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



↳ PROLOG
  ↳ PrologToPiTRSProof

member2(X, .2(X, underscore)).
member2(X, .2(underscore1, Xs)) :- member2(X, Xs).


With regard to the inferred argument filtering the predicates were used in the following modes:
member2: (f,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


member_2_in_ag2(X, ._22(X, underscore)) -> member_2_out_ag2(X, ._22(X, underscore))
member_2_in_ag2(X, ._22(underscore1, Xs)) -> if_member_2_in_1_ag4(X, underscore1, Xs, member_2_in_ag2(X, Xs))
if_member_2_in_1_ag4(X, underscore1, Xs, member_2_out_ag2(X, Xs)) -> member_2_out_ag2(X, ._22(underscore1, Xs))

The argument filtering Pi contains the following mapping:
member_2_in_ag2(x1, x2)  =  member_2_in_ag1(x2)
._22(x1, x2)  =  ._22(x1, x2)
member_2_out_ag2(x1, x2)  =  member_2_out_ag1(x1)
if_member_2_in_1_ag4(x1, x2, x3, x4)  =  if_member_2_in_1_ag1(x4)

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:

member_2_in_ag2(X, ._22(X, underscore)) -> member_2_out_ag2(X, ._22(X, underscore))
member_2_in_ag2(X, ._22(underscore1, Xs)) -> if_member_2_in_1_ag4(X, underscore1, Xs, member_2_in_ag2(X, Xs))
if_member_2_in_1_ag4(X, underscore1, Xs, member_2_out_ag2(X, Xs)) -> member_2_out_ag2(X, ._22(underscore1, Xs))

The argument filtering Pi contains the following mapping:
member_2_in_ag2(x1, x2)  =  member_2_in_ag1(x2)
._22(x1, x2)  =  ._22(x1, x2)
member_2_out_ag2(x1, x2)  =  member_2_out_ag1(x1)
if_member_2_in_1_ag4(x1, x2, x3, x4)  =  if_member_2_in_1_ag1(x4)


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

MEMBER_2_IN_AG2(X, ._22(underscore1, Xs)) -> IF_MEMBER_2_IN_1_AG4(X, underscore1, Xs, member_2_in_ag2(X, Xs))
MEMBER_2_IN_AG2(X, ._22(underscore1, Xs)) -> MEMBER_2_IN_AG2(X, Xs)

The TRS R consists of the following rules:

member_2_in_ag2(X, ._22(X, underscore)) -> member_2_out_ag2(X, ._22(X, underscore))
member_2_in_ag2(X, ._22(underscore1, Xs)) -> if_member_2_in_1_ag4(X, underscore1, Xs, member_2_in_ag2(X, Xs))
if_member_2_in_1_ag4(X, underscore1, Xs, member_2_out_ag2(X, Xs)) -> member_2_out_ag2(X, ._22(underscore1, Xs))

The argument filtering Pi contains the following mapping:
member_2_in_ag2(x1, x2)  =  member_2_in_ag1(x2)
._22(x1, x2)  =  ._22(x1, x2)
member_2_out_ag2(x1, x2)  =  member_2_out_ag1(x1)
if_member_2_in_1_ag4(x1, x2, x3, x4)  =  if_member_2_in_1_ag1(x4)
IF_MEMBER_2_IN_1_AG4(x1, x2, x3, x4)  =  IF_MEMBER_2_IN_1_AG1(x4)
MEMBER_2_IN_AG2(x1, x2)  =  MEMBER_2_IN_AG1(x2)

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:

MEMBER_2_IN_AG2(X, ._22(underscore1, Xs)) -> IF_MEMBER_2_IN_1_AG4(X, underscore1, Xs, member_2_in_ag2(X, Xs))
MEMBER_2_IN_AG2(X, ._22(underscore1, Xs)) -> MEMBER_2_IN_AG2(X, Xs)

The TRS R consists of the following rules:

member_2_in_ag2(X, ._22(X, underscore)) -> member_2_out_ag2(X, ._22(X, underscore))
member_2_in_ag2(X, ._22(underscore1, Xs)) -> if_member_2_in_1_ag4(X, underscore1, Xs, member_2_in_ag2(X, Xs))
if_member_2_in_1_ag4(X, underscore1, Xs, member_2_out_ag2(X, Xs)) -> member_2_out_ag2(X, ._22(underscore1, Xs))

The argument filtering Pi contains the following mapping:
member_2_in_ag2(x1, x2)  =  member_2_in_ag1(x2)
._22(x1, x2)  =  ._22(x1, x2)
member_2_out_ag2(x1, x2)  =  member_2_out_ag1(x1)
if_member_2_in_1_ag4(x1, x2, x3, x4)  =  if_member_2_in_1_ag1(x4)
IF_MEMBER_2_IN_1_AG4(x1, x2, x3, x4)  =  IF_MEMBER_2_IN_1_AG1(x4)
MEMBER_2_IN_AG2(x1, x2)  =  MEMBER_2_IN_AG1(x2)

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:

MEMBER_2_IN_AG2(X, ._22(underscore1, Xs)) -> MEMBER_2_IN_AG2(X, Xs)

The TRS R consists of the following rules:

member_2_in_ag2(X, ._22(X, underscore)) -> member_2_out_ag2(X, ._22(X, underscore))
member_2_in_ag2(X, ._22(underscore1, Xs)) -> if_member_2_in_1_ag4(X, underscore1, Xs, member_2_in_ag2(X, Xs))
if_member_2_in_1_ag4(X, underscore1, Xs, member_2_out_ag2(X, Xs)) -> member_2_out_ag2(X, ._22(underscore1, Xs))

The argument filtering Pi contains the following mapping:
member_2_in_ag2(x1, x2)  =  member_2_in_ag1(x2)
._22(x1, x2)  =  ._22(x1, x2)
member_2_out_ag2(x1, x2)  =  member_2_out_ag1(x1)
if_member_2_in_1_ag4(x1, x2, x3, x4)  =  if_member_2_in_1_ag1(x4)
MEMBER_2_IN_AG2(x1, x2)  =  MEMBER_2_IN_AG1(x2)

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:

MEMBER_2_IN_AG2(X, ._22(underscore1, Xs)) -> MEMBER_2_IN_AG2(X, Xs)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._22(x1, x2)
MEMBER_2_IN_AG2(x1, x2)  =  MEMBER_2_IN_AG1(x2)

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:

MEMBER_2_IN_AG1(._22(underscore1, Xs)) -> MEMBER_2_IN_AG1(Xs)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {MEMBER_2_IN_AG1}.
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: