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



PROLOG
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

reverse2(X1s, X2s) :- reverse3(X1s, {}0, X2s).
reverse3({}0, Xs, Xs).
reverse3(.2(X, X1s), X2s, Ys) :- reverse3(X1s, .2(X, X2s), Ys).


With regard to the inferred argument filtering the predicates were used in the following modes:
reverse2: (f,b)
reverse3: (f,b,b) (f,f,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(X1s, X2s) -> if_reverse_2_in_1_ag3(X1s, X2s, reverse_3_in_agg3(X1s, []_0, X2s))
reverse_3_in_agg3([]_0, Xs, Xs) -> reverse_3_out_agg3([]_0, Xs, Xs)
reverse_3_in_agg3(._22(X, X1s), X2s, Ys) -> if_reverse_3_in_1_agg5(X, X1s, X2s, Ys, reverse_3_in_aag3(X1s, ._22(X, X2s), Ys))
reverse_3_in_aag3([]_0, Xs, Xs) -> reverse_3_out_aag3([]_0, Xs, Xs)
reverse_3_in_aag3(._22(X, X1s), X2s, Ys) -> if_reverse_3_in_1_aag5(X, X1s, X2s, Ys, reverse_3_in_aag3(X1s, ._22(X, X2s), Ys))
if_reverse_3_in_1_aag5(X, X1s, X2s, Ys, reverse_3_out_aag3(X1s, ._22(X, X2s), Ys)) -> reverse_3_out_aag3(._22(X, X1s), X2s, Ys)
if_reverse_3_in_1_agg5(X, X1s, X2s, Ys, reverse_3_out_aag3(X1s, ._22(X, X2s), Ys)) -> reverse_3_out_agg3(._22(X, X1s), X2s, Ys)
if_reverse_2_in_1_ag3(X1s, X2s, reverse_3_out_agg3(X1s, []_0, X2s)) -> reverse_2_out_ag2(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_2_in_ag2(x1, x2)  =  reverse_2_in_ag1(x2)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_reverse_2_in_1_ag3(x1, x2, x3)  =  if_reverse_2_in_1_ag1(x3)
reverse_3_in_agg3(x1, x2, x3)  =  reverse_3_in_agg2(x2, x3)
reverse_3_out_agg3(x1, x2, x3)  =  reverse_3_out_agg1(x1)
if_reverse_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_reverse_3_in_1_agg1(x5)
reverse_3_in_aag3(x1, x2, x3)  =  reverse_3_in_aag1(x3)
reverse_3_out_aag3(x1, x2, x3)  =  reverse_3_out_aag2(x1, x2)
if_reverse_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_reverse_3_in_1_aag1(x5)
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(X1s, X2s) -> if_reverse_2_in_1_ag3(X1s, X2s, reverse_3_in_agg3(X1s, []_0, X2s))
reverse_3_in_agg3([]_0, Xs, Xs) -> reverse_3_out_agg3([]_0, Xs, Xs)
reverse_3_in_agg3(._22(X, X1s), X2s, Ys) -> if_reverse_3_in_1_agg5(X, X1s, X2s, Ys, reverse_3_in_aag3(X1s, ._22(X, X2s), Ys))
reverse_3_in_aag3([]_0, Xs, Xs) -> reverse_3_out_aag3([]_0, Xs, Xs)
reverse_3_in_aag3(._22(X, X1s), X2s, Ys) -> if_reverse_3_in_1_aag5(X, X1s, X2s, Ys, reverse_3_in_aag3(X1s, ._22(X, X2s), Ys))
if_reverse_3_in_1_aag5(X, X1s, X2s, Ys, reverse_3_out_aag3(X1s, ._22(X, X2s), Ys)) -> reverse_3_out_aag3(._22(X, X1s), X2s, Ys)
if_reverse_3_in_1_agg5(X, X1s, X2s, Ys, reverse_3_out_aag3(X1s, ._22(X, X2s), Ys)) -> reverse_3_out_agg3(._22(X, X1s), X2s, Ys)
if_reverse_2_in_1_ag3(X1s, X2s, reverse_3_out_agg3(X1s, []_0, X2s)) -> reverse_2_out_ag2(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_2_in_ag2(x1, x2)  =  reverse_2_in_ag1(x2)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_reverse_2_in_1_ag3(x1, x2, x3)  =  if_reverse_2_in_1_ag1(x3)
reverse_3_in_agg3(x1, x2, x3)  =  reverse_3_in_agg2(x2, x3)
reverse_3_out_agg3(x1, x2, x3)  =  reverse_3_out_agg1(x1)
if_reverse_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_reverse_3_in_1_agg1(x5)
reverse_3_in_aag3(x1, x2, x3)  =  reverse_3_in_aag1(x3)
reverse_3_out_aag3(x1, x2, x3)  =  reverse_3_out_aag2(x1, x2)
if_reverse_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_reverse_3_in_1_aag1(x5)
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(X1s, X2s) -> IF_REVERSE_2_IN_1_AG3(X1s, X2s, reverse_3_in_agg3(X1s, []_0, X2s))
REVERSE_2_IN_AG2(X1s, X2s) -> REVERSE_3_IN_AGG3(X1s, []_0, X2s)
REVERSE_3_IN_AGG3(._22(X, X1s), X2s, Ys) -> IF_REVERSE_3_IN_1_AGG5(X, X1s, X2s, Ys, reverse_3_in_aag3(X1s, ._22(X, X2s), Ys))
REVERSE_3_IN_AGG3(._22(X, X1s), X2s, Ys) -> REVERSE_3_IN_AAG3(X1s, ._22(X, X2s), Ys)
REVERSE_3_IN_AAG3(._22(X, X1s), X2s, Ys) -> IF_REVERSE_3_IN_1_AAG5(X, X1s, X2s, Ys, reverse_3_in_aag3(X1s, ._22(X, X2s), Ys))
REVERSE_3_IN_AAG3(._22(X, X1s), X2s, Ys) -> REVERSE_3_IN_AAG3(X1s, ._22(X, X2s), Ys)

