Left Termination of the query pattern member_in_2(a, g) w.r.t. the given Prolog program could successfully be proven:

0 Prolog
↳1 PrologToPrologProblemTransformerProof (⇐)
↳2 Prolog
↳3 PrologToPiTRSProof (⇐)
↳4 PiTRS
↳5 DependencyPairsProof (⇔)
↳6 PiDP
↳7 DependencyGraphProof (⇔)
↳8 PiDP
↳9 UsableRulesProof (⇔)
↳10 PiDP
↳11 PiDPToQDPProof (⇐)
↳12 QDP
↳13 QDPSizeChangeProof (⇔)
↳14 YES

(0) Obligation:

Clauses:

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

Queries:

member(a,g).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

(2) Obligation:

Clauses:

member1(T5, .(T5, T6)).
member1(T22, .(T11, .(T22, T23))).
member1(T33, .(T11, .(T31, T32))) :- member1(T33, T32).
member1(T53, .(T42, .(T53, T54))).
member1(T64, .(T42, .(T62, T63))) :- member1(T64, T63).

Queries:

member1(a,g).

(3) PrologToPiTRSProof (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: (f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

member1_in_ag(T5, .(T5, T6)) → member1_out_ag(T5, .(T5, T6))
member1_in_ag(T22, .(T11, .(T22, T23))) → member1_out_ag(T22, .(T11, .(T22, T23)))
member1_in_ag(T33, .(T11, .(T31, T32))) → U1_ag(T33, T11, T31, T32, member1_in_ag(T33, T32))
member1_in_ag(T64, .(T42, .(T62, T63))) → U2_ag(T64, T42, T62, T63, member1_in_ag(T64, T63))
U2_ag(T64, T42, T62, T63, member1_out_ag(T64, T63)) → member1_out_ag(T64, .(T42, .(T62, T63)))
U1_ag(T33, T11, T31, T32, member1_out_ag(T33, T32)) → member1_out_ag(T33, .(T11, .(T31, T32)))

The argument filtering Pi contains the following mapping:
member1_in_ag(x1, x2)  =  member1_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
member1_out_ag(x1, x2)  =  member1_out_ag(x1, x2)
U1_ag(x1, x2, x3, x4, x5)  =  U1_ag(x2, x3, x4, x5)
U2_ag(x1, x2, x3, x4, x5)  =  U2_ag(x2, x3, x4, x5)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

member1_in_ag(T5, .(T5, T6)) → member1_out_ag(T5, .(T5, T6))
member1_in_ag(T22, .(T11, .(T22, T23))) → member1_out_ag(T22, .(T11, .(T22, T23)))
member1_in_ag(T33, .(T11, .(T31, T32))) → U1_ag(T33, T11, T31, T32, member1_in_ag(T33, T32))
member1_in_ag(T64, .(T42, .(T62, T63))) → U2_ag(T64, T42, T62, T63, member1_in_ag(T64, T63))
U2_ag(T64, T42, T62, T63, member1_out_ag(T64, T63)) → member1_out_ag(T64, .(T42, .(T62, T63)))
U1_ag(T33, T11, T31, T32, member1_out_ag(T33, T32)) → member1_out_ag(T33, .(T11, .(T31, T32)))

The argument filtering Pi contains the following mapping:
member1_in_ag(x1, x2)  =  member1_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
member1_out_ag(x1, x2)  =  member1_out_ag(x1, x2)
U1_ag(x1, x2, x3, x4, x5)  =  U1_ag(x2, x3, x4, x5)
U2_ag(x1, x2, x3, x4, x5)  =  U2_ag(x2, x3, x4, x5)

(5) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

MEMBER1_IN_AG(T33, .(T11, .(T31, T32))) → U1_AG(T33, T11, T31, T32, member1_in_ag(T33, T32))
MEMBER1_IN_AG(T33, .(T11, .(T31, T32))) → MEMBER1_IN_AG(T33, T32)
MEMBER1_IN_AG(T64, .(T42, .(T62, T63))) → U2_AG(T64, T42, T62, T63, member1_in_ag(T64, T63))

The TRS R consists of the following rules:

member1_in_ag(T5, .(T5, T6)) → member1_out_ag(T5, .(T5, T6))
member1_in_ag(T22, .(T11, .(T22, T23))) → member1_out_ag(T22, .(T11, .(T22, T23)))
member1_in_ag(T33, .(T11, .(T31, T32))) → U1_ag(T33, T11, T31, T32, member1_in_ag(T33, T32))
member1_in_ag(T64, .(T42, .(T62, T63))) → U2_ag(T64, T42, T62, T63, member1_in_ag(T64, T63))
U2_ag(T64, T42, T62, T63, member1_out_ag(T64, T63)) → member1_out_ag(T64, .(T42, .(T62, T63)))
U1_ag(T33, T11, T31, T32, member1_out_ag(T33, T32)) → member1_out_ag(T33, .(T11, .(T31, T32)))

The argument filtering Pi contains the following mapping:
member1_in_ag(x1, x2)  =  member1_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
member1_out_ag(x1, x2)  =  member1_out_ag(x1, x2)
U1_ag(x1, x2, x3, x4, x5)  =  U1_ag(x2, x3, x4, x5)
U2_ag(x1, x2, x3, x4, x5)  =  U2_ag(x2, x3, x4, x5)
MEMBER1_IN_AG(x1, x2)  =  MEMBER1_IN_AG(x2)
U1_AG(x1, x2, x3, x4, x5)  =  U1_AG(x2, x3, x4, x5)
U2_AG(x1, x2, x3, x4, x5)  =  U2_AG(x2, x3, x4, x5)

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

(6) Obligation:

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

MEMBER1_IN_AG(T33, .(T11, .(T31, T32))) → U1_AG(T33, T11, T31, T32, member1_in_ag(T33, T32))
MEMBER1_IN_AG(T33, .(T11, .(T31, T32))) → MEMBER1_IN_AG(T33, T32)
MEMBER1_IN_AG(T64, .(T42, .(T62, T63))) → U2_AG(T64, T42, T62, T63, member1_in_ag(T64, T63))

The TRS R consists of the following rules:

member1_in_ag(T5, .(T5, T6)) → member1_out_ag(T5, .(T5, T6))
member1_in_ag(T22, .(T11, .(T22, T23))) → member1_out_ag(T22, .(T11, .(T22, T23)))
member1_in_ag(T33, .(T11, .(T31, T32))) → U1_ag(T33, T11, T31, T32, member1_in_ag(T33, T32))
member1_in_ag(T64, .(T42, .(T62, T63))) → U2_ag(T64, T42, T62, T63, member1_in_ag(T64, T63))
U2_ag(T64, T42, T62, T63, member1_out_ag(T64, T63)) → member1_out_ag(T64, .(T42, .(T62, T63)))
U1_ag(T33, T11, T31, T32, member1_out_ag(T33, T32)) → member1_out_ag(T33, .(T11, .(T31, T32)))

The argument filtering Pi contains the following mapping:
member1_in_ag(x1, x2)  =  member1_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
member1_out_ag(x1, x2)  =  member1_out_ag(x1, x2)
U1_ag(x1, x2, x3, x4, x5)  =  U1_ag(x2, x3, x4, x5)
U2_ag(x1, x2, x3, x4, x5)  =  U2_ag(x2, x3, x4, x5)
MEMBER1_IN_AG(x1, x2)  =  MEMBER1_IN_AG(x2)
U1_AG(x1, x2, x3, x4, x5)  =  U1_AG(x2, x3, x4, x5)
U2_AG(x1, x2, x3, x4, x5)  =  U2_AG(x2, x3, x4, x5)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(8) Obligation:

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

MEMBER1_IN_AG(T33, .(T11, .(T31, T32))) → MEMBER1_IN_AG(T33, T32)

The TRS R consists of the following rules:

member1_in_ag(T5, .(T5, T6)) → member1_out_ag(T5, .(T5, T6))
member1_in_ag(T22, .(T11, .(T22, T23))) → member1_out_ag(T22, .(T11, .(T22, T23)))
member1_in_ag(T33, .(T11, .(T31, T32))) → U1_ag(T33, T11, T31, T32, member1_in_ag(T33, T32))
member1_in_ag(T64, .(T42, .(T62, T63))) → U2_ag(T64, T42, T62, T63, member1_in_ag(T64, T63))
U2_ag(T64, T42, T62, T63, member1_out_ag(T64, T63)) → member1_out_ag(T64, .(T42, .(T62, T63)))
U1_ag(T33, T11, T31, T32, member1_out_ag(T33, T32)) → member1_out_ag(T33, .(T11, .(T31, T32)))

The argument filtering Pi contains the following mapping:
member1_in_ag(x1, x2)  =  member1_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
member1_out_ag(x1, x2)  =  member1_out_ag(x1, x2)
U1_ag(x1, x2, x3, x4, x5)  =  U1_ag(x2, x3, x4, x5)
U2_ag(x1, x2, x3, x4, x5)  =  U2_ag(x2, x3, x4, x5)
MEMBER1_IN_AG(x1, x2)  =  MEMBER1_IN_AG(x2)

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

(9) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(10) Obligation:

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

MEMBER1_IN_AG(T33, .(T11, .(T31, T32))) → MEMBER1_IN_AG(T33, T32)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
MEMBER1_IN_AG(x1, x2)  =  MEMBER1_IN_AG(x2)

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

(11) PiDPToQDPProof (SOUND transformation)

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

(12) Obligation:

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

MEMBER1_IN_AG(.(T11, .(T31, T32))) → MEMBER1_IN_AG(T32)

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

(13) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • MEMBER1_IN_AG(.(T11, .(T31, T32))) → MEMBER1_IN_AG(T32)
    The graph contains the following edges 1 > 1

(14) YES