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



PROLOG
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

suffix2(Xs, Ys) :- app3(underscore, Xs, Ys).
app3({}0, X, X).
app3(.2(X, Xs), Ys, .2(X, Zs)) :- app3(Xs, Ys, Zs).


With regard to the inferred argument filtering the predicates were used in the following modes:
suffix2: (b,f)
app3: (f,b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


suffix_2_in_ga2(Xs, Ys) -> if_suffix_2_in_1_ga3(Xs, Ys, app_3_in_aga3(underscore, Xs, Ys))
app_3_in_aga3([]_0, X, X) -> app_3_out_aga3([]_0, X, X)
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))
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_suffix_2_in_1_ga3(Xs, Ys, app_3_out_aga3(underscore, Xs, Ys)) -> suffix_2_out_ga2(Xs, Ys)

The argument filtering Pi contains the following mapping:
suffix_2_in_ga2(x1, x2)  =  suffix_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
if_suffix_2_in_1_ga3(x1, x2, x3)  =  if_suffix_2_in_1_ga1(x3)
app_3_in_aga3(x1, x2, x3)  =  app_3_in_aga1(x2)
app_3_out_aga3(x1, x2, x3)  =  app_3_out_aga2(x1, x3)
if_app_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_aga1(x5)
suffix_2_out_ga2(x1, x2)  =  suffix_2_out_ga1(x2)

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:

suffix_2_in_ga2(Xs, Ys) -> if_suffix_2_in_1_ga3(Xs, Ys, app_3_in_aga3(underscore, Xs, Ys))
app_3_in_aga3([]_0, X, X) -> app_3_out_aga3([]_0, X, X)
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))
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_suffix_2_in_1_ga3(Xs, Ys, app_3_out_aga3(underscore, Xs, Ys)) -> suffix_2_out_ga2(Xs, Ys)

The argument filtering Pi contains the following mapping:
suffix_2_in_ga2(x1, x2)  =  suffix_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
if_suffix_2_in_1_ga3(x1, x2, x3)  =  if_suffix_2_in_1_ga1(x3)
app_3_in_aga3(x1, x2, x3)  =  app_3_in_aga1(x2)
app_3_out_aga3(x1, x2, x3)  =  app_3_out_aga2(x1, x3)
if_app_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_aga1(x5)
suffix_2_out_ga2(x1, x2)  =  suffix_2_out_ga1(x2)


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

SUFFIX_2_IN_GA2(Xs, Ys) -> IF_SUFFIX_2_IN_1_GA3(Xs, Ys, app_3_in_aga3(underscore, Xs, Ys))
SUFFIX_2_IN_GA2(Xs, Ys) -> APP_3_IN_AGA3(underscore, Xs, 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)

The TRS R consists of the following rules:

suffix_2_in_ga2(Xs, Ys) -> if_suffix_2_in_1_ga3(Xs, Ys, app_3_in_aga3(underscore, Xs, Ys))
app_3_in_aga3([]_0, X, X) -> app_3_out_aga3([]_0, X, X)
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))
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_suffix_2_in_1_ga3(Xs, Ys, app_3_out_aga3(underscore, Xs, Ys)) -> suffix_2_out_ga2(Xs, Ys)

The argument filtering Pi contains the following mapping:
suffix_2_in_ga2(x1, x2)  =  suffix_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
if_suffix_2_in_1_ga3(x1, x2, x3)  =  if_suffix_2_in_1_ga1(x3)
app_3_in_aga3(x1, x2, x3)  =  app_3_in_aga1(x2)
app_3_out_aga3(x1, x2, x3)  =  app_3_out_aga2(x1, x3)
if_app_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_aga1(x5)
suffix_2_out_ga2(x1, x2)  =  suffix_2_out_ga1(x2)
SUFFIX_2_IN_GA2(x1, x2)  =  SUFFIX_2_IN_GA1(x1)
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_SUFFIX_2_IN_1_GA3(x1, x2, x3)  =  IF_SUFFIX_2_IN_1_GA1(x3)

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:

SUFFIX_2_IN_GA2(Xs, Ys) -> IF_SUFFIX_2_IN_1_GA3(Xs, Ys, app_3_in_aga3(underscore, Xs, Ys))
SUFFIX_2_IN_GA2(Xs, Ys) -> APP_3_IN_AGA3(underscore, Xs, 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)

The TRS R consists of the following rules:

suffix_2_in_ga2(Xs, Ys) -> if_suffix_2_in_1_ga3(Xs, Ys, app_3_in_aga3(underscore, Xs, Ys))
app_3_in_aga3([]_0, X, X) -> app_3_out_aga3([]_0, X, X)
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))
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_suffix_2_in_1_ga3(Xs, Ys, app_3_out_aga3(underscore, Xs, Ys)) -> suffix_2_out_ga2(Xs, Ys)

The argument filtering Pi contains the following mapping:
suffix_2_in_ga2(x1, x2)  =  suffix_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
if_suffix_2_in_1_ga3(x1, x2, x3)  =  if_suffix_2_in_1_ga1(x3)
app_3_in_aga3(x1, x2, x3)  =  app_3_in_aga1(x2)
app_3_out_aga3(x1, x2, x3)  =  app_3_out_aga2(x1, x3)
if_app_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_aga1(x5)
suffix_2_out_ga2(x1, x2)  =  suffix_2_out_ga1(x2)
SUFFIX_2_IN_GA2(x1, x2)  =  SUFFIX_2_IN_GA1(x1)
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_SUFFIX_2_IN_1_GA3(x1, x2, x3)  =  IF_SUFFIX_2_IN_1_GA1(x3)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 1 SCC with 3 less nodes.

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

suffix_2_in_ga2(Xs, Ys) -> if_suffix_2_in_1_ga3(Xs, Ys, app_3_in_aga3(underscore, Xs, Ys))
app_3_in_aga3([]_0, X, X) -> app_3_out_aga3([]_0, X, X)
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))
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_suffix_2_in_1_ga3(Xs, Ys, app_3_out_aga3(underscore, Xs, Ys)) -> suffix_2_out_ga2(Xs, Ys)

The argument filtering Pi contains the following mapping:
suffix_2_in_ga2(x1, x2)  =  suffix_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
if_suffix_2_in_1_ga3(x1, x2, x3)  =  if_suffix_2_in_1_ga1(x3)
app_3_in_aga3(x1, x2, x3)  =  app_3_in_aga1(x2)
app_3_out_aga3(x1, x2, x3)  =  app_3_out_aga2(x1, x3)
if_app_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_aga1(x5)
suffix_2_out_ga2(x1, x2)  =  suffix_2_out_ga1(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
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
PiDP
                  ↳ PiDPToQDPProof
  ↳ 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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
QDP
  ↳ 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}.
With regard to the inferred argument filtering the predicates were used in the following modes:
suffix2: (b,f)
app3: (f,b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

suffix_2_in_ga2(Xs, Ys) -> if_suffix_2_in_1_ga3(Xs, Ys, app_3_in_aga3(underscore, Xs, Ys))
app_3_in_aga3([]_0, X, X) -> app_3_out_aga3([]_0, X, X)
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))
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_suffix_2_in_1_ga3(Xs, Ys, app_3_out_aga3(underscore, Xs, Ys)) -> suffix_2_out_ga2(Xs, Ys)

The argument filtering Pi contains the following mapping:
suffix_2_in_ga2(x1, x2)  =  suffix_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
if_suffix_2_in_1_ga3(x1, x2, x3)  =  if_suffix_2_in_1_ga2(x1, x3)
app_3_in_aga3(x1, x2, x3)  =  app_3_in_aga1(x2)
app_3_out_aga3(x1, x2, x3)  =  app_3_out_aga3(x1, x2, x3)
if_app_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_aga2(x3, x5)
suffix_2_out_ga2(x1, x2)  =  suffix_2_out_ga2(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:

suffix_2_in_ga2(Xs, Ys) -> if_suffix_2_in_1_ga3(Xs, Ys, app_3_in_aga3(underscore, Xs, Ys))
app_3_in_aga3([]_0, X, X) -> app_3_out_aga3([]_0, X, X)
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))
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_suffix_2_in_1_ga3(Xs, Ys, app_3_out_aga3(underscore, Xs, Ys)) -> suffix_2_out_ga2(Xs, Ys)

The argument filtering Pi contains the following mapping:
suffix_2_in_ga2(x1, x2)  =  suffix_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
if_suffix_2_in_1_ga3(x1, x2, x3)  =  if_suffix_2_in_1_ga2(x1, x3)
app_3_in_aga3(x1, x2, x3)  =  app_3_in_aga1(x2)
app_3_out_aga3(x1, x2, x3)  =  app_3_out_aga3(x1, x2, x3)
if_app_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_aga2(x3, x5)
suffix_2_out_ga2(x1, x2)  =  suffix_2_out_ga2(x1, x2)


