An Example Proof using Identities of Regular Expressions

[@spoidermon4323]

no need to use E.R=R just see 1+00*1 take 1 out or common it becomes (E+00*)1 by {(P+Q)R=PR+QR}

[@Apoorvpandey]

good observation

[@itsmeakash_]

Good

[@shraddhatiwari2500]

Ya exactly .......

[@lehaswinipalla3365]

he did the same thing,just explained that 1 is E.R

[@divyaanshuagrawal3383]

Yeah that makes much more sense now! Thanks!

[@ashadas724]

My cllg teacher is also teaching me the same as this automata playlist I think my teacher is also learning from this neso academy 😂😂😂😂

[@ruthtamiru577]

yea mine too🤣🤣🤣

[@sakshidherange7554]

🤣😂

[@preetijha6968]

Yes .🤣🤣

[@sameen_17]

sameeeee🤣🤣💔

[@mesumhussain3566]

same bro xd

[@sanjeevkumarsingh4198]

in 4th line if you multiply with E then it should be like this E(1+00*1)(0+10*1)* but here we only need to do is take 1 as common so the 5th step will come out.

[@technicalspecial2770]

actually you are tracking 1 commen in the 5th steep, not the [€+00*] you are talking about. the process is correct but not the explanation

[@DolaLado]

Yes. That was confusing.

[@parwindersingh9151]

Thanks

[@tobito8897]

thanks for clearing the confusion

[@annawickramasinghe2983]

Thanks for clearing the confusion

[@zeeshanpatel5587]

Thanks, I was so confused

[@s0ge770]

4 hours left for exam Only hope is above

[@lakshaytalkstocomputer6649]

SSAAAAAMMMMEEEEEEE!!!!!!!!!!!!!!!!!!!!!!!!!

[@kadirdeniz2570]

Did you pass the exam ?

[@prateeksengar6634]

How did it go?

[@mahadevhatti5228]

@@lakshaytalkstocomputer6649 Is that " SAME " or " SHAME " ?

[@lakshaytalkstocomputer6649]

@@mahadevhatti5228 same bro, never shame

[@Tia-sy6zu]

Can anyone reconfirm my logic so in the first step we are taking (1+00*1) as common and in algebra we would have taken 1 to get the string again but here we used E or epsilon so we are basically trying to say that the common we took it has . operator of concatenation in it because e concatenated with the term would give the original term not multiplication as it seems while taking common

[@federicobianchi5191]

how do you take epsilon and 00* out? I can't get the formal logic of it

[@SandipHodkhasa]

from (1+00*1) , if you take out 1 common you are left with (1)(epsilon + 00*)

[@Harshal102]

Sandip Hodkhasa thank you so much.

[@javaexpertsa8947]

Thank you so much, i was searching the whole internet, because i couldn't understand this step. That make's sense now to me.

[@curias7]

where is the rule about that?

[@zyro9922]

@@SandipHodkhasa Thanks.

[@yashpandey1372]

In my university 4th sem exam same question come. And five questions exactly same of pumping lemma, melay machine of 2'complement , nfa to dfa

[@deepanshukumar5407]

Sir your teching method is very sime and intership

[@shaswatsoni3228]

please explain the 5th step?

[@AbhishekNigam]

Thanks for the video!

[@keshavraghav3896]

4:10 , yaha 1 ko common liya hai...

[@singh_1057]

Ya exactly

[@chabotaluputa7665]

Your amazing❤❤

[@vepakommavamsi]

Can this is also called as equivalence of two regular expression?

[@BetioLopezMenendez4]

Hey! Hope you be ok, I have a question, so you said that E.R = R, why E.1=E?

[@anshulgangwal4391]

Think of the '.' operator as string concatenation. Joining a null string(E) to another string(R) gives you R itself.

[@ChristianBurnsShafer]

He didn't use E.1=E, rather he factored out the 1 using identity (12). You are correct in believing E.1=1.

[@MrKyuubiJesus]

Well, he did factor it out, but regex is not commutative (ab != ba). To see this we can use the LHS to obtain a language that includes all of LSH. L = 10(0|1)* includes LSH. 01 is a string in RHS and not in L. Thus the languages are not equivalent.

[@dhanushsivajaya1356]

Thankyou sir

[@vepakommavamsi]

Can this is also called as equivalence of two regular expression

[@sainikhilkura5585]

(a*ab+ba)*a*=(a+ab+ba)* solve ths que usng the identitties

[@Explore-Fashioninlife]

Sir please make a another video with more examples. Please😫🙏🙏💓😫🙏🙏💓😫🙏🙏💓😫🙏🙏💓😫🙏🙏💓😫🙏🙏💓😫🙏

[@technicalspecial2770]

in 4th steep you miss to put the brackek

[@kadambalapavan2280]

in step -5 which one are we taking common E+00* or 1 ?

[@chetanbudhathoki1033]

he took 1 common

[@peacecop]

How? in THE leder Side, THE furst Can behöver 1, in THE right Side 0. THE expressionism Åre not equal

[@preetijha6968]

(a*ab+ba)*a* = (a+ab+ba)*. Prove this solution plzzz ..

[@supersakib62]

Where can I get the slides? Anybody has them??

[@rupakdutta8575]

I was confused at E+00* part after that I realized 1+1 = 1 because R+R = R

[@HamzaShahzadEB_]

Where are the questions like write an re for the set of string that contains only a and b ? Where are these types of questions?????

[@nasikchanana9658]

Loved u video lexture

[@whateverbehappy5206]

thnq sir

[@sameerkhatri997]

Where dose the one go

[@sajidsiddiqui4652]

45 min left for the exam 😢😢

[@ssm840]

is ε+r=r?? can anyone clarify?

[@LadderVictims]

if r doesnt contain e then r + e is not r

[@karthikmandelli1612]

6 hrs to go for a exam

[@MyAsdfqwe]

feels like gibberish with imaginary methods.

[@N12458]

imagination is more important than knowledge-Rick Astley

[@AhamedKabeer-wn1jb]

Thank you..

[@pashamshyamsundharreddy4438]

sir,€.1=1,,1+1 remains know sir taking common

[@bimalendusarma3289]

have confusing in that part....can anyone explain this.

[@bimalendusarma3289]

got it..

[@abdelrhmanahmed1378]

explain please

[@ChristianBurnsShafer]

He factored out the 1 using identity (12).

[@neiljohn2637]

49

[@marcaironcantal515]

Hello Guys

[@sia2225]

step 4 is not logically correct

[@omergmail9295]

confused

[@vepakommavamsi]

Plz reply

[@shahzadabbasi5225]

LOL. I think the logics you're following here are useless.