Left Termination of the query pattern append3(f,f,f,b) w.r.t. the given Prolog program could not be shown:



↳ PROLOG
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

append3({}0, L, L).
append3(.2(H, L1), L2, .2(H, L3)) :- append3(L1, L2, L3).
append34(A, B, C, D) :- append3(A, B, E), append3(E, C, D).


With regard to the inferred argument filtering the predicates were used in the following modes:
append34: (f,f,f,b)
append3: (f,f,f) (f,f,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


append3_4_in_aaag4(A, B, C, D) -> if_append3_4_in_1_aaag5(A, B, C, D, append_3_in_aaa3(A, B, E))
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))
if_append3_4_in_1_aaag5(A, B, C, D, append_3_out_aaa3(A, B, E)) -> if_append3_4_in_2_aaag6(A, B, C, D, E, append_3_in_aag3(E, C, D))
append_3_in_aag3([]_0, L, L) -> append_3_out_aag3([]_0, L, L)
append_3_in_aag3(._22(H, L1), L2, ._22(H, L3)) -> if_append_3_in_1_aag5(H, L1, L2, L3, append_3_in_aag3(L1, L2, L3))
if_append_3_in_1_aag5(H, L1, L2, L3, append_3_out_aag3(L1, L2, L3)) -> append_3_out_aag3(._22(H, L1), L2, ._22(H, L3))
if_append3_4_in_2_aaag6(A, B, C, D, E, append_3_out_aag3(E, C, D)) -> append3_4_out_aaag4(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_4_in_aaag4(x1, x2, x3, x4)  =  append3_4_in_aaag1(x4)
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
if_append3_4_in_1_aaag5(x1, x2, x3, x4, x5)  =  if_append3_4_in_1_aaag2(x4, x5)
append_3_in_aaa3(x1, x2, x3)  =  append_3_in_aaa
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_append3_4_in_2_aaag6(x1, x2, x3, x4, x5, x6)  =  if_append3_4_in_2_aaag2(x1, x6)
append_3_in_aag3(x1, x2, x3)  =  append_3_in_aag1(x3)
append_3_out_aag3(x1, x2, x3)  =  append_3_out_aag2(x1, x2)
if_append_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_aag1(x5)
append3_4_out_aaag4(x1, x2, x3, x4)  =  append3_4_out_aaag2(x1, x3)

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:

append3_4_in_aaag4(A, B, C, D) -> if_append3_4_in_1_aaag5(A, B, C, D, append_3_in_aaa3(A, B, E))
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))
if_append3_4_in_1_aaag5(A, B, C, D, append_3_out_aaa3(A, B, E)) -> if_append3_4_in_2_aaag6(A, B, C, D, E, append_3_in_aag3(E, C, D))
append_3_in_aag3([]_0, L, L) -> append_3_out_aag3([]_0, L, L)
append_3_in_aag3(._22(H, L1), L2, ._22(H, L3)) -> if_append_3_in_1_aag5(H, L1, L2, L3, append_3_in_aag3(L1, L2, L3))
if_append_3_in_1_aag5(H, L1, L2, L3, append_3_out_aag3(L1, L2, L3)) -> append_3_out_aag3(._22(H, L1), L2, ._22(H, L3))
if_append3_4_in_2_aaag6(A, B, C, D, E, append_3_out_aag3(E, C, D)) -> append3_4_out_aaag4(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_4_in_aaag4(x1, x2, x3, x4)  =  append3_4_in_aaag1(x4)
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
if_append3_4_in_1_aaag5(x1, x2, x3, x4, x5)  =  if_append3_4_in_1_aaag2(x4, x5)
append_3_in_aaa3(x1, x2, x3)  =  append_3_in_aaa
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_append3_4_in_2_aaag6(x1, x2, x3, x4, x5, x6)  =  if_append3_4_in_2_aaag2(x1, x6)
append_3_in_aag3(x1, x2, x3)  =  append_3_in_aag1(x3)
append_3_out_aag3(x1, x2, x3)  =  append_3_out_aag2(x1, x2)
if_append_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_aag1(x5)
append3_4_out_aaag4(x1, x2, x3, x4)  =  append3_4_out_aaag2(x1, x3)


