Left Termination of the query pattern app2(f,b,b) w.r.t. the given Prolog program could successfully be proven:



↳ PROLOG
  ↳ UnrequestedClauseRemoverProof

app13(.2(X, Xs), Ys, .2(X, Zs)) :- app13(Xs, Ys, Zs).
app13({}0, Ys, Ys).
app23(.2(X, Xs), Ys, .2(X, Zs)) :- app23(Xs, Ys, Zs).
app23({}0, Ys, Ys).


The clauses

app13(.2(X, Xs), Ys, .2(X, Zs)) :- app13(Xs, Ys, Zs).
app13({}0, Ys, Ys).

can be ignored, as they are not needed by any of the given querys.

Deleting these clauses results in the following prolog program:

app23(.2(X, Xs), Ys, .2(X, Zs)) :- app23(Xs, Ys, Zs).
app23({}0, Ys, Ys).



↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof

app23(.2(X, Xs), Ys, .2(X, Zs)) :- app23(Xs, Ys, Zs).
app23({}0, Ys, Ys).


With regard to the inferred argument filtering the predicates were used in the following modes:
app23: (f,b,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


app2_3_in_agg3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app2_3_in_1_agg5(X, Xs, Ys, Zs, app2_3_in_agg3(Xs, Ys, Zs))
app2_3_in_agg3([]_0, Ys, Ys) -> app2_3_out_agg3([]_0, Ys, Ys)
if_app2_3_in_1_agg5(X, Xs, Ys, Zs, app2_3_out_agg3(Xs, Ys, Zs)) -> app2_3_out_agg3(._22(X, Xs), Ys, ._22(X, Zs))

The argument filtering Pi contains the following mapping:
app2_3_in_agg3(x1, x2, x3)  =  app2_3_in_agg2(x2, x3)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_app2_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_app2_3_in_1_agg2(x1, x5)
app2_3_out_agg3(x1, x2, x3)  =  app2_3_out_agg1(x1)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG



↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
        ↳ PiTRS
          ↳ DependencyPairsProof

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

app2_3_in_agg3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app2_3_in_1_agg5(X, Xs, Ys, Zs, app2_3_in_agg3(Xs, Ys, Zs))
app2_3_in_agg3([]_0, Ys, Ys) -> app2_3_out_agg3([]_0, Ys, Ys)
if_app2_3_in_1_agg5(X, Xs, Ys, Zs, app2_3_out_agg3(Xs, Ys, Zs)) -> app2_3_out_agg3(._22(X, Xs), Ys, ._22(X, Zs))

The argument filtering Pi contains the following mapping:
app2_3_in_agg3(x1, x2, x3)  =  app2_3_in_agg2(x2, x3)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_app2_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_app2_3_in_1_agg2(x1, x5)
app2_3_out_agg3(x1, x2, x3)  =  app2_3_out_agg1(x1)


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

APP2_3_IN_AGG3(._22(X, Xs), Ys, ._22(X, Zs)) -> IF_APP2_3_IN_1_AGG5(X, Xs, Ys, Zs, app2_3_in_agg3(Xs, Ys, Zs))
APP2_3_IN_AGG3(._22(X, Xs), Ys, ._22(X, Zs)) -> APP2_3_IN_AGG3(Xs, Ys, Zs)

The TRS R consists of the following rules:

app2_3_in_agg3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app2_3_in_1_agg5(X, Xs, Ys, Zs, app2_3_in_agg3(Xs, Ys, Zs))
app2_3_in_agg3([]_0, Ys, Ys) -> app2_3_out_agg3([]_0, Ys, Ys)
if_app2_3_in_1_agg5(X, Xs, Ys, Zs, app2_3_out_agg3(Xs, Ys, Zs)) -> app2_3_out_agg3(._22(X, Xs), Ys, ._22(X, Zs))

The argument filtering Pi contains the following mapping:
app2_3_in_agg3(x1, x2, x3)  =  app2_3_in_agg2(x2, x3)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_app2_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_app2_3_in_1_agg2(x1, x5)
app2_3_out_agg3(x1, x2, x3)  =  app2_3_out_agg1(x1)
APP2_3_IN_AGG3(x1, x2, x3)  =  APP2_3_IN_AGG2(x2, x3)
IF_APP2_3_IN_1_AGG5(x1, x2, x3, x4, x5)  =  IF_APP2_3_IN_1_AGG2(x1, x5)

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

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

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

APP2_3_IN_AGG3(._22(X, Xs), Ys, ._22(X, Zs)) -> IF_APP2_3_IN_1_AGG5(X, Xs, Ys, Zs, app2_3_in_agg3(Xs, Ys, Zs))
APP2_3_IN_AGG3(._22(X, Xs), Ys, ._22(X, Zs)) -> APP2_3_IN_AGG3(Xs, Ys, Zs)

The TRS R consists of the following rules:

app2_3_in_agg3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app2_3_in_1_agg5(X, Xs, Ys, Zs, app2_3_in_agg3(Xs, Ys, Zs))
app2_3_in_agg3([]_0, Ys, Ys) -> app2_3_out_agg3([]_0, Ys, Ys)
if_app2_3_in_1_agg5(X, Xs, Ys, Zs, app2_3_out_agg3(Xs, Ys, Zs)) -> app2_3_out_agg3(._22(X, Xs), Ys, ._22(X, Zs))

The argument filtering Pi contains the following mapping:
app2_3_in_agg3(x1, x2, x3)  =  app2_3_in_agg2(x2, x3)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_app2_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_app2_3_in_1_agg2(x1, x5)
app2_3_out_agg3(x1, x2, x3)  =  app2_3_out_agg1(x1)
APP2_3_IN_AGG3(x1, x2, x3)  =  APP2_3_IN_AGG2(x2, x3)
IF_APP2_3_IN_1_AGG5(x1, x2, x3, x4, x5)  =  IF_APP2_3_IN_1_AGG2(x1, x5)

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

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

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

APP2_3_IN_AGG3(._22(X, Xs), Ys, ._22(X, Zs)) -> APP2_3_IN_AGG3(Xs, Ys, Zs)

The TRS R consists of the following rules:

app2_3_in_agg3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app2_3_in_1_agg5(X, Xs, Ys, Zs, app2_3_in_agg3(Xs, Ys, Zs))
app2_3_in_agg3([]_0, Ys, Ys) -> app2_3_out_agg3([]_0, Ys, Ys)
if_app2_3_in_1_agg5(X, Xs, Ys, Zs, app2_3_out_agg3(Xs, Ys, Zs)) -> app2_3_out_agg3(._22(X, Xs), Ys, ._22(X, Zs))

The argument filtering Pi contains the following mapping:
app2_3_in_agg3(x1, x2, x3)  =  app2_3_in_agg2(x2, x3)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_app2_3_in_1_agg5(x1, x2, x3, x4, x5)  =  if_app2_3_in_1_agg2(x1, x5)
app2_3_out_agg3(x1, x2, x3)  =  app2_3_out_agg1(x1)
APP2_3_IN_AGG3(x1, x2, x3)  =  APP2_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
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
        ↳ PiTRS
          ↳ DependencyPairsProof
            ↳ PiDP
              ↳ DependencyGraphProof
                ↳ PiDP
                  ↳ UsableRulesProof
                    ↳ PiDP
                      ↳ PiDPToQDPProof

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

APP2_3_IN_AGG3(._22(X, Xs), Ys, ._22(X, Zs)) -> APP2_3_IN_AGG3(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._22(x1, x2)
APP2_3_IN_AGG3(x1, x2, x3)  =  APP2_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
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
        ↳ PiTRS
          ↳ DependencyPairsProof
            ↳ PiDP
              ↳ DependencyGraphProof
                ↳ PiDP
                  ↳ UsableRulesProof
                    ↳ PiDP
                      ↳ PiDPToQDPProof
                        ↳ QDP
                          ↳ QDPSizeChangeProof

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

APP2_3_IN_AGG2(Ys, ._22(X, Zs)) -> APP2_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 {APP2_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: