Left Termination of the query pattern
subset(g,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)).
subset(.(X, Xs), Ys) :- ','(member(X, Ys), subset(Xs, Ys)).
subset([], Ys).
member1(X, .(Y, Xs)) :- member1(X, Xs).
member1(X, .(X, Xs)).
subset1(.(X, Xs), Ys) :- ','(member1(X, Ys), subset1(Xs, Ys)).
subset1([], Ys).

Query: subset(g,g)

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: subset(T1, T2)\n\nT1 is ground\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: subset(T1, T2), SCOPE: 1, CLAUSE: 2 | subset(T1, T2), SCOPE: 1, CLAUSE: 3\n\nT1 is ground\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: ','(member(T6, T8), subset(T7, T8)) | subset(.(T6, T7), T8), SCOPE: 1, CLAUSE: 3\n\nT6 is ground\nT7 is ground\nT8 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: subset(T1, T2), SCOPE: 1, CLAUSE: 3\n\nT1 is ground\nT2 is ground\nX4 is free\nX5 is free\nX6 is free\nsubset(T1, T2) !=? subset(.(X4, X5), X6)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: ','(member(T6, T8), subset(T7, T8)), SCOPE: 2, CLAUSE: 0 | ','(member(T6, T8), subset(T7, T8)), SCOPE: 2, CLAUSE: 1 | ?_2 | subset(.(T6, T7), T8), SCOPE: 1, CLAUSE: 3\n\nT6 is ground\nT7 is ground\nT8 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: ','(member(T6, T8), subset(T7, T8)), SCOPE: 2, CLAUSE: 0\n\nT6 is ground\nT7 is ground\nT8 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: ','(member(T6, T8), subset(T7, T8)), SCOPE: 2, CLAUSE: 1 | ?_2 | subset(.(T6, T7), T8), SCOPE: 1, CLAUSE: 3\n\nT6 is ground\nT7 is ground\nT8 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: ','(member(T21, T23), subset(T7, .(T22, T23)))\n\nT7 is ground\nT21 is ground\nT22 is ground\nT23 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: member(T21, T23)\n\nT21 is ground\nT23 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: subset(T7, .(T22, T23))\n\nT7 is ground\nT22 is ground\nT23 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: member(T21, T23), SCOPE: 3, CLAUSE: 0 | member(T21, T23), SCOPE: 3, CLAUSE: 1\n\nT21 is ground\nT23 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: member(T21, T23), SCOPE: 3, CLAUSE: 0\n\nT21 is ground\nT23 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: member(T21, T23), SCOPE: 3, CLAUSE: 1\n\nT21 is ground\nT23 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: member(T42, T44)\n\nT42 is ground\nT44 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: ','(member(T6, T8), subset(T7, T8)), SCOPE: 2, CLAUSE: 1\n\nT6 is ground\nT7 is ground\nT8 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: ?_2 | subset(.(T6, T7), T8), SCOPE: 1, CLAUSE: 3\n\nT6 is ground\nT7 is ground\nT8 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: subset(T7, .(T66, T67))\n\nT7 is ground\nT66 is ground\nT67 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: subset(.(T6, T7), T8), SCOPE: 1, CLAUSE: 3\n\nT6 is ground\nT7 is ground\nT8 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 29 [arrowhead = none ];
29 -> 2;

30 [label="EVAL with clause\nsubset(.(X4, X5), X6) :- ','(member(X4, X6), subset(X5, X6)).\nand substitutionX4 -> T6,\nX5 -> T7,\nT1 -> .(T6, T7),\nT2 -> T8,\nX6 -> T8", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 30 [arrowhead = none ];
30 -> 3;

31 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 31 [arrowhead = none ];
31 -> 4;

32 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 32 [arrowhead = none ];
32 -> 5;

33 [label="EVAL with clause\nsubset([], X70).\nand substitutionT1 -> [],\nT2 -> T73,\nX70 -> T73", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 33 [arrowhead = none ];
33 -> 26;

34 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 34 [arrowhead = none ];
34 -> 27;

35 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
5 -> 35 [arrowhead = none ];
35 -> 6;

36 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
5 -> 36 [arrowhead = none ];
36 -> 7;

