Left Termination proof of /tmp/tmpoPIF

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

0 Prolog

↳1 PrologToDTProblemTransformerProof (⇒, 83 ms)

↳2 TRIPLES

↳3 TriplesToPiDPProof (⇒, 223 ms)

↳4 PiDP

↳5 DependencyGraphProof (⇔, 0 ms)

↳6 PiDP

↳7 PiDPToQDPProof (⇔, 42 ms)

↳8 QDP

↳9 QDPOrderProof (⇔, 493 ms)

↳10 QDP

↳11 DependencyGraphProof (⇔, 1 ms)

↳12 QDP

↳13 QDPOrderProof (⇔, 282 ms)

↳14 QDP

↳15 DependencyGraphProof (⇔, 0 ms)

↳16 QDP

↳17 UsableRulesProof (⇔, 0 ms)

↳18 QDP

↳19 QReductionProof (⇔, 0 ms)

↳20 QDP

↳21 UsableRulesReductionPairsProof (⇔, 0 ms)

↳22 QDP

↳23 PisEmptyProof (⇔, 0 ms)

↳24 YES

(0) Obligation:

Clauses:

p(cons(X, nil)).
p(cons(s(s(X)), cons(Y, Xs))) :- ','(p(cons(X, cons(Y, Xs))), p(cons(s(s(s(s(Y)))), Xs))).
p(cons(0, Xs)) :- p(Xs).

