Left Termination of the query pattern append3(f,f,b) w.r.t. the given Prolog program could successfully be proven:



↳ PROLOG
  ↳ UnrequestedClauseRemoverProof

append13({}0, X, X).
append13(.2(X, Y), U, .2(X, Z)) :- append13(Y, U, Z).
append23({}0, X, X).
append23(.2(X, Y), U, .2(X, Z)) :- append23(Y, U, Z).
append33({}0, X, X).
append33(.2(X, Y), U, .2(X, Z)) :- append33(Y, U, Z).


The clauses

append13({}0, X, X).
append13(.2(X, Y), U, .2(X, Z)) :- append13(Y, U, Z).
append23({}0, X, X).
append23(.2(X, Y), U, .2(X, Z)) :- append23(Y, U, Z).

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

Deleting these clauses results in the following prolog program:

append33({}0, X, X).
append33(.2(X, Y), U, .2(X, Z)) :- append33(Y, U, Z).



↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof

append33({}0, X, X).
append33(.2(X, Y), U, .2(X, Z)) :- append33(Y, U, Z).


With regard to the inferred argument filtering the predicates were used in the following modes:
append33: (f,f,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


append3_3_in_aag3([]_0, X, X) -> append3_3_out_aag3([]_0, X, X)
append3_3_in_aag3(._22(X, Y), U, ._22(X, Z)) -> if_append3_3_in_1_aag5(X, Y, U, Z, append3_3_in_aag3(Y, U, Z))
if_append3_3_in_1_aag5(X, Y, U, Z, append3_3_out_aag3(Y, U, Z)) -> append3_3_out_aag3(._22(X, Y), U, ._22(X, Z))

The argument filtering Pi contains the following mapping:
append3_3_in_aag3(x1, x2, x3)  =  append3_3_in_aag1(x3)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
append3_3_out_aag3(x1, x2, x3)  =  append3_3_out_aag2(x1, x2)
if_append3_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_append3_3_in_1_aag2(x1, x5)

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:

append3_3_in_aag3([]_0, X, X) -> append3_3_out_aag3([]_0, X, X)
append3_3_in_aag3(._22(X, Y), U, ._22(X, Z)) -> if_append3_3_in_1_aag5(X, Y, U, Z, append3_3_in_aag3(Y, U, Z))
if_append3_3_in_1_aag5(X, Y, U, Z, append3_3_out_aag3(Y, U, Z)) -> append3_3_out_aag3(._22(X, Y), U, ._22(X, Z))

The argument filtering Pi contains the following mapping:
append3_3_in_aag3(x1, x2, x3)  =  append3_3_in_aag1(x3)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
append3_3_out_aag3(x1, x2, x3)  =  append3_3_out_aag2(x1, x2)
if_append3_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_append3_3_in_1_aag2(x1, x5)


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

APPEND3_3_IN_AAG3(._22(X, Y), U, ._22(X, Z)) -> IF_APPEND3_3_IN_1_AAG5(X, Y, U, Z, append3_3_in_aag3(Y, U, Z))
APPEND3_3_IN_AAG3(._22(X, Y), U, ._22(X, Z)) -> APPEND3_3_IN_AAG3(Y, U, Z)

The TRS R consists of the following rules:

append3_3_in_aag3([]_0, X, X) -> append3_3_out_aag3([]_0, X, X)
append3_3_in_aag3(._22(X, Y), U, ._22(X, Z)) -> if_append3_3_in_1_aag5(X, Y, U, Z, append3_3_in_aag3(Y, U, Z))
if_append3_3_in_1_aag5(X, Y, U, Z, append3_3_out_aag3(Y, U, Z)) -> append3_3_out_aag3(._22(X, Y), U, ._22(X, Z))

