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



↳ PROLOG
  ↳ PrologToPiTRSProof

append3({}0, Ys, Ys).
append3(.2(X, Xs), Ys, .2(X, Zs)) :- append3(Xs, Ys, Zs).
sublist2(X, Y) :- append3(P, underscore, Y), append3(underscore1, X, P).


With regard to the inferred argument filtering the predicates were used in the following modes:
sublist2: (f,b)
append3: (f,f,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


sublist_2_in_ag2(X, Y) -> if_sublist_2_in_1_ag3(X, Y, append_3_in_aag3(P, underscore, Y))
append_3_in_aag3([]_0, Ys, Ys) -> append_3_out_aag3([]_0, Ys, Ys)
append_3_in_aag3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_append_3_in_1_aag5(X, Xs, Ys, Zs, append_3_in_aag3(Xs, Ys, Zs))
if_append_3_in_1_aag5(X, Xs, Ys, Zs, append_3_out_aag3(Xs, Ys, Zs)) -> append_3_out_aag3(._22(X, Xs), Ys, ._22(X, Zs))
if_sublist_2_in_1_ag3(X, Y, append_3_out_aag3(P, underscore, Y)) -> if_sublist_2_in_2_ag4(X, Y, P, append_3_in_aag3(underscore1, X, P))
if_sublist_2_in_2_ag4(X, Y, P, append_3_out_aag3(underscore1, X, P)) -> sublist_2_out_ag2(X, Y)

The argument filtering Pi contains the following mapping:
sublist_2_in_ag2(x1, x2)  =  sublist_2_in_ag1(x2)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_sublist_2_in_1_ag3(x1, x2, x3)  =  if_sublist_2_in_1_ag1(x3)
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_aag2(x1, x5)
if_sublist_2_in_2_ag4(x1, x2, x3, x4)  =  if_sublist_2_in_2_ag1(x4)
sublist_2_out_ag2(x1, x2)  =  sublist_2_out_ag1(x1)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG



↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof

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

sublist_2_in_ag2(X, Y) -> if_sublist_2_in_1_ag3(X, Y, append_3_in_aag3(P, underscore, Y))
append_3_in_aag3([]_0, Ys, Ys) -> append_3_out_aag3([]_0, Ys, Ys)
append_3_in_aag3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_append_3_in_1_aag5(X, Xs, Ys, Zs, append_3_in_aag3(Xs, Ys, Zs))
if_append_3_in_1_aag5(X, Xs, Ys, Zs, append_3_out_aag3(Xs, Ys, Zs)) -> append_3_out_aag3(._22(X, Xs), Ys, ._22(X, Zs))
if_sublist_2_in_1_ag3(X, Y, append_3_out_aag3(P, underscore, Y)) -> if_sublist_2_in_2_ag4(X, Y, P, append_3_in_aag3(underscore1, X, P))
if_sublist_2_in_2_ag4(X, Y, P, append_3_out_aag3(underscore1, X, P)) -> sublist_2_out_ag2(X, Y)

The argument filtering Pi contains the following mapping:
sublist_2_in_ag2(x1, x2)  =  sublist_2_in_ag1(x2)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_sublist_2_in_1_ag3(x1, x2, x3)  =  if_sublist_2_in_1_ag1(x3)
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_aag2(x1, x5)
if_sublist_2_in_2_ag4(x1, x2, x3, x4)  =  if_sublist_2_in_2_ag1(x4)
sublist_2_out_ag2(x1, x2)  =  sublist_2_out_ag1(x1)


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

SUBLIST_2_IN_AG2(X, Y) -> IF_SUBLIST_2_IN_1_AG3(X, Y, append_3_in_aag3(P, underscore, Y))
SUBLIST_2_IN_AG2(X, Y) -> APPEND_3_IN_AAG3(P, underscore, Y)
APPEND_3_IN_AAG3(._22(X, Xs), Ys, ._22(X, Zs)) -> IF_APPEND_3_IN_1_AAG5(X, Xs, Ys, Zs, append_3_in_aag3(Xs, Ys, Zs))
APPEND_3_IN_AAG3(._22(X, Xs), Ys, ._22(X, Zs)) -> APPEND_3_IN_AAG3(Xs, Ys, Zs)
IF_SUBLIST_2_IN_1_AG3(X, Y, append_3_out_aag3(P, underscore, Y)) -> IF_SUBLIST_2_IN_2_AG4(X, Y, P, append_3_in_aag3(underscore1, X, P))
IF_SUBLIST_2_IN_1_AG3(X, Y, append_3_out_aag3(P, underscore, Y)) -> APPEND_3_IN_AAG3(underscore1, X, P)