Query: p(g)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: p(T1)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: p(T1), SCOPE: 1, CLAUSE: 0 | p(T1), SCOPE: 1, CLAUSE: 1 | p(T1), SCOPE: 1, CLAUSE: 2\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: true | p(cons(T3, nil)), SCOPE: 1, CLAUSE: 1 | p(cons(T3, nil)), SCOPE: 1, CLAUSE: 2\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: p(T1), SCOPE: 1, CLAUSE: 1 | p(T1), SCOPE: 1, CLAUSE: 2\n\nT1 is ground\nX2 is free\np(T1) !=? p(cons(X2, nil))", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: p(cons(T3, nil)), SCOPE: 1, CLAUSE: 1 | p(cons(T3, nil)), SCOPE: 1, CLAUSE: 2\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: p(cons(T3, nil)), SCOPE: 1, CLAUSE: 2\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: p(nil)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: p(nil), SCOPE: 2, CLAUSE: 0 | p(nil), SCOPE: 2, CLAUSE: 1 | p(nil), SCOPE: 2, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: p(nil), SCOPE: 2, CLAUSE: 1 | p(nil), SCOPE: 2, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: p(nil), SCOPE: 2, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(p(cons(T8, cons(T9, T10))), p(cons(s(s(s(s(T9)))), T10))) | p(cons(s(s(T8)), cons(T9, T10))), SCOPE: 1, CLAUSE: 2\n\nT8 is ground\nT9 is ground\nT10 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: p(T1), SCOPE: 1, CLAUSE: 2\n\nT1 is ground\nX2 is free\nX16 is free\nX17 is free\nX18 is free\np(T1) !=? p(cons(X2, nil))\np(T1) !=? p(cons(s(s(X16)), cons(X17, X18)))", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: ','(p(cons(T8, cons(T9, T10))), p(cons(s(s(s(s(T9)))), T10))), SCOPE: 3, CLAUSE: 0 | ','(p(cons(T8, cons(T9, T10))), p(cons(s(s(s(s(T9)))), T10))), SCOPE: 3, CLAUSE: 1 | ','(p(cons(T8, cons(T9, T10))), p(cons(s(s(s(s(T9)))), T10))), SCOPE: 3, CLAUSE: 2 | ?_3 | p(cons(s(s(T8)), cons(T9, T10))), SCOPE: 1, CLAUSE: 2\n\nT8 is ground\nT9 is ground\nT10 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: ','(p(cons(T8, cons(T9, T10))), p(cons(s(s(s(s(T9)))), T10))), SCOPE: 3, CLAUSE: 1 | ','(p(cons(T8, cons(T9, T10))), p(cons(s(s(s(s(T9)))), T10))), SCOPE: 3, CLAUSE: 2 | ?_3 | p(cons(s(s(T8)), cons(T9, T10))), SCOPE: 1, CLAUSE: 2\n\nT8 is ground\nT9 is ground\nT10 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: ','(p(cons(T8, cons(T9, T10))), p(cons(s(s(s(s(T9)))), T10))), SCOPE: 3, CLAUSE: 1\n\nT8 is ground\nT9 is ground\nT10 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: ','(p(cons(T8, cons(T9, T10))), p(cons(s(s(s(s(T9)))), T10))), SCOPE: 3, CLAUSE: 2 | ?_3 | p(cons(s(s(T8)), cons(T9, T10))), SCOPE: 1, CLAUSE: 2\n\nT8 is ground\nT9 is ground\nT10 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: ','(','(p(cons(T24, cons(T25, T26))), p(cons(s(s(s(s(T25)))), T26))), p(cons(s(s(s(s(T25)))), T26)))\n\nT24 is ground\nT25 is ground\nT26 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: p(cons(T24, cons(T25, T26)))\n\nT24 is ground\nT25 is ground\nT26 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: ','(p(cons(s(s(s(s(T25)))), T26)), p(cons(s(s(s(s(T25)))), T26)))\n\nT25 is ground\nT26 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: p(cons(s(s(s(s(T25)))), T26))\n\nT25 is ground\nT26 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: p(cons(s(s(s(s(T25)))), T26))\n\nT25 is ground\nT26 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: ','(p(cons(T8, cons(T9, T10))), p(cons(s(s(s(s(T9)))), T10))), SCOPE: 3, CLAUSE: 2\n\nT8 is ground\nT9 is ground\nT10 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: ?_3 | p(cons(s(s(T8)), cons(T9, T10))), SCOPE: 1, CLAUSE: 2\n\nT8 is ground\nT9 is ground\nT10 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: ','(p(cons(T45, T46)), p(cons(s(s(s(s(T45)))), T46)))\n\nT45 is ground\nT46 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="29: p(cons(T45, T46))\n\nT45 is ground\nT46 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
30 [label="30: p(cons(s(s(s(s(T45)))), T46))\n\nT45 is ground\nT46 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
31 [label="31: p(cons(s(s(T8)), cons(T9, T10))), SCOPE: 1, CLAUSE: 2\n\nT8 is ground\nT9 is ground\nT10 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
32 [label="32: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
33 [label="33: p(T54)\n\nT54 is ground\nX2 is free\np(cons(0, T54)) !=? p(cons(X2, nil))", fontsize=16, style=filled, fillcolor=lightyellow];
34 [label="34: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
35 [label="35: p(T54), SCOPE: 4, CLAUSE: 0 | p(T54), SCOPE: 4, CLAUSE: 1 | p(T54), SCOPE: 4, CLAUSE: 2\n\nT54 is ground\nX2 is free\np(cons(0, T54)) !=? p(cons(X2, nil))", fontsize=16, style=filled, fillcolor=lightyellow];
36 [label="36: p(T54), SCOPE: 4, CLAUSE: 0\n\nT54 is ground\nX2 is free\np(cons(0, T54)) !=? p(cons(X2, nil))", fontsize=16, style=filled, fillcolor=lightyellow];
37 [label="37: p(T54), SCOPE: 4, CLAUSE: 1 | p(T54), SCOPE: 4, CLAUSE: 2\n\nT54 is ground\nX2 is free\np(cons(0, T54)) !=? p(cons(X2, nil))", fontsize=16, style=filled, fillcolor=lightyellow];
38 [label="38: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
39 [label="39: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
40 [label="40: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
41 [label="41: p(T54), SCOPE: 4, CLAUSE: 1\n\nT54 is ground\nX2 is free\np(cons(0, T54)) !=? p(cons(X2, nil))", fontsize=16, style=filled, fillcolor=lightyellow];
42 [label="42: p(T54), SCOPE: 4, CLAUSE: 2\n\nT54 is ground\nX2 is free\np(cons(0, T54)) !=? p(cons(X2, nil))", fontsize=16, style=filled, fillcolor=lightyellow];
43 [label="43: ','(p(cons(T72, cons(T73, T74))), p(cons(s(s(s(s(T73)))), T74)))\n\nT72 is ground\nT73 is ground\nT74 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
44 [label="44: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
45 [label="45: p(cons(T72, cons(T73, T74)))\n\nT72 is ground\nT73 is ground\nT74 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
46 [label="46: p(cons(s(s(s(s(T73)))), T74))\n\nT73 is ground\nT74 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
47 [label="47: p(T83)\n\nT83 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
48 [label="48: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
49 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 49 [arrowhead = none ];
49 -> 2;

50 [label="EVAL with clause\np(cons(X2, nil)).\nand substitutionX2 -> T3,\nT1 -> cons(T3, nil)", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 50 [arrowhead = none ];
50 -> 3;

51 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 51 [arrowhead = none ];
51 -> 4;

52 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 52 [arrowhead = none ];
52 -> 5;

53 [label="EVAL with clause\np(cons(s(s(X16)), cons(X17, X18))) :- ','(p(cons(X16, cons(X17, X18))), p(cons(s(s(s(s(X17)))), X18))).\nand substitutionX16 -> T8,\nX17 -> T9,\nX18 -> T10,\nT1 -> cons(s(s(T8)), cons(T9, T10))", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 53 [arrowhead = none ];
53 -> 13;

54 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 54 [arrowhead = none ];
54 -> 14;

55 [label="BACKTRACK\nfor clause: p(cons(s(s(X)), cons(Y, Xs))) :- ','(p(cons(X, cons(Y, Xs))), p(cons(s(s(s(s(Y)))), Xs)))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 55 [arrowhead = none ];
55 -> 6;

56 [label="EVAL with clause\np(cons(0, X7)) :- p(X7).\nand substitutionT3 -> 0,\nX7 -> nil", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 56 [arrowhead = none ];
56 -> 7;

57 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 57 [arrowhead = none ];
57 -> 8;

58 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 58 [arrowhead = none ];
58 -> 9;

59 [label="BACKTRACK\nfor clause: p(cons(X, nil))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 59 [arrowhead = none ];
59 -> 10;

60 [label="BACKTRACK\nfor clause: p(cons(s(s(X)), cons(Y, Xs))) :- ','(p(cons(X, cons(Y, Xs))), p(cons(s(s(s(s(Y)))), Xs)))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 60 [arrowhead = none ];
60 -> 11;

61 [label="BACKTRACK\nfor clause: p(cons(0, Xs)) :- p(Xs)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 61 [arrowhead = none ];
61 -> 12;

62 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
13 -> 62 [arrowhead = none ];
62 -> 15;

63 [label="EVAL with clause\np(cons(0, X58)) :- p(X58).\nand substitutionX58 -> T54,\nT1 -> cons(0, T54)", fontsize=14, style = filled, fillcolor = lightblue];
14 -> 63 [arrowhead = none ];
63 -> 33;

64 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
14 -> 64 [arrowhead = none ];
64 -> 34;

65 [label="BACKTRACK\nfor clause: p(cons(X, nil))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 65 [arrowhead = none ];
65 -> 16;

66 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
16 -> 66 [arrowhead = none ];
66 -> 17;

67 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
16 -> 67 [arrowhead = none ];
67 -> 18;

68 [label="EVAL with clause\np(cons(s(s(X32)), cons(X33, X34))) :- ','(p(cons(X32, cons(X33, X34))), p(cons(s(s(s(s(X33)))), X34))).\nand substitutionX32 -> T24,\nT8 -> s(s(T24)),\nT9 -> T25,\nX33 -> T25,\nT10 -> T26,\nX34 -> T26", fontsize=14, style = filled, fillcolor = lightblue];
17 -> 68 [arrowhead = none ];
68 -> 19;

69 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 69 [arrowhead = none ];
69 -> 20;

70 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
18 -> 70 [arrowhead = none ];
70 -> 25;

71 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
18 -> 71 [arrowhead = none ];
71 -> 26;

72 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
19 -> 72 [arrowhead = none ];
72 -> 21;

73 [label="SPLIT 2\nnew knowledge:\nT24 is ground\nT25 is ground\nT26 is ground", fontsize=14, style = filled, fillcolor = violet];
19 -> 73 [arrowhead = none ];
73 -> 22;

74 [label="INSTANCE with matching:\nT1 -> cons(T24, cons(T25, T26))", fontsize=14, style = filled, fillcolor = lightgrey];
21 -> 74 [arrowhead = none , style=dashed];
74 -> 1[style=dashed];

75 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
22 -> 75 [arrowhead = none ];
75 -> 23;

76 [label="SPLIT 2\nnew knowledge:\nT25 is ground\nT26 is ground", fontsize=14, style = filled, fillcolor = violet];
22 -> 76 [arrowhead = none ];
76 -> 24;

77 [label="INSTANCE with matching:\nT1 -> cons(s(s(s(s(T25)))), T26)", fontsize=14, style = filled, fillcolor = lightgrey];
23 -> 77 [arrowhead = none , style=dashed];
77 -> 1[style=dashed];

78 [label="INSTANCE with matching:\nT1 -> cons(s(s(s(s(T25)))), T26)", fontsize=14, style = filled, fillcolor = lightgrey];
24 -> 78 [arrowhead = none , style=dashed];
78 -> 1[style=dashed];

79 [label="EVAL with clause\np(cons(0, X49)) :- p(X49).\nand substitutionT8 -> 0,\nT9 -> T45,\nT10 -> T46,\nX49 -> cons(T45, T46)", fontsize=14, style = filled, fillcolor = lightblue];
25 -> 79 [arrowhead = none ];
79 -> 27;

80 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
25 -> 80 [arrowhead = none ];
80 -> 28;

81 [label="FAILURE", fontsize=14, style = filled, fillcolor = lightgreen];
26 -> 81 [arrowhead = none ];
81 -> 31;

82 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
27 -> 82 [arrowhead = none ];
82 -> 29;

83 [label="SPLIT 2\nnew knowledge:\nT45 is ground\nT46 is ground", fontsize=14, style = filled, fillcolor = violet];
27 -> 83 [arrowhead = none ];
83 -> 30;

84 [label="INSTANCE with matching:\nT1 -> cons(T45, T46)", fontsize=14, style = filled, fillcolor = lightgrey];
29 -> 84 [arrowhead = none , style=dashed];
84 -> 1[style=dashed];

85 [label="INSTANCE with matching:\nT1 -> cons(s(s(s(s(T45)))), T46)", fontsize=14, style = filled, fillcolor = lightgrey];
30 -> 85 [arrowhead = none , style=dashed];
85 -> 1[style=dashed];

86 [label="BACKTRACK\nfor clause: p(cons(0, Xs)) :- p(Xs)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
31 -> 86 [arrowhead = none ];
86 -> 32;

87 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
33 -> 87 [arrowhead = none ];
87 -> 35;

88 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
35 -> 88 [arrowhead = none ];
88 -> 36;

89 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
35 -> 89 [arrowhead = none ];
89 -> 37;

90 [label="EVAL with clause\np(cons(X63, nil)).\nand substitutionX63 -> T59,\nT54 -> cons(T59, nil)", fontsize=14, style = filled, fillcolor = lightblue];
36 -> 90 [arrowhead = none ];
90 -> 38;

91 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
36 -> 91 [arrowhead = none ];
91 -> 39;

92 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
37 -> 92 [arrowhead = none ];
92 -> 41;

93 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
37 -> 93 [arrowhead = none ];
93 -> 42;

94 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
38 -> 94 [arrowhead = none ];
94 -> 40;

95 [label="EVAL with clause\np(cons(s(s(X76)), cons(X77, X78))) :- ','(p(cons(X76, cons(X77, X78))), p(cons(s(s(s(s(X77)))), X78))).\nand substitutionX76 -> T72,\nX77 -> T73,\nX78 -> T74,\nT54 -> cons(s(s(T72)), cons(T73, T74))", fontsize=14, style = filled, fillcolor = lightblue];
41 -> 95 [arrowhead = none ];
95 -> 43;

96 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
41 -> 96 [arrowhead = none ];
96 -> 44;

97 [label="EVAL with clause\np(cons(0, X87)) :- p(X87).\nand substitutionX87 -> T83,\nT54 -> cons(0, T83)", fontsize=14, style = filled, fillcolor = lightblue];
42 -> 97 [arrowhead = none ];
97 -> 47;

98 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
42 -> 98 [arrowhead = none ];
98 -> 48;

99 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
43 -> 99 [arrowhead = none ];
99 -> 45;

100 [label="SPLIT 2\nnew knowledge:\nT72 is ground\nT73 is ground\nT74 is ground", fontsize=14, style = filled, fillcolor = violet];
43 -> 100 [arrowhead = none ];
100 -> 46;

101 [label="INSTANCE with matching:\nT1 -> cons(T72, cons(T73, T74))", fontsize=14, style = filled, fillcolor = lightgrey];
45 -> 101 [arrowhead = none , style=dashed];
101 -> 1[style=dashed];

102 [label="INSTANCE with matching:\nT1 -> cons(s(s(s(s(T73)))), T74)", fontsize=14, style = filled, fillcolor = lightgrey];
46 -> 102 [arrowhead = none , style=dashed];
102 -> 1[style=dashed];

103 [label="INSTANCE with matching:\nT1 -> T83", fontsize=14, style = filled, fillcolor = lightgrey];
47 -> 103 [arrowhead = none , style=dashed];
103 -> 1[style=dashed];

}

(2) Obligation:

Triples:

pA(cons(s(s(s(s(X1)))), cons(X2, X3))) :- pA(cons(X1, cons(X2, X3))).
pA(cons(s(s(s(s(X1)))), cons(X2, X3))) :- ','(pcA(cons(X1, cons(X2, X3))), pA(cons(s(s(s(s(X2)))), X3))).
pA(cons(s(s(s(s(X1)))), cons(X2, X3))) :- ','(pcA(cons(X1, cons(X2, X3))), ','(pcA(cons(s(s(s(s(X2)))), X3)), pA(cons(s(s(s(s(X2)))), X3)))).
pA(cons(s(s(0)), cons(X1, X2))) :- pA(cons(X1, X2)).
pA(cons(s(s(0)), cons(X1, X2))) :- ','(pcA(cons(X1, X2)), pA(cons(s(s(s(s(X1)))), X2))).
pA(cons(0, cons(s(s(X1)), cons(X2, X3)))) :- pA(cons(X1, cons(X2, X3))).
pA(cons(0, cons(s(s(X1)), cons(X2, X3)))) :- ','(pcA(cons(X1, cons(X2, X3))), pA(cons(s(s(s(s(X2)))), X3))).
pA(cons(0, cons(0, X1))) :- pA(X1).

Clauses:

pcA(cons(X1, nil)).
pcA(cons(s(s(s(s(X1)))), cons(X2, X3))) :- ','(pcA(cons(X1, cons(X2, X3))), ','(pcA(cons(s(s(s(s(X2)))), X3)), pcA(cons(s(s(s(s(X2)))), X3)))).
pcA(cons(s(s(0)), cons(X1, X2))) :- ','(pcA(cons(X1, X2)), pcA(cons(s(s(s(s(X1)))), X2))).
pcA(cons(0, cons(X1