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 PrologToPiTRSProof (⇐)
    ↳2 PiTRS
    ↳3 DependencyPairsProof (⇔)
    ↳4 PiDP
    ↳5 DependencyGraphProof (⇔)
    ↳6 PiDP
    ↳7 UsableRulesProof (⇔)
    ↳8 PiDP
    ↳9 PiDPToQDPProof (⇐)
    ↳10 QDP
  ↳11 PrologToPiTRSProof (⇐)
    ↳12 PiTRS
    ↳13 DependencyPairsProof (⇔)
    ↳14 PiDP
    ↳15 DependencyGraphProof (⇔)
    ↳16 PiDP
    ↳17 UsableRulesProof (⇔)
    ↳18 PiDP
    ↳19 PiDPToQDPProof (⇐)
    ↳20 QDP
    ↳21 QDPSizeChangeProof (⇔)
    ↳22 TRUE

(0) Obligation:

Clauses:

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

Queries:

member(a,g).

(1) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
member_in: (f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

member_in_ag(X, .(X, X1)) → member_out_ag(X, .(X, X1))
member_in_ag(X, .(X2, Xs)) → U1_ag(X, X2, Xs, member_in_ag(X, Xs))
U1_ag(X, X2, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(X2, Xs))

The argument filtering Pi contains the following mapping:
member_in_ag(x1, x2)  =  member_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
member_out_ag(x1, x2)  =  member_out_ag(x1)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x4)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(2) Obligation:

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

member_in_ag(X, .(X, X1)) → member_out_ag(X, .(X, X1))
member_in_ag(X, .(X2, Xs)) → U1_ag(X, X2, Xs, member_in_ag(X, Xs))
U1_ag(X, X2, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(X2, Xs))

The argument filtering Pi contains the following mapping:
member_in_ag(x1, x2)  =  member_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
member_out_ag(x1, x2)  =  member_out_ag(x1)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x4)

(3) 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:

MEMBER_IN_AG(X, .(X2, Xs)) → U1_AG(X, X2, Xs, member_in_ag(X, Xs))
MEMBER_IN_AG(X, .(X2, Xs)) → MEMBER_IN_AG(X, Xs)

The TRS R consists of the following rules:

member_in_ag(X, .(X, X1)) → member_out_ag(X, .(X, X1))
member_in_ag(X, .(X2, Xs)) → U1_ag(X, X2, Xs, member_in_ag(X, Xs))
U1_ag(X, X2, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(X2, Xs))

The argument filtering Pi contains the following mapping:
member_in_ag(x1, x2)  =  member_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
member_out_ag(x1, x2)  =  member_out_ag(x1)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x4)
MEMBER_IN_AG(x1, x2)  =  MEMBER_IN_AG(x2)
U1_AG(x1, x2, x3, x4)  =  U1_AG(x4)

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

(4) Obligation:

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

MEMBER_IN_AG(X, .(X2, Xs)) → U1_AG(X, X2, Xs, member_in_ag(X, Xs))
MEMBER_IN_AG(X, .(X2, Xs)) → MEMBER_IN_AG(X, Xs)

The TRS R consists of the following rules:

member_in_ag(X, .(X, X1)) → member_out_ag(X, .(X, X1))
member_in_ag(X, .(X2, Xs)) → U1_ag(X, X2, Xs, member_in_ag(X, Xs))
U1_ag(X, X2, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(X2, Xs))

The argument filtering Pi contains the following mapping:
member_in_ag(x1, x2)  =  member_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
member_out_ag(x1, x2)  =  member_out_ag(x1)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x4)
MEMBER_IN_AG(x1, x2)  =  MEMBER_IN_AG(x2)
U1_AG(x1, x2, x3, x4)  =  U1_AG(x4)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Obligation:

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

MEMBER_IN_AG(X, .(X2, Xs)) → MEMBER_IN_AG(X, Xs)

The TRS R consists of the following rules:

member_in_ag(X, .(X, X1)) → member_out_ag(X, .(X, X1))
member_in_ag(X, .(X2, Xs)) → U1_ag(X, X2, Xs, member_in_ag(X, Xs))
U1_ag(X, X2, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(X2, Xs))

The argument filtering Pi contains the following mapping:
member_in_ag(x1, x2)  =  member_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
member_out_ag(x1, x2)  =  member_out_ag(x1)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x4)
MEMBER_IN_AG(x1, x2)  =  MEMBER_IN_AG(x2)

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

(7) UsableRulesProof (EQUIVALENT transformation)

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

(8) Obligation:

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

MEMBER_IN_AG(X, .(X2, Xs)) → MEMBER_IN_AG(X, Xs)

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

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

(9) PiDPToQDPProof (SOUND transformation)

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

(10) Obligation:

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

MEMBER_IN_AG(.(X2, Xs)) → MEMBER_IN_AG(Xs)

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

(11) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
member_in: (f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

member_in_ag(X, .(X, X1)) → member_out_ag(X, .(X, X1))
member_in_ag(X, .(X2, Xs)) → U1_ag(X, X2, Xs, member_in_ag(X, Xs))
U1_ag(X, X2, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(X2, Xs))

The argument filtering Pi contains the following mapping:
member_in_ag(x1, x2)  =  member_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
member_out_ag(x1, x2)  =  member_out_ag(x1, x2)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x2, x3, x4)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(12) Obligation:

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

member_in_ag(X, .(X, X1)) → member_out_ag(X, .(X, X1))
member_in_ag(X, .(X2, Xs)) → U1_ag(X, X2, Xs, member_in_ag(X, Xs))
U1_ag(X, X2, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(X2, Xs))

The argument filtering Pi contains the following mapping:
member_in_ag(x1, x2)  =  member_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
member_out_ag(x1, x2)  =  member_out_ag(x1, x2)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x2, x3, x4)

(13) 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:

MEMBER_IN_AG(X, .(X2, Xs)) → U1_AG(X, X2, Xs, member_in_ag(X, Xs))
MEMBER_IN_AG(X, .(X2, Xs)) → MEMBER_IN_AG(X, Xs)

The TRS R consists of the following rules:

member_in_ag(X, .(X, X1)) → member_out_ag(X, .(X, X1))
member_in_ag(X, .(X2, Xs)) → U1_ag(X, X2, Xs, member_in_ag(X, Xs))
U1_ag(X, X2, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(X2, Xs))

The argument filtering Pi contains the following mapping:
member_in_ag(x1, x2)  =  member_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
member_out_ag(x1, x2)  =  member_out_ag(x1, x2)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x2, x3, x4)
MEMBER_IN_AG(x1, x2)  =  MEMBER_IN_AG(x2)
U1_AG(x1, x2, x3, x4)  =  U1_AG(x2, x3, x4)

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

(14) Obligation:

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

MEMBER_IN_AG(X, .(X2, Xs)) → U1_AG(X, X2, Xs, member_in_ag(X, Xs))
MEMBER_IN_AG(X, .(X2, Xs)) → MEMBER_IN_AG(X, Xs)

The TRS R consists of the following rules:

member_in_ag(X, .(X, X1)) → member_out_ag(X, .(X, X1))
member_in_ag(X, .(X2, Xs)) → U1_ag(X, X2, Xs, member_in_ag(X, Xs))
U1_ag(X, X2, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(X2, Xs))

The argument filtering Pi contains the following mapping:
member_in_ag(x1, x2)  =  member_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
member_out_ag(x1, x2)  =  member_out_ag(x1, x2)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x2, x3, x4)
MEMBER_IN_AG(x1, x2)  =  MEMBER_IN_AG(x2)
U1_AG(x1, x2, x3, x4)  =  U1_AG(x2, x3, x4)

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

(15) DependencyGraphProof (EQUIVALENT transformation)

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

(16) Obligation:

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

MEMBER_IN_AG(X, .(X2, Xs)) → MEMBER_IN_AG(X, Xs)

The TRS R consists of the following rules:

member_in_ag(X, .(X, X1)) → member_out_ag(X, .(X, X1))
member_in_ag(X, .(X2, Xs)) → U1_ag(X, X2, Xs, member_in_ag(X, Xs))
U1_ag(X, X2, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(X2, Xs))

The argument filtering Pi contains the following mapping:
member_in_ag(x1, x2)  =  member_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
member_out_ag(x1, x2)  =  member_out_ag(x1, x2)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x2, x3, x4)
MEMBER_IN_AG(x1, x2)  =  MEMBER_IN_AG(x2)

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

(17) UsableRulesProof (EQUIVALENT transformation)

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

(18) Obligation:

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

MEMBER_IN_AG(X, .(X2, Xs)) → MEMBER_IN_AG(X, Xs)

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

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

(19) PiDPToQDPProof (SOUND transformation)

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

(20) Obligation:

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

MEMBER_IN_AG(.(X2, Xs)) → MEMBER_IN_AG(Xs)

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

(21) 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:

  • MEMBER_IN_AG(.(X2, Xs)) → MEMBER_IN_AG(Xs)
    The graph contains the following edges 1 > 1

(22) TRUE