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



↳ PROLOG
  ↳ UnrequestedClauseRemoverProof

append3({}0, L, L).
append3(.2(H, L1), L2, .2(H, L3)) :- append3(L1, L2, L3).
append13({}0, L, L).
append13(.2(H, L1), L2, .2(H, L3)) :- append13(L1, L2, L3).


The clauses

append13({}0, L, L).
append13(.2(H, L1), L2, .2(H, L3)) :- append13(L1, L2, L3).

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

Deleting these clauses results in the following prolog program:

append3({}0, L, L).
append3(.2(H, L1), L2, .2(H, L3)) :- append3(L1, L2, L3).



↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
      ↳ PrologToPiTRSProof

append3({}0, L, L).
append3(.2(H, L1), L2, .2(H, L3)) :- append3(L1, L2, L3).


With regard to the inferred argument filtering the predicates were used in the following modes:
append3: (f,f,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


append_3_in_aaa3([]_0, L, L) -> append_3_out_aaa3([]_0, L, L)
append_3_in_aaa3(._22(H, L1), L2, ._22(H, L3)) -> if_append_3_in_1_aaa5(H, L1, L2, L3, append_3_in_aaa3(L1, L2, L3))
if_append_3_in_1_aaa5(H, L1, L2, L3, append_3_out_aaa3(L1, L2, L3)) -> append_3_out_aaa3(._22(H, L1), L2, ._22(H, L3))

The argument filtering Pi contains the following mapping:
append_3_in_aaa3(x1, x2, x3)  =  append_3_in_aaa
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
append_3_out_aaa3(x1, x2, x3)  =  append_3_out_aaa1(x1)
if_append_3_in_1_aaa5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_aaa1(x5)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG



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

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

append_3_in_aaa3([]_0, L, L) -> append_3_out_aaa3([]_0, L, L)
append_3_in_aaa3(._22(H, L1), L2, ._22(H, L3)) -> if_append_3_in_1_aaa5(H, L1, L2, L3, append_3_in_aaa3(L1, L2, L3))
if_append_3_in_1_aaa5(H, L1, L2, L3, append_3_out_aaa3(L1, L2, L3)) -> append_3_out_aaa3(._22(H, L1), L2, ._22(H, L3))

The argument filtering Pi contains the following mapping:
append_3_in_aaa3(x1, x2, x3)  =  append_3_in_aaa
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
append_3_out_aaa3(x1, x2, x3)  =  append_3_out_aaa1(x1)
if_append_3_in_1_aaa5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_aaa1(x5)


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

APPEND_3_IN_AAA3(._22(H, L1), L2, ._22(H, L3)) -> IF_APPEND_3_IN_1_AAA5(H, L1, L2, L3, append_3_in_aaa3(L1, L2, L3))
APPEND_3_IN_AAA3(._22(H, L1), L2, ._22(H, L3)) -> APPEND_3_IN_AAA3(L1, L2, L3)

The TRS R consists of the following rules:

append_3_in_aaa3([]_0, L, L) -> append_3_out_aaa3([]_0, L, L)
append_3_in_aaa3(._22(H, L1), L2, ._22(H, L3)) -> if_append_3_in_1_aaa5(H, L1, L2, L3, append_3_in_aaa3(L1, L2, L3))
if_append_3_in_1_aaa5(H, L1, L2, L3, append_3_out_aaa3(L1, L2, L3)) -> append_3_out_aaa3(._22(H, L1), L2, ._22(H, L3))

The argument filtering Pi contains the following mapping:
append_3_in_aaa3(x1, x2, x3)  =  append_3_in_aaa
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
append_3_out_aaa3(x1, x2, x3)  =  append_3_out_aaa1(x1)
if_append_3_in_1_aaa5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_aaa1(x5)
IF_APPEND_3_IN_1_AAA5(x1, x2, x3, x4, x5)  =  IF_APPEND_3_IN_1_AAA1(x5)
APPEND_3_IN_AAA3(x1, x2, x3)  =  APPEND_3_IN_AAA

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

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

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

APPEND_3_IN_AAA3(._22(H, L1), L2, ._22(H, L3)) -> IF_APPEND_3_IN_1_AAA5(H, L1, L2, L3, append_3_in_aaa3(L1, L2, L3))
APPEND_3_IN_AAA3(._22(H, L1), L2, ._22(H, L3)) -> APPEND_3_IN_AAA3(L1, L2, L3)