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

APPEND3_4_IN_AAAG4(A, B, C, D) -> IF_APPEND3_4_IN_1_AAAG5(A, B, C, D, append_3_in_aaa3(A, B, E))
APPEND3_4_IN_AAAG4(A, B, C, D) -> APPEND_3_IN_AAA3(A, B, E)
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)
IF_APPEND3_4_IN_1_AAAG5(A, B, C, D, append_3_out_aaa3(A, B, E)) -> IF_APPEND3_4_IN_2_AAAG6(A, B, C, D, E, append_3_in_aag3(E, C, D))
IF_APPEND3_4_IN_1_AAAG5(A, B, C, D, append_3_out_aaa3(A, B, E)) -> APPEND_3_IN_AAG3(E, C, D)
APPEND_3_IN_AAG3(._22(H, L1), L2, ._22(H, L3)) -> IF_APPEND_3_IN_1_AAG5(H, L1, L2, L3, append_3_in_aag3(L1, L2, L3))
APPEND_3_IN_AAG3(._22(H, L1), L2, ._22(H, L3)) -> APPEND_3_IN_AAG3(L1, L2, L3)

The TRS R consists of the following rules:

append3_4_in_aaag4(A, B, C, D) -> if_append3_4_in_1_aaag5(A, B, C, D, append_3_in_aaa3(A, B, E))
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))
if_append3_4_in_1_aaag5(A, B, C, D, append_3_out_aaa3(A, B, E)) -> if_append3_4_in_2_aaag6(A, B, C, D, E, append_3_in_aag3(E, C, D))
append_3_in_aag3([]_0, L, L) -> append_3_out_aag3([]_0, L, L)
append_3_in_aag3(._22(H, L1), L2, ._22(H, L3)) -> if_append_3_in_1_aag5(H, L1, L2, L3, append_3_in_aag3(L1, L2, L3))
if_append_3_in_1_aag5(H, L1, L2, L3, append_3_out_aag3(L1, L2, L3)) -> append_3_out_aag3(._22(H, L1), L2, ._22(H, L3))
if_append3_4_in_2_aaag6(A, B, C, D, E, append_3_out_aag3(E, C, D)) -> append3_4_out_aaag4(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_4_in_aaag4(x1, x2, x3, x4)  =  append3_4_in_aaag1(x4)
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
if_append3_4_in_1_aaag5(x1, x2, x3, x4, x5)  =  if_append3_4_in_1_aaag2(x4, x5)
append_3_in_aaa3(x1, x2, x3)  =  append_3_in_aaa
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_append3_4_in_2_aaag6(x1, x2, x3, x4, x5, x6)  =  if_append3_4_in_2_aaag2(x1, x6)
append_3_in_aag3(x1, x2, x3)  =  append_3_in_aag1(x3)
append_3_out_aag3(x1, x2, x3)  =  append_3_out_aag2(x1, x2)
if_append_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_aag1(x5)
append3_4_out_aaag4(x1, x2, x3, x4)  =  append3_4_out_aaag2(x1, x3)
IF_APPEND_3_IN_1_AAA5(x1, x2, x3, x4, x5)  =  IF_APPEND_3_IN_1_AAA1(x5)
IF_APPEND3_4_IN_2_AAAG6(x1, x2, x3, x4, x5, x6)  =  IF_APPEND3_4_IN_2_AAAG2(x1, x6)
IF_APPEND_3_IN_1_AAG5(x1, x2, x3, x4, x5)  =  IF_APPEND_3_IN_1_AAG1(x5)
APPEND_3_IN_AAG3(x1, x2, x3)  =  APPEND_3_IN_AAG1(x3)
APPEND3_4_IN_AAAG4(x1, x2, x3, x4)  =  APPEND3_4_IN_AAAG1(x4)
APPEND_3_IN_AAA3(x1, x2, x3)  =  APPEND_3_IN_AAA
IF_APPEND3_4_IN_1_AAAG5(x1, x2, x3, x4, x5)  =  IF_APPEND3_4_IN_1_AAAG2(x4, x5)

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:

APPEND3_4_IN_AAAG4(A, B, C, D) -> IF_APPEND3_4_IN_1_AAAG5(A, B, C, D, append_3_in_aaa3(A, B, E))
APPEND3_4_IN_AAAG4(A, B, C, D) -> APPEND_3_IN_AAA3(A, B, E)
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)
IF_APPEND3_4_IN_1_AAAG5(A, B, C, D, append_3_out_aaa3(A, B, E)) -> IF_APPEND3_4_IN_2_AAAG6(A, B, C, D, E, append_3_in_aag3(E, C, D))
IF_APPEND3_4_IN_1_AAAG5(A, B, C, D, append_3_out_aaa3(A, B, E)) -> APPEND_3_IN_AAG3(E, C, D)
APPEND_3_IN_AAG3(._22(H, L1), L2, ._22(H, L3)) -> IF_APPEND_3_IN_1_AAG5(H, L1, L2, L3, append_3_in_aag3(L1, L2, L3))
APPEND_3_IN_AAG3(._22(H, L1), L2, ._22(H, L3)) -> APPEND_3_IN_AAG3(L1, L2, L3)

The TRS R consists of the following rules:

append3_4_in_aaag4(A, B, C, D) -> if_append3_4_in_1_aaag5(A, B, C, D, append_3_in_aaa3(A, B, E))
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))
if_append3_4_in_1_aaag5(A, B, C, D, append_3_out_aaa3(A, B, E)) -> if_append3_4_in_2_aaag6(A, B, C, D, E, append_3_in_aag3(E, C, D))
append_3_in_aag3([]_0, L, L) -> append_3_out_aag3([]_0, L, L)
append_3_in_aag3(._22(H, L1), L2, ._22(H, L3)) -> if_append_3_in_1_aag5(H, L1, L2, L3, append_3_in_aag3(L1, L2, L3))
if_append_3_in_1_aag5(H, L1, L2, L3, append_3_out_aag3(L1, L2, L3)) -> append_3_out_aag3(._22(H, L1), L2, ._22(H, L3))
if_append3_4_in_2_aaag6(A, B, C, D, E, append_3_out_aag3(E, C, D)) -> append3_4_out_aaag4(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_4_in_aaag4(x1, x2, x3, x4)  =  append3_4_in_aaag1(x4)
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
if_append3_4_in_1_aaag5(x1, x2, x3, x4, x5)  =  if_append3_4_in_1_aaag2(x4, x5)
append_3_in_aaa3(x1, x2, x3)  =  append_3_in_aaa
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_append3_4_in_2_aaag6(x1, x2, x3, x4, x5, x6)  =  if_append3_4_in_2_aaag2(x1, x6)
append_3_in_aag3(x1, x2, x3)  =  append_3_in_aag1(x3)
append_3_out_aag3(x1, x2, x3)  =  append_3_out_aag2(x1, x2)
if_append_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_aag1(x5)
append3_4_out_aaag4(x1, x2, x3, x4)  =  append3_4_out_aaag2(x1, x3)
IF_APPEND_3_IN_1_AAA5(x1, x2, x3, x4, x5)  =  IF_APPEND_3_IN_1_AAA1(x5)
IF_APPEND3_4_IN_2_AAAG6(x1, x2, x3, x4, x5, x6)  =  IF_APPEND3_4_IN_2_AAAG2(x1, x6)
IF_APPEND_3_IN_1_AAG5(x1, x2, x3, x4, x5)  =  IF_APPEND_3_IN_1_AAG1(x5)
APPEND_3_IN_AAG3(x1, x2, x3)  =  APPEND_3_IN_AAG1(x3)
APPEND3_4_IN_AAAG4(x1, x2, x3, x4)  =  APPEND3_4_IN_AAAG1(x4)
APPEND_3_IN_AAA3(x1, x2, x3)  =  APPEND_3_IN_AAA
IF_APPEND3_4_IN_1_AAAG5(x1, x2, x3, x4, x5)  =  IF_APPEND3_4_IN_1_AAAG2(x4, x5)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 2 SCCs with 6 less nodes.

