Answer :
The statement is disproven as it is possible for L₁ U L₂ to be regular while L₁ and L₂ are non-regular languages.
A regular language can be recognized by a finite automaton, whereas a non-regular language cannot. To prove the given statement, consider the following:
- It is possible for the union of two non-regular languages to be regular. For instance, let L₁ = {aⁿbⁿ | n ≥ 0} and L₂ = {bⁿaⁿ | n ≥ 0}. Both L₁ and L₂ are non-regular as they cannot be recognized by finite automata.
- However, L₁ U L₂ = {aⁿbⁿ | n ≥ 0} U {bⁿaⁿ | n ≥ 0} can be regular if n = 1, since the finite automaton can distinguish between strings 'ab' and 'ba'.
- More importantly, there are specific constructions where the union remains non-regular, highlighting that the regularity of the union is not guaranteed simply by the non-regularity of component languages.
Thus, the statement is disproven as it is possible for L₁ U L₂ to be regular even if L₁ and L₂ are non-regular.