The TRS R consists of the following rules:

append_3_in_aaa3([]_0, L, L) -> append_3_out_aaa3([]_0, L, L)
append_3_in_aaa3(._22(H, L1), L2, ._22(H, L3)) -> if_append_3_in_1_aaa5(H, L1, L2, L3, append_3_in_aaa3(L1, L2, L3))
if_append_3_in_1_aaa5(H, L1, L2, L3, append_3_out_aaa3(L1, L2, L3)) -> append_3_out_aaa3(._22(H, L1), L2, ._22(H, L3))

The argument filtering Pi contains the following mapping:
append_3_in_aaa3(x1, x2, x3)  =  append_3_in_aaa
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
append_3_out_aaa3(x1, x2, x3)  =  append_3_out_aaa1(x1)
if_append_3_in_1_aaa5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_aaa1(x5)
IF_APPEND_3_IN_1_AAA5(x1, x2, x3, x4, x5)  =  IF_APPEND_3_IN_1_AAA1(x5)
APPEND_3_IN_AAA3(x1, x2, x3)  =  APPEND_3_IN_AAA

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
      ↳ PrologToPiTRSProof

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

APPEND_3_IN_AAA3(._22(H, L1), L2, ._22(H, L3)) -> APPEND_3_IN_AAA3(L1, L2, L3)

The TRS R consists of the following rules:

append_3_in_aaa3([]_0, L, L) -> append_3_out_aaa3([]_0, L, L)
append_3_in_aaa3(._22(H, L1), L2, ._22(H, L3)) -> if_append_3_in_1_aaa5(H, L1, L2, L3, append_3_in_aaa3(L1, L2, L3))
if_append_3_in_1_aaa5(H, L1, L2, L3, append_3_out_aaa3(L1, L2, L3)) -> append_3_out_aaa3(._22(H, L1), L2, ._22(H, L3))

The argument filtering Pi contains the following mapping:
append_3_in_aaa3(x1, x2, x3)  =  append_3_in_aaa
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
append_3_out_aaa3(x1, x2, x3)  =  append_3_out_aaa1(x1)
if_append_3_in_1_aaa5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_aaa1(x5)
APPEND_3_IN_AAA3(x1, x2, x3)  =  APPEND_3_IN_AAA

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
      ↳ PrologToPiTRSProof

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

APPEND_3_IN_AAA3(._22(H, L1), L2, ._22(H, L3)) -> APPEND_3_IN_AAA3(L1, L2, L3)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._21(x2)
APPEND_3_IN_AAA3(x1, x2, x3)  =  APPEND_3_IN_AAA

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
      ↳ PrologToPiTRSProof

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

APPEND_3_IN_AAA -> APPEND_3_IN_AAA

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {APPEND_3_IN_AAA}.
With regard to the inferred argument filtering the predicates were used in the following modes:
append3: (f,f,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

append_3_in_aaa3([]_0, L, L) -> append_3_out_aaa3([]_0, L, L)
append_3_in_aaa3(._22(H, L1), L2, ._22(H, L3)) -> if_append_3_in_1_aaa5(H, L1, L2, L3, append_3_in_aaa3(L1, L2, L3))
if_append_3_in_1_aaa5(H, L1, L2, L3, append_3_out_aaa3(L1, L2, L3)) -> append_3_out_aaa3(._22(H, L1), L2, ._22(H, L3))

The argument filtering Pi contains the following mapping:
append_3_in_aaa3(x1, x2, x3)  =  append_3_in_aaa
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
append_3_out_aaa3(x1, x2, x3)  =  append_3_out_aaa1(x1)
if_append_3_in_1_aaa5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_aaa1(x5)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG



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

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

append_3_in_aaa3([]_0, L, L) -> append_3_out_aaa3([]_0, L, L)
append_3_in_aaa3(._22(H, L1), L2, ._22(H, L3)) -> if_append_3_in_1_aaa5(H, L1, L2, L3, append_3_in_aaa3(L1, L2, L3))
if_append_3_in_1_aaa5(H, L1, L2, L3, append_3_out_aaa3(L1, L2, L3)) -> append_3_out_aaa3(._22(H, L1), L2, ._22(H, L3))

The argument filtering Pi contains the following mapping:
append_3_in_aaa3(x1, x2, x3)  =  append_3_in_aaa
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
append_3_out_aaa3(x1, x2, x3)  =  append_3_out_aaa1(x1)
if_append_3_in_1_aaa5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_aaa1(x5)


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

APPEND_3_IN_AAA3(._22(H, L1), L2, ._22(H, L3)) -> IF_APPEND_3_IN_1_AAA5(H, L1, L2, L3, append_3_in_aaa3(L1, L2, L3))
APPEND_3_IN_AAA3(._22(H, L1), L2, ._22(H, L3)) -> APPEND_3_IN_AAA3(L1, L2, L3)

The TRS R consists of the following rules:

append_3_in_aaa3([]_0, L, L) -> append_3_out_aaa3([]_0, L, L)
append_3_in_aaa3(._22(H, L1), L2, ._22(H, L3)) -> if_append_3_in_1_aaa5(H, L1, L2, L3, append_3_in_aaa3(L1, L2, L3))
if_append_3_in_1_aaa5(H, L1, L2, L3, append_3_out_aaa3(L1, L2, L3)) -> append_3_out_aaa3(._22(H, L1), L2, ._22(H, L3))

The argument filtering Pi contains the following mapping:
append_3_in_aaa3(x1, x2, x3)  =  append_3_in_aaa
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
append_3_out_aaa3(x1, x2, x3)  =  append_3_out_aaa1(x1)
if_append_3_in_1_aaa5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_aaa1(x5)
IF_APPEND_3_IN_1_AAA5(x1, x2, x3, x4, x5)  =  IF_APPEND_3_IN_1_AAA1(x5)
APPEND_3_IN_AAA3(x1, x2, x3)  =  APPEND_3_IN_AAA

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

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

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

APPEND_3_IN_AAA3(._22(H, L1), L2, ._22(H, L3)) -> IF_APPEND_3_IN_1_AAA5(H, L1, L2, L3, append_3_in_aaa3(L1, L2, L3))
APPEND_3_IN_AAA3(._22(H, L1), L2, ._22(H, L3)) -> APPEND_3_IN_AAA3(L1, L2, L3)

The TRS R consists of the following rules:

append_3_in_aaa3([]_0, L, L) -> append_3_out_aaa3([]_0, L, L)
append_3_in_aaa3(._22(H, L1), L2, ._22(H, L3)) -> if_append_3_in_1_aaa5(H, L1, L2, L3, append_3_in_aaa3(L1, L2, L3))
if_append_3_in_1_aaa5(H, L1, L2, L3, append_3_out_aaa3(L1, L2, L3)) -> append_3_out_aaa3(._22(H, L1), L2, ._22(H, L3))

The argument filtering Pi contains the following mapping:
append_3_in_aaa3(x1, x2, x3)  =  append_3_in_aaa
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
append_3_out_aaa3(x1, x2, x3)  =  append_3_out_aaa1(x1)
if_append_3_in_1_aaa5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_aaa1(x5)
IF_APPEND_3_IN_1_AAA5(x1, x2, x3, x4, x5)  =  IF_APPEND_3_IN_1_AAA1(x5)
APPEND_3_IN_AAA3(x1, x2, x3)  =  APPEND_3_IN_AAA

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

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

APPEND_3_IN_AAA3(._22(H, L1), L2, ._22(H, L3)) -> APPEND_3_IN_AAA3(L1, L2, L3)

The TRS R consists of the following rules:

append_3_in_aaa3([]_0, L, L) -> append_3_out_aaa3([]_0, L, L)
append_3_in_aaa3(._22(H, L1), L2, ._22(H, L3)) -> if_append_3_in_1_aaa5(H, L1, L2, L3, append_3_in_aaa3(L1, L2, L3))
if_append_3_in_1_aaa5(H, L1, L2, L3, append_3_out_aaa3(L1, L2, L3)) -> append_3_out_aaa3(._22(H, L1), L2, ._22(H, L3))

The argument filtering Pi contains the following mapping:
append_3_in_aaa3(x1, x2, x3)  =  append_3_in_aaa
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
append_3_out_aaa3(x1, x2, x3)  =  append_3_out_aaa1(x1)
if_append_3_in_1_aaa5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_aaa1(x5)
APPEND_3_IN_AAA3(x1, x2, x3)  =  APPEND_3_IN_AAA

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

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

APPEND_3_IN_AAA3(._22(H, L1), L2, ._22(H, L3)) -> APPEND_3_IN_AAA3(L1, L2, L3)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._21(x2)
APPEND_3_IN_AAA3(x1, x2, x3)  =  APPEND_3_IN_AAA

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

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

APPEND_3_IN_AAA -> APPEND_3_IN_AAA

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {APPEND_3_IN_AAA}.