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

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

APPEND_3_IN_AAG3(._22(H, L1), L2, ._22(H, L3)) -> APPEND_3_IN_AAG3(L1, L2, L3)

The TRS R consists of the following rules:

append3_4_in_aaag4(A, B, C, D) -> if_append3_4_in_1_aaag5(A, B, C, D, append_3_in_aaa3(A, B, E))
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))
if_append3_4_in_1_aaag5(A, B, C, D, append_3_out_aaa3(A, B, E)) -> if_append3_4_in_2_aaag6(A, B, C, D, E, append_3_in_aag3(E, C, D))
append_3_in_aag3([]_0, L, L) -> append_3_out_aag3([]_0, L, L)
append_3_in_aag3(._22(H, L1), L2, ._22(H, L3)) -> if_append_3_in_1_aag5(H, L1, L2, L3, append_3_in_aag3(L1, L2, L3))
if_append_3_in_1_aag5(H, L1, L2, L3, append_3_out_aag3(L1, L2, L3)) -> append_3_out_aag3(._22(H, L1), L2, ._22(H, L3))
if_append3_4_in_2_aaag6(A, B, C, D, E, append_3_out_aag3(E, C, D)) -> append3_4_out_aaag4(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_4_in_aaag4(x1, x2, x3, x4)  =  append3_4_in_aaag1(x4)
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
if_append3_4_in_1_aaag5(x1, x2, x3, x4, x5)  =  if_append3_4_in_1_aaag2(x4, x5)
append_3_in_aaa3(x1, x2, x3)  =  append_3_in_aaa
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_append3_4_in_2_aaag6(x1, x2, x3, x4, x5, x6)  =  if_append3_4_in_2_aaag2(x1, x6)
append_3_in_aag3(x1, x2, x3)  =  append_3_in_aag1(x3)
append_3_out_aag3(x1, x2, x3)  =  append_3_out_aag2(x1, x2)
if_append_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_aag1(x5)
append3_4_out_aaag4(x1, x2, x3, x4)  =  append3_4_out_aaag2(x1, x3)
APPEND_3_IN_AAG3(x1, x2, x3)  =  APPEND_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
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

APPEND_3_IN_AAG3(._22(H, L1), L2, ._22(H, L3)) -> APPEND_3_IN_AAG3(L1, L2, L3)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._21(x2)
APPEND_3_IN_AAG3(x1, x2, x3)  =  APPEND_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
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

APPEND_3_IN_AAG1(._21(L3)) -> APPEND_3_IN_AAG1(L3)

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_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: 



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

append3_4_in_aaag4(A, B, C, D) -> if_append3_4_in_1_aaag5(A, B, C, D, append_3_in_aaa3(A, B, E))
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))
if_append3_4_in_1_aaag5(A, B, C, D, append_3_out_aaa3(A, B, E)) -> if_append3_4_in_2_aaag6(A, B, C, D, E, append_3_in_aag3(E, C, D))
append_3_in_aag3([]_0, L, L) -> append_3_out_aag3([]_0, L, L)
append_3_in_aag3(._22(H, L1), L2, ._22(H, L3)) -> if_append_3_in_1_aag5(H, L1, L2, L3, append_3_in_aag3(L1, L2, L3))
if_append_3_in_1_aag5(H, L1, L2, L3, append_3_out_aag3(L1, L2, L3)) -> append_3_out_aag3(._22(H, L1), L2, ._22(H, L3))
if_append3_4_in_2_aaag6(A, B, C, D, E, append_3_out_aag3(E, C, D)) -> append3_4_out_aaag4(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_4_in_aaag4(x1, x2, x3, x4)  =  append3_4_in_aaag1(x4)
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
if_append3_4_in_1_aaag5(x1, x2, x3, x4, x5)  =  if_append3_4_in_1_aaag2(x4, x5)
append_3_in_aaa3(x1, x2, x3)  =  append_3_in_aaa
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_append3_4_in_2_aaag6(x1, x2, x3, x4, x5, x6)  =  if_append3_4_in_2_aaag2(x1, x6)
append_3_in_aag3(x1, x2, x3)  =  append_3_in_aag1(x3)
append_3_out_aag3(x1, x2, x3)  =  append_3_out_aag2(x1, x2)
if_append_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_aag1(x5)
append3_4_out_aaag4(x1, x2, x3, x4)  =  append3_4_out_aaag2(x1, x3)
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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ 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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ 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:
append34: (f,f,f,b)
append3: (f,f,f) (f,f,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

append3_4_in_aaag4(A, B, C, D) -> if_append3_4_in_1_aaag5(A, B, C, D, append_3_in_aaa3(A, B, E))
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))
if_append3_4_in_1_aaag5(A, B, C, D, append_3_out_aaa3(A, B, E)) -> if_append3_4_in_2_aaag6(A, B, C, D, E, append_3_in_aag3(E, C, D))
append_3_in_aag3([]_0, L, L) -> append_3_out_aag3([]_0, L, L)
append_3_in_aag3(._22(H, L1), L2, ._22(H, L3)) -> if_append_3_in_1_aag5(H, L1, L2, L3, append_3_in_aag3(L1, L2, L3))
if_append_3_in_1_aag5(H, L1, L2, L3, append_3_out_aag3(L1, L2, L3)) -> append_3_out_aag3(._22(H, L1), L2, ._22(H, L3))
if_append3_4_in_2_aaag6(A, B, C, D, E, append_3_out_aag3(E, C, D)) -> append3_4_out_aaag4(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_4_in_aaag4(x1, x2, x3, x4)  =  append3_4_in_aaag1(x4)
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
if_append3_4_in_1_aaag5(x1, x2, x3, x4, x5)  =  if_append3_4_in_1_aaag2(x4, x5)
append_3_in_aaa3(x1, x2, x3)  =  append_3_in_aaa
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_append3_4_in_2_aaag6(x1, x2, x3, x4, x5, x6)  =  if_append3_4_in_2_aaag3(x1, x4, x6)
append_3_in_aag3(x1, x2, x3)  =  append_3_in_aag1(x3)
append_3_out_aag3(x1, x2, x3)  =  append_3_out_aag3(x1, x2, x3)
if_append_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_aag2(x4, x5)
append3_4_out_aaag4(x1, x2, x3, x4)  =  append3_4_out_aaag3(x1, x3, x4)

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:

append3_4_in_aaag4(A, B, C, D) -> if_append3_4_in_1_aaag5(A, B, C, D, append_3_in_aaa3(A, B, E))
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))
if_append3_4_in_1_aaag5(A, B, C, D, append_3_out_aaa3(A, B, E)) -> if_append3_4_in_2_aaag6(A, B, C, D, E, append_3_in_aag3(E, C, D))
append_3_in_aag3([]_0, L, L) -> append_3_out_aag3([]_0, L, L)
append_3_in_aag3(._22(H, L1), L2, ._22(H, L3)) -> if_append_3_in_1_aag5(H, L1, L2, L3, append_3_in_aag3(L1, L2, L3))
if_append_3_in_1_aag5(H, L1, L2, L3, append_3_out_aag3(L1, L2, L3)) -> append_3_out_aag3(._22(H, L1), L2, ._22(H, L3))
if_append3_4_in_2_aaag6(A, B, C, D, E, append_3_out_aag3(E, C, D)) -> append3_4_out_aaag4(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_4_in_aaag4(x1, x2, x3, x4)  =  append3_4_in_aaag1(x4)
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
if_append3_4_in_1_aaag5(x1, x2, x3, x4, x5)  =  if_append3_4_in_1_aaag2(x4, x5)
append_3_in_aaa3(x1, x2, x3)  =  append_3_in_aaa
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_append3_4_in_2_aaag6(x1, x2, x3, x4, x5, x6)  =  if_append3_4_in_2_aaag3(x1, x4, x6)
append_3_in_aag3(x1, x2, x3)  =  append_3_in_aag1(x3)
append_3_out_aag3(x1, x2, x3)  =  append_3_out_aag3(x1, x2, x3)
if_append_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_aag2(x4, x5)
append3_4_out_aaag4(x1, x2, x3, x4)  =  append3_4_out_aaag3(x1, x3, x4)


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

APPEND3_4_IN_AAAG4(A, B, C, D) -> IF_APPEND3_4_IN_1_AAAG5(A, B, C, D, append_3_in_aaa3(A, B, E))
APPEND3_4_IN_AAAG4(A, B, C, D) -> APPEND_3_IN_AAA3(A, B, E)
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)
IF_APPEND3_4_IN_1_AAAG5(A, B, C, D, append_3_out_aaa3(A, B, E)) -> IF_APPEND3_4_IN_2_AAAG6(A, B, C, D, E, append_3_in_aag3(E, C, D))
IF_APPEND3_4_IN_1_AAAG5(A, B, C, D, append_3_out_aaa3(A, B, E)) -> APPEND_3_IN_AAG3(E, C, D)
APPEND_3_IN_AAG3(._22(H, L1), L2, ._22(H, L3)) -> IF_APPEND_3_IN_1_AAG5(H, L1, L2, L3, append_3_in_aag3(L1, L2, L3))
APPEND_3_IN_AAG3(._22(H, L1), L2, ._22(H, L3)) -> APPEND_3_IN_AAG3(L1, L2, L3)

The TRS R consists of the following rules:

append3_4_in_aaag4(A, B, C, D) -> if_append3_4_in_1_aaag5(A, B, C, D, append_3_in_aaa3(A, B, E))
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))
if_append3_4_in_1_aaag5(A, B, C, D, append_3_out_aaa3(A, B, E)) -> if_append3_4_in_2_aaag6(A, B, C, D, E, append_3_in_aag3(E, C, D))
append_3_in_aag3([]_0, L, L) -> append_3_out_aag3([]_0, L, L)
append_3_in_aag3(._22(H, L1), L2, ._22(H, L3)) -> if_append_3_in_1_aag5(H, L1, L2, L3, append_3_in_aag3(L1, L2, L3))
if_append_3_in_1_aag5(H, L1, L2, L3, append_3_out_aag3(L1, L2, L3)) -> append_3_out_aag3(._22(H, L1), L2, ._22(H, L3))
if_append3_4_in_2_aaag6(A, B, C, D, E, append_3_out_aag3(E, C, D)) -> append3_4_out_aaag4(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_4_in_aaag4(x1, x2, x3, x4)  =  append3_4_in_aaag1(x4)
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
if_append3_4_in_1_aaag5(x1, x2, x3, x4, x5)  =  if_append3_4_in_1_aaag2(x4, x5)
append_3_in_aaa3(x1, x2, x3)  =  append_3_in_aaa
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_append3_4_in_2_aaag6(x1, x2, x3, x4, x5, x6)  =  if_append3_4_in_2_aaag3(x1, x4, x6)
append_3_in_aag3(x1, x2, x3)  =  append_3_in_aag1(x3)
append_3_out_aag3(x1, x2, x3)  =  append_3_out_aag3(x1, x2, x3)
if_append_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_aag2(x4, x5)
append3_4_out_aaag4(x1, x2, x3, x4)  =  append3_4_out_aaag3(x1, x3, x4)
IF_APPEND_3_IN_1_AAA5(x1, x2, x3, x4, x5)  =  IF_APPEND_3_IN_1_AAA1(x5)
IF_APPEND3_4_IN_2_AAAG6(x1, x2, x3, x4, x5, x6)  =  IF_APPEND3_4_IN_2_AAAG3(x1, x4, x6)
IF_APPEND_3_IN_1_AAG5(x1, x2, x3, x4, x5)  =  IF_APPEND_3_IN_1_AAG2(x4, x5)
APPEND_3_IN_AAG3(x1, x2, x3)  =  APPEND_3_IN_AAG1(x3)
APPEND3_4_IN_AAAG4(x1, x2, x3, x4)  =  APPEND3_4_IN_AAAG1(x4)
APPEND_3_IN_AAA3(x1, x2, x3)  =  APPEND_3_IN_AAA
IF_APPEND3_4_IN_1_AAAG5(x1, x2, x3, x4, x5)  =  IF_APPEND3_4_IN_1_AAAG2(x4, x5)

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:

