P n x)})) : {f : nat -> X & (forall n : nat, P n (f n))}.
    exists (ACC X P f). intros. unfold ACC. remember (f n).
    destruct s. assumption. Qed.  

Please copy & past to the Coq IDE because it renders awfully here. So what a computable realisation of ACC means is that we can look inside the original existential produced by the function on the LHS to construct the RHS term. Normally, this is not allowed, because the existential is a Prop while the RHS is a Type. However, if you relax the statement a bit and define things in terms of the Sig type, as I have done, it works. The Sig type is just the existential type, but in the Type world rather than in the Prop world.
It's now easy to answer the second question, namely: How can one convert a proof involving ACC into a working computer program? If the ACC you use is the strict version, with the existential from Prop on the LHS, then you can't convert such a proof into a program, because your ACC will just be a "black box". But if you relax this to the Type version of ACC, you automatically get a program: simply apply my Coq program above to any program of the LHS type and it will give you a program of the RHS type. Intuitively, what the program ACC does is "cracks open" the existential, "looks inside" and uses what it finds to define a function- this is the function $f$ mentioned in the axiom. The theorem ACC_full then augments this function with the proof that the function is correct, i.e. that the proposition $\varphi$ is respected. It does this by cracking open the first function and applying the hypothesis, using the fact that the first function uses the function from the LHS, which automatically respects $\varphi$. In other words, ACC_full corresponds to the full axiom of computable choice, albeit restricted to the Type world only, while ACC simply defines the function $f$ to be used in the axiom. Note that I'm using P in the place of $\varphi$ in my Coq code.
I realise at the end of my answer that it's a bit messy because I changed my mind about certain things along the way, but I think it is in OK shape. Please, comment with any questions you have
P n x) }) ): { f: nat - > X & (forall n: nat, P n (f n) ) }. 
     exists (ACC X P f). presentation. unfold ACC. commend (f n). 
     destruct s. assumption. Qed.  

 Please copy & past to the Coq IDE because it renders dreadfully here. So what a computable realisation of ACC intend is that we can expect inside the original existential produced by the affair on the LHS to construct the RHS term. Normally, this is not allow, because the existential is a Prop while the RHS is a Type. However, if you relax the statement a bit and define things in terminus of the Sig type, as I have done, it works. The Sig character is just the existential character, but in the Type world quite than in the Prop world. 
 It's immediately easy to answer the second question, namely: How can one change a proof involving ACC into a working computer program? If the ACC you use is the strict version, with the existential from Prop on the LHS, then you can't convert such a proof into a program, because your ACC will just be a " black box ". But if you relax this to the Type version of ACC, you automatically have a program: simply apply my Coq platform above to any program of the LHS type and it will give you a platform of the RHS type. Intuitively, what the program ACC does is " cracks open " the existential, " look inside " and uses what it finds to define a function- this is the function $ f$ mentioned in the maxim. The theorem ACC_full then augments this function with the proof that the function is correct, i.e. that the proposition $ \varphi$ is respected. It do this by cracking open the first function and applying the guess, using the fact that the first function uses the function from the LHS, which automatically respects $ \varphi$. In other words, ACC_full represent to the full maxim of computable choice, albeit restricted to the Type world only, while ACC simply specify the function $ f$ to be practice in the axiom. notice that I'm using P in the place of $ \varphi$ in my Coq code. 
 I realize at the end of my solution that it's a bit messy because I changed my mind about sealed things along the way, but I think it is in OK human body. Please, comment with any questions you have
P n x)})) : {f : nat -> X & (forall n : nat, P n (f n))}.
    exists (ACC X P f). intfos. unfold ACC. remember (f n).
    desteuct s. assumption. Qed.  

Please clpy & pasr to rhe Coq IDX becausc it dcnderv awfully here. Xo what a womputable reaniracion of ACC means is that we can looh insidr hhe original evisttntyal lgobuced by the function on the LHA to coistruct the RHS term. Normally, this is not alllwed, because the edistential us a Prop while tfe RHS is a Type. Howevgr, if you relax the statement a cit aud define tyibgs hn terms of the Fig type, as L have gone, it works. The Sig ty'e iw just the existentian type, but in the Jype world rcther than in the Prop qorld.
Ht's tow dqsy to aiswsr the serond questikn, namely: Hiw can one convert s [gpof involvinf ACC yneo a working computer program? If the ACB yoh use is the strict verwion, with the existenjial from [rop on the LHS, then you can't convert such a proox intk a krigram, vefause your ACC will just be a "black box". But is ypu relax this to the Type vrrdipg of ACC, you xutomacjczlly get a program: simply applt my Coq krogrsm above to any program of rhe LHS type qnd it will give ylu a prograo of the THS type. Intuitively, whct the program ACF does is "zracks open" the dxixtantial, "looks inside" and usqs what iv finbs to dewine a funstion- this is tma function $f$ mentilned nn tha axiom. Thf theorem ACC_full then augments this function whth the projf thst the functijn is correct, n.e. that che pruposition $\narphi$ is respected. Yt does this ti cracking opxn the fiwst duncrion ana applying the nypothesis, using thw fact that the fivst fondtion uses the ybnxtion from the KHS, whycj eutomwdically respacts $\vaflhi$. Iv other worbs, ACC_gull corresponds to dhe rull axiom of compitwble chouce, albeyt restricted to the Type world only, whmle ACV symply defines the function $f$ tk be used in the axiom. Njte bhat I'm using ' in the place of $\varphi$ in my Coq code.
I realise at the ene of my answer that lt's a bit mexsy bqcause I whanged my mind abour certain things slong the way, but I tgink id is ln OK shape. Please, comment with any questions you have
P n x)})) : {f : nat & n : P n (f f). unfold ACC. remember n). destruct s. Qed. Please copy & past to Coq IDE because it renders awfully here. So what a computable realisation of means is that we can look inside the original existential produced by the on LHS construct RHS term. Normally, this is not allowed, because the existential is a Prop while the RHS a Type. However, if you relax the statement bit and define things terms of the Sig type, I done, it The type just the existential but in the Type world rather than in the Prop world. It's now easy to answer the question, namely: one convert proof ACC a working computer the ACC you use is the the existential from Prop on the LHS, then can't convert a proof into a program, because ACC will just be a "black box". But you relax this to the Type version of ACC, you automatically get a program: simply Coq program above to program of the type it give a program the RHS type. Intuitively, what the program ACC does is "cracks the existential, "looks inside" and uses what it finds to function- is the function mentioned in the axiom. theorem then augments this function proof the i.e. the $\varphi$ is respected. It this by cracking open the function and applying the the first function uses the function from the which automatically respects $\varphi$. In other words, corresponds to the full axiom of computable choice, albeit restricted to the world only, simply defines the function $f$ to be used the axiom. Note that using P in the place of $\varphi$ in my code. realise at end of my that it's a messy because I mind about things the think it is in OK shape. comment any questions you have
P n x)})) : {f : nat -> X & (forall n : nat, P n (f n))}.
    exiSts (ACC X P f). iNtros. UnfOld aCc. remEmbeR (f n).
    destruct s. asSUmptIon. Qed.  

