Left Termination of the query pattern reverse(f,b) w.r.t. the given Prolog program could not be shown:



↳ PROLOG
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

app3(.2(X, Xs), Ys, .2(X, Zs)) :- app3(Xs, Ys, Zs).
app3({}0, Ys, Ys).
reverse2(.2(X, Xs), Ys) :- reverse2(Xs, Zs), app3(Zs, .2(X, {}0), Ys).
reverse2({}0, {}0).


With regard to the inferred argument filtering the predicates were used in the following modes:
reverse2: (f,b) (f,f)
app3: (f,b,f) (f,b,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


reverse_2_in_ag2(._22(X, Xs), Ys) -> if_reverse_2_in_1_ag4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
reverse_2_in_aa2(._22(X, Xs), Ys) -> if_reverse_2_in_1_aa4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
reverse_2_in_aa2([]_0, []_0) -> reverse_2_out_aa2([]_0, []_0)
if_reverse_2_in_1_aa4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> if_reverse_2_in_2_aa5(X, Xs, Ys, Zs, app_3_in_aga3(Zs, ._22(X, []_0), Ys))
app_3_in_aga3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_aga5(X, Xs, Ys, Zs, app_3_in_aga3(Xs, Ys, Zs))
app_3_in_aga3([]_0, Ys, Ys) -> app_3_out_aga3([]_0, Ys, Ys)
if_app_3_in_1_aga5(X, Xs, Ys, Zs, app_3_out_aga3(Xs, Ys, Zs)) -> app_3_out_aga3(._22(X, Xs), Ys, ._22(X, Zs))
if_reverse_2_in_2_aa5(X, Xs, Ys, Zs, app_3_out_aga3(Zs, ._22(X, []_0), Ys)) -> reverse_2_out_aa2(._22(X, Xs), Ys)
if_reverse_2_in_1_ag4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> if_reverse_2_in_2_ag5(X, Xs, Ys, Zs, app_3_in_agg3(Zs, ._22(X, []_0), Ys))
app_3_in_agg3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_agg5(X, Xs, Ys, Zs, app_3_in_agg3(Xs, Ys, Zs))
app_3_in_agg3([]_0, Ys, Ys) -> app_3_out_agg3([]_0, Ys, Ys)
if_app_3_in_1_agg5(X, Xs, Ys, Zs, app_3_out_agg3(Xs, Ys, Zs)) -> app_3_out_agg3(._22(X, Xs), Ys, ._22(X, Zs))
if_reverse_2_in_2_ag5(X, Xs, Ys, Zs, app_3_out_agg3(Zs, ._22(X, []_0), Ys)) -> reverse_2_out_ag2(._22(X, Xs), Ys)
reverse_2_in_ag2([]_0, []_0) -> reverse_2_out_ag2([]_0, []_0)

The argument filtering Pi contains the following mapping:
reverse_2_in_ag2(x1, x2)  =  reverse_2_in_ag1(x2)
._22(x1, x2)  =  ._21(x2)
[]_0  =  []_0
if_reverse_2_in_1_ag4(x1, x2, x3, x4)  =  if_reverse_2_in_1_ag2(x3, x4)
reverse_2_in_aa2(x1, x2)  =  reverse_2_in_aa
if_reverse_2_in_1_aa4(x1, x2, x3, x4)  =  if_reverse_2_in_1_aa1(x4)
reverse_2_out_aa2(x1, x2)  =  reverse_2_out_aa1(x1)
if_reverse_2_in_2_aa5(x1, x2, x3, x4, x5)  =  if_reverse_2_in_2_aa2(x2, x5)
app_3_in_aga3(x1, x2, x3)  =  app_3_in_aga1(x2)
if_app_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_aga1(x5)
app_3_out_aga3(x1, x2, x3)  =  app_3_out_aga2(x1, x3)
if_reverse_2_in_2_ag5(x1, x2, x3, x4, x5)  =  if_reverse_2_in_2_ag2(x2, x5)
app_3_in_agg3(x1, x2, x3)  =  app_3_in_agg2(x2, x3)
if_app_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_agg1(x5)
app_3_out_agg3(x1, x2, x3)  =  app_3_out_agg1(x1)
reverse_2_out_ag2(x1, x2)  =  reverse_2_out_ag1(x1)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG



↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
  ↳ PrologToPiTRSProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

reverse_2_in_ag2(._22(X, Xs), Ys) -> if_reverse_2_in_1_ag4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
reverse_2_in_aa2(._22(X, Xs), Ys) -> if_reverse_2_in_1_aa4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
reverse_2_in_aa2([]_0, []_0) -> reverse_2_out_aa2([]_0, []_0)
if_reverse_2_in_1_aa4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> if_reverse_2_in_2_aa5(X, Xs, Ys, Zs, app_3_in_aga3(Zs, ._22(X, []_0), Ys))
app_3_in_aga3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_aga5(X, Xs, Ys, Zs, app_3_in_aga3(Xs, Ys, Zs))
app_3_in_aga3([]_0, Ys, Ys) -> app_3_out_aga3([]_0, Ys, Ys)
if_app_3_in_1_aga5(X, Xs, Ys, Zs, app_3_out_aga3(Xs, Ys, Zs)) -> app_3_out_aga3(._22(X, Xs), Ys, ._22(X, Zs))
if_reverse_2_in_2_aa5(X, Xs, Ys, Zs, app_3_out_aga3(Zs, ._22(X, []_0), Ys)) -> reverse_2_out_aa2(._22(X, Xs), Ys)
if_reverse_2_in_1_ag4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> if_reverse_2_in_2_ag5(X, Xs, Ys, Zs, app_3_in_agg3(Zs, ._22(X, []_0), Ys))
app_3_in_agg3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_agg5(X, Xs, Ys, Zs, app_3_in_agg3(Xs, Ys, Zs))
app_3_in_agg3([]_0, Ys, Ys) -> app_3_out_agg3([]_0, Ys, Ys)
if_app_3_in_1_agg5(X, Xs, Ys, Zs, app_3_out_agg3(Xs, Ys, Zs)) -> app_3_out_agg3(._22(X, Xs), Ys, ._22(X, Zs))
if_reverse_2_in_2_ag5(X, Xs, Ys, Zs, app_3_out_agg3(Zs, ._22(X, []_0), Ys)) -> reverse_2_out_ag2(._22(X, Xs), Ys)
reverse_2_in_ag2([]_0, []_0) -> reverse_2_out_ag2([]_0, []_0)

The argument filtering Pi contains the following mapping:
reverse_2_in_ag2(x1, x2)  =  reverse_2_in_ag1(x2)
._22(x1, x2)  =  ._21(x2)
[]_0  =  []_0
if_reverse_2_in_1_ag4(x1, x2, x3, x4)  =  if_reverse_2_in_1_ag2(x3, x4)
reverse_2_in_aa2(x1, x2)  =  reverse_2_in_aa
if_reverse_2_in_1_aa4(x1, x2, x3, x4)  =  if_reverse_2_in_1_aa1(x4)
reverse_2_out_aa2(x1, x2)  =  reverse_2_out_aa1(x1)
if_reverse_2_in_2_aa5(x1, x2, x3, x4, x5)  =  if_reverse_2_in_2_aa2(x2, x5)
app_3_in_aga3(x1, x2, x3)  =  app_3_in_aga1(x2)
if_app_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_aga1(x5)
app_3_out_aga3(x1, x2, x3)  =  app_3_out_aga2(x1, x3)
if_reverse_2_in_2_ag5(x1, x2, x3, x4, x5)  =  if_reverse_2_in_2_ag2(x2, x5)
app_3_in_agg3(x1, x2, x3)  =  app_3_in_agg2(x2, x3)
if_app_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_agg1(x5)
app_3_out_agg3(x1, x2, x3)  =  app_3_out_agg1(x1)
reverse_2_out_ag2(x1, x2)  =  reverse_2_out_ag1(x1)


Pi DP problem:
The TRS P consists of the following rules:

REVERSE_2_IN_AG2(._22(X, Xs), Ys) -> IF_REVERSE_2_IN_1_AG4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
REVERSE_2_IN_AG2(._22(X, Xs), Ys) -> REVERSE_2_IN_AA2(Xs, Zs)
REVERSE_2_IN_AA2(._22(X, Xs), Ys) -> IF_REVERSE_2_IN_1_AA4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
REVERSE_2_IN_AA2(._22(X, Xs), Ys) -> REVERSE_2_IN_AA2(Xs, Zs)
IF_REVERSE_2_IN_1_AA4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> IF_REVERSE_2_IN_2_AA5(X, Xs, Ys, Zs, app_3_in_aga3(Zs, ._22(X, []_0), Ys))
IF_REVERSE_2_IN_1_AA4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> APP_3_IN_AGA3(Zs, ._22(X, []_0), Ys)
APP_3_IN_AGA3(._22(X, Xs), Ys, ._22(X, Zs)) -> IF_APP_3_IN_1_AGA5(X, Xs, Ys, Zs, app_3_in_aga3(Xs, Ys, Zs))
APP_3_IN_AGA3(._22(X, Xs), Ys, ._22(X, Zs)) -> APP_3_IN_AGA3(Xs, Ys, Zs)
IF_REVERSE_2_IN_1_AG4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> IF_REVERSE_2_IN_2_AG5(X, Xs, Ys, Zs, app_3_in_agg3(Zs, ._22(X, []_0), Ys))
IF_REVERSE_2_IN_1_AG4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> APP_3_IN_AGG3(Zs, ._22(X, []_0), Ys)
APP_3_IN_AGG3(._22(X, Xs), Ys, ._22(X, Zs)) -> IF_APP_3_IN_1_AGG5(X, Xs, Ys, Zs, app_3_in_agg3(Xs, Ys, Zs))
APP_3_IN_AGG3(._22(X, Xs), Ys, ._22(X, Zs)) -> APP_3_IN_AGG3(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_2_in_ag2(._22(X, Xs), Ys) -> if_reverse_2_in_1_ag4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
reverse_2_in_aa2(._22(X, Xs), Ys) -> if_reverse_2_in_1_aa4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
reverse_2_in_aa2([]_0, []_0) -> reverse_2_out_aa2([]_0, []_0)
if_reverse_2_in_1_aa4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> if_reverse_2_in_2_aa5(X, Xs, Ys, Zs, app_3_in_aga3(Zs, ._22(X, []_0), Ys))
app_3_in_aga3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_aga5(X, Xs, Ys, Zs, app_3_in_aga3(Xs, Ys, Zs))
app_3_in_aga3([]_0, Ys, Ys) -> app_3_out_aga3([]_0, Ys, Ys)
if_app_3_in_1_aga5(X, Xs, Ys, Zs, app_3_out_aga3(Xs, Ys, Zs)) -> app_3_out_aga3(._22(X, Xs), Ys, ._22(X, Zs))
if_reverse_2_in_2_aa5(X, Xs, Ys, Zs, app_3_out_aga3(Zs, ._22(X, []_0), Ys)) -> reverse_2_out_aa2(._22(X, Xs), Ys)
if_reverse_2_in_1_ag4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> if_reverse_2_in_2_ag5(X, Xs, Ys, Zs, app_3_in_agg3(Zs, ._22(X, []_0), Ys))
app_3_in_agg3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_agg5(X, Xs, Ys, Zs, app_3_in_agg3(Xs, Ys, Zs))
app_3_in_agg3([]_0, Ys, Ys) -> app_3_out_agg3([]_0, Ys, Ys)
if_app_3_in_1_agg5(X, Xs, Ys, Zs, app_3_out_agg3(Xs, Ys, Zs)) -> app_3_out_agg3(._22(X, Xs), Ys, ._22(X, Zs))
if_reverse_2_in_2_ag5(X, Xs, Ys, Zs, app_3_out_agg3(Zs, ._22(X, []_0), Ys)) -> reverse_2_out_ag2(._22(X, Xs), Ys)
reverse_2_in_ag2([]_0, []_0) -> reverse_2_out_ag2([]_0, []_0)

The argument filtering Pi contains the following mapping:
reverse_2_in_ag2(x1, x2)  =  reverse_2_in_ag1(x2)
._22(x1, x2)  =  ._21(x2)
[]_0  =  []_0
if_reverse_2_in_1_ag4(x1, x2, x3, x4)  =  if_reverse_2_in_1_ag2(x3, x4)
reverse_2_in_aa2(x1, x2)  =  reverse_2_in_aa
if_reverse_2_in_1_aa4(x1, x2, x3, x4)  =  if_reverse_2_in_1_aa1(x4)
reverse_2_out_aa2(x1, x2)  =  reverse_2_out_aa1(x1)
if_reverse_2_in_2_aa5(x1, x2, x3, x4, x5)  =  if_reverse_2_in_2_aa2(x2, x5)
app_3_in_aga3(x1, x2, x3)  =  app_3_in_aga1(x2)
if_app_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_aga1(x5)
app_3_out_aga3(x1, x2, x3)  =  app_3_out_aga2(x1, x3)
if_reverse_2_in_2_ag5(x1, x2, x3, x4, x5)  =  if_reverse_2_in_2_ag2(x2, x5)
app_3_in_agg3(x1, x2, x3)  =  app_3_in_agg2(x2, x3)
if_app_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_agg1(x5)
app_3_out_agg3(x1, x2, x3)  =  app_3_out_agg1(x1)
reverse_2_out_ag2(x1, x2)  =  reverse_2_out_ag1(x1)
APP_3_IN_AGG3(x1, x2, x3)  =  APP_3_IN_AGG2(x2, x3)
APP_3_IN_AGA3(x1, x2, x3)  =  APP_3_IN_AGA1(x2)
IF_APP_3_IN_1_AGA5(x1, x2, x3, x4, x5)  =  IF_APP_3_IN_1_AGA1(x5)
IF_REVERSE_2_IN_1_AA4(x1, x2, x3, x4)  =  IF_REVERSE_2_IN_1_AA1(x4)
REVERSE_2_IN_AA2(x1, x2)  =  REVERSE_2_IN_AA
IF_APP_3_IN_1_AGG5(x1, x2, x3, x4, x5)  =  IF_APP_3_IN_1_AGG1(x5)
IF_REVERSE_2_IN_2_AG5(x1, x2, x3, x4, x5)  =  IF_REVERSE_2_IN_2_AG2(x2, x5)
REVERSE_2_IN_AG2(x1, x2)  =  REVERSE_2_IN_AG1(x2)
IF_REVERSE_2_IN_2_AA5(x1, x2, x3, x4, x5)  =  IF_REVERSE_2_IN_2_AA2(x2, x5)
IF_REVERSE_2_IN_1_AG4(x1, x2, x3, x4)  =  IF_REVERSE_2_IN_1_AG2(x3, x4)

We have to consider all (P,R,Pi)-chains

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
  ↳ PrologToPiTRSProof

Pi DP problem:
The TRS P consists of the following rules:

REVERSE_2_IN_AG2(._22(X, Xs), Ys) -> IF_REVERSE_2_IN_1_AG4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
REVERSE_2_IN_AG2(._22(X, Xs), Ys) -> REVERSE_2_IN_AA2(Xs, Zs)
REVERSE_2_IN_AA2(._22(X, Xs), Ys) -> IF_REVERSE_2_IN_1_AA4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
REVERSE_2_IN_AA2(._22(X, Xs), Ys) -> REVERSE_2_IN_AA2(Xs, Zs)
IF_REVERSE_2_IN_1_AA4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> IF_REVERSE_2_IN_2_AA5(X, Xs, Ys, Zs, app_3_in_aga3(Zs, ._22(X, []_0), Ys))
IF_REVERSE_2_IN_1_AA4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> APP_3_IN_AGA3(Zs, ._22(X, []_0), Ys)
APP_3_IN_AGA3(._22(X, Xs), Ys, ._22(X, Zs)) -> IF_APP_3_IN_1_AGA5(X, Xs, Ys, Zs, app_3_in_aga3(Xs, Ys, Zs))
APP_3_IN_AGA3(._22(X, Xs), Ys, ._22(X, Zs)) -> APP_3_IN_AGA3(Xs, Ys, Zs)
IF_REVERSE_2_IN_1_AG4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> IF_REVERSE_2_IN_2_AG5(X, Xs, Ys, Zs, app_3_in_agg3(Zs, ._22(X, []_0), Ys))
IF_REVERSE_2_IN_1_AG4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> APP_3_IN_AGG3(Zs, ._22(X, []_0), Ys)
APP_3_IN_AGG3(._22(X, Xs), Ys, ._22(X, Zs)) -> IF_APP_3_IN_1_AGG5(X, Xs, Ys, Zs, app_3_in_agg3(Xs, Ys, Zs))
APP_3_IN_AGG3(._22(X, Xs), Ys, ._22(X, Zs)) -> APP_3_IN_AGG3(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_2_in_ag2(._22(X, Xs), Ys) -> if_reverse_2_in_1_ag4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
reverse_2_in_aa2(._22(X, Xs), Ys) -> if_reverse_2_in_1_aa4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
reverse_2_in_aa2([]_0, []_0) -> reverse_2_out_aa2([]_0, []_0)
if_reverse_2_in_1_aa4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> if_reverse_2_in_2_aa5(X, Xs, Ys, Zs, app_3_in_aga3(Zs, ._22(X, []_0), Ys))
app_3_in_aga3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_aga5(X, Xs, Ys, Zs, app_3_in_aga3(Xs, Ys, Zs))
app_3_in_aga3([]_0, Ys, Ys) -> app_3_out_aga3([]_0, Ys, Ys)
if_app_3_in_1_aga5(X, Xs, Ys, Zs, app_3_out_aga3(Xs, Ys, Zs)) -> app_3_out_aga3(._22(X, Xs), Ys, ._22(X, Zs))
if_reverse_2_in_2_aa5(X, Xs, Ys, Zs, app_3_out_aga3(Zs, ._22(X, []_0), Ys)) -> reverse_2_out_aa2(._22(X, Xs), Ys)
if_reverse_2_in_1_ag4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> if_reverse_2_in_2_ag5(X, Xs, Ys, Zs, app_3_in_agg3(Zs, ._22(X, []_0), Ys))
app_3_in_agg3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_agg5(X, Xs, Ys, Zs, app_3_in_agg3(Xs, Ys, Zs))
app_3_in_agg3([]_0, Ys, Ys) -> app_3_out_agg3([]_0, Ys, Ys)
if_app_3_in_1_agg5(X, Xs, Ys, Zs, app_3_out_agg3(Xs, Ys, Zs)) -> app_3_out_agg3(._22(X, Xs), Ys, ._22(X, Zs))
if_reverse_2_in_2_ag5(X, Xs, Ys, Zs, app_3_out_agg3(Zs, ._22(X, []_0), Ys)) -> reverse_2_out_ag2(._22(X, Xs), Ys)
reverse_2_in_ag2([]_0, []_0) -> reverse_2_out_ag2([]_0, []_0)

The argument filtering Pi contains the following mapping:
reverse_2_in_ag2(x1, x2)  =  reverse_2_in_ag1(x2)
._22(x1, x2)  =  ._21(x2)
[]_0  =  []_0
if_reverse_2_in_1_ag4(x1, x2, x3, x4)  =  if_reverse_2_in_1_ag2(x3, x4)
reverse_2_in_aa2(x1, x2)  =  reverse_2_in_aa
if_reverse_2_in_1_aa4(x1, x2, x3, x4)  =  if_reverse_2_in_1_aa1(x4)
reverse_2_out_aa2(x1, x2)  =  reverse_2_out_aa1(x1)
if_reverse_2_in_2_aa5(x1, x2, x3, x4, x5)  =  if_reverse_2_in_2_aa2(x2, x5)
app_3_in_aga3(x1, x2, x3)  =  app_3_in_aga1(x2)
if_app_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_aga1(x5)
app_3_out_aga3(x1, x2, x3)  =  app_3_out_aga2(x1, x3)
if_reverse_2_in_2_ag5(x1, x2, x3, x4, x5)  =  if_reverse_2_in_2_ag2(x2, x5)
app_3_in_agg3(x1, x2, x3)  =  app_3_in_agg2(x2, x3)
if_app_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_agg1(x5)
app_3_out_agg3(x1, x2, x3)  =  app_3_out_agg1(x1)
reverse_2_out_ag2(x1, x2)  =  reverse_2_out_ag1(x1)
APP_3_IN_AGG3(x1, x2, x3)  =  APP_3_IN_AGG2(x2, x3)
APP_3_IN_AGA3(x1, x2, x3)  =  APP_3_IN_AGA1(x2)
IF_APP_3_IN_1_AGA5(x1, x2, x3, x4, x5)  =  IF_APP_3_IN_1_AGA1(x5)
IF_REVERSE_2_IN_1_AA4(x1, x2, x3, x4)  =  IF_REVERSE_2_IN_1_AA1(x4)
REVERSE_2_IN_AA2(x1, x2)  =  REVERSE_2_IN_AA
IF_APP_3_IN_1_AGG5(x1, x2, x3, x4, x5)  =  IF_APP_3_IN_1_AGG1(x5)
IF_REVERSE_2_IN_2_AG5(x1, x2, x3, x4, x5)  =  IF_REVERSE_2_IN_2_AG2(x2, x5)
REVERSE_2_IN_AG2(x1, x2)  =  REVERSE_2_IN_AG1(x2)
IF_REVERSE_2_IN_2_AA5(x1, x2, x3, x4, x5)  =  IF_REVERSE_2_IN_2_AA2(x2, x5)
IF_REVERSE_2_IN_1_AG4(x1, x2, x3, x4)  =  IF_REVERSE_2_IN_1_AG2(x3, x4)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 3 SCCs with 9 less nodes.

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
              ↳ PiDP
              ↳ PiDP
  ↳ PrologToPiTRSProof

Pi DP problem:
The TRS P consists of the following rules:

APP_3_IN_AGG3(._22(X, Xs), Ys, ._22(X, Zs)) -> APP_3_IN_AGG3(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_2_in_ag2(._22(X, Xs), Ys) -> if_reverse_2_in_1_ag4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
reverse_2_in_aa2(._22(X, Xs), Ys) -> if_reverse_2_in_1_aa4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
reverse_2_in_aa2([]_0, []_0) -> reverse_2_out_aa2([]_0, []_0)
if_reverse_2_in_1_aa4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> if_reverse_2_in_2_aa5(X, Xs, Ys, Zs, app_3_in_aga3(Zs, ._22(X, []_0), Ys))
app_3_in_aga3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_aga5(X, Xs, Ys, Zs, app_3_in_aga3(Xs, Ys, Zs))
app_3_in_aga3([]_0, Ys, Ys) -> app_3_out_aga3([]_0, Ys, Ys)
if_app_3_in_1_aga5(X, Xs, Ys, Zs, app_3_out_aga3(Xs, Ys, Zs)) -> app_3_out_aga3(._22(X, Xs), Ys, ._22(X, Zs))
if_reverse_2_in_2_aa5(X, Xs, Ys, Zs, app_3_out_aga3(Zs, ._22(X, []_0), Ys)) -> reverse_2_out_aa2(._22(X, Xs), Ys)
if_reverse_2_in_1_ag4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> if_reverse_2_in_2_ag5(X, Xs, Ys, Zs, app_3_in_agg3(Zs, ._22(X, []_0), Ys))
app_3_in_agg3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_agg5(X, Xs, Ys, Zs, app_3_in_agg3(Xs, Ys, Zs))
app_3_in_agg3([]_0, Ys, Ys) -> app_3_out_agg3([]_0, Ys, Ys)
if_app_3_in_1_agg5(X, Xs, Ys, Zs, app_3_out_agg3(Xs, Ys, Zs)) -> app_3_out_agg3(._22(X, Xs), Ys, ._22(X, Zs))
if_reverse_2_in_2_ag5(X, Xs, Ys, Zs, app_3_out_agg3(Zs, ._22(X, []_0), Ys)) -> reverse_2_out_ag2(._22(X, Xs), Ys)
reverse_2_in_ag2([]_0, []_0) -> reverse_2_out_ag2([]_0, []_0)

The argument filtering Pi contains the following mapping:
reverse_2_in_ag2(x1, x2)  =  reverse_2_in_ag1(x2)
._22(x1, x2)  =  ._21(x2)
[]_0  =  []_0
if_reverse_2_in_1_ag4(x1, x2, x3, x4)  =  if_reverse_2_in_1_ag2(x3, x4)
reverse_2_in_aa2(x1, x2)  =  reverse_2_in_aa
if_reverse_2_in_1_aa4(x1, x2, x3, x4)  =  if_reverse_2_in_1_aa1(x4)
reverse_2_out_aa2(x1, x2)  =  reverse_2_out_aa1(x1)
if_reverse_2_in_2_aa5(x1, x2, x3, x4, x5)  =  if_reverse_2_in_2_aa2(x2, x5)
app_3_in_aga3(x1, x2, x3)  =  app_3_in_aga1(x2)
if_app_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_aga1(x5)
app_3_out_aga3(x1, x2, x3)  =  app_3_out_aga2(x1, x3)
if_reverse_2_in_2_ag5(x1, x2, x3, x4, x5)  =  if_reverse_2_in_2_ag2(x2, x5)
app_3_in_agg3(x1, x2, x3)  =  app_3_in_agg2(x2, x3)
if_app_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_agg1(x5)
app_3_out_agg3(x1, x2, x3)  =  app_3_out_agg1(x1)
reverse_2_out_ag2(x1, x2)  =  reverse_2_out_ag1(x1)
APP_3_IN_AGG3(x1, x2, x3)  =  APP_3_IN_AGG2(x2, x3)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting we can delete all non-usable rules from R.

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP
              ↳ PiDP
  ↳ PrologToPiTRSProof

Pi DP problem:
The TRS P consists of the following rules:

APP_3_IN_AGG3(._22(X, Xs), Ys, ._22(X, Zs)) -> APP_3_IN_AGG3(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._21(x2)
APP_3_IN_AGG3(x1, x2, x3)  =  APP_3_IN_AGG2(x2, x3)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem by application of Pi.

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
              ↳ PiDP
              ↳ PiDP
  ↳ PrologToPiTRSProof

Q DP problem:
The TRS P consists of the following rules:

APP_3_IN_AGG2(Ys, ._21(Zs)) -> APP_3_IN_AGG2(Ys, Zs)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {APP_3_IN_AGG2}.
By using the subterm criterion together with the size-change analysis we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: 



↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

Pi DP problem:
The TRS P consists of the following rules:

APP_3_IN_AGA3(._22(X, Xs), Ys, ._22(X, Zs)) -> APP_3_IN_AGA3(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_2_in_ag2(._22(X, Xs), Ys) -> if_reverse_2_in_1_ag4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
reverse_2_in_aa2(._22(X, Xs), Ys) -> if_reverse_2_in_1_aa4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
reverse_2_in_aa2([]_0, []_0) -> reverse_2_out_aa2([]_0, []_0)
if_reverse_2_in_1_aa4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> if_reverse_2_in_2_aa5(X, Xs, Ys, Zs, app_3_in_aga3(Zs, ._22(X, []_0), Ys))
app_3_in_aga3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_aga5(X, Xs, Ys, Zs, app_3_in_aga3(Xs, Ys, Zs))
app_3_in_aga3([]_0, Ys, Ys) -> app_3_out_aga3([]_0, Ys, Ys)
if_app_3_in_1_aga5(X, Xs, Ys, Zs, app_3_out_aga3(Xs, Ys, Zs)) -> app_3_out_aga3(._22(X, Xs), Ys, ._22(X, Zs))
if_reverse_2_in_2_aa5(X, Xs, Ys, Zs, app_3_out_aga3(Zs, ._22(X, []_0), Ys)) -> reverse_2_out_aa2(._22(X, Xs), Ys)
if_reverse_2_in_1_ag4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> if_reverse_2_in_2_ag5(X, Xs, Ys, Zs, app_3_in_agg3(Zs, ._22(X, []_0), Ys))
app_3_in_agg3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_agg5(X, Xs, Ys, Zs, app_3_in_agg3(Xs, Ys, Zs))
app_3_in_agg3([]_0, Ys, Ys) -> app_3_out_agg3([]_0, Ys, Ys)
if_app_3_in_1_agg5(X, Xs, Ys, Zs, app_3_out_agg3(Xs, Ys, Zs)) -> app_3_out_agg3(._22(X, Xs), Ys, ._22(X, Zs))
if_reverse_2_in_2_ag5(X, Xs, Ys, Zs, app_3_out_agg3(Zs, ._22(X, []_0), Ys)) -> reverse_2_out_ag2(._22(X, Xs), Ys)
reverse_2_in_ag2([]_0, []_0) -> reverse_2_out_ag2([]_0, []_0)

The argument filtering Pi contains the following mapping:
reverse_2_in_ag2(x1, x2)  =  reverse_2_in_ag1(x2)
._22(x1, x2)  =  ._21(x2)
[]_0  =  []_0
if_reverse_2_in_1_ag4(x1, x2, x3, x4)  =  if_reverse_2_in_1_ag2(x3, x4)
reverse_2_in_aa2(x1, x2)  =  reverse_2_in_aa
if_reverse_2_in_1_aa4(x1, x2, x3, x4)  =  if_reverse_2_in_1_aa1(x4)
reverse_2_out_aa2(x1, x2)  =  reverse_2_out_aa1(x1)
if_reverse_2_in_2_aa5(x1, x2, x3, x4, x5)  =  if_reverse_2_in_2_aa2(x2, x5)
app_3_in_aga3(x1, x2, x3)  =  app_3_in_aga1(x2)
if_app_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_aga1(x5)
app_3_out_aga3(x1, x2, x3)  =  app_3_out_aga2(x1, x3)
if_reverse_2_in_2_ag5(x1, x2, x3, x4, x5)  =  if_reverse_2_in_2_ag2(x2, x5)
app_3_in_agg3(x1, x2, x3)  =  app_3_in_agg2(x2, x3)
if_app_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_agg1(x5)
app_3_out_agg3(x1, x2, x3)  =  app_3_out_agg1(x1)
reverse_2_out_ag2(x1, x2)  =  reverse_2_out_ag1(x1)
APP_3_IN_AGA3(x1, x2, x3)  =  APP_3_IN_AGA1(x2)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting we can delete all non-usable rules from R.

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

Pi DP problem:
The TRS P consists of the following rules:

APP_3_IN_AGA3(._22(X, Xs), Ys, ._22(X, Zs)) -> APP_3_IN_AGA3(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._21(x2)
APP_3_IN_AGA3(x1, x2, x3)  =  APP_3_IN_AGA1(x2)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem by application of Pi.

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
              ↳ PiDP
  ↳ PrologToPiTRSProof

Q DP problem:
The TRS P consists of the following rules:

APP_3_IN_AGA1(Ys) -> APP_3_IN_AGA1(Ys)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {APP_3_IN_AGA1}.

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
  ↳ PrologToPiTRSProof

Pi DP problem:
The TRS P consists of the following rules:

REVERSE_2_IN_AA2(._22(X, Xs), Ys) -> REVERSE_2_IN_AA2(Xs, Zs)

The TRS R consists of the following rules:

reverse_2_in_ag2(._22(X, Xs), Ys) -> if_reverse_2_in_1_ag4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
reverse_2_in_aa2(._22(X, Xs), Ys) -> if_reverse_2_in_1_aa4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
reverse_2_in_aa2([]_0, []_0) -> reverse_2_out_aa2([]_0, []_0)
if_reverse_2_in_1_aa4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> if_reverse_2_in_2_aa5(X, Xs, Ys, Zs, app_3_in_aga3(Zs, ._22(X, []_0), Ys))
app_3_in_aga3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_aga5(X, Xs, Ys, Zs, app_3_in_aga3(Xs, Ys, Zs))
app_3_in_aga3([]_0, Ys, Ys) -> app_3_out_aga3([]_0, Ys, Ys)
if_app_3_in_1_aga5(X, Xs, Ys, Zs, app_3_out_aga3(Xs, Ys, Zs)) -> app_3_out_aga3(._22(X, Xs), Ys, ._22(X, Zs))
if_reverse_2_in_2_aa5(X, Xs, Ys, Zs, app_3_out_aga3(Zs, ._22(X, []_0), Ys)) -> reverse_2_out_aa2(._22(X, Xs), Ys)
if_reverse_2_in_1_ag4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> if_reverse_2_in_2_ag5(X, Xs, Ys, Zs, app_3_in_agg3(Zs, ._22(X, []_0), Ys))
app_3_in_agg3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_agg5(X, Xs, Ys, Zs, app_3_in_agg3(Xs, Ys, Zs))
app_3_in_agg3([]_0, Ys, Ys) -> app_3_out_agg3([]_0, Ys, Ys)
if_app_3_in_1_agg5(X, Xs, Ys, Zs, app_3_out_agg3(Xs, Ys, Zs)) -> app_3_out_agg3(._22(X, Xs), Ys, ._22(X, Zs))
if_reverse_2_in_2_ag5(X, Xs, Ys, Zs, app_3_out_agg3(Zs, ._22(X, []_0), Ys)) -> reverse_2_out_ag2(._22(X, Xs), Ys)
reverse_2_in_ag2([]_0, []_0) -> reverse_2_out_ag2([]_0, []_0)

The argument filtering Pi contains the following mapping:
reverse_2_in_ag2(x1, x2)  =  reverse_2_in_ag1(x2)
._22(x1, x2)  =  ._21(x2)
[]_0  =  []_0
if_reverse_2_in_1_ag4(x1, x2, x3, x4)  =  if_reverse_2_in_1_ag2(x3, x4)
reverse_2_in_aa2(x1, x2)  =  reverse_2_in_aa
if_reverse_2_in_1_aa4(x1, x2, x3, x4)  =  if_reverse_2_in_1_aa1(x4)
reverse_2_out_aa2(x1, x2)  =  reverse_2_out_aa1(x1)
if_reverse_2_in_2_aa5(x1, x2, x3, x4, x5)  =  if_reverse_2_in_2_aa2(x2, x5)
app_3_in_aga3(x1, x2, x3)  =  app_3_in_aga1(x2)
if_app_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_aga1(x5)
app_3_out_aga3(x1, x2, x3)  =  app_3_out_aga2(x1, x3)
if_reverse_2_in_2_ag5(x1, x2, x3, x4, x5)  =  if_reverse_2_in_2_ag2(x2, x5)
app_3_in_agg3(x1, x2, x3)  =  app_3_in_agg2(x2, x3)
if_app_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_agg1(x5)
app_3_out_agg3(x1, x2, x3)  =  app_3_out_agg1(x1)
reverse_2_out_ag2(x1, x2)  =  reverse_2_out_ag1(x1)
REVERSE_2_IN_AA2(x1, x2)  =  REVERSE_2_IN_AA

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting we can delete all non-usable rules from R.

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
  ↳ PrologToPiTRSProof

Pi DP problem:
The TRS P consists of the following rules:

REVERSE_2_IN_AA2(._22(X, Xs), Ys) -> REVERSE_2_IN_AA2(Xs, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._21(x2)
REVERSE_2_IN_AA2(x1, x2)  =  REVERSE_2_IN_AA

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem by application of Pi.

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
  ↳ PrologToPiTRSProof

Q DP problem:
The TRS P consists of the following rules:

REVERSE_2_IN_AA -> REVERSE_2_IN_AA

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {REVERSE_2_IN_AA}.
With regard to the inferred argument filtering the predicates were used in the following modes:
reverse2: (f,b) (f,f)
app3: (f,b,f) (f,b,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

reverse_2_in_ag2(._22(X, Xs), Ys) -> if_reverse_2_in_1_ag4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
reverse_2_in_aa2(._22(X, Xs), Ys) -> if_reverse_2_in_1_aa4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
reverse_2_in_aa2([]_0, []_0) -> reverse_2_out_aa2([]_0, []_0)
if_reverse_2_in_1_aa4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> if_reverse_2_in_2_aa5(X, Xs, Ys, Zs, app_3_in_aga3(Zs, ._22(X, []_0), Ys))
app_3_in_aga3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_aga5(X, Xs, Ys, Zs, app_3_in_aga3(Xs, Ys, Zs))
app_3_in_aga3([]_0, Ys, Ys) -> app_3_out_aga3([]_0, Ys, Ys)
if_app_3_in_1_aga5(X, Xs, Ys, Zs, app_3_out_aga3(Xs, Ys, Zs)) -> app_3_out_aga3(._22(X, Xs), Ys, ._22(X, Zs))
if_reverse_2_in_2_aa5(X, Xs, Ys, Zs, app_3_out_aga3(Zs, ._22(X, []_0), Ys)) -> reverse_2_out_aa2(._22(X, Xs), Ys)
if_reverse_2_in_1_ag4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> if_reverse_2_in_2_ag5(X, Xs, Ys, Zs, app_3_in_agg3(Zs, ._22(X, []_0), Ys))
app_3_in_agg3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_agg5(X, Xs, Ys, Zs, app_3_in_agg3(Xs, Ys, Zs))
app_3_in_agg3([]_0, Ys, Ys) -> app_3_out_agg3([]_0, Ys, Ys)
if_app_3_in_1_agg5(X, Xs, Ys, Zs, app_3_out_agg3(Xs, Ys, Zs)) -> app_3_out_agg3(._22(X, Xs), Ys, ._22(X, Zs))
if_reverse_2_in_2_ag5(X, Xs, Ys, Zs, app_3_out_agg3(Zs, ._22(X, []_0), Ys)) -> reverse_2_out_ag2(._22(X, Xs), Ys)
reverse_2_in_ag2([]_0, []_0) -> reverse_2_out_ag2([]_0, []_0)

The argument filtering Pi contains the following mapping:
reverse_2_in_ag2(x1, x2)  =  reverse_2_in_ag1(x2)
._22(x1, x2)  =  ._21(x2)
[]_0  =  []_0
if_reverse_2_in_1_ag4(x1, x2, x3, x4)  =  if_reverse_2_in_1_ag2(x3, x4)
reverse_2_in_aa2(x1, x2)  =  reverse_2_in_aa
if_reverse_2_in_1_aa4(x1, x2, x3, x4)  =  if_reverse_2_in_1_aa1(x4)
reverse_2_out_aa2(x1, x2)  =  reverse_2_out_aa1(x1)
if_reverse_2_in_2_aa5(x1, x2, x3, x4, x5)  =  if_reverse_2_in_2_aa2(x2, x5)
app_3_in_aga3(x1, x2, x3)  =  app_3_in_aga1(x2)
if_app_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_aga2(x3, x5)
app_3_out_aga3(x1, x2, x3)  =  app_3_out_aga3(x1, x2, x3)
if_reverse_2_in_2_ag5(x1, x2, x3, x4, x5)  =  if_reverse_2_in_2_ag3(x2, x3, x5)
app_3_in_agg3(x1, x2, x3)  =  app_3_in_agg2(x2, x3)
if_app_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_agg3(x3, x4, x5)
app_3_out_agg3(x1, x2, x3)  =  app_3_out_agg3(x1, x2, x3)
reverse_2_out_ag2(x1, x2)  =  reverse_2_out_ag2(x1, x2)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG



↳ PROLOG
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

reverse_2_in_ag2(._22(X, Xs), Ys) -> if_reverse_2_in_1_ag4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
reverse_2_in_aa2(._22(X, Xs), Ys) -> if_reverse_2_in_1_aa4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
reverse_2_in_aa2([]_0, []_0) -> reverse_2_out_aa2([]_0, []_0)
if_reverse_2_in_1_aa4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> if_reverse_2_in_2_aa5(X, Xs, Ys, Zs, app_3_in_aga3(Zs, ._22(X, []_0), Ys))
app_3_in_aga3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_aga5(X, Xs, Ys, Zs, app_3_in_aga3(Xs, Ys, Zs))
app_3_in_aga3([]_0, Ys, Ys) -> app_3_out_aga3([]_0, Ys, Ys)
if_app_3_in_1_aga5(X, Xs, Ys, Zs, app_3_out_aga3(Xs, Ys, Zs)) -> app_3_out_aga3(._22(X, Xs), Ys, ._22(X, Zs))
if_reverse_2_in_2_aa5(X, Xs, Ys, Zs, app_3_out_aga3(Zs, ._22(X, []_0), Ys)) -> reverse_2_out_aa2(._22(X, Xs), Ys)
if_reverse_2_in_1_ag4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> if_reverse_2_in_2_ag5(X, Xs, Ys, Zs, app_3_in_agg3(Zs, ._22(X, []_0), Ys))
app_3_in_agg3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_agg5(X, Xs, Ys, Zs, app_3_in_agg3(Xs, Ys, Zs))
app_3_in_agg3([]_0, Ys, Ys) -> app_3_out_agg3([]_0, Ys, Ys)
if_app_3_in_1_agg5(X, Xs, Ys, Zs, app_3_out_agg3(Xs, Ys, Zs)) -> app_3_out_agg3(._22(X, Xs), Ys, ._22(X, Zs))
if_reverse_2_in_2_ag5(X, Xs, Ys, Zs, app_3_out_agg3(Zs, ._22(X, []_0), Ys)) -> reverse_2_out_ag2(._22(X, Xs), Ys)
reverse_2_in_ag2([]_0, []_0) -> reverse_2_out_ag2([]_0, []_0)

The argument filtering Pi contains the following mapping:
reverse_2_in_ag2(x1, x2)  =  reverse_2_in_ag1(x2)
._22(x1, x2)  =  ._21(x2)
[]_0  =  []_0
if_reverse_2_in_1_ag4(x1, x2, x3, x4)  =  if_reverse_2_in_1_ag2(x3, x4)
reverse_2_in_aa2(x1, x2)  =  reverse_2_in_aa
if_reverse_2_in_1_aa4(x1, x2, x3, x4)  =  if_reverse_2_in_1_aa1(x4)
reverse_2_out_aa2(x1, x2)  =  reverse_2_out_aa1(x1)
if_reverse_2_in_2_aa5(x1, x2, x3, x4, x5)  =  if_reverse_2_in_2_aa2(x2, x5)
app_3_in_aga3(x1, x2, x3)  =  app_3_in_aga1(x2)
if_app_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_aga2(x3, x5)
app_3_out_aga3(x1, x2, x3)  =  app_3_out_aga3(x1, x2, x3)
if_reverse_2_in_2_ag5(x1, x2, x3, x4, x5)  =  if_reverse_2_in_2_ag3(x2, x3, x5)
app_3_in_agg3(x1, x2, x3)  =  app_3_in_agg2(x2, x3)
if_app_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_agg3(x3, x4, x5)
app_3_out_agg3(x1, x2, x3)  =  app_3_out_agg3(x1, x2, x3)
reverse_2_out_ag2(x1, x2)  =  reverse_2_out_ag2(x1, x2)


Pi DP problem:
The TRS P consists of the following rules:

REVERSE_2_IN_AG2(._22(X, Xs), Ys) -> IF_REVERSE_2_IN_1_AG4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
REVERSE_2_IN_AG2(._22(X, Xs), Ys) -> REVERSE_2_IN_AA2(Xs, Zs)
REVERSE_2_IN_AA2(._22(X, Xs), Ys) -> IF_REVERSE_2_IN_1_AA4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
REVERSE_2_IN_AA2(._22(X, Xs), Ys) -> REVERSE_2_IN_AA2(Xs, Zs)
IF_REVERSE_2_IN_1_AA4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> IF_REVERSE_2_IN_2_AA5(X, Xs, Ys, Zs, app_3_in_aga3(Zs, ._22(X, []_0), Ys))
IF_REVERSE_2_IN_1_AA4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> APP_3_IN_AGA3(Zs, ._22(X, []_0), Ys)
APP_3_IN_AGA3(._22(X, Xs), Ys, ._22(X, Zs)) -> IF_APP_3_IN_1_AGA5(X, Xs, Ys, Zs, app_3_in_aga3(Xs, Ys, Zs))
APP_3_IN_AGA3(._22(X, Xs), Ys, ._22(X, Zs)) -> APP_3_IN_AGA3(Xs, Ys, Zs)
IF_REVERSE_2_IN_1_AG4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> IF_REVERSE_2_IN_2_AG5(X, Xs, Ys, Zs, app_3_in_agg3(Zs, ._22(X, []_0), Ys))
IF_REVERSE_2_IN_1_AG4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> APP_3_IN_AGG3(Zs, ._22(X, []_0), Ys)
APP_3_IN_AGG3(._22(X, Xs), Ys, ._22(X, Zs)) -> IF_APP_3_IN_1_AGG5(X, Xs, Ys, Zs, app_3_in_agg3(Xs, Ys, Zs))
APP_3_IN_AGG3(._22(X, Xs), Ys, ._22(X, Zs)) -> APP_3_IN_AGG3(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_2_in_ag2(._22(X, Xs), Ys) -> if_reverse_2_in_1_ag4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
reverse_2_in_aa2(._22(X, Xs), Ys) -> if_reverse_2_in_1_aa4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
reverse_2_in_aa2([]_0, []_0) -> reverse_2_out_aa2([]_0, []_0)
if_reverse_2_in_1_aa4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> if_reverse_2_in_2_aa5(X, Xs, Ys, Zs, app_3_in_aga3(Zs, ._22(X, []_0), Ys))
app_3_in_aga3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_aga5(X, Xs, Ys, Zs, app_3_in_aga3(Xs, Ys, Zs))
app_3_in_aga3([]_0, Ys, Ys) -> app_3_out_aga3([]_0, Ys, Ys)
if_app_3_in_1_aga5(X, Xs, Ys, Zs, app_3_out_aga3(Xs, Ys, Zs)) -> app_3_out_aga3(._22(X, Xs), Ys, ._22(X, Zs))
if_reverse_2_in_2_aa5(X, Xs, Ys, Zs, app_3_out_aga3(Zs, ._22(X, []_0), Ys)) -> reverse_2_out_aa2(._22(X, Xs), Ys)
if_reverse_2_in_1_ag4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> if_reverse_2_in_2_ag5(X, Xs, Ys, Zs, app_3_in_agg3(Zs, ._22(X, []_0), Ys))
app_3_in_agg3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_agg5(X, Xs, Ys, Zs, app_3_in_agg3(Xs, Ys, Zs))
app_3_in_agg3([]_0, Ys, Ys) -> app_3_out_agg3([]_0, Ys, Ys)
if_app_3_in_1_agg5(X, Xs, Ys, Zs, app_3_out_agg3(Xs, Ys, Zs)) -> app_3_out_agg3(._22(X, Xs), Ys, ._22(X, Zs))
if_reverse_2_in_2_ag5(X, Xs, Ys, Zs, app_3_out_agg3(Zs, ._22(X, []_0), Ys)) -> reverse_2_out_ag2(._22(X, Xs), Ys)
reverse_2_in_ag2([]_0, []_0) -> reverse_2_out_ag2([]_0, []_0)

The argument filtering Pi contains the following mapping:
reverse_2_in_ag2(x1, x2)  =  reverse_2_in_ag1(x2)
._22(x1, x2)  =  ._21(x2)
[]_0  =  []_0
if_reverse_2_in_1_ag4(x1, x2, x3, x4)  =  if_reverse_2_in_1_ag2(x3, x4)
reverse_2_in_aa2(x1, x2)  =  reverse_2_in_aa
if_reverse_2_in_1_aa4(x1, x2, x3, x4)  =  if_reverse_2_in_1_aa1(x4)
reverse_2_out_aa2(x1, x2)  =  reverse_2_out_aa1(x1)
if_reverse_2_in_2_aa5(x1, x2, x3, x4, x5)  =  if_reverse_2_in_2_aa2(x2, x5)
app_3_in_aga3(x1, x2, x3)  =  app_3_in_aga1(x2)
if_app_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_aga2(x3, x5)
app_3_out_aga3(x1, x2, x3)  =  app_3_out_aga3(x1, x2, x3)
if_reverse_2_in_2_ag5(x1, x2, x3, x4, x5)  =  if_reverse_2_in_2_ag3(x2, x3, x5)
app_3_in_agg3(x1, x2, x3)  =  app_3_in_agg2(x2, x3)
if_app_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_agg3(x3, x4, x5)
app_3_out_agg3(x1, x2, x3)  =  app_3_out_agg3(x1, x2, x3)
reverse_2_out_ag2(x1, x2)  =  reverse_2_out_ag2(x1, x2)
APP_3_IN_AGG3(x1, x2, x3)  =  APP_3_IN_AGG2(x2, x3)
APP_3_IN_AGA3(x1, x2, x3)  =  APP_3_IN_AGA1(x2)
IF_APP_3_IN_1_AGA5(x1, x2, x3, x4, x5)  =  IF_APP_3_IN_1_AGA2(x3, x5)
IF_REVERSE_2_IN_1_AA4(x1, x2, x3, x4)  =  IF_REVERSE_2_IN_1_AA1(x4)
REVERSE_2_IN_AA2(x1, x2)  =  REVERSE_2_IN_AA
IF_APP_3_IN_1_AGG5(x1, x2, x3, x4, x5)  =  IF_APP_3_IN_1_AGG3(x3, x4, x5)
IF_REVERSE_2_IN_2_AG5(x1, x2, x3, x4, x5)  =  IF_REVERSE_2_IN_2_AG3(x2, x3, x5)
REVERSE_2_IN_AG2(x1, x2)  =  REVERSE_2_IN_AG1(x2)
IF_REVERSE_2_IN_2_AA5(x1, x2, x3, x4, x5)  =  IF_REVERSE_2_IN_2_AA2(x2, x5)
IF_REVERSE_2_IN_1_AG4(x1, x2, x3, x4)  =  IF_REVERSE_2_IN_1_AG2(x3, x4)

We have to consider all (P,R,Pi)-chains

↳ PROLOG
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof

Pi DP problem:
The TRS P consists of the following rules:

REVERSE_2_IN_AG2(._22(X, Xs), Ys) -> IF_REVERSE_2_IN_1_AG4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
REVERSE_2_IN_AG2(._22(X, Xs), Ys) -> REVERSE_2_IN_AA2(Xs, Zs)
REVERSE_2_IN_AA2(._22(X, Xs), Ys) -> IF_REVERSE_2_IN_1_AA4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
REVERSE_2_IN_AA2(._22(X, Xs), Ys) -> REVERSE_2_IN_AA2(Xs, Zs)
IF_REVERSE_2_IN_1_AA4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> IF_REVERSE_2_IN_2_AA5(X, Xs, Ys, Zs, app_3_in_aga3(Zs, ._22(X, []_0), Ys))
IF_REVERSE_2_IN_1_AA4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> APP_3_IN_AGA3(Zs, ._22(X, []_0), Ys)
APP_3_IN_AGA3(._22(X, Xs), Ys, ._22(X, Zs)) -> IF_APP_3_IN_1_AGA5(X, Xs, Ys, Zs, app_3_in_aga3(Xs, Ys, Zs))
APP_3_IN_AGA3(._22(X, Xs), Ys, ._22(X, Zs)) -> APP_3_IN_AGA3(Xs, Ys, Zs)
IF_REVERSE_2_IN_1_AG4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> IF_REVERSE_2_IN_2_AG5(X, Xs, Ys, Zs, app_3_in_agg3(Zs, ._22(X, []_0), Ys))
IF_REVERSE_2_IN_1_AG4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> APP_3_IN_AGG3(Zs, ._22(X, []_0), Ys)
APP_3_IN_AGG3(._22(X, Xs), Ys, ._22(X, Zs)) -> IF_APP_3_IN_1_AGG5(X, Xs, Ys, Zs, app_3_in_agg3(Xs, Ys, Zs))
APP_3_IN_AGG3(._22(X, Xs), Ys, ._22(X, Zs)) -> APP_3_IN_AGG3(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_2_in_ag2(._22(X, Xs), Ys) -> if_reverse_2_in_1_ag4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
reverse_2_in_aa2(._22(X, Xs), Ys) -> if_reverse_2_in_1_aa4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
reverse_2_in_aa2([]_0, []_0) -> reverse_2_out_aa2([]_0, []_0)
if_reverse_2_in_1_aa4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> if_reverse_2_in_2_aa5(X, Xs, Ys, Zs, app_3_in_aga3(Zs, ._22(X, []_0), Ys))
app_3_in_aga3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_aga5(X, Xs, Ys, Zs, app_3_in_aga3(Xs, Ys, Zs))
app_3_in_aga3([]_0, Ys, Ys) -> app_3_out_aga3([]_0, Ys, Ys)
if_app_3_in_1_aga5(X, Xs, Ys, Zs, app_3_out_aga3(Xs, Ys, Zs)) -> app_3_out_aga3(._22(X, Xs), Ys, ._22(X, Zs))
if_reverse_2_in_2_aa5(X, Xs, Ys, Zs, app_3_out_aga3(Zs, ._22(X, []_0), Ys)) -> reverse_2_out_aa2(._22(X, Xs), Ys)
if_reverse_2_in_1_ag4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> if_reverse_2_in_2_ag5(X, Xs, Ys, Zs, app_3_in_agg3(Zs, ._22(X, []_0), Ys))
app_3_in_agg3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_agg5(X, Xs, Ys, Zs, app_3_in_agg3(Xs, Ys, Zs))
app_3_in_agg3([]_0, Ys, Ys) -> app_3_out_agg3([]_0, Ys, Ys)
if_app_3_in_1_agg5(X, Xs, Ys, Zs, app_3_out_agg3(Xs, Ys, Zs)) -> app_3_out_agg3(._22(X, Xs), Ys, ._22(X, Zs))
if_reverse_2_in_2_ag5(X, Xs, Ys, Zs, app_3_out_agg3(Zs, ._22(X, []_0), Ys)) -> reverse_2_out_ag2(._22(X, Xs), Ys)
reverse_2_in_ag2([]_0, []_0) -> reverse_2_out_ag2([]_0, []_0)

The argument filtering Pi contains the following mapping:
reverse_2_in_ag2(x1, x2)  =  reverse_2_in_ag1(x2)
._22(x1, x2)  =  ._21(x2)
[]_0  =  []_0
if_reverse_2_in_1_ag4(x1, x2, x3, x4)  =  if_reverse_2_in_1_ag2(x3, x4)
reverse_2_in_aa2(x1, x2)  =  reverse_2_in_aa
if_reverse_2_in_1_aa4(x1, x2, x3, x4)  =  if_reverse_2_in_1_aa1(x4)
reverse_2_out_aa2(x1, x2)  =  reverse_2_out_aa1(x1)
if_reverse_2_in_2_aa5(x1, x2, x3, x4, x5)  =  if_reverse_2_in_2_aa2(x2, x5)
app_3_in_aga3(x1, x2, x3)  =  app_3_in_aga1(x2)
if_app_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_aga2(x3, x5)
app_3_out_aga3(x1, x2, x3)  =  app_3_out_aga3(x1, x2, x3)
if_reverse_2_in_2_ag5(x1, x2, x3, x4, x5)  =  if_reverse_2_in_2_ag3(x2, x3, x5)
app_3_in_agg3(x1, x2, x3)  =  app_3_in_agg2(x2, x3)
if_app_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_agg3(x3, x4, x5)
app_3_out_agg3(x1, x2, x3)  =  app_3_out_agg3(x1, x2, x3)
reverse_2_out_ag2(x1, x2)  =  reverse_2_out_ag2(x1, x2)
APP_3_IN_AGG3(x1, x2, x3)  =  APP_3_IN_AGG2(x2, x3)
APP_3_IN_AGA3(x1, x2, x3)  =  APP_3_IN_AGA1(x2)
IF_APP_3_IN_1_AGA5(x1, x2, x3, x4, x5)  =  IF_APP_3_IN_1_AGA2(x3, x5)
IF_REVERSE_2_IN_1_AA4(x1, x2, x3, x4)  =  IF_REVERSE_2_IN_1_AA1(x4)
REVERSE_2_IN_AA2(x1, x2)  =  REVERSE_2_IN_AA
IF_APP_3_IN_1_AGG5(x1, x2, x3, x4, x5)  =  IF_APP_3_IN_1_AGG3(x3, x4, x5)
IF_REVERSE_2_IN_2_AG5(x1, x2, x3, x4, x5)  =  IF_REVERSE_2_IN_2_AG3(x2, x3, x5)
REVERSE_2_IN_AG2(x1, x2)  =  REVERSE_2_IN_AG1(x2)
IF_REVERSE_2_IN_2_AA5(x1, x2, x3, x4, x5)  =  IF_REVERSE_2_IN_2_AA2(x2, x5)
IF_REVERSE_2_IN_1_AG4(x1, x2, x3, x4)  =  IF_REVERSE_2_IN_1_AG2(x3, x4)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 3 SCCs with 9 less nodes.

↳ PROLOG
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
              ↳ PiDP
              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

APP_3_IN_AGG3(._22(X, Xs), Ys, ._22(X, Zs)) -> APP_3_IN_AGG3(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_2_in_ag2(._22(X, Xs), Ys) -> if_reverse_2_in_1_ag4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
reverse_2_in_aa2(._22(X, Xs), Ys) -> if_reverse_2_in_1_aa4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
reverse_2_in_aa2([]_0, []_0) -> reverse_2_out_aa2([]_0, []_0)
if_reverse_2_in_1_aa4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> if_reverse_2_in_2_aa5(X, Xs, Ys, Zs, app_3_in_aga3(Zs, ._22(X, []_0), Ys))
app_3_in_aga3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_aga5(X, Xs, Ys, Zs, app_3_in_aga3(Xs, Ys, Zs))
app_3_in_aga3([]_0, Ys, Ys) -> app_3_out_aga3([]_0, Ys, Ys)
if_app_3_in_1_aga5(X, Xs, Ys, Zs, app_3_out_aga3(Xs, Ys, Zs)) -> app_3_out_aga3(._22(X, Xs), Ys, ._22(X, Zs))
if_reverse_2_in_2_aa5(X, Xs, Ys, Zs, app_3_out_aga3(Zs, ._22(X, []_0), Ys)) -> reverse_2_out_aa2(._22(X, Xs), Ys)
if_reverse_2_in_1_ag4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> if_reverse_2_in_2_ag5(X, Xs, Ys, Zs, app_3_in_agg3(Zs, ._22(X, []_0), Ys))
app_3_in_agg3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_agg5(X, Xs, Ys, Zs, app_3_in_agg3(Xs, Ys, Zs))
app_3_in_agg3([]_0, Ys, Ys) -> app_3_out_agg3([]_0, Ys, Ys)
if_app_3_in_1_agg5(X, Xs, Ys, Zs, app_3_out_agg3(Xs, Ys, Zs)) -> app_3_out_agg3(._22(X, Xs), Ys, ._22(X, Zs))
if_reverse_2_in_2_ag5(X, Xs, Ys, Zs, app_3_out_agg3(Zs, ._22(X, []_0), Ys)) -> reverse_2_out_ag2(._22(X, Xs), Ys)
reverse_2_in_ag2([]_0, []_0) -> reverse_2_out_ag2([]_0, []_0)

The argument filtering Pi contains the following mapping:
reverse_2_in_ag2(x1, x2)  =  reverse_2_in_ag1(x2)
._22(x1, x2)  =  ._21(x2)
[]_0  =  []_0
if_reverse_2_in_1_ag4(x1, x2, x3, x4)  =  if_reverse_2_in_1_ag2(x3, x4)
reverse_2_in_aa2(x1, x2)  =  reverse_2_in_aa
if_reverse_2_in_1_aa4(x1, x2, x3, x4)  =  if_reverse_2_in_1_aa1(x4)
reverse_2_out_aa2(x1, x2)  =  reverse_2_out_aa1(x1)
if_reverse_2_in_2_aa5(x1, x2, x3, x4, x5)  =  if_reverse_2_in_2_aa2(x2, x5)
app_3_in_aga3(x1, x2, x3)  =  app_3_in_aga1(x2)
if_app_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_aga2(x3, x5)
app_3_out_aga3(x1, x2, x3)  =  app_3_out_aga3(x1, x2, x3)
if_reverse_2_in_2_ag5(x1, x2, x3, x4, x5)  =  if_reverse_2_in_2_ag3(x2, x3, x5)
app_3_in_agg3(x1, x2, x3)  =  app_3_in_agg2(x2, x3)
if_app_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_agg3(x3, x4, x5)
app_3_out_agg3(x1, x2, x3)  =  app_3_out_agg3(x1, x2, x3)
reverse_2_out_ag2(x1, x2)  =  reverse_2_out_ag2(x1, x2)
APP_3_IN_AGG3(x1, x2, x3)  =  APP_3_IN_AGG2(x2, x3)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting we can delete all non-usable rules from R.

↳ PROLOG
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP
              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

APP_3_IN_AGG3(._22(X, Xs), Ys, ._22(X, Zs)) -> APP_3_IN_AGG3(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._21(x2)
APP_3_IN_AGG3(x1, x2, x3)  =  APP_3_IN_AGG2(x2, x3)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem by application of Pi.

↳ PROLOG
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
              ↳ PiDP
              ↳ PiDP

Q DP problem:
The TRS P consists of the following rules:

APP_3_IN_AGG2(Ys, ._21(Zs)) -> APP_3_IN_AGG2(Ys, Zs)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {APP_3_IN_AGG2}.
By using the subterm criterion together with the size-change analysis we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: 



↳ PROLOG
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

APP_3_IN_AGA3(._22(X, Xs), Ys, ._22(X, Zs)) -> APP_3_IN_AGA3(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_2_in_ag2(._22(X, Xs), Ys) -> if_reverse_2_in_1_ag4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
reverse_2_in_aa2(._22(X, Xs), Ys) -> if_reverse_2_in_1_aa4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
reverse_2_in_aa2([]_0, []_0) -> reverse_2_out_aa2([]_0, []_0)
if_reverse_2_in_1_aa4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> if_reverse_2_in_2_aa5(X, Xs, Ys, Zs, app_3_in_aga3(Zs, ._22(X, []_0), Ys))
app_3_in_aga3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_aga5(X, Xs, Ys, Zs, app_3_in_aga3(Xs, Ys, Zs))
app_3_in_aga3([]_0, Ys, Ys) -> app_3_out_aga3([]_0, Ys, Ys)
if_app_3_in_1_aga5(X, Xs, Ys, Zs, app_3_out_aga3(Xs, Ys, Zs)) -> app_3_out_aga3(._22(X, Xs), Ys, ._22(X, Zs))
if_reverse_2_in_2_aa5(X, Xs, Ys, Zs, app_3_out_aga3(Zs, ._22(X, []_0), Ys)) -> reverse_2_out_aa2(._22(X, Xs), Ys)
if_reverse_2_in_1_ag4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> if_reverse_2_in_2_ag5(X, Xs, Ys, Zs, app_3_in_agg3(Zs, ._22(X, []_0), Ys))
app_3_in_agg3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_agg5(X, Xs, Ys, Zs, app_3_in_agg3(Xs, Ys, Zs))
app_3_in_agg3([]_0, Ys, Ys) -> app_3_out_agg3([]_0, Ys, Ys)
if_app_3_in_1_agg5(X, Xs, Ys, Zs, app_3_out_agg3(Xs, Ys, Zs)) -> app_3_out_agg3(._22(X, Xs), Ys, ._22(X, Zs))
if_reverse_2_in_2_ag5(X, Xs, Ys, Zs, app_3_out_agg3(Zs, ._22(X, []_0), Ys)) -> reverse_2_out_ag2(._22(X, Xs), Ys)
reverse_2_in_ag2([]_0, []_0) -> reverse_2_out_ag2([]_0, []_0)

The argument filtering Pi contains the following mapping:
reverse_2_in_ag2(x1, x2)  =  reverse_2_in_ag1(x2)
._22(x1, x2)  =  ._21(x2)
[]_0  =  []_0
if_reverse_2_in_1_ag4(x1, x2, x3, x4)  =  if_reverse_2_in_1_ag2(x3, x4)
reverse_2_in_aa2(x1, x2)  =  reverse_2_in_aa
if_reverse_2_in_1_aa4(x1, x2, x3, x4)  =  if_reverse_2_in_1_aa1(x4)
reverse_2_out_aa2(x1, x2)  =  reverse_2_out_aa1(x1)
if_reverse_2_in_2_aa5(x1, x2, x3, x4, x5)  =  if_reverse_2_in_2_aa2(x2, x5)
app_3_in_aga3(x1, x2, x3)  =  app_3_in_aga1(x2)
if_app_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_aga2(x3, x5)
app_3_out_aga3(x1, x2, x3)  =  app_3_out_aga3(x1, x2, x3)
if_reverse_2_in_2_ag5(x1, x2, x3, x4, x5)  =  if_reverse_2_in_2_ag3(x2, x3, x5)
app_3_in_agg3(x1, x2, x3)  =  app_3_in_agg2(x2, x3)
if_app_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_agg3(x3, x4, x5)
app_3_out_agg3(x1, x2, x3)  =  app_3_out_agg3(x1, x2, x3)
reverse_2_out_ag2(x1, x2)  =  reverse_2_out_ag2(x1, x2)
APP_3_IN_AGA3(x1, x2, x3)  =  APP_3_IN_AGA1(x2)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting we can delete all non-usable rules from R.

↳ PROLOG
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

APP_3_IN_AGA3(._22(X, Xs), Ys, ._22(X, Zs)) -> APP_3_IN_AGA3(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._21(x2)
APP_3_IN_AGA3(x1, x2, x3)  =  APP_3_IN_AGA1(x2)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem by application of Pi.

↳ PROLOG
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
              ↳ PiDP

Q DP problem:
The TRS P consists of the following rules:

APP_3_IN_AGA1(Ys) -> APP_3_IN_AGA1(Ys)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {APP_3_IN_AGA1}.

↳ PROLOG
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof

Pi DP problem:
The TRS P consists of the following rules:

REVERSE_2_IN_AA2(._22(X, Xs), Ys) -> REVERSE_2_IN_AA2(Xs, Zs)

The TRS R consists of the following rules:

reverse_2_in_ag2(._22(X, Xs), Ys) -> if_reverse_2_in_1_ag4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
reverse_2_in_aa2(._22(X, Xs), Ys) -> if_reverse_2_in_1_aa4(X, Xs, Ys, reverse_2_in_aa2(Xs, Zs))
reverse_2_in_aa2([]_0, []_0) -> reverse_2_out_aa2([]_0, []_0)
if_reverse_2_in_1_aa4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> if_reverse_2_in_2_aa5(X, Xs, Ys, Zs, app_3_in_aga3(Zs, ._22(X, []_0), Ys))
app_3_in_aga3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_aga5(X, Xs, Ys, Zs, app_3_in_aga3(Xs, Ys, Zs))
app_3_in_aga3([]_0, Ys, Ys) -> app_3_out_aga3([]_0, Ys, Ys)
if_app_3_in_1_aga5(X, Xs, Ys, Zs, app_3_out_aga3(Xs, Ys, Zs)) -> app_3_out_aga3(._22(X, Xs), Ys, ._22(X, Zs))
if_reverse_2_in_2_aa5(X, Xs, Ys, Zs, app_3_out_aga3(Zs, ._22(X, []_0), Ys)) -> reverse_2_out_aa2(._22(X, Xs), Ys)
if_reverse_2_in_1_ag4(X, Xs, Ys, reverse_2_out_aa2(Xs, Zs)) -> if_reverse_2_in_2_ag5(X, Xs, Ys, Zs, app_3_in_agg3(Zs, ._22(X, []_0), Ys))
app_3_in_agg3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_agg5(X, Xs, Ys, Zs, app_3_in_agg3(Xs, Ys, Zs))
app_3_in_agg3([]_0, Ys, Ys) -> app_3_out_agg3([]_0, Ys, Ys)
if_app_3_in_1_agg5(X, Xs, Ys, Zs, app_3_out_agg3(Xs, Ys, Zs)) -> app_3_out_agg3(._22(X, Xs), Ys, ._22(X, Zs))
if_reverse_2_in_2_ag5(X, Xs, Ys, Zs, app_3_out_agg3(Zs, ._22(X, []_0), Ys)) -> reverse_2_out_ag2(._22(X, Xs), Ys)
reverse_2_in_ag2([]_0, []_0) -> reverse_2_out_ag2([]_0, []_0)

The argument filtering Pi contains the following mapping:
reverse_2_in_ag2(x1, x2)  =  reverse_2_in_ag1(x2)
._22(x1, x2)  =  ._21(x2)
[]_0  =  []_0
if_reverse_2_in_1_ag4(x1, x2, x3, x4)  =  if_reverse_2_in_1_ag2(x3, x4)
reverse_2_in_aa2(x1, x2)  =  reverse_2_in_aa
if_reverse_2_in_1_aa4(x1, x2, x3, x4)  =  if_reverse_2_in_1_aa1(x4)
reverse_2_out_aa2(x1, x2)  =  reverse_2_out_aa1(x1)
if_reverse_2_in_2_aa5(x1, x2, x3, x4, x5)  =  if_reverse_2_in_2_aa2(x2, x5)
app_3_in_aga3(x1, x2, x3)  =  app_3_in_aga1(x2)
if_app_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_aga2(x3, x5)
app_3_out_aga3(x1, x2, x3)  =  app_3_out_aga3(x1, x2, x3)
if_reverse_2_in_2_ag5(x1, x2, x3, x4, x5)  =  if_reverse_2_in_2_ag3(x2, x3, x5)
app_3_in_agg3(x1, x2, x3)  =  app_3_in_agg2(x2, x3)
if_app_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_agg3(x3, x4, x5)
app_3_out_agg3(x1, x2, x3)  =  app_3_out_agg3(x1, x2, x3)
reverse_2_out_ag2(x1, x2)  =  reverse_2_out_ag2(x1, x2)
REVERSE_2_IN_AA2(x1, x2)  =  REVERSE_2_IN_AA

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting we can delete all non-usable rules from R.

↳ PROLOG
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof

Pi DP problem:
The TRS P consists of the following rules:

REVERSE_2_IN_AA2(._22(X, Xs), Ys) -> REVERSE_2_IN_AA2(Xs, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._21(x2)
REVERSE_2_IN_AA2(x1, x2)  =  REVERSE_2_IN_AA

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem by application of Pi.

↳ PROLOG
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

REVERSE_2_IN_AA -> REVERSE_2_IN_AA

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {REVERSE_2_IN_AA}.