APPEND3_4_IN_AAAG4(A, B, C, D) -> IF_APPEND3_4_IN_1_AAAG5(A, B, C, D, append_3_in_aaa3(A, B, E))
APPEND3_4_IN_AAAG4(A, B, C, D) -> APPEND_3_IN_AAA3(A, B, E)
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)
IF_APPEND3_4_IN_1_AAAG5(A, B, C, D, append_3_out_aaa3(A, B, E)) -> IF_APPEND3_4_IN_2_AAAG6(A, B, C, D, E, append_3_in_aag3(E, C, D))
IF_APPEND3_4_IN_1_AAAG5(A, B, C, D, append_3_out_aaa3(A, B, E)) -> APPEND_3_IN_AAG3(E, C, D)
APPEND_3_IN_AAG3(._22(H, L1), L2, ._22(H, L3)) -> IF_APPEND_3_IN_1_AAG5(H, L1, L2, L3, append_3_in_aag3(L1, L2, L3))
APPEND_3_IN_AAG3(._22(H, L1), L2, ._22(H, L3)) -> APPEND_3_IN_AAG3(L1, L2, L3)

The TRS R consists of the following rules:

append3_4_in_aaag4(A, B, C, D) -> if_append3_4_in_1_aaag5(A, B, C, D, append_3_in_aaa3(A, B, E))
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))
if_append3_4_in_1_aaag5(A, B, C, D, append_3_out_aaa3(A, B, E)) -> if_append3_4_in_2_aaag6(A, B, C, D, E, append_3_in_aag3(E, C, D))
append_3_in_aag3([]_0, L, L) -> append_3_out_aag3([]_0, L, L)
append_3_in_aag3(._22(H, L1), L2, ._22(H, L3)) -> if_append_3_in_1_aag5(H, L1, L2, L3, append_3_in_aag3(L1, L2, L3))
if_append_3_in_1_aag5(H, L1, L2, L3, append_3_out_aag3(L1, L2, L3)) -> append_3_out_aag3(._22(H, L1), L2, ._22(H, L3))
if_append3_4_in_2_aaag6(A, B, C, D, E, append_3_out_aag3(E, C, D)) -> append3_4_out_aaag4(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_4_in_aaag4(x1, x2, x3, x4)  =  append3_4_in_aaag1(x4)
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
if_append3_4_in_1_aaag5(x1, x2, x3, x4, x5)  =  if_append3_4_in_1_aaag2(x4, x5)
append_3_in_aaa3(x1, x2, x3)  =  append_3_in_aaa
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_append3_4_in_2_aaag6(x1, x2, x3, x4, x5, x6)  =  if_append3_4_in_2_aaag3(x1, x4, x6)
append_3_in_aag3(x1, x2, x3)  =  append_3_in_aag1(x3)
append_3_out_aag3(x1, x2, x3)  =  append_3_out_aag3(x1, x2, x3)
if_append_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_aag2(x4, x5)
append3_4_out_aaag4(x1, x2, x3, x4)  =  append3_4_out_aaag3(x1, x3, x4)
IF_APPEND_3_IN_1_AAA5(x1, x2, x3, x4, x5)  =  IF_APPEND_3_IN_1_AAA1(x5)
IF_APPEND3_4_IN_2_AAAG6(x1, x2, x3, x4, x5, x6)  =  IF_APPEND3_4_IN_2_AAAG3(x1, x4, x6)
IF_APPEND_3_IN_1_AAG5(x1, x2, x3, x4, x5)  =  IF_APPEND_3_IN_1_AAG2(x4, x5)
APPEND_3_IN_AAG3(x1, x2, x3)  =  APPEND_3_IN_AAG1(x3)
APPEND3_4_IN_AAAG4(x1, x2, x3, x4)  =  APPEND3_4_IN_AAAG1(x4)
APPEND_3_IN_AAA3(x1, x2, x3)  =  APPEND_3_IN_AAA
IF_APPEND3_4_IN_1_AAAG5(x1, x2, x3, x4, x5)  =  IF_APPEND3_4_IN_1_AAAG2(x4, x5)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 2 SCCs with 6 less nodes.

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

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

APPEND_3_IN_AAG3(._22(H, L1), L2, ._22(H, L3)) -> APPEND_3_IN_AAG3(L1, L2, L3)

The TRS R consists of the following rules:

append3_4_in_aaag4(A, B, C, D) -> if_append3_4_in_1_aaag5(A, B, C, D, append_3_in_aaa3(A, B, E))
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))
if_append3_4_in_1_aaag5(A, B, C, D, append_3_out_aaa3(A, B, E)) -> if_append3_4_in_2_aaag6(A, B, C, D, E, append_3_in_aag3(E, C, D))
append_3_in_aag3([]_0, L, L) -> append_3_out_aag3([]_0, L, L)
append_3_in_aag3(._22(H, L1), L2, ._22(H, L3)) -> if_append_3_in_1_aag5(H, L1, L2, L3, append_3_in_aag3(L1, L2, L3))
if_append_3_in_1_aag5(H, L1, L2, L3, append_3_out_aag3(L1, L2, L3)) -> append_3_out_aag3(._22(H, L1), L2, ._22(H, L3))
if_append3_4_in_2_aaag6(A, B, C, D, E, append_3_out_aag3(E, C, D)) -> append3_4_out_aaag4(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_4_in_aaag4(x1, x2, x3, x4)  =  append3_4_in_aaag1(x4)
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
if_append3_4_in_1_aaag5(x1, x2, x3, x4, x5)  =  if_append3_4_in_1_aaag2(x4, x5)
append_3_in_aaa3(x1, x2, x3)  =  append_3_in_aaa
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_append3_4_in_2_aaag6(x1, x2, x3, x4, x5, x6)  =  if_append3_4_in_2_aaag3(x1, x4, x6)
append_3_in_aag3(x1, x2, x3)  =  append_3_in_aag1(x3)
append_3_out_aag3(x1, x2, x3)  =  append_3_out_aag3(x1, x2, x3)
if_append_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_aag2(x4, x5)
append3_4_out_aaag4(x1, x2, x3, x4)  =  append3_4_out_aaag3(x1, x3, x4)
APPEND_3_IN_AAG3(x1, x2, x3)  =  APPEND_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
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
              ↳ PiDP

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

APPEND_3_IN_AAG3(._22(H, L1), L2, ._22(H, L3)) -> APPEND_3_IN_AAG3(L1, L2, L3)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._21(x2)
APPEND_3_IN_AAG3(x1, x2, x3)  =  APPEND_3_IN_AAG1(x3)

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

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

append3_4_in_aaag4(A, B, C, D) -> if_append3_4_in_1_aaag5(A, B, C, D, append_3_in_aaa3(A, B, E))
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))
if_append3_4_in_1_aaag5(A, B, C, D, append_3_out_aaa3(A, B, E)) -> if_append3_4_in_2_aaag6(A, B, C, D, E, append_3_in_aag3(E, C, D))
append_3_in_aag3([]_0, L, L) -> append_3_out_aag3([]_0, L, L)
append_3_in_aag3(._22(H, L1), L2, ._22(H, L3)) -> if_append_3_in_1_aag5(H, L1, L2, L3, append_3_in_aag3(L1, L2, L3))
if_append_3_in_1_aag5(H, L1, L2, L3, append_3_out_aag3(L1, L2, L3)) -> append_3_out_aag3(._22(H, L1), L2, ._22(H, L3))
if_append3_4_in_2_aaag6(A, B, C, D, E, append_3_out_aag3(E, C, D)) -> append3_4_out_aaag4(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_4_in_aaag4(x1, x2, x3, x4)  =  append3_4_in_aaag1(x4)
[]_0  =  []_0
._22(x1, x2)  =  ._21(x2)
if_append3_4_in_1_aaag5(x1, x2, x3, x4, x5)  =  if_append3_4_in_1_aaag2(x4, x5)
append_3_in_aaa3(x1, x2, x3)  =  append_3_in_aaa
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_append3_4_in_2_aaag6(x1, x2, x3, x4, x5, x6)  =  if_append3_4_in_2_aaag3(x1, x4, x6)
append_3_in_aag3(x1, x2, x3)  =  append_3_in_aag1(x3)
append_3_out_aag3(x1, x2, x3)  =  append_3_out_aag3(x1, x2, x3)
if_append_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_aag2(x4, x5)
append3_4_out_aaag4(x1, x2, x3, x4)  =  append3_4_out_aaag3(x1, x3, x4)
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
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP

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