this were not the case; then equivalence classes 1, 2, …, p contain only c[1], c[2], …, c[p] strings. But then only c[1] + c[2] + … + c[p] strings are in an equivalence class. This number is finite so we must have missed some strings. The claim is therefore true, having been proven by contradiction.
