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

(0) Obligation:

Clauses:

member(X, .(Y, Xs)) :- member(X, Xs).
member(X, .(X, Xs)).

Query: member(a,g)

(1) PrologToPiTRSProof (SOUND transformation)

We use the technique of [TOCL09]. 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, .(Y, Xs)) → U1_ag(X, Y, Xs, member_in_ag(X, Xs))
member_in_ag(X, .(X, Xs)) → member_out_ag(X, .(X, Xs))
U1_ag(X, Y, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(Y, Xs))

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

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, .(Y, Xs)) → U1_ag(X, Y, Xs, member_in_ag(X, Xs))
member_in_ag(X, .(X, Xs)) → member_out_ag(X, .(X, Xs))
U1_ag(X, Y, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(Y, Xs))

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

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

The TRS R consists of the following rules:

member_in_ag(X, .(Y, Xs)) → U1_ag(X, Y, Xs, member_in_ag(X, Xs))
member_in_ag(X, .(X, Xs)) → member_out_ag(X, .(X, Xs))
U1_ag(X, Y, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(Y, Xs))

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

The TRS R consists of the following rules:

member_in_ag(X, .(Y, Xs)) → U1_ag(X, Y, Xs, member_in_ag(X, Xs))
member_in_ag(X, .(X, Xs)) → member_out_ag(X, .(X, Xs))
U1_ag(X, Y, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(Y, Xs))

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

The TRS R consists of the following rules:

member_in_ag(X, .(Y, Xs)) → U1_ag(X, Y, Xs, member_in_ag(X, Xs))
member_in_ag(X, .(X, Xs)) → member_out_ag(X, .(X, Xs))
U1_ag(X, Y, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(Y, Xs))

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

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

(11) 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(.(Y, Xs)) → MEMBER_IN_AG(Xs)
  The graph contains the following edges 1 > 1

(12) YES