37 [label="EVAL with clause\nmember(X19, .(X20, X21)) :- member(X19, X21).\nand substitutionT6 -> T21,\nX19 -> T21,\nX20 -> T22,\nX21 -> T23,\nT8 -> .(T22, T23)", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 37 [arrowhead = none ];
37 -> 8;

38 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 38 [arrowhead = none ];
38 -> 9;

39 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
7 -> 39 [arrowhead = none ];
39 -> 20;

40 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
7 -> 40 [arrowhead = none ];
40 -> 21;

41 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
8 -> 41 [arrowhead = none ];
41 -> 10;

42 [label="SPLIT 2\nnew knowledge:\nT21 is ground\nT23 is ground", fontsize=14, style = filled, fillcolor = violet];
8 -> 42 [arrowhead = none ];
42 -> 11;

43 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
10 -> 43 [arrowhead = none ];
43 -> 12;

44 [label="INSTANCE with matching:\nT1 -> T7\nT2 -> .(T22, T23)", fontsize=14, style = filled, fillcolor = lightgrey];
11 -> 44 [arrowhead = none , style=dashed];
44 -> 1[style=dashed];

45 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
12 -> 45 [arrowhead = none ];
45 -> 13;

46 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
12 -> 46 [arrowhead = none ];
46 -> 14;

47 [label="EVAL with clause\nmember(X40, .(X41, X42)) :- member(X40, X42).\nand substitutionT21 -> T42,\nX40 -> T42,\nX41 -> T43,\nX42 -> T44,\nT23 -> .(T43, T44)", fontsize=14, style = filled, fillcolor = lightblue];
13 -> 47 [arrowhead = none ];
47 -> 15;

48 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
13 -> 48 [arrowhead = none ];
48 -> 16;

49 [label="EVAL with clause\nmember(X50, .(X50, X51)).\nand substitutionT21 -> T52,\nX50 -> T52,\nX51 -> T53,\nT23 -> .(T52, T53)", fontsize=14, style = filled, fillcolor = lightblue];
14 -> 49 [arrowhead = none ];
49 -> 17;

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

51 [label="INSTANCE with matching:\nT21 -> T42\nT23 -> T44", fontsize=14, style = filled, fillcolor = lightgrey];
15 -> 51 [arrowhead = none , style=dashed];
51 -> 10[style=dashed];

52 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 52 [arrowhead = none ];
52 -> 19;

53 [label="EVAL with clause\nmember(X63, .(X63, X64)).\nand substitutionT6 -> T66,\nX63 -> T66,\nX64 -> T67,\nT8 -> .(T66, T67)", fontsize=14, style = filled, fillcolor = lightblue];
20 -> 53 [arrowhead = none ];
53 -> 22;

54 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
20 -> 54 [arrowhead = none ];
54 -> 23;

55 [label="FAILURE", fontsize=14, style = filled, fillcolor = lightgreen];
21 -> 55 [arrowhead = none ];
55 -> 24;

56 [label="INSTANCE with matching:\nT1 -> T7\nT2 -> .(T66, T67)", fontsize=14, style = filled, fillcolor = lightgrey];
22 -> 56 [arrowhead = none , style=dashed];
56 -> 1[style=dashed];

57 [label="BACKTRACK\nfor clause: subset([], Ys)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
24 -> 57 [arrowhead = none ];
57 -> 25;

58 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
26 -> 58 [arrowhead = none ];
58 -> 28;

}

(2) Obligation:

Clauses:

memberA(T42, .(T43, T44)) :- memberA(T42, T44).
memberA(T52, .(T52, T53)).
subsetB(.(T21, T7), .(T22, T23)) :- memberA(T21, T23).
subsetB(.(T21, T7), .(T22, T23)) :- ','(memberA(T21, T23), subsetB(T7, .(T22, T23))).
subsetB(.(T66, T7), .(T66, T67)) :- subsetB(T7, .(T66, T67)).
subsetB([], T73).

Query: subsetB(g,g)

(3) PrologToPiTRSProof (SOUND transformation)

We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
subsetB_in: (b,b)
memberA_in: (b,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

subsetB_in_gg(.(T21, T7), .(T22, T23)) → U2_gg(T21, T7, T22, T23, memberA_in_gg(T21, T23))
memberA_in_gg(T42, .(T43, T44)) → U1_gg(T42, T43, T44, memberA_in_gg(T42, T44))
memberA_in_gg(T52, .(T52, T53)) → memberA_out_gg(T52, .(T52, T53))
U1_gg(T42, T43, T44, memberA_out_gg(T42, T44)) → memberA_out_gg(T42, .(T43, T44))
U2_gg(T21, T7, T22, T23, memberA_out_gg(T21, T23)) → subsetB_out_gg(.(T21, T7), .(T22, T23))
U2_gg(T21, T7, T22, T23, memberA_out_gg(T21, T23)) → U3_gg(T21, T7, T22, T23, subsetB_in_gg(T7, .(T22, T23)))
subsetB_in_gg(.(T66, T7), .(T66, T67)) → U4_gg(T66, T7, T67, subsetB_in_gg(T7, .(T66, T67)))
subsetB_in_gg([], T73) → subsetB_out_gg([], T73)
U4_gg(T66, T7, T67, subsetB_out_gg(T7, .(T66, T67))) → subsetB_out_gg(.(T66, T7), .(T66, T67))
U3_gg(T21, T7, T22, T23, subsetB_out_gg(T7, .(T22, T23))) → subsetB_out_gg(.(T21, T7), .(T22, T23))

The argument filtering Pi contains the following mapping:
subsetB_in_gg(x1, x2)  =  subsetB_in_gg(x1, x2)
.(x1, x2)  =  .(x1, x2)
U2_gg(x1, x2, x3, x4, x5)  =  U2_gg(x2, x3, x4, x5)
memberA_in_gg(x1, x2)  =  memberA_in_gg(x1, x2)
U1_gg(x1, x2, x3, x4)  =  U1_gg(x4)
memberA_out_gg(x1, x2)  =  memberA_out_gg
subsetB_out_gg(x1, x2)  =  subsetB_out_gg
U3_gg(x1, x2, x3, x4, x5)  =  U3_gg(x5)
U4_gg(x1, x2, x3, x4)  =  U4_gg(x4)
[]  =  []

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

subsetB_in_gg(.(T21, T7), .(T22, T23)) → U2_gg(T21, T7, T22, T23, memberA_in_gg(T21, T23))
memberA_in_gg(T42, .(T43, T44)) → U1_gg(T42, T43, T44, memberA_in_gg(T42, T44))
memberA_in_gg(T52, .(T52, T53)) → memberA_out_gg(T52, .(T52, T53))
U1_gg(T42, T43, T44, memberA_out_gg(T42, T44)) → memberA_out_gg(T42, .(T43, T44))
U2_gg(T21, T7, T22, T23, memberA_out_gg(T21, T23)) → subsetB_out_gg(.(T21, T7), .(T22, T23))
U2_gg(T21, T7, T22, T23, memberA_out_gg(T21, T23)) → U3_gg(T21, T7, T22, T23, subsetB_in_gg(T7, .(T22, T23)))
subsetB_in_gg(.(T66, T7), .(T66, T67)) → U4_gg(T66, T7, T67, subsetB_in_gg(T7, .(T66, T67)))
subsetB_in_gg([], T73) → subsetB_out_gg([], T73)
U4_gg(T66, T7, T67, subsetB_out_gg(T7, .(T66, T67))) → subsetB_out_gg(.(T66, T7), .(T66, T67))
U3_gg(T21, T7, T22, T23, subsetB_out_gg(T7, .(T22, T23))) → subsetB_out_gg(.(T21, T7), .(T22, T23))

The argument filtering Pi contains the following mapping:
subsetB_in_gg(x1, x2)  =  subsetB_in_gg(x1, x2)
.(x1, x2)  =  .(x1, x2)
U2_gg(x1, x2, x3, x4, x5)  =  U2_gg(x2, x3, x4, x5)
memberA_in_gg(x1, x2)  =  memberA_in_gg(x1, x2)
U1_gg(x1, x2, x3, x4)  =  U1_gg(x4)
memberA_out_gg(x1, x2)  =  memberA_out_gg
subsetB_out_gg(x1, x2)  =  subsetB_out_gg
U3_gg(x1, x2, x3, x4, x5)  =  U3_gg(x5)
U4_gg(x1, x2, x3, x4)  =  U4_gg(x4)
[]  =  []

(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:

SUBSETB_IN_GG(.(T21, T7), .(T22, T23)) → U2_GG(T21, T7, T22, T23, memberA_in_gg(T21, T23))
SUBSETB_IN_GG(.(T21, T7), .(T22, T23)) → MEMBERA_IN_GG(T21, T23)
MEMBERA_IN_GG(T42, .(T43, T44)) → U1_GG(T42, T43, T44, memberA_in_gg(T42, T44))
MEMBERA_IN_GG(T42, .(T43, T44)) → MEMBERA_IN_GG(T42, T44)
U2_GG(T21, T7, T22, T23, memberA_out_gg(T21, T23)) → U3_GG(T21, T7, T22, T23, subsetB_in_gg(T7, .(T22, T23)))
U2_GG(T21, T7, T22, T23, memberA_out_gg(T21, T23)) → SUBSETB_IN_GG(T7, .(T22, T23))
SUBSETB_IN_GG(.(T66, T7), .(T66, T67)) → U4_GG(T66, T7, T67, subsetB_in_gg(T7, .(T66, T67)))
SUBSETB_IN_GG(.(T66, T7), .(T66, T67)) → SUBSETB_IN_GG(T7, .(T66, T67))

The TRS R consists of the following rules:

subsetB_in_gg(.(T21, T7), .(T22, T23)) → U2_gg(T21, T7, T22, T23, memberA_in_gg(T21, T23))
memberA_in_gg(T42, .(T43, T44)) → U1_gg(T42, T43, T44, memberA_in_gg(T42, T44))
memberA_in_gg(T52, .(T52, T53)) → memberA_out_gg(T52, .(T52, T53))
U1_gg(T42, T43, T44, memberA_out_gg(T42, T44)) → memberA_out_gg(T42, .(T43, T44))
U2_gg(T21, T7, T22, T23, memberA_out_gg(T21, T23)) → subsetB_out_gg(.(T21, T7), .(T22, T23))
U2_gg(T21, T7, T22, T23, memberA_out_gg(T21, T23)) → U3_gg(T21, T7, T22, T23, subsetB_in_gg(T7, .(T22, T23)))
subsetB_in_gg(.(T66, T7), .(T66, T67)) → U4_gg(T66, T7, T67, subsetB_in_gg(T7, .(T66, T67)))
subsetB_in_gg([], T73) → subsetB_out_gg([], T73)
U4_gg(T66, T7, T67, subsetB_out_gg(T7, .(T66, T67))) → subsetB_out_gg(.(T66, T7), .(T66, T67))
U3_gg(T21, T7, T22, T23, subsetB_out_gg(T7, .(T22, T23))) → subsetB_out_gg(.(T21, T7), .(T22, T23))

The argument filtering Pi contains the following mapping:
subsetB_in_gg(x1, x2)  =  subsetB_in_gg(x1, x2)
.(x1, x2)  =  .(x1, x2)
U2_gg(x1, x2, x3, x4, x5)  =  U2_gg(x2, x3, x4, x5)
memberA_in_gg(x1, x2)  =  memberA_in_gg(x1, x2)
U1_gg(x1, x2, x3, x4)  =  U1_gg(x4)
memberA_out_gg(x1, x2)  =  memberA_out_gg
subsetB_out_gg(x1, x2)  =  subsetB_out_gg
U3_gg(x1, x2, x3, x4, x5)  =  U3_gg(x5)
U4_gg(x1, x2, x3, x4)  =  U4_gg(x4)
[]  =  []
SUBSETB_IN_GG(x1, x2)  =  SUBSETB_IN_GG(x1, x2)
U2_GG(x1, x2, x3, x4, x5)  =  U2_GG(x2, x3, x4, x5)
MEMBERA_IN_GG(x1, x2)  =  MEMBERA_IN_GG(x1, x2)
U1_GG(x1, x2, x3, x4)  =  U1_GG(x4)
U3_GG(x1, x2, x3, x4, x5)  =  U3_GG(x5)
U4_GG(x1, x2, x3, x4)  =  U4_GG(x4)

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

(6) Obligation:

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

SUBSETB_IN_GG(.(T21, T7), .(T22, T23)) → U2_GG(T21, T7, T22, T23, memberA_in_gg(T21, T23))
SUBSETB_IN_GG(.(T21, T7), .(T22, T23)) → MEMBERA_IN_GG(T21, T23)
MEMBERA_IN_GG(T42, .(T43, T44)) → U1_GG(T42, T43, T44, memberA_in_gg(T42, T44))
MEMBERA_IN_GG(T42, .(T43, T44)) → MEMBERA_IN_GG(T42, T44)
U2_GG(T21, T7, T22, T23, memberA_out_gg(T21, T23)) → U3_GG(T21, T7, T22, T23, subsetB_in_gg(T7, .(T22, T23)))
U2_GG(T21, T7, T22, T23, memberA_out_gg(T21, T23)) → SUBSETB_IN_GG(T7, .(T22, T23))
SUBSETB_IN_GG(.(T66, T7), .(T66, T67)) → U4_GG(T66, T7, T67, subsetB_in_gg(T7, .(T66, T67)))
SUBSETB_IN_GG(.(T66, T7), .(T66, T67)) → SUBSETB_IN_GG(T7, .(T66, T67))

The TRS R consists of the following rules:

subsetB_in_gg(.(T21, T7), .(T22, T23)) → U2_gg(T21, T7, T22, T23, memberA_in_gg(T21, T23))
memberA_in_gg(T42, .(T43, T44)) → U1_gg(T42, T43, T44, memberA_in_gg(T42, T44))
memberA_in_gg(T52, .(T52, T53)) → memberA_out_gg(T52, .(T52, T53))
U1_gg(T42, T43, T44, memberA_out_gg(T42, T44)) → memberA_out_gg(T42, .(T43, T44))
U2_gg(T21, T7, T22, T23, memberA_out_gg(T21, T23)) → subsetB_out_gg(.(T21, T7), .(T22, T23))
U2_gg(T21, T7, T22, T23, memberA_out_gg(T21, T23)) → U3_gg(T21, T7, T22, T23, subsetB_in_gg(T7, .(T22, T23)))
subsetB_in_gg(.(T66, T7), .(T66, T67)) → U4_gg(T66, T7, T67, subsetB_in_gg(T7, .(T66, T67)))
subsetB_in_gg([], T73) → subsetB_out_gg([], T73)
U4_gg(T66, T7, T67, subsetB_out_gg(T7, .(T66, T67))) → subsetB_out_gg(.(T66, T7), .(T66, T67))
U3_gg(T21, T7, T22, T23, subsetB_out_gg(T7, .(T22, T23))) → subsetB_out_gg(.(T21, T7), .(T22, T23))

The argument filtering Pi contains the following mapping:
subsetB_in_gg(x1, x2)  =  subsetB_in_gg(x1, x2)
.(x1, x2)  =  .(x1, x2)
U2_gg(x1, x2, x3, x4, x5)  =  U2_gg(x2, x3, x4, x5)
memberA_in_gg(x1, x2)  =  memberA_in_gg(x1, x2)
U1_gg(x1, x2, x3, x4)  =  U1_gg(x4)
memberA_out_gg(x1, x2)  =  memberA_out_gg
subsetB_out_gg(x1, x2)  =  subsetB_out_gg
U3_gg(x1, x2, x3, x4, x5)  =  U3_gg(x5)
U4_gg(x1, x2, x3, x4)  =  U4_gg(x4)
[]  =  []
SUBSETB_IN_GG(x1, x2)  =  SUBSETB_IN_GG(x1, x2)
U2_GG(x1, x2, x3, x4, x5)  =  U2_GG(x2, x3, x4, x5)
MEMBERA_IN_GG(x1, x2)  =  MEMBERA_IN_GG(x1, x2)
U1_GG(x1, x2, x3, x4)  =  U1_GG(x4)
U3_GG(x1, x2, x3, x4, x5)  =  U3_GG(x5)
U4_GG(x1, x2, x3, x4)  =  U4_GG(x4)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 4 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

MEMBERA_IN_GG(T42, .(T43, T44)) → MEMBERA_IN_GG(T42, T44)

The TRS R consists of the following rules:

subsetB_in_gg(.(T21, T7), .(T22, T23)) → U2_gg(T21, T7, T22, T23, memberA_in_gg(T21, T23))
memberA_in_gg(T42, .(T43, T44)) → U1_gg(T42, T43, T44, memberA_in_gg(T42, T44))
memberA_in_gg(T52, .(T52, T53)) → memberA_out_gg(T52, .(T52, T53))
U1_gg(T42, T43, T44, memberA_out_gg(T42, T44)) → memberA_out_gg(T42, .(T43, T44))
U2_gg(T21, T7, T22, T23, memberA_out_gg(T21, T23)) → subsetB_out_gg(.(T21, T7), .(T22, T23))
U2_gg(T21, T7, T22, T23, memberA_out_gg(T21, T23)) → U3_gg(T21, T7, T22, T23, subsetB_in_gg(T7, .(T22, T23)))
subsetB_in_gg(.(T66, T7), .(T66, T67)) → U4_gg(T66, T7, T67, subsetB_in_gg(T7, .(T66, T67)))
subsetB_in_gg([], T73) → subsetB_out_gg([], T73)
U4_gg(T66, T7, T67, subsetB_out_gg(T7, .(T66, T67))) → subsetB_out_gg(.(T66, T7), .(T66, T67))
U3_gg(T21, T7, T22, T23, subsetB_out_gg(T7, .(T22, T23))) → subsetB_out_gg(.(T21, T7), .(T22, T23))

The argument filtering Pi contains the following mapping:
subsetB_in_gg(x1, x2)  =  subsetB_in_gg(x1, x2)
.(x1, x2)  =  .(x1, x2)
U2_gg(x1, x2, x3, x4, x5)  =  U2_gg(x2, x3, x4, x5)
memberA_in_gg(x1, x2)  =  memberA_in_gg(x1, x2)
U1_gg(x1, x2, x3, x4)  =  U1_gg(x4)
memberA_out_gg(x1, x2)  =  memberA_out_gg
subsetB_out_gg(x1, x2)  =  subsetB_out_gg
U3_gg(x1, x2, x3, x4, x5)  =  U3_gg(x5)
U4_gg(x1, x2, x3, x4)  =  U4_gg(x4)
[]  =  []
MEMBERA_IN_GG(x1, x2)  =  MEMBERA_IN_GG(x1, x2)

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

(10) UsableRulesProof (EQUIVALENT transformation)

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

(11) Obligation:

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

MEMBERA_IN_GG(T42, .(T43, T44)) → MEMBERA_IN_GG(T42, T44)

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

(12) PiDPToQDPProof (EQUIVALENT transformation)

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

(13) Obligation:

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

MEMBERA_IN_GG(T42, .(T43, T44)) → MEMBERA_IN_GG(T42, T44)

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

(14) 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_GG(T42, .(T43, T44)) → MEMBERA_IN_GG(T42, T44)
  The graph contains the following edges 1 >= 1, 2 > 2

(15) YES

(16) Obligation:

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

U2_GG(T21, T7, T22, T23, memberA_out_gg(T21, T23)) → SUBSETB_IN_GG(T7, .(T22, T23))
SUBSETB_IN_GG(.(T21, T7), .(T22, T23)) → U2_GG(T21, T7, T22, T23, memberA_in_gg(T21, T23))
SUBSETB_IN_GG(.(T66, T7), .(T66, T67)) → SUBSETB_IN_GG(T7, .(T66, T67))

The TRS R consists of the following rules:

subsetB_in_gg(.(T21, T7), .(T22, T23)) → U2_gg(T21, T7, T22, T23, memberA_in_gg(T21, T23))
memberA_in_gg(T42, .(T43, T44)) → U1_gg(T42, T43, T44, memberA_in_gg(T42, T44))
memberA_in_gg(T52, .(T52, T53)) → memberA_out_gg(T52, .(T52, T53))
U1_gg(T42, T43, T44, memberA_out_gg(T42, T44)) → memberA_out_gg(T42, .(T43, T44))
U2_gg(T21, T7, T22, T23, memberA_out_gg(T21, T23)) → subsetB_out_gg(.(T21, T7), .(T22, T23))
U2_gg(T21, T7, T22, T23, memberA_out_gg(T21, T23)) → U3_gg(T21, T7, T22, T23, subsetB_in_gg(T7, .(T22, T23)))
subsetB_in_gg(.(T66, T7), .(T66, T67)) → U4_gg(T66, T7, T67, subsetB_in_gg(T7, .(T66, T67)))
subsetB_in_gg([], T73) → subsetB_out_gg([], T73)
U4_gg(T66, T7, T67, subsetB_out_gg(T7, .(T66, T67))) → subsetB_out_gg(.(T66, T7), .(T66, T67))
U3_gg(T21, T7, T22, T23, subsetB_out_gg(T7, .(T22, T23))) → subsetB_out_gg(.(T21, T7), .(T22, T23))

The argument filtering Pi contains the following mapping:
subsetB_in_gg(x1, x2)  =  subsetB_in_gg(x1, x2)
.(x1, x2)  =  .(x1, x2)
U2_gg(x1, x2, x3, x4, x5)  =  U2_gg(x2, x3, x4, x5)
memberA_in_gg(x1, x2)  =  memberA_in_gg(x1, x2)
U1_gg(x1, x2, x3, x4)  =  U1_gg(x4)
memberA_out_gg(x1, x2)  =  memberA_out_gg
subsetB_out_gg(x1, x2)  =  subsetB_out_gg
U3_gg(x1, x2, x3, x4, x5)  =  U3_gg(x5)
U4_gg(x1, x2, x3, x4)  =  U4_gg(x4)
[]  =  []
SUBSETB_IN_GG(x1, x2)  =  SUBSETB_IN_GG(x1, x2)
U2_GG(x1, x2, x3, x4, x5)  =  U2_GG(x2, x3, x4, x5)

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:

U2_GG(T21, T7, T22, T23, memberA_out_gg(T21, T23)) → SUBSETB_IN_GG(T7, .(T22, T23))
SUBSETB_IN_GG(.(T21, T7), .(T22, T23)) → U2_GG(T21, T7, T22, T23, memberA_in_gg(T21, T23))
SUBSETB_IN_GG(.(T66, T7), .(T66, T67)) → SUBSETB_IN_GG(T7, .(T66, T67))

The TRS R consists of the following rules:

memberA_in_gg(T42, .(T43, T44)) → U1_gg(T42, T43, T44, memberA_in_gg(T42, T44))
memberA_in_gg(T52, .(T52, T53)) → memberA_out_gg(T52, .(T52, T53))
U1_gg(T42, T43, T44, memberA_out_gg(T42, T44)) → memberA_out_gg(T42, .(T43, T44))

The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
memberA_in_gg(x1, x2)  =  memberA_in_gg(x1, x2)
U1_gg(x1, x2, x3, x4)  =  U1_gg(x4)
memberA_out_gg(x1, x2)  =  memberA_out_gg
SUBSETB_IN_GG(x1, x2)  =  SUBSETB_IN_GG(x1, x2)
U2_GG(x1, x2, x3, x4, x5)  =  U2_GG(x2, x3, x4, x5)

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:

U2_GG(T7, T22, T23, memberA_out_gg) → SUBSETB_IN_GG(T7, .(T22, T23))
SUBSETB_IN_GG(.(T21, T7), .(T22, T23)) → U2_GG(T7, T22, T23, memberA_in_gg(T21, T23))
SUBSETB_IN_GG(.(T66, T7), .(T66, T67)) → SUBSETB_IN_GG(T7, .(T66, T67))

The TRS R consists of the following rules:

memberA_in_gg(T42, .(T43, T44)) → U1_gg(memberA_in_gg(T42, T44))
memberA_in_gg(T52, .(T52, T53)) → memberA_out_gg
U1_gg(memberA_out_gg) → memberA_out_gg

The set Q consists of the following terms:

memberA_in_gg(x0, x1)
U1_gg(x0)

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:

• SUBSETB_IN_GG(.(T21, T7), .(T22, T23)) → U2_GG(T7, T22, T23, memberA_in_gg(T21, T23))
  The graph contains the following edges 1 > 1, 2 > 2, 2 > 3

• SUBSETB_IN_GG(.(T66, T7), .(T66, T67)) → SUBSETB_IN_GG(T7, .(T66, T67))
  The graph contains the following edges 1 > 1, 2 >= 2

• U2_GG(T7, T22, T23, memberA_out_gg) → SUBSETB_IN_GG(T7, .(T22, T23))
  The graph contains the following edges 1 >= 1

(22) YES