The TRS R consists of the following rules:

reverse_2_in_ag2(X1s, X2s) -> if_reverse_2_in_1_ag3(X1s, X2s, reverse_3_in_agg3(X1s, []_0, X2s))
reverse_3_in_agg3([]_0, Xs, Xs) -> reverse_3_out_agg3([]_0, Xs, Xs)
reverse_3_in_agg3(._22(X, X1s), X2s, Ys) -> if_reverse_3_in_1_agg5(X, X1s, X2s, Ys, reverse_3_in_aag3(X1s, ._22(X, X2s), Ys))
reverse_3_in_aag3([]_0, Xs, Xs) -> reverse_3_out_aag3([]_0, Xs, Xs)
reverse_3_in_aag3(._22(X, X1s), X2s, Ys) -> if_reverse_3_in_1_aag5(X, X1s, X2s, Ys, reverse_3_in_aag3(X1s, ._22(X, X2s), Ys))
if_reverse_3_in_1_aag5(X, X1s, X2s, Ys, reverse_3_out_aag3(X1s, ._22(X, X2s), Ys)) -> reverse_3_out_aag3(._22(X, X1s), X2s, Ys)
if_reverse_3_in_1_agg5(X, X1s, X2s, Ys, reverse_3_out_aag3(X1s, ._22(X, X2s), Ys)) -> reverse_3_out_agg3(._22(X, X1s), X2s, Ys)
if_reverse_2_in_1_ag3(X1s, X2s, reverse_3_out_agg3(X1s, []_0, X2s)) -> reverse_2_out_ag2(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_2_in_ag2(x1, x2)  =  reverse_2_in_ag1(x2)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_reverse_2_in_1_ag3(x1, x2, x3)  =  if_reverse_2_in_1_ag1(x3)
reverse_3_in_agg3(x1, x2, x3)  =  reverse_3_in_agg2(x2, x3)
reverse_3_out_agg3(x1, x2, x3)  =  reverse_3_out_agg1(x1)
if_reverse_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_reverse_3_in_1_agg1(x5)
reverse_3_in_aag3(x1, x2, x3)  =  reverse_3_in_aag1(x3)
reverse_3_out_aag3(x1, x2, x3)  =  reverse_3_out_aag2(x1, x2)
if_reverse_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_reverse_3_in_1_aag1(x5)
reverse_2_out_ag2(x1, x2)  =  reverse_2_out_ag1(x1)
IF_REVERSE_3_IN_1_AAG5(x1, x2, x3, x4, x5)  =  IF_REVERSE_3_IN_1_AAG1(x5)
REVERSE_3_IN_AAG3(x1, x2, x3)  =  REVERSE_3_IN_AAG1(x3)
IF_REVERSE_3_IN_1_AGG5(x1, x2, x3, x4, x5)  =  IF_REVERSE_3_IN_1_AGG1(x5)
IF_REVERSE_2_IN_1_AG3(x1, x2, x3)  =  IF_REVERSE_2_IN_1_AG1(x3)
REVERSE_3_IN_AGG3(x1, x2, x3)  =  REVERSE_3_IN_AGG2(x2, x3)
REVERSE_2_IN_AG2(x1, x2)  =  REVERSE_2_IN_AG1(x2)