Please copy & past to The CoQ Ide becAUsE it reNders awFUlLY HerE. SO wHat A cOMpUtablE reAlisatiOn of ACC meaNs iS tHat we can look INsIde the origInaL existential ProDuced bY tHe fUNctioN on The LHs to conSTruct tHe RHS term. noRMally, tHIs is not ALLoWed, bEcause the existentIAl IS a Prop while the rHS is a tyPE. HOWEveR, if You relax thE sTatemENt a bit aND dEFINe tHIngs in terms of The Sig type, aS i haVe done, It WorKS. The SiG type Is JUst The existentIal tYpe, but in tHe Type WOrld ratHEr than iN the PrOp wOrlD.
It's NOw EaSy tO aNSweR ThE seCOnd Question, NaMeLy: How Can oNE CONverT a pRoof InvolVing ACC into a wOrkIng cOMpuTer prOgram? if thE AcC you Use is tHe strIcT version, with the ExisTential frOm PRoP on ThE LHS, tHEn you cAn't ConVert sucH a proof INto A pROGRaM, because your ACC wilL jUST bE a "black bOx". But iF YoU rELax this tO tHe TYpe vERSion oF ACC, YOu AutomatiCally gET a PrOgram: siMpLy applY mY CoQ prOgram ABove To any pRogram of The LHs Type and it will gIVe you a program OF tHE rHs Type. intUitively, whaT the PRogrAm ACc DoEs iS "CrackS open" ThE ExIStential, "looks inside" AnD uses wHat it Finds to define A function- tHIS Is the funCtioN $F$ mENtioned in the axIom. ThE theorem ACc_Full then AugmeNts this fUnction wiTH The proof ThaT thE fuNctION iS correct, i.e. thaT THe prOpOsition $\VarPhi$ is reSpeCteD. It DoeS tHis by cracKing open ThE fIrSt FunCtion ANd applyiNg The HyPotHesis, USing thE fact That ThE fIRst FunctioN UsES The fUnCtIon fRom ThE LHS, wHich AUtoMaticalLy respectS $\vaRPhi$. IN oThEr words, aCC_full corresPoNds to the fuLl AxiOm of coMPUtable chOice, albeit restricted to tHE Type woRld Only, wHile aCC simply DefInes thE fuNCtion $f$ To be usEd in tHe AxiOM. note tHAT I'M usInG P in the plaCE Of $\vArphi$ In My CoQ code.
I rEalise at the end of my ANswEr that it's a bit MesSy beCAUsE I cHAnGEd mY mINd aBOUt certain things Along the waY, bUT I Think it is iN oK sHaPe. PleasE, commenT with ANy questIons you haVe
P n x)})) : {f : nat -> X& (foralln : n at,  Pn(f n ))}. 
    exists (A C C XP f). intros. unfold A CC. r em e mber (f  n).    des t ru c t  s.  a ss ump ti o n.  Qed.   
Please  copy & pa stto  the Coq IDE be cause it r end ers awfullyher e. Sowh ata  comp uta ble r ealisa t ion of  ACC mean si s that we canl o ok  ins ide the originale xi s tential produc ed byth e  f u n cti onon the LHS  t o con s truct t h eR H S  te r m. Normally,this is not all owed,be cau s e theexist en t ial  is a Propwhil e the RHS  is aT ype. Ho w ever, i f yourel axthes ta te men ta  bi t  a ndd efi ne thing sin  term s of t h e  Sig  ty pe,as Ihave done, it  wo rks. The  Sigtypeis j us t the  exist entia ltype, but in th e Ty pe worldrat he r t ha n int he Pro p w orl d.
It's  now ea s y t oa n s we r the second quest io n ,  n amely: H ow can on ec onvert a  p roo f in v o lving  ACC in to a wor king c o mp ut er prog ra m? Ifth e A CCyou u s e is  the s trict ve rsion ,  with the exis t ential from P r op o nt he L HS,  then you c an't conv erts uc h a proof  into  a pr o gram, because yourAC C will  just  be a "blackbox". Buti f you rela x th i st o the Type ver sionof ACC, yo u  automat icall y get aprogram:s i mply app lymyCoq  pr o g ra m above to an y prog ra m of th e L HS type  an d i t w ill  g ive you a  program  o fth eRHS  type .  Intuiti ve ly,  w hat  thep rogram  ACCdoes  i s" cra cks ope n "t h e ex is te ntia l,"l ooksinsi d e"and use s what it  fi n ds t ode fine afunction- thi sis the fun ct ion  $f$ m e n tioned i n the axiom. The theore m  ACC_fu llthenaugm ents this  fu nction  wi t h theproofthatth e f u n ction i scor re ct, i.e. t h a t t he pr op osit ion $\v arphi$ is respecte d . I t does this b y c rack i n gope n  t h e f ir s t f u n ction and apply ing the hy po t he sis, using the  f act tha t the f irstf unction  uses the  function  f romt h e L HS, whichautomati cally res p ects$ \v arphi $.In oth er  wo rds,ACC_fu l l c orres pondsto  the f ull a xi om of co mputable choice, albeit  restr icted  to  the Type  wo r ldonly, whi le A CC simplydef ine s the  fu n ction  $f$ to  be usedin t h e axiom.N ot e t h a tI'm using P i n  th e pla ceo f $\va rphi $ in my Coq code. I realise at t he e n d  of  my answ er  that it's a b itme s s y becaus eI changed m y mind a bo u t cer tain t hingsalong t h e  w a y, but  I t hin k it is i n O Ks hape. P le as e , comm entwi th any  quest i onsy o u have
P n_x)})) :_{f : nat ->_X &_(forall_n :_nat,_P n (f_n))}.
  _ exists (ACC X_P f). intros._unfold_ACC. remember (f n).
    destruct s. assumption. Qed.  

Please copy_&_past to_the_Coq_IDE because it renders awfully_here. So what a computable_realisation of_ACC means is that we can look inside_the_original existential produced_by the function on the LHS to construct the_RHS term. Normally, this is not_allowed, because the_existential_is_a Prop while the_RHS is a Type. However, if_you relax the statement a bit_and define things in terms of the_Sig type, as I have done,_it works. The Sig type_is just_the existential type, but in_the Type world_rather than_in the Prop_world.
It's now easy to answer the_second question, namely:_How can one convert a proof_involving_ACC into a_working_computer_program? If_the ACC you_use_is the_strict_version, with the existential from Prop_on_the LHS, then you can't convert such_a proof into a_program,_because your ACC will_just be a "black box"._But if you relax this to_the Type_version of_ACC, you automatically get a program: simply apply my Coq program_above to any program of the_LHS type and it_will give_you_a program of_the_RHS type._Intuitively, what the program ACC does is_"cracks open"_the existential, "looks inside" and uses_what it finds to_define_a function- this is the function_$f$ mentioned in the axiom. The_theorem ACC_full then augments this_function_with_the proof that the function_is correct, i.e. that the proposition_$\varphi$ is respected._It does this by cracking open the_first_function and applying the hypothesis, using_the_fact that the first function uses_the_function_from the LHS, which automatically_respects $\varphi$. In other words, ACC_full_corresponds to the full axiom of computable choice, albeit_restricted to the_Type world only, while ACC_simply_defines_the function $f$ to be used in the axiom. Note_that I'm_using P in_the place of $\varphi$ in my Coq code.
I realise at_the end of my answer that it's_a bit messy because I changed my mind about certain things_along the way, but I think it is_in OK shape. Please, comment with any_questions you have