The argument filtering Pi contains the following mapping:
append3_3_in_aag3(x1, x2, x3)  =  append3_3_in_aag1(x3)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
append3_3_out_aag3(x1, x2, x3)  =  append3_3_out_aag2(x1, x2)
if_append3_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_append3_3_in_1_aag2(x1, x5)
IF_APPEND3_3_IN_1_AAG5(x1, x2, x3, x4, x5)  =  IF_APPEND3_3_IN_1_AAG2(x1, x5)
APPEND3_3_IN_AAG3(x1, x2, x3)  =  APPEND3_3_IN_AAG1(x3)

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:

APPEND3_3_IN_AAG3(._22(X, Y), U, ._22(X, Z)) -> IF_APPEND3_3_IN_1_AAG5(X, Y, U, Z, append3_3_in_aag3(Y, U, Z))
APPEND3_3_IN_AAG3(._22(X, Y), U, ._22(X, Z)) -> APPEND3_3_IN_AAG3(Y, U, Z)

The TRS R consists of the following rules:

append3_3_in_aag3([]_0, X, X) -> append3_3_out_aag3([]_0, X, X)
append3_3_in_aag3(._22(X, Y), U, ._22(X, Z)) -> if_append3_3_in_1_aag5(X, Y, U, Z, append3_3_in_aag3(Y, U, Z))
if_append3_3_in_1_aag5(X, Y, U, Z, append3_3_out_aag3(Y, U, Z)) -> append3_3_out_aag3(._22(X, Y), U, ._22(X, Z))

The argument filtering Pi contains the following mapping:
append3_3_in_aag3(x1, x2, x3)  =  append3_3_in_aag1(x3)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
append3_3_out_aag3(x1, x2, x3)  =  append3_3_out_aag2(x1, x2)
if_append3_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_append3_3_in_1_aag2(x1, x5)
IF_APPEND3_3_IN_1_AAG5(x1, x2, x3, x4, x5)  =  IF_APPEND3_3_IN_1_AAG2(x1, x5)
APPEND3_3_IN_AAG3(x1, x2, x3)  =  APPEND3_3_IN_AAG1(x3)

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:

APPEND3_3_IN_AAG3(._22(X, Y), U, ._22(X, Z)) -> APPEND3_3_IN_AAG3(Y, U, Z)

The TRS R consists of the following rules:

append3_3_in_aag3([]_0, X, X) -> append3_3_out_aag3([]_0, X, X)
append3_3_in_aag3(._22(X, Y), U, ._22(X, Z)) -> if_append3_3_in_1_aag5(X, Y, U, Z, append3_3_in_aag3(Y, U, Z))
if_append3_3_in_1_aag5(X, Y, U, Z, append3_3_out_aag3(Y, U, Z)) -> append3_3_out_aag3(._22(X, Y), U, ._22(X, Z))

The argument filtering Pi contains the following mapping:
append3_3_in_aag3(x1, x2, x3)  =  append3_3_in_aag1(x3)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
append3_3_out_aag3(x1, x2, x3)  =  append3_3_out_aag2(x1, x2)
if_append3_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_append3_3_in_1_aag2(x1, x5)
APPEND3_3_IN_AAG3(x1, x2, x3)  =  APPEND3_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
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
        ↳ PiTRS
          ↳ DependencyPairsProof
            ↳ PiDP
              ↳ DependencyGraphProof
                ↳ PiDP
                  ↳ UsableRulesProof
                    ↳ PiDP
                      ↳ PiDPToQDPProof

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

APPEND3_3_IN_AAG3(._22(X, Y), U, ._22(X, Z)) -> APPEND3_3_IN_AAG3(Y, U, Z)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._22(x1, x2)
APPEND3_3_IN_AAG3(x1, x2, x3)  =  APPEND3_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
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
        ↳ PiTRS
          ↳ DependencyPairsProof
            ↳ PiDP
              ↳ DependencyGraphProof
                ↳ PiDP
                  ↳ UsableRulesProof
                    ↳ PiDP
                      ↳ PiDPToQDPProof
                        ↳ QDP
                          ↳ QDPSizeChangeProof

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

APPEND3_3_IN_AAG1(._22(X, Z)) -> APPEND3_3_IN_AAG1(Z)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {APPEND3_3_IN_AAG1}.
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: