Left Termination of the query pattern member_in_2(g, a) w.r.t. the given Prolog program could not be shown:

(0) Obligation:

Clauses:

member(X, .(X, X1)).
member(X, .(X2, Xs)) :- member(X, Xs).

Queries:

member(g,a).

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph.

(2) Obligation:

Triples:

member1(T30, .(T11, .(T31, T33))) :- member1(T30, T33).
member1(T61, .(T42, .(T62, T64))) :- member1(T61, T64).

Clauses:

memberc1(T5, .(T5, T6)).
memberc1(T22, .(T11, .(T22, T23))).
memberc1(T30, .(T11, .(T31, T33))) :- memberc1(T30, T33).
memberc1(T53, .(T42, .(T53, T54))).
memberc1(T61, .(T42, .(T62, T64))) :- memberc1(T61, T64).

Afs:

member1(x1, x2)  =  member1(x1)

(3) TriplesToPiDPProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
member1_in: (b,f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

MEMBER1_IN_GA(T30, .(T11, .(T31, T33))) → U1_GA(T30, T11, T31, T33, member1_in_ga(T30, T33))
MEMBER1_IN_GA(T30, .(T11, .(T31, T33))) → MEMBER1_IN_GA(T30, T33)
MEMBER1_IN_GA(T61, .(T42, .(T62, T64))) → U2_GA(T61, T42, T62, T64, member1_in_ga(T61, T64))

R is empty.
The argument filtering Pi contains the following mapping:
member1_in_ga(x1, x2)  =  member1_in_ga(x1)
.(x1, x2)  =  .(x2)
MEMBER1_IN_GA(x1, x2)  =  MEMBER1_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5)  =  U1_GA(x1, x5)
U2_GA(x1, x2, x3, x4, x5)  =  U2_GA(x1, x5)

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

Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES

(4) Obligation:

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

MEMBER1_IN_GA(T30, .(T11, .(T31, T33))) → U1_GA(T30, T11, T31, T33, member1_in_ga(T30, T33))
MEMBER1_IN_GA(T30, .(T11, .(T31, T33))) → MEMBER1_IN_GA(T30, T33)
MEMBER1_IN_GA(T61, .(T42, .(T62, T64))) → U2_GA(T61, T42, T62, T64, member1_in_ga(T61, T64))

R is empty.
The argument filtering Pi contains the following mapping:
member1_in_ga(x1, x2)  =  member1_in_ga(x1)
.(x1, x2)  =  .(x2)
MEMBER1_IN_GA(x1, x2)  =  MEMBER1_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5)  =  U1_GA(x1, x5)
U2_GA(x1, x2, x3, x4, x5)  =  U2_GA(x1, x5)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 2 less nodes.

(6) Obligation:

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

MEMBER1_IN_GA(T30, .(T11, .(T31, T33))) → MEMBER1_IN_GA(T30, T33)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x2)
MEMBER1_IN_GA(x1, x2)  =  MEMBER1_IN_GA(x1)

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

(7) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(8) Obligation:

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

MEMBER1_IN_GA(T30) → MEMBER1_IN_GA(T30)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(9) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = MEMBER1_IN_GA(T30) evaluates to t =MEMBER1_IN_GA(T30)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

• Semiunifier: [ ]
• Matcher: [ ]

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Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from MEMBER1_IN_GA(T30) to MEMBER1_IN_GA(T30).

(10) NO