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(X1s, X2s) -> IF_REVERSE_2_IN_1_AG3(X1s, X2s, reverse_3_in_agg3(X1s, []_0, X2s))
REVERSE_2_IN_AG2(X1s, X2s) -> REVERSE_3_IN_AGG3(X1s, []_0, X2s)
REVERSE_3_IN_AGG3(._22(X, X1s), X2s, Ys) -> IF_REVERSE_3_IN_1_AGG5(X, X1s, X2s, Ys, reverse_3_in_aag3(X1s, ._22(X, X2s), Ys))
REVERSE_3_IN_AGG3(._22(X, X1s), X2s, Ys) -> REVERSE_3_IN_AAG3(X1s, ._22(X, X2s), Ys)
REVERSE_3_IN_AAG3(._22(X, X1s), X2s, Ys) -> IF_REVERSE_3_IN_1_AAG5(X, X1s, X2s, Ys, reverse_3_in_aag3(X1s, ._22(X, X2s), Ys))
REVERSE_3_IN_AAG3(._22(X, X1s), X2s, Ys) -> REVERSE_3_IN_AAG3(X1s, ._22(X, X2s), Ys)

The TRS R consists of the following rules:

reverse_2_in_ag2(X1s, X2s) -> if_reverse_2_in_1_ag3(X1s, X2s, reverse_3_in_agg3(X1s, []_0, X2s))
reverse_3_in_agg3([]_0, Xs, Xs) -> reverse_3_out_agg3([]_0, Xs, Xs)
reverse_3_in_agg3(._22(X, X1s), X2s, Ys) -> if_reverse_3_in_1_agg5(X, X1s, X2s, Ys, reverse_3_in_aag3(X1s, ._22(X, X2s), Ys))
reverse_3_in_aag3([]_0, Xs, Xs) -> reverse_3_out_aag3([]_0, Xs, Xs)
reverse_3_in_aag3(._22(X, X1s), X2s, Ys) -> if_reverse_3_in_1_aag5(X, X1s, X2s, Ys, reverse_3_in_aag3(X1s, ._22(X, X2s), Ys))
if_reverse_3_in_1_aag5(X, X1s, X2s, Ys, reverse_3_out_aag3(X1s, ._22(X, X2s), Ys)) -> reverse_3_out_aag3(._22(X, X1s), X2s, Ys)
if_reverse_3_in_1_agg5(X, X1s, X2s, Ys, reverse_3_out_aag3(X1s, ._22(X, X2s), Ys)) -> reverse_3_out_agg3(._22(X, X1s), X2s, Ys)
if_reverse_2_in_1_ag3(X1s, X2s, reverse_3_out_agg3(X1s, []_0, X2s)) -> reverse_2_out_ag2(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_2_in_ag2(x1, x2)  =  reverse_2_in_ag1(x2)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_reverse_2_in_1_ag3(x1, x2, x3)  =  if_reverse_2_in_1_ag1(x3)
reverse_3_in_agg3(x1, x2, x3)  =  reverse_3_in_agg2(x2, x3)
reverse_3_out_agg3(x1, x2, x3)  =  reverse_3_out_agg1(x1)
if_reverse_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_reverse_3_in_1_agg1(x5)
reverse_3_in_aag3(x1, x2, x3)  =  reverse_3_in_aag1(x3)
reverse_3_out_aag3(x1, x2, x3)  =  reverse_3_out_aag2(x1, x2)
if_reverse_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_reverse_3_in_1_aag1(x5)
reverse_2_out_ag2(x1, x2)  =  reverse_2_out_ag1(x1)
IF_REVERSE_3_IN_1_AAG5(x1, x2, x3, x4, x5)  =  IF_REVERSE_3_IN_1_AAG1(x5)
REVERSE_3_IN_AAG3(x1, x2, x3)  =  REVERSE_3_IN_AAG1(x3)
IF_REVERSE_3_IN_1_AGG5(x1, x2, x3, x4, x5)  =  IF_REVERSE_3_IN_1_AGG1(x5)
IF_REVERSE_2_IN_1_AG3(x1, x2, x3)  =  IF_REVERSE_2_IN_1_AG1(x3)
REVERSE_3_IN_AGG3(x1, x2, x3)  =  REVERSE_3_IN_AGG2(x2, x3)
REVERSE_2_IN_AG2(x1, x2)  =  REVERSE_2_IN_AG1(x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 1 SCC with 5 less nodes.

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

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

REVERSE_3_IN_AAG3(._22(X, X1s), X2s, Ys) -> REVERSE_3_IN_AAG3(X1s, ._22(X, X2s), Ys)

The TRS R consists of the following rules:

reverse_2_in_ag2(X1s, X2s) -> if_reverse_2_in_1_ag3(X1s, X2s, reverse_3_in_agg3(X1s, []_0, X2s))
reverse_3_in_agg3([]_0, Xs, Xs) -> reverse_3_out_agg3([]_0, Xs, Xs)
reverse_3_in_agg3(._22(X, X1s), X2s, Ys) -> if_reverse_3_in_1_agg5(X, X1s, X2s, Ys, reverse_3_in_aag3(X1s, ._22(X, X2s), Ys))
reverse_3_in_aag3([]_0, Xs, Xs) -> reverse_3_out_aag3([]_0, Xs, Xs)
reverse_3_in_aag3(._22(X, X1s), X2s, Ys) -> if_reverse_3_in_1_aag5(X, X1s, X2s, Ys, reverse_3_in_aag3(X1s, ._22(X, X2s), Ys))
if_reverse_3_in_1_aag5(X, X1s, X2s, Ys, reverse_3_out_aag3(X1s, ._22(X, X2s), Ys)) -> reverse_3_out_aag3(._22(X, X1s), X2s, Ys)
if_reverse_3_in_1_agg5(X, X1s, X2s, Ys, reverse_3_out_aag3(X1s, ._22(X, X2s), Ys)) -> reverse_3_out_agg3(._22(X, X1s), X2s, Ys)
if_reverse_2_in_1_ag3(X1s, X2s, reverse_3_out_agg3(X1s, []_0, X2s)) -> reverse_2_out_ag2(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_2_in_ag2(x1, x2)  =  reverse_2_in_ag1(x2)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_reverse_2_in_1_ag3(x1, x2, x3)  =  if_reverse_2_in_1_ag1(x3)
reverse_3_in_agg3(x1, x2, x3)  =  reverse_3_in_agg2(x2, x3)
reverse_3_out_agg3(x1, x2, x3)  =  reverse_3_out_agg1(x1)
if_reverse_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_reverse_3_in_1_agg1(x5)
reverse_3_in_aag3(x1, x2, x3)  =  reverse_3_in_aag1(x3)
reverse_3_out_aag3(x1, x2, x3)  =  reverse_3_out_aag2(x1, x2)
if_reverse_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_reverse_3_in_1_aag1(x5)
reverse_2_out_ag2(x1, x2)  =  reverse_2_out_ag1(x1)
REVERSE_3_IN_AAG3(x1, x2, x3)  =  REVERSE_3_IN_AAG1(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
            ↳ PiDP
              ↳ UsableRulesProof
PiDP
                  ↳ PiDPToQDPProof
  ↳ PrologToPiTRSProof

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

REVERSE_3_IN_AAG3(._22(X, X1s), X2s, Ys) -> REVERSE_3_IN_AAG3(X1s, ._22(X, X2s), Ys)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._22(x1, x2)
REVERSE_3_IN_AAG3(x1, x2, x3)  =  REVERSE_3_IN_AAG1(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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
QDP
  ↳ PrologToPiTRSProof

