Skip to main content

Timeline for Regular language with pumping lemma

Current License: CC BY-SA 3.0

4 events
when toggle format what by license comment
Sep 23, 2016 at 18:13 history edited Hendrik Jan CC BY-SA 3.0
empty language gives me an headache.
Sep 23, 2016 at 18:01 comment added Hendrik Jan @JanDvorak Exactly. When replacing in the automaton for the concatenation all transitions from one alphabet by $\epsilon$ transitions one obtains a finite state automaton for the other half. If needed these $\epsilon$ can be removed by standard techniques. (Just removing the transitions leaves us with the task of deciding where to begin/end.)
Sep 23, 2016 at 16:51 comment added John Dvorak To prove the theorem, consider a FSM recognizing L1L2 and remove from it the transition states corresponding to one of L1 or L2?
Sep 23, 2016 at 13:00 history answered Hendrik Jan CC BY-SA 3.0