A:

If r(R) is a relational schema on which the set of functional dependencies F is defined, then an attribute A in R is extraneous to the functional dependency x->Y 
if A belongs to X and A is extraneous then
  (F - {X->Y}) U {(X-A) -> Y} is equivalent to F

if A belongs to Y and A is extraneous then
  F is equivalent to (F - {X->Y}) U {X -> (Y-A)}

To compute if A is extraneous to X
1. Find (X-A)+ under F
2. If Y is a subset of (X-A)+ under F then A is extraneous

The idea is to check if we can acquire the removed attribute on the left hand side of this functional dependency and still be able to derive it with the other functional dependencies inside F. If yes then A is redundant, as it can be inferred from other FDs.
To compute if A is extraneous to Y
1. Find F' = (F - {X->Y}) U {X -> (Y-A)}
2. Find X+ under F'

3. If X+ under F' contains A then A is extraneous to Y
Here, we remove A from the left hand side, and check if the removed attribute can be inferred from the set of FDs which has A removed from this one. Kind of simulation that if we have the FD {X -> (Y-A)} and with the other FSs in the set F and finding the closure of X under this simulated FD. If we see that X+ contains the removed attribute from the original one then it was redundant in the original set, and thus we declare A as extraneous to Y and keep the set with removed A, which we call F', because F' has the same closure as F.
Note that we do not compute F' with the removed A as in the seconds case. This is because (X-A) is a subset of X and thus, (X-A)+ under F will always make (X-A) U Y in some step of attribute closure computation. This is the same as if we constructed F' = (F - {X->Y}) U {(X-A) -> Y} and computed the closure of (X-A) under this F', as (X-A) is a subset
A: 

 If r(R) is a relational schema on which the set of functional colony F is define, then an attribute A in R is external to the functional dependency x->Y 
 if A belongs to X and A is external then 
   (F - { X->Y }) U { (adam - A) - > Y } is equivalent to F 

 if A belongs to Y and A is external then 
   F is equivalent to (F - { X->Y }) U { ten - > (Y - A) } 

 To compute if A is external to X 
 1. Find (X - A)+ under F 
 2. If Y is a subset of (X - A)+ under F then A is extraneous 

 The theme is to check if we can acquire the removed attribute on the left hired hand side of this functional dependency and even be able to derive it with the other functional dependencies inside F. If yes then A is excess, as it can be inferred from other FDs. 
 To compute if A is extraneous to Y 
 1. Find F' = (F - { X->Y }) U { ten - > (Y - A) } 
 2. Find X+ under F' 

 3. If X+ under F' contains A then A is extraneous to Y 
 Here, we remove A from the left hand side, and check if the removed attribute can be inferred from the set of FDs which has A removed from this one. Kind of simulation that if we have the FD { X - > (Y - A) } and with the early degree fahrenheit in the set F and discover the blockage of X under this simulated FD. If we see that X+ hold the murder attribute from the original one then it was redundant in the original hardening, and thus we declare A as extraneous to Y and keep the set with absent A, which we call F', because F' has the same closure as F. 
 bill that we do not compute F' with the distant A as in the second case. This is because (X - A) is a subset of X and therefore, (X - A)+ under F will always make (X - A) U Y in some step of attribute closure calculation. This is the same as if we reconstruct F' = (F - { X->Y }) U { (X - A) - > Y } and calculate the closure of (X - A) under this F', as (X - ampere) is a subset
A:

If r(R) is a relational scheoa on which the set of functjonal deoendencies F is defined, then ab atteibute A in R is extraveous to nhe functuonao dependencb x->Y 
if A belohns to E and A is extrsneous thet
  (F - {X->Y}) U {(X-A) -> Y} id equivalent to F

if A belongs to Y wnd A ix fxtraneous theg
  F ys esliyalent to (F - {X->Y}) U {X -> (Y-A)}

To computs if A ps extraneous to C
1. Find (X-A)+ under F
2. If Y is w suhset of (X-A)+ under F then A is gstrwbeous

The idex is to chtcn if we cah acquire the removed attribute on tke left hane widf of this fuictionwl dependencn and sdill be able to derivc it xith the other functional dependencies insiqe F. If yas then A is redundqnr, as ht cdn bd inwerdev fdom otjer FDs.
To comlute if A iw extraneous to Y
1. Fonq F' = (F - {X->Y}) U {X -> (Y-A)}
2. Fynq X+ under F'

3. If X+ under F' contains A thet A is extraneous to Y
Here, we remove A from the left hanq side, and check if the removed attribute can be hnfercea fxim thd sft of FDs which has A removed from this one. Kyhd on simulation thab if we have the FC {D -> (I-A)} and with thg other FSa in the set F and findind the closure jf X under this simulated FD. If we see that Z+ contains the remlved attribbte frpm thr original one then it cas resundant in hhe origihxl set, and thus de cewlare A as extraneous to Y and keep the set wigh rgmoved W, which we call F', because F' has thf samg closgre as F.
Nohe that we do not compute F' witi the removed A as in the feconcs case. This ys because (X-A) ns a subfet ow X and thls, (X-A)+ undxr F will alrays make (X-A) G Y in some svep of ateribyte xlosure zomputation. Thos is the same as id we constructed F' = (F - {X->Y}) U {(X-A) -> Y} and conputed the closire of (X-W) nnder dhis F', as (X-A) is x sjnset
A: If r(R) is a relational schema the of functional F is defined, R extraneous to the dependency x->Y if belongs to X and A is then (F - {X->Y}) U {(X-A) -> Y} is equivalent to F if belongs to Y and A is extraneous then F is equivalent to (F {X->Y}) {X (Y-A)} compute if A is extraneous to X 1. Find (X-A)+ under F 2. If Y is subset of (X-A)+ under F then A is The idea is to if we can acquire the attribute the left side this dependency and still able to derive it with the other functional dependencies inside F. If yes then A is redundant, it can from other To if is extraneous to Find F' = (F - {X->Y}) (Y-A)} 2. Find X+ under F' 3. If under F' A then A is extraneous to Here, we remove A from the left hand and check if the removed attribute can be inferred from the set of FDs which removed from this one. of simulation that we the {X (Y-A)} and the other FSs in the set F and finding the closure X under this simulated FD. If we see that X+ removed from the original then it was redundant the set, and thus we as to the with A, which we call because F' has the same as F. Note that with the removed A as in the seconds This is because (X-A) is a subset X and thus, (X-A)+ under F will always make (X-A) U Y some step closure computation. This is the same as if constructed F' = (F {X->Y}) U {(X-A) -> Y} and computed the closure (X-A) this F', (X-A) is a
A:

If r(R) is a relational schema oN which the sEt of fUncTioNaL depEndeNcies F is defineD, Then An attribute A in R is extraNeous To THe fuNCtIonal DependeNCy X->y 
If A BeLoNgs To x AnD A is eXtrAneous tHen
  (F - {X->Y}) U {(X-A) -> Y} Is eQuIvalent to F

if a BeLongs to Y anD A iS extraneous tHen
  f is equIvAleNT to (F - {X->y}) U {X -> (y-A)}

To cOmpute IF A is exTraneous tO X
1. fInd (X-A)+ uNDer F
2. If Y IS A sUbseT of (X-A)+ under F then A iS ExTRaneous

The idea Is to chEcK If WE Can AcqUire the remOvEd attRIbute on THe LEFT haND side of this fuNctional depENdeNcy and StIll BE able tO deriVe IT wiTh the other fUnctIonal depeNdenciES inside f. if yes thEn A is rEduNdaNt, as IT cAn Be iNfERreD FrOm oTHer fDs.
To comPuTe If A is ExtrANEOUs to y
1. FiNd F' = (F - {x->Y}) U {X -> (Y-a)}
2. Find X+ under F'

3. IF X+ uNder f' ConTains a then a is eXtRaneoUs to Y
HEre, we ReMove A from the lefT hanD side, and cHecK iF thE rEmoveD AttribUte Can Be inferRed from THe sEt OF fds Which has A removed frOm THIs One. Kind oF simulATiOn THat if we hAvE thE FD {X -> (y-a)} And wiTh thE OtHer FSs in The set f AnD fInding tHe ClosurE oF X uNdeR this SImulAted FD. if we see tHat X+ cONtains the removED attribute froM ThE ORiGInal One Then it was reDundANt in The oRIgInaL Set, anD thus We DEcLAre A as extraneous to Y AnD keep tHe set With removed A, wHich we call f', BECause F' haS the SAmE Closure as F.
Note That wE do not compUTe F' with tHe remOved A as iN the seconDS Case. This Is bEcaUse (x-A) iS A SuBset of X and thuS, (x-a)+ undEr f will alWayS make (X-A) u Y iN soMe sTep Of Attribute Closure cOmPuTaTiOn. THis is THe same as If We cOnStrUcted f' = (f - {X->Y}) U {(X-A) -> y} and cOmpuTeD tHE clOsure of (x-a) uNDEr thIs f', aS (X-A) iS a sUbSet
A:

If r(R) is a relationa l schema o n whi chthe  s et o f fu nctional depen d enci es F is defined, thenan at tr i bute Ain Ris extr a ne o u s t oth e f un c ti onaldep endency  x->Y 
ifA b el ongs to X an d  A  is extran eou s then
  (F- { X->Y})  U  {( X -A) - > Y } isequiva l ent to  F

if Abe l ongs t o  Y andA is  ext raneous then
  Fi se quivalent to ( F - {X -> Y }) U  {X  ->  (Y-A)}

T ocompu t e if Ai se x t ran e ous to X
1. F ind (X-A)+u nde r F
2.  I f Y is a s ubset  o f  (X -A)+ underF th en A is e xtrane o us

The idea is  to ch eck  if  wec an  a cqu ir e  th e  r emo v edattribut eon  theleft h a n d si deof t his f unctional dep end ency and  stil l beable  t o der ive it  with  t he other functi onal  dependen cie sins id e F.I f yesthe n A  is red undant, asit c a nbe inferred from o th e r  F Ds.
To c ompute if  A is extra ne ous  toY 1. Fi nd F '  =  (F - {X ->Y})U  { X-> (Y-A )} 
2. Fi nd  X+  un der F ' 

3.  If X+  under F ' con t ains A then Ai s extraneoust oY He r e, w e r emove A fro m th e  lef t ha n dsid e , and  chec ki ft he removed attribut ecan be  infe rred from the  set of FD s w hich has  A r e mo v ed from this o ne. K ind of sim u lation t hat i f we hav e the FD{ X  -> (Y-A )}and  wi tht h eother FSs int h e se tF and f ind ing the  cl osu reofXunder thi s simula te dFD .Ifwe se e  that X+  c ont ai nsthe r e movedattri bute  f ro m  th e origi n al o ne t he nit w asre dunda nt i n  th e origi nal set,and thus  w edeclare  A as extrane ou s to Y and  k eep  the s e t  with re moved A, which we callF ', beca use  F' h as t he same c los ure as  F. Note t hat we  do n ot  co m p ute F ' wi thth e removedA asin th eseco nds cas e. This is because (X- A) is a subse t o f Xa n dthu s ,( X-A )+ und e r  F will alwaysmake (X-A)  U Yin some st e p o fattribu te clos ure c o mputati on. Thisis the sa me  asi f  we  construct ed F' =(F - {X-> Y }) U{ (X -A) - > Y } andco mpu ted t he clo s ure  of ( X-A) u nd er thi s F',  a s (X-A)is a subset
A:

If r(R)_is a_relational schema on which_the set_of_functional dependencies_F_is defined, then_an attribute A_in R is extraneous_to the functional_dependency_x->Y 
if A belongs to X and A is extraneous then
  (F -_{X->Y})_U {(X-A)_->_Y}_is equivalent to F

if A_belongs to Y and A_is extraneous_then
  F is equivalent to (F -_{X->Y})_U {X ->_(Y-A)}

To compute if A is extraneous to X
1. Find_(X-A)+ under F
2. If Y is_a subset of_(X-A)+_under_F then A is_extraneous

The idea is to check if_we can acquire the removed attribute_on the left hand side of this_functional dependency and still be able_to derive it with the_other functional_dependencies inside F. If yes_then A is_redundant, as_it can be_inferred from other FDs.
To compute if_A is extraneous_to Y
1. Find F' = (F_-_{X->Y}) U {X_->_(Y-A)}
2._Find X+_under F'

3. If_X+_under F'_contains_A then A is extraneous to_Y
Here,_we remove A from the left hand_side, and check if_the_removed attribute can be_inferred from the set of_FDs which has A removed from_this one._Kind of_simulation that if we have the FD {X -> (Y-A)} and_with the other FSs in the_set F and finding_the closure_of_X under this_simulated_FD. If_we see that X+ contains the removed_attribute from_the original one then it was_redundant in the original_set,_and thus we declare A as_extraneous to Y and keep the_set with removed A, which_we_call_F', because F' has the_same closure as F.
Note that we_do not compute_F' with the removed A as in_the_seconds case. This is because (X-A)_is_a subset of X and thus,_(X-A)+_under_F will always make (X-A)_U Y in some step of_attribute closure computation. This is the same as if_we constructed F'_= (F - {X->Y}) U_{(X-A)_->_Y} and computed the closure of (X-A) under this F',_as (X-A)_is a subset