Proof of (w * w^R)^R = w*w^R

Question

Question: if w exist in T*, prove that (w * w^R)^R = w * w^R

Hi, I am new to theories of algorithms, and I am having trouble understanding how to prove this, if someone can hint me towards the right direction that would be much appreciated.

Note: R means the string is reversed, ex: (abc)^R = cba

Also Note: * means concatenation so (abc * def) = abcdef

It follows from 2 lemmas: 1) `(s^R)^R = s` (for any `s` in `T*`). And 2) `(u*v)^R = v^R * u^R` for any `u,v` in `T*`. Can you prove these two lemmas and then show how the result follows? — John Coleman, Jan 17 '17 at 03:51
Not a programming question - try http://cstheory.stackexchange.com ? — Paul R, Jan 17 '17 at 07:58

Dmitry Bychenko · Answer 1 · 2017-01-17T12:09:30.903

In general case, try using group theory (you even have a hint: "Also Note: * means concatenation so (abc * def) = abcdef"):

  a, b, c, ...          - (characters)    - group's elements (generators in fact)
  a * b       == ab     - (concatenation) - group's operation
  ε                     - (empty string)  - group's 1 
  a..z**-1    == z..a   - rule for the item reversing; a**-1 == a

So far, so good, string reversing is **-1 operation, for any ab...yz item we have:

  (ab...yz)((ab..yz)^R) == ab..yz * zy ..ba == ab..yzzy..ba

    since zz == z * z == ε (z == z**-1) 

  we have

    ab..yzzy..ba == 
    ab..yy..ba == 
    ab..ba ==
    ε

Your theorem is quite easy then: change string reversing to **-1 and have

  (w * w**-1)**-1=(w**-1)**-1 * w**-1 == w * w**-1

For this particular case the group we've built may be an overkill, however, it can be very useful when'll be solving the similar problems.

Quibble: this is a semigroup rather than a group. – John Coleman Jan 17 '17 at 12:01 — John Coleman, Jan 17 '17 at 12:01

Proof of (w * w^R)^R = w*w^R

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