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

REVERSE_3_IN_AAG1(Ys) -> REVERSE_3_IN_AAG1(Ys)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {REVERSE_3_IN_AAG1}.
With regard to the inferred argument filtering the predicates were used in the following modes:
reverse2: (f,b)
reverse3: (f,b,b) (f,f,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(X1s, X2s) -> if_reverse_2_in_1_ag3(X1s, X2s, reverse_3_in_agg3(X1s, []_0, X2s))
reverse_3_in_agg3([]_0, Xs, Xs) -> reverse_3_out_agg3([]_0, Xs, Xs)
reverse_3_in_agg3(._22(X, X1s), X2s, Ys) -> if_reverse_3_in_1_agg5(X, X1s, X2s, Ys, reverse_3_in_aag3(X1s, ._22(X, X2s), Ys))
reverse_3_in_aag3([]_0, Xs, Xs) -> reverse_3_out_aag3([]_0, Xs, Xs)
reverse_3_in_aag3(._22(X, X1s), X2s, Ys) -> if_reverse_3_in_1_aag5(X, X1s, X2s, Ys, reverse_3_in_aag3(X1s, ._22(X, X2s), Ys))
if_reverse_3_in_1_aag5(X, X1s, X2s, Ys, reverse_3_out_aag3(X1s, ._22(X, X2s), Ys)) -> reverse_3_out_aag3(._22(X, X1s), X2s, Ys)
if_reverse_3_in_1_agg5(X, X1s, X2s, Ys, reverse_3_out_aag3(X1s, ._22(X, X2s), Ys)) -> reverse_3_out_agg3(._22(X, X1s), X2s, Ys)
if_reverse_2_in_1_ag3(X1s, X2s, reverse_3_out_agg3(X1s, []_0, X2s)) -> reverse_2_out_ag2(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_2_in_ag2(x1, x2)  =  reverse_2_in_ag1(x2)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_reverse_2_in_1_ag3(x1, x2, x3)  =  if_reverse_2_in_1_ag2(x2, x3)
reverse_3_in_agg3(x1, x2, x3)  =  reverse_3_in_agg2(x2, x3)
reverse_3_out_agg3(x1, x2, x3)  =  reverse_3_out_agg3(x1, x2, x3)
if_reverse_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_reverse_3_in_1_agg3(x3, x4, x5)
reverse_3_in_aag3(x1, x2, x3)  =  reverse_3_in_aag1(x3)
reverse_3_out_aag3(x1, x2, x3)  =  reverse_3_out_aag3(x1, x2, x3)
if_reverse_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_reverse_3_in_1_aag2(x4, x5)
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(X1s, X2s) -> if_reverse_2_in_1_ag3(X1s, X2s, reverse_3_in_agg3(X1s, []_0, X2s))
reverse_3_in_agg3([]_0, Xs, Xs) -> reverse_3_out_agg3([]_0, Xs, Xs)
reverse_3_in_agg3(._22(X, X1s), X2s, Ys) -> if_reverse_3_in_1_agg5(X, X1s, X2s, Ys, reverse_3_in_aag3(X1s, ._22(X, X2s), Ys))
reverse_3_in_aag3([]_0, Xs, Xs) -> reverse_3_out_aag3([]_0, Xs, Xs)
reverse_3_in_aag3(._22(X, X1s), X2s, Ys) -> if_reverse_3_in_1_aag5(X, X1s, X2s, Ys, reverse_3_in_aag3(X1s, ._22(X, X2s), Ys))
if_reverse_3_in_1_aag5(X, X1s, X2s, Ys, reverse_3_out_aag3(X1s, ._22(X, X2s), Ys)) -> reverse_3_out_aag3(._22(X, X1s), X2s, Ys)
if_reverse_3_in_1_agg5(X, X1s, X2s, Ys, reverse_3_out_aag3(X1s, ._22(X, X2s), Ys)) -> reverse_3_out_agg3(._22(X, X1s), X2s, Ys)
if_reverse_2_in_1_ag3(X1s, X2s, reverse_3_out_agg3(X1s, []_0, X2s)) -> reverse_2_out_ag2(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_2_in_ag2(x1, x2)  =  reverse_2_in_ag1(x2)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_reverse_2_in_1_ag3(x1, x2, x3)  =  if_reverse_2_in_1_ag2(x2, x3)
reverse_3_in_agg3(x1, x2, x3)  =  reverse_3_in_agg2(x2, x3)
reverse_3_out_agg3(x1, x2, x3)  =  reverse_3_out_agg3(x1, x2, x3)
if_reverse_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_reverse_3_in_1_agg3(x3, x4, x5)
reverse_3_in_aag3(x1, x2, x3)  =  reverse_3_in_aag1(x3)
reverse_3_out_aag3(x1, x2, x3)  =  reverse_3_out_aag3(x1, x2, x3)
if_reverse_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_reverse_3_in_1_aag2(x4, x5)
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(X1s, X2s) -> IF_REVERSE_2_IN_1_AG3(X1s, X2s, reverse_3_in_agg3(X1s, []_0, X2s))
REVERSE_2_IN_AG2(X1s, X2s) -> REVERSE_3_IN_AGG3(X1s, []_0, X2s)
REVERSE_3_IN_AGG3(._22(X, X1s), X2s, Ys) -> IF_REVERSE_3_IN_1_AGG5(X, X1s, X2s, Ys, reverse_3_in_aag3(X1s, ._22(X, X2s), Ys))
REVERSE_3_IN_AGG3(._22(X, X1s), X2s, Ys) -> REVERSE_3_IN_AAG3(X1s, ._22(X, X2s), Ys)
REVERSE_3_IN_AAG3(._22(X, X1s), X2s, Ys) -> IF_REVERSE_3_IN_1_AAG5(X, X1s, X2s, Ys, reverse_3_in_aag3(X1s, ._22(X, X2s), Ys))
REVERSE_3_IN_AAG3(._22(X, X1s), X2s, Ys) -> REVERSE_3_IN_AAG3(X1s, ._22(X, X2s), Ys)

The TRS R consists of the following rules:

reverse_2_in_ag2(X1s, X2s) -> if_reverse_2_in_1_ag3(X1s, X2s, reverse_3_in_agg3(X1s, []_0, X2s))
reverse_3_in_agg3([]_0, Xs, Xs) -> reverse_3_out_agg3([]_0, Xs, Xs)
reverse_3_in_agg3(._22(X, X1s), X2s, Ys) -> if_reverse_3_in_1_agg5(X, X1s, X2s, Ys, reverse_3_in_aag3(X1s, ._22(X, X2s), Ys))
reverse_3_in_aag3([]_0, Xs, Xs) -> reverse_3_out_aag3([]_0, Xs, Xs)
reverse_3_in_aag3(._22(X, X1s), X2s, Ys) -> if_reverse_3_in_1_aag5(X, X1s, X2s, Ys, reverse_3_in_aag3(X1s, ._22(X, X2s), Ys))
if_reverse_3_in_1_aag5(X, X1s, X2s, Ys, reverse_3_out_aag3(X1s, ._22(X, X2s), Ys)) -> reverse_3_out_aag3(._22(X, X1s), X2s, Ys)
if_reverse_3_in_1_agg5(X, X1s, X2s, Ys, reverse_3_out_aag3(X1s, ._22(X, X2s), Ys)) -> reverse_3_out_agg3(._22(X, X1s), X2s, Ys)
if_reverse_2_in_1_ag3(X1s, X2s, reverse_3_out_agg3(X1s, []_0, X2s)) -> reverse_2_out_ag2(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_2_in_ag2(x1, x2)  =  reverse_2_in_ag1(x2)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_reverse_2_in_1_ag3(x1, x2, x3)  =  if_reverse_2_in_1_ag2(x2, x3)
reverse_3_in_agg3(x1, x2, x3)  =  reverse_3_in_agg2(x2, x3)
reverse_3_out_agg3(x1, x2, x3)  =  reverse_3_out_agg3(x1, x2, x3)
if_reverse_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_reverse_3_in_1_agg3(x3, x4, x5)
reverse_3_in_aag3(x1, x2, x3)  =  reverse_3_in_aag1(x3)
reverse_3_out_aag3(x1, x2, x3)  =  reverse_3_out_aag3(x1, x2, x3)
if_reverse_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_reverse_3_in_1_aag2(x4, x5)
reverse_2_out_ag2(x1, x2)  =  reverse_2_out_ag2(x1, x2)
IF_REVERSE_3_IN_1_AAG5(x1, x2, x3, x4, x5)  =  IF_REVERSE_3_IN_1_AAG2(x4, x5)
REVERSE_3_IN_AAG3(x1, x2, x3)  =  REVERSE_3_IN_AAG1(x3)
IF_REVERSE_3_IN_1_AGG5(x1, x2, x3, x4, x5)  =  IF_REVERSE_3_IN_1_AGG3(x3, x4, x5)
IF_REVERSE_2_IN_1_AG3(x1, x2, x3)  =  IF_REVERSE_2_IN_1_AG2(x2, x3)
REVERSE_3_IN_AGG3(x1, x2, x3)  =  REVERSE_3_IN_AGG2(x2, x3)
REVERSE_2_IN_AG2(x1, x2)  =  REVERSE_2_IN_AG1(x2)

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(X1s, X2s) -> IF_REVERSE_2_IN_1_AG3(X1s, X2s, reverse_3_in_agg3(X1s, []_0, X2s))
REVERSE_2_IN_AG2(X1s, X2s) -> REVERSE_3_IN_AGG3(X1s, []_0, X2s)
REVERSE_3_IN_AGG3(._22(X, X1s), X2s, Ys) -> IF_REVERSE_3_IN_1_AGG5(X, X1s, X2s, Ys, reverse_3_in_aag3(X1s, ._22(X, X2s), Ys))
REVERSE_3_IN_AGG3(._22(X, X1s), X2s, Ys) -> REVERSE_3_IN_AAG3(X1s, ._22(X, X2s), Ys)
REVERSE_3_IN_AAG3(._22(X, X1s), X2s, Ys) -> IF_REVERSE_3_IN_1_AAG5(X, X1s, X2s, Ys, reverse_3_in_aag3(X1s, ._22(X, X2s), Ys))
REVERSE_3_IN_AAG3(._22(X, X1s), X2s, Ys) -> REVERSE_3_IN_AAG3(X1s, ._22(X, X2s), Ys)

The TRS R consists of the following rules:

reverse_2_in_ag2(X1s, X2s) -> if_reverse_2_in_1_ag3(X1s, X2s, reverse_3_in_agg3(X1s, []_0, X2s))
reverse_3_in_agg3([]_0, Xs, Xs) -> reverse_3_out_agg3([]_0, Xs, Xs)
reverse_3_in_agg3(._22(X, X1s), X2s, Ys) -> if_reverse_3_in_1_agg5(X, X1s, X2s, Ys, reverse_3_in_aag3(X1s, ._22(X, X2s), Ys))
reverse_3_in_aag3([]_0, Xs, Xs) -> reverse_3_out_aag3([]_0, Xs, Xs)
reverse_3_in_aag3(._22(X, X1s), X2s, Ys) -> if_reverse_3_in_1_aag5(X, X1s, X2s, Ys, reverse_3_in_aag3(X1s, ._22(X, X2s), Ys))
if_reverse_3_in_1_aag5(X, X1s, X2s, Ys, reverse_3_out_aag3(X1s, ._22(X, X2s), Ys)) -> reverse_3_out_aag3(._22(X, X1s), X2s, Ys)
if_reverse_3_in_1_agg5(X, X1s, X2s, Ys, reverse_3_out_aag3(X1s, ._22(X, X2s), Ys)) -> reverse_3_out_agg3(._22(X, X1s), X2s, Ys)
if_reverse_2_in_1_ag3(X1s, X2s, reverse_3_out_agg3(X1s, []_0, X2s)) -> reverse_2_out_ag2(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_2_in_ag2(x1, x2)  =  reverse_2_in_ag1(x2)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_reverse_2_in_1_ag3(x1, x2, x3)  =  if_reverse_2_in_1_ag2(x2, x3)
reverse_3_in_agg3(x1, x2, x3)  =  reverse_3_in_agg2(x2, x3)
reverse_3_out_agg3(x1, x2, x3)  =  reverse_3_out_agg3(x1, x2, x3)
if_reverse_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_reverse_3_in_1_agg3(x3, x4, x5)
reverse_3_in_aag3(x1, x2, x3)  =  reverse_3_in_aag1(x3)
reverse_3_out_aag3(x1, x2, x3)  =  reverse_3_out_aag3(x1, x2, x3)
if_reverse_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_reverse_3_in_1_aag2(x4, x5)
reverse_2_out_ag2(x1, x2)  =  reverse_2_out_ag2(x1, x2)
IF_REVERSE_3_IN_1_AAG5(x1, x2, x3, x4, x5)  =  IF_REVERSE_3_IN_1_AAG2(x4, x5)
REVERSE_3_IN_AAG3(x1, x2, x3)  =  REVERSE_3_IN_AAG1(x3)
IF_REVERSE_3_IN_1_AGG5(x1, x2, x3, x4, x5)  =  IF_REVERSE_3_IN_1_AGG3(x3, x4, x5)
IF_REVERSE_2_IN_1_AG3(x1, x2, x3)  =  IF_REVERSE_2_IN_1_AG2(x2, x3)
REVERSE_3_IN_AGG3(x1, x2, x3)  =  REVERSE_3_IN_AGG2(x2, x3)
REVERSE_2_IN_AG2(x1, x2)  =  REVERSE_2_IN_AG1(x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 1 SCC with 5 less nodes.

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

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

REVERSE_3_IN_AAG3(._22(X, X1s), X2s, Ys) -> REVERSE_3_IN_AAG3(X1s, ._22(X, X2s), Ys)

The TRS R consists of the following rules:

reverse_2_in_ag2(X1s, X2s) -> if_reverse_2_in_1_ag3(X1s, X2s, reverse_3_in_agg3(X1s, []_0, X2s))
reverse_3_in_agg3([]_0, Xs, Xs) -> reverse_3_out_agg3([]_0, Xs, Xs)
reverse_3_in_agg3(._22(X, X1s), X2s, Ys) -> if_reverse_3_in_1_agg5(X, X1s, X2s, Ys, reverse_3_in_aag3(X1s, ._22(X, X2s), Ys))
reverse_3_in_aag3([]_0, Xs, Xs) -> reverse_3_out_aag3([]_0, Xs, Xs)
reverse_3_in_aag3(._22(X, X1s), X2s, Ys) -> if_reverse_3_in_1_aag5(X, X1s, X2s, Ys, reverse_3_in_aag3(X1s, ._22(X, X2s), Ys))
if_reverse_3_in_1_aag5(X, X1s, X2s, Ys, reverse_3_out_aag3(X1s, ._22(X, X2s), Ys)) -> reverse_3_out_aag3(._22(X, X1s), X2s, Ys)
if_reverse_3_in_1_agg5(X, X1s, X2s, Ys, reverse_3_out_aag3(X1s, ._22(X, X2s), Ys)) -> reverse_3_out_agg3(._22(X, X1s), X2s, Ys)
if_reverse_2_in_1_ag3(X1s, X2s, reverse_3_out_agg3(X1s, []_0, X2s)) -> reverse_2_out_ag2(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_2_in_ag2(x1, x2)  =  reverse_2_in_ag1(x2)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_reverse_2_in_1_ag3(x1, x2, x3)  =  if_reverse_2_in_1_ag2(x2, x3)
reverse_3_in_agg3(x1, x2, x3)  =  reverse_3_in_agg2(x2, x3)
reverse_3_out_agg3(x1, x2, x3)  =  reverse_3_out_agg3(x1, x2, x3)
if_reverse_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_reverse_3_in_1_agg3(x3, x4, x5)
reverse_3_in_aag3(x1, x2, x3)  =  reverse_3_in_aag1(x3)
reverse_3_out_aag3(x1, x2, x3)  =  reverse_3_out_aag3(x1, x2, x3)
if_reverse_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_reverse_3_in_1_aag2(x4, x5)
reverse_2_out_ag2(x1, x2)  =  reverse_2_out_ag2(x1, x2)
REVERSE_3_IN_AAG3(x1, x2, x3)  =  REVERSE_3_IN_AAG1(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
            ↳ PiDP
              ↳ UsableRulesProof
PiDP
                  ↳ PiDPToQDPProof

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

REVERSE_3_IN_AAG3(._22(X, X1s), X2s, Ys) -> REVERSE_3_IN_AAG3(X1s, ._22(X, X2s), Ys)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._22(x1, x2)
REVERSE_3_IN_AAG3(x1, x2, x3)  =  REVERSE_3_IN_AAG1(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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
QDP

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

REVERSE_3_IN_AAG1(Ys) -> REVERSE_3_IN_AAG1(Ys)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {REVERSE_3_IN_AAG1}.