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



↳ PROLOG
  ↳ PrologToPiTRSProof

list1({}0).
list1(.2(underscore, Ts)) :- list1(Ts).


With regard to the inferred argument filtering the predicates were used in the following modes:
list1: (b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


list_1_in_g1([]_0) -> list_1_out_g1([]_0)
list_1_in_g1(._22(underscore, Ts)) -> if_list_1_in_1_g3(underscore, Ts, list_1_in_g1(Ts))
if_list_1_in_1_g3(underscore, Ts, list_1_out_g1(Ts)) -> list_1_out_g1(._22(underscore, Ts))

The argument filtering Pi contains the following mapping:
list_1_in_g1(x1)  =  list_1_in_g1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
list_1_out_g1(x1)  =  list_1_out_g
if_list_1_in_1_g3(x1, x2, x3)  =  if_list_1_in_1_g1(x3)

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:

list_1_in_g1([]_0) -> list_1_out_g1([]_0)
list_1_in_g1(._22(underscore, Ts)) -> if_list_1_in_1_g3(underscore, Ts, list_1_in_g1(Ts))
if_list_1_in_1_g3(underscore, Ts, list_1_out_g1(Ts)) -> list_1_out_g1(._22(underscore, Ts))

The argument filtering Pi contains the following mapping:
list_1_in_g1(x1)  =  list_1_in_g1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
list_1_out_g1(x1)  =  list_1_out_g
if_list_1_in_1_g3(x1, x2, x3)  =  if_list_1_in_1_g1(x3)


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

LIST_1_IN_G1(._22(underscore, Ts)) -> IF_LIST_1_IN_1_G3(underscore, Ts, list_1_in_g1(Ts))
LIST_1_IN_G1(._22(underscore, Ts)) -> LIST_1_IN_G1(Ts)

The TRS R consists of the following rules:

list_1_in_g1([]_0) -> list_1_out_g1([]_0)
list_1_in_g1(._22(underscore, Ts)) -> if_list_1_in_1_g3(underscore, Ts, list_1_in_g1(Ts))
if_list_1_in_1_g3(underscore, Ts, list_1_out_g1(Ts)) -> list_1_out_g1(._22(underscore, Ts))

The argument filtering Pi contains the following mapping:
list_1_in_g1(x1)  =  list_1_in_g1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
list_1_out_g1(x1)  =  list_1_out_g
if_list_1_in_1_g3(x1, x2, x3)  =  if_list_1_in_1_g1(x3)
IF_LIST_1_IN_1_G3(x1, x2, x3)  =  IF_LIST_1_IN_1_G1(x3)
LIST_1_IN_G1(x1)  =  LIST_1_IN_G1(x1)

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:

LIST_1_IN_G1(._22(underscore, Ts)) -> IF_LIST_1_IN_1_G3(underscore, Ts, list_1_in_g1(Ts))
LIST_1_IN_G1(._22(underscore, Ts)) -> LIST_1_IN_G1(Ts)

The TRS R consists of the following rules:

list_1_in_g1([]_0) -> list_1_out_g1([]_0)
list_1_in_g1(._22(underscore, Ts)) -> if_list_1_in_1_g3(underscore, Ts, list_1_in_g1(Ts))
if_list_1_in_1_g3(underscore, Ts, list_1_out_g1(Ts)) -> list_1_out_g1(._22(underscore, Ts))

The argument filtering Pi contains the following mapping:
list_1_in_g1(x1)  =  list_1_in_g1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
list_1_out_g1(x1)  =  list_1_out_g
if_list_1_in_1_g3(x1, x2, x3)  =  if_list_1_in_1_g1(x3)
IF_LIST_1_IN_1_G3(x1, x2, x3)  =  IF_LIST_1_IN_1_G1(x3)
LIST_1_IN_G1(x1)  =  LIST_1_IN_G1(x1)

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

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

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

LIST_1_IN_G1(._22(underscore, Ts)) -> LIST_1_IN_G1(Ts)

The TRS R consists of the following rules:

list_1_in_g1([]_0) -> list_1_out_g1([]_0)
list_1_in_g1(._22(underscore, Ts)) -> if_list_1_in_1_g3(underscore, Ts, list_1_in_g1(Ts))
if_list_1_in_1_g3(underscore, Ts, list_1_out_g1(Ts)) -> list_1_out_g1(._22(underscore, Ts))

The argument filtering Pi contains the following mapping:
list_1_in_g1(x1)  =  list_1_in_g1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
list_1_out_g1(x1)  =  list_1_out_g
if_list_1_in_1_g3(x1, x2, x3)  =  if_list_1_in_1_g1(x3)
LIST_1_IN_G1(x1)  =  LIST_1_IN_G1(x1)

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:

LIST_1_IN_G1(._22(underscore, Ts)) -> LIST_1_IN_G1(Ts)

R is empty.
Pi is empty.
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:

LIST_1_IN_G1(._22(underscore, Ts)) -> LIST_1_IN_G1(Ts)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {LIST_1_IN_G1}.
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: