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) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
2 [label="A: member(T1, T2)\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
5 [label="5: member(T1, T2), SCOPE: 1, CLAUSE: 0 | member(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: member(T9, T8) | member(T1, .(T7, T8)), SCOPE: 1, CLAUSE: 1\n\nT7 is ground\nT8 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: member(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT2 is ground\nX4 is free\nX5 is free\nX6 is free\nmember(T1, T2) !=? member(X4, .(X5, X6))", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: member(T9, T8), SCOPE: 2, CLAUSE: 0 | member(T9, T8), SCOPE: 2, CLAUSE: 1 | ?_2 | member(T1, .(T7, T8)), SCOPE: 1, CLAUSE: 1\n\nT7 is ground\nT8 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: member(T9, T8), SCOPE: 2, CLAUSE: 0\n\nT8 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: member(T9, T8), SCOPE: 2, CLAUSE: 1 | ?_2 | member(T1, .(T7, T8)), SCOPE: 1, CLAUSE: 1\n\nT7 is ground\nT8 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: member(T25, T24)\n\nT24 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: member(T9, T8), SCOPE: 2, CLAUSE: 1\n\nT8 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: ?_2 | member(T1, .(T7, T8)), SCOPE: 1, CLAUSE: 1\n\nT7 is ground\nT8 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
36 [label="36: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
38 [label="38: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
39 [label="39: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
41 [label="41: member(T1, .(T7, T8)), SCOPE: 1, CLAUSE: 1\n\nT7 is ground\nT8 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
44 [label="44: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
45 [label="45: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
46 [label="46: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
47 [label="47: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
48 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
2 -> 48 [arrowhead = none ];
48 -> 5;

49 [label="EVAL with clause\nmember(X4, .(X5, X6)) :- member(X4, X6).\nand substitutionT1 -> T9,\nX4 -> T9,\nX5 -> T7,\nX6 -> T8,\nT2 -> .(T7, T8),\nT6 -> T9", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 49 [arrowhead = none ];
49 -> 11;

50 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 50 [arrowhead = none ];
50 -> 18;

51 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
11 -> 51 [arrowhead = none ];
51 -> 20;

52 [label="BACKTRACK\nfor clause: member(X, .(X, Xs))\nwith clash: (member(T1, T2), member(X4, .(X5, X6)))", fontsize=14, style = filled, fillcolor = lightgreen];
18 -> 52 [arrowhead = none ];
52 -> 47;

53 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
20 -> 53 [arrowhead = none ];
53 -> 21;

54 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
20 -> 54 [arrowhead = none ];
54 -> 22;

55 [label="EVAL with clause\nmember(X19, .(X20, X21)) :- member(X19, X21).\nand substitutionT9 -> T25,\nX19 -> T25,\nX20 -> T23,\nX21 -> T24,\nT8 -> .(T23, T24),\nT22 -> T25", fontsize=14, style = filled, fillcolor = lightblue];
21 -> 55 [arrowhead = none ];
55 -> 23;

56 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
21 -> 56 [arrowhead = none ];
56 -> 24;

57 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
22 -> 57 [arrowhead = none ];
57 -> 27;

58 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
22 -> 58 [arrowhead = none ];
58 -> 28;

59 [label="INSTANCE with matching:\nT1 -> T25\nT2 -> T24", fontsize=14, style = filled, fillcolor = lightgrey];
23 -> 59 [arrowhead = none , style=dashed];
59 -> 2[style=dashed];

60 [label="EVAL with clause\nmember(X33, .(X33, X34)).\nand substitutionT9 -> T37,\nX33 -> T37,\nX34 -> T38,\nT8 -> .(T37, T38)", fontsize=14, style = filled, fillcolor = lightblue];
27 -> 60 [arrowhead = none ];
60 -> 36;

61 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
27 -> 61 [arrowhead = none ];
61 -> 38;

62 [label="FAILURE", fontsize=14, style = filled, fillcolor = lightgreen];
28 -> 62 [arrowhead = none ];
62 -> 41;

63 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
36 -> 63 [arrowhead = none ];
63 -> 39;

64 [label="EVAL with clause\nmember(X39, .(X39, X40)).\nand substitutionT1 -> T43,\nX39 -> T43,\nT7 -> T43,\nT8 -> T44,\nX40 -> T44", fontsize=14, style = filled, fillcolor = lightblue];
41 -> 64 [arrowhead = none ];
64 -> 44;

65 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
41 -> 65 [arrowhead = none ];
65 -> 45;

66 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
44 -> 66 [arrowhead = none ];
66 -> 46;

}

(2) Obligation:

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

memberA_in_ag(T25, .(T7, .(T23, T24))) → U1_ag(T25, T7, T23, T24, memberA_in_ag(T25, T24))
memberA_in_ag(T37, .(T7, .(T37, T38))) → memberA_out_ag(T37, .(T7, .(T37, T38)))
memberA_in_ag(T43, .(T43, T44)) → memberA_out_ag(T43, .(T43, T44))
U1_ag(T25, T7, T23, T24, memberA_out_ag(T25, T24)) → memberA_out_ag(T25, .(T7, .(T23, T24)))

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

(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:

MEMBERA_IN_AG(T25, .(T7, .(T23, T24))) → U1_AG(T25, T7, T23, T24, memberA_in_ag(T25, T24))
MEMBERA_IN_AG(T25, .(T7, .(T23, T24))) → MEMBERA_IN_AG(T25, T24)

The TRS R consists of the following rules:

memberA_in_ag(T25, .(T7, .(T23, T24))) → U1_ag(T25, T7, T23, T24, memberA_in_ag(T25, T24))
memberA_in_ag(T37, .(T7, .(T37, T38))) → memberA_out_ag(T37, .(T7, .(T37, T38)))
memberA_in_ag(T43, .(T43, T44)) → memberA_out_ag(T43, .(T43, T44))
U1_ag(T25, T7, T23, T24, memberA_out_ag(T25, T24)) → memberA_out_ag(T25, .(T7, .(T23, T24)))

The argument filtering Pi contains the following mapping:
memberA_in_ag(x1, x2)  =  memberA_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
U1_ag(x1, x2, x3, x4, x5)  =  U1_ag(x2, x3, x4, x5)
memberA_out_ag(x1, x2)  =  memberA_out_ag(x1, x2)
MEMBERA_IN_AG(x1, x2)  =  MEMBERA_IN_AG(x2)
U1_AG(x1, x2, x3, x4, x5)  =  U1_AG(x2, x3, x4, x5)

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

(4) Obligation:

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

MEMBERA_IN_AG(T25, .(T7, .(T23, T24))) → U1_AG(T25, T7, T23, T24, memberA_in_ag(T25, T24))
MEMBERA_IN_AG(T25, .(T7, .(T23, T24))) → MEMBERA_IN_AG(T25, T24)

The TRS R consists of the following rules:

memberA_in_ag(T25, .(T7, .(T23, T24))) → U1_ag(T25, T7, T23, T24, memberA_in_ag(T25, T24))
memberA_in_ag(T37, .(T7, .(T37, T38))) → memberA_out_ag(T37, .(T7, .(T37, T38)))
memberA_in_ag(T43, .(T43, T44)) → memberA_out_ag(T43, .(T43, T44))
U1_ag(T25, T7, T23, T24, memberA_out_ag(T25, T24)) → memberA_out_ag(T25, .(T7, .(T23, T24)))

The argument filtering Pi contains the following mapping:
memberA_in_ag(x1, x2)  =  memberA_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
U1_ag(x1, x2, x3, x4, x5)  =  U1_ag(x2, x3, x4, x5)
memberA_out_ag(x1, x2)  =  memberA_out_ag(x1, x2)
MEMBERA_IN_AG(x1, x2)  =  MEMBERA_IN_AG(x2)
U1_AG(x1, x2, x3, x4, x5)  =  U1_AG(x2, x3, x4, 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 1 less node.

(6) Obligation:

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

MEMBERA_IN_AG(T25, .(T7, .(T23, T24))) → MEMBERA_IN_AG(T25, T24)

The TRS R consists of the following rules:

memberA_in_ag(T25, .(T7, .(T23, T24))) → U1_ag(T25, T7, T23, T24, memberA_in_ag(T25, T24))
memberA_in_ag(T37, .(T7, .(T37, T38))) → memberA_out_ag(T37, .(T7, .(T37, T38)))
memberA_in_ag(T43, .(T43, T44)) → memberA_out_ag(T43, .(T43, T44))
U1_ag(T25, T7, T23, T24, memberA_out_ag(T25, T24)) → memberA_out_ag(T25, .(T7, .(T23, T24)))

The argument filtering Pi contains the following mapping:
memberA_in_ag(x1, x2)  =  memberA_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
U1_ag(x1, x2, x3, x4, x5)  =  U1_ag(x2, x3, x4, x5)
memberA_out_ag(x1, x2)  =  memberA_out_ag(x1, x2)
MEMBERA_IN_AG(x1, x2)  =  MEMBERA_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:

MEMBERA_IN_AG(T25, .(T7, .(T23, T24))) → MEMBERA_IN_AG(T25, T24)

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

MEMBERA_IN_AG(.(T7, .(T23, T24))) → MEMBERA_IN_AG(T24)

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:

• MEMBERA_IN_AG(.(T7, .(T23, T24))) → MEMBERA_IN_AG(T24)
  The graph contains the following edges 1 > 1

(12) YES