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

SUFFIX_2_IN_GA2(Xs, Ys) -> IF_SUFFIX_2_IN_1_GA3(Xs, Ys, app_3_in_aga3(underscore, Xs, Ys))
SUFFIX_2_IN_GA2(Xs, Ys) -> APP_3_IN_AGA3(underscore, Xs, 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)

The TRS R consists of the following rules:

suffix_2_in_ga2(Xs, Ys) -> if_suffix_2_in_1_ga3(Xs, Ys, app_3_in_aga3(underscore, Xs, Ys))
app_3_in_aga3([]_0, X, X) -> app_3_out_aga3([]_0, X, X)
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))
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_suffix_2_in_1_ga3(Xs, Ys, app_3_out_aga3(underscore, Xs, Ys)) -> suffix_2_out_ga2(Xs, Ys)

The argument filtering Pi contains the following mapping:
suffix_2_in_ga2(x1, x2)  =  suffix_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
if_suffix_2_in_1_ga3(x1, x2, x3)  =  if_suffix_2_in_1_ga2(x1, x3)
app_3_in_aga3(x1, x2, x3)  =  app_3_in_aga1(x2)
app_3_out_aga3(x1, x2, x3)  =  app_3_out_aga3(x1, x2, x3)
if_app_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_aga2(x3, x5)
suffix_2_out_ga2(x1, x2)  =  suffix_2_out_ga2(x1, x2)
SUFFIX_2_IN_GA2(x1, x2)  =  SUFFIX_2_IN_GA1(x1)
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_SUFFIX_2_IN_1_GA3(x1, x2, x3)  =  IF_SUFFIX_2_IN_1_GA2(x1, x3)

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:

SUFFIX_2_IN_GA2(Xs, Ys) -> IF_SUFFIX_2_IN_1_GA3(Xs, Ys, app_3_in_aga3(underscore, Xs, Ys))
SUFFIX_2_IN_GA2(Xs, Ys) -> APP_3_IN_AGA3(underscore, Xs, 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)

The TRS R consists of the following rules:

suffix_2_in_ga2(Xs, Ys) -> if_suffix_2_in_1_ga3(Xs, Ys, app_3_in_aga3(underscore, Xs, Ys))
app_3_in_aga3([]_0, X, X) -> app_3_out_aga3([]_0, X, X)
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))
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_suffix_2_in_1_ga3(Xs, Ys, app_3_out_aga3(underscore, Xs, Ys)) -> suffix_2_out_ga2(Xs, Ys)

The argument filtering Pi contains the following mapping:
suffix_2_in_ga2(x1, x2)  =  suffix_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
if_suffix_2_in_1_ga3(x1, x2, x3)  =  if_suffix_2_in_1_ga2(x1, x3)
app_3_in_aga3(x1, x2, x3)  =  app_3_in_aga1(x2)
app_3_out_aga3(x1, x2, x3)  =  app_3_out_aga3(x1, x2, x3)
if_app_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_aga2(x3, x5)
suffix_2_out_ga2(x1, x2)  =  suffix_2_out_ga2(x1, x2)
SUFFIX_2_IN_GA2(x1, x2)  =  SUFFIX_2_IN_GA1(x1)
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_SUFFIX_2_IN_1_GA3(x1, x2, x3)  =  IF_SUFFIX_2_IN_1_GA2(x1, x3)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 1 SCC with 3 less nodes.

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

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:

suffix_2_in_ga2(Xs, Ys) -> if_suffix_2_in_1_ga3(Xs, Ys, app_3_in_aga3(underscore, Xs, Ys))
app_3_in_aga3([]_0, X, X) -> app_3_out_aga3([]_0, X, X)
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))
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_suffix_2_in_1_ga3(Xs, Ys, app_3_out_aga3(underscore, Xs, Ys)) -> suffix_2_out_ga2(Xs, Ys)

The argument filtering Pi contains the following mapping:
suffix_2_in_ga2(x1, x2)  =  suffix_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
if_suffix_2_in_1_ga3(x1, x2, x3)  =  if_suffix_2_in_1_ga2(x1, x3)
app_3_in_aga3(x1, x2, x3)  =  app_3_in_aga1(x2)
app_3_out_aga3(x1, x2, x3)  =  app_3_out_aga3(x1, x2, x3)
if_app_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_aga2(x3, x5)
suffix_2_out_ga2(x1, x2)  =  suffix_2_out_ga2(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
            ↳ PiDP
              ↳ UsableRulesProof
PiDP
                  ↳ PiDPToQDPProof

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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
QDP

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}.