The TRS R consists of the following rules:

sublist_2_in_ag2(X, Y) -> if_sublist_2_in_1_ag3(X, Y, append_3_in_aag3(P, underscore, Y))
append_3_in_aag3([]_0, Ys, Ys) -> append_3_out_aag3([]_0, Ys, Ys)
append_3_in_aag3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_append_3_in_1_aag5(X, Xs, Ys, Zs, append_3_in_aag3(Xs, Ys, Zs))
if_append_3_in_1_aag5(X, Xs, Ys, Zs, append_3_out_aag3(Xs, Ys, Zs)) -> append_3_out_aag3(._22(X, Xs), Ys, ._22(X, Zs))
if_sublist_2_in_1_ag3(X, Y, append_3_out_aag3(P, underscore, Y)) -> if_sublist_2_in_2_ag4(X, Y, P, append_3_in_aag3(underscore1, X, P))
if_sublist_2_in_2_ag4(X, Y, P, append_3_out_aag3(underscore1, X, P)) -> sublist_2_out_ag2(X, Y)

The argument filtering Pi contains the following mapping:
sublist_2_in_ag2(x1, x2)  =  sublist_2_in_ag1(x2)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_sublist_2_in_1_ag3(x1, x2, x3)  =  if_sublist_2_in_1_ag1(x3)
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_aag2(x1, x5)
if_sublist_2_in_2_ag4(x1, x2, x3, x4)  =  if_sublist_2_in_2_ag1(x4)
sublist_2_out_ag2(x1, x2)  =  sublist_2_out_ag1(x1)
SUBLIST_2_IN_AG2(x1, x2)  =  SUBLIST_2_IN_AG1(x2)
IF_APPEND_3_IN_1_AAG5(x1, x2, x3, x4, x5)  =  IF_APPEND_3_IN_1_AAG2(x1, x5)
APPEND_3_IN_AAG3(x1, x2, x3)  =  APPEND_3_IN_AAG1(x3)
IF_SUBLIST_2_IN_2_AG4(x1, x2, x3, x4)  =  IF_SUBLIST_2_IN_2_AG1(x4)
IF_SUBLIST_2_IN_1_AG3(x1, x2, x3)  =  IF_SUBLIST_2_IN_1_AG1(x3)

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

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

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

SUBLIST_2_IN_AG2(X, Y) -> IF_SUBLIST_2_IN_1_AG3(X, Y, append_3_in_aag3(P, underscore, Y))
SUBLIST_2_IN_AG2(X, Y) -> APPEND_3_IN_AAG3(P, underscore, Y)
APPEND_3_IN_AAG3(._22(X, Xs), Ys, ._22(X, Zs)) -> IF_APPEND_3_IN_1_AAG5(X, Xs, Ys, Zs, append_3_in_aag3(Xs, Ys, Zs))
APPEND_3_IN_AAG3(._22(X, Xs), Ys, ._22(X, Zs)) -> APPEND_3_IN_AAG3(Xs, Ys, Zs)
IF_SUBLIST_2_IN_1_AG3(X, Y, append_3_out_aag3(P, underscore, Y)) -> IF_SUBLIST_2_IN_2_AG4(X, Y, P, append_3_in_aag3(underscore1, X, P))
IF_SUBLIST_2_IN_1_AG3(X, Y, append_3_out_aag3(P, underscore, Y)) -> APPEND_3_IN_AAG3(underscore1, X, P)

The TRS R consists of the following rules:

sublist_2_in_ag2(X, Y) -> if_sublist_2_in_1_ag3(X, Y, append_3_in_aag3(P, underscore, Y))
append_3_in_aag3([]_0, Ys, Ys) -> append_3_out_aag3([]_0, Ys, Ys)
append_3_in_aag3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_append_3_in_1_aag5(X, Xs, Ys, Zs, append_3_in_aag3(Xs, Ys, Zs))
if_append_3_in_1_aag5(X, Xs, Ys, Zs, append_3_out_aag3(Xs, Ys, Zs)) -> append_3_out_aag3(._22(X, Xs), Ys, ._22(X, Zs))
if_sublist_2_in_1_ag3(X, Y, append_3_out_aag3(P, underscore, Y)) -> if_sublist_2_in_2_ag4(X, Y, P, append_3_in_aag3(underscore1, X, P))
if_sublist_2_in_2_ag4(X, Y, P, append_3_out_aag3(underscore1, X, P)) -> sublist_2_out_ag2(X, Y)

The argument filtering Pi contains the following mapping:
sublist_2_in_ag2(x1, x2)  =  sublist_2_in_ag1(x2)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_sublist_2_in_1_ag3(x1, x2, x3)  =  if_sublist_2_in_1_ag1(x3)
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_aag2(x1, x5)
if_sublist_2_in_2_ag4(x1, x2, x3, x4)  =  if_sublist_2_in_2_ag1(x4)
sublist_2_out_ag2(x1, x2)  =  sublist_2_out_ag1(x1)
SUBLIST_2_IN_AG2(x1, x2)  =  SUBLIST_2_IN_AG1(x2)
IF_APPEND_3_IN_1_AAG5(x1, x2, x3, x4, x5)  =  IF_APPEND_3_IN_1_AAG2(x1, x5)
APPEND_3_IN_AAG3(x1, x2, x3)  =  APPEND_3_IN_AAG1(x3)
IF_SUBLIST_2_IN_2_AG4(x1, x2, x3, x4)  =  IF_SUBLIST_2_IN_2_AG1(x4)
IF_SUBLIST_2_IN_1_AG3(x1, x2, x3)  =  IF_SUBLIST_2_IN_1_AG1(x3)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 1 SCC with 5 less nodes.

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

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

APPEND_3_IN_AAG3(._22(X, Xs), Ys, ._22(X, Zs)) -> APPEND_3_IN_AAG3(Xs, Ys, Zs)

The TRS R consists of the following rules:

sublist_2_in_ag2(X, Y) -> if_sublist_2_in_1_ag3(X, Y, append_3_in_aag3(P, underscore, Y))
append_3_in_aag3([]_0, Ys, Ys) -> append_3_out_aag3([]_0, Ys, Ys)
append_3_in_aag3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_append_3_in_1_aag5(X, Xs, Ys, Zs, append_3_in_aag3(Xs, Ys, Zs))
if_append_3_in_1_aag5(X, Xs, Ys, Zs, append_3_out_aag3(Xs, Ys, Zs)) -> append_3_out_aag3(._22(X, Xs), Ys, ._22(X, Zs))
if_sublist_2_in_1_ag3(X, Y, append_3_out_aag3(P, underscore, Y)) -> if_sublist_2_in_2_ag4(X, Y, P, append_3_in_aag3(underscore1, X, P))
if_sublist_2_in_2_ag4(X, Y, P, append_3_out_aag3(underscore1, X, P)) -> sublist_2_out_ag2(X, Y)

The argument filtering Pi contains the following mapping:
sublist_2_in_ag2(x1, x2)  =  sublist_2_in_ag1(x2)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_sublist_2_in_1_ag3(x1, x2, x3)  =  if_sublist_2_in_1_ag1(x3)
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_aag2(x1, x5)
if_sublist_2_in_2_ag4(x1, x2, x3, x4)  =  if_sublist_2_in_2_ag1(x4)
sublist_2_out_ag2(x1, x2)  =  sublist_2_out_ag1(x1)
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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof

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

APPEND_3_IN_AAG3(._22(X, Xs), Ys, ._22(X, Zs)) -> APPEND_3_IN_AAG3(Xs, Ys, Zs)

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

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

APPEND_3_IN_AAG1(._22(X, Zs)) -> APPEND_3_IN_AAG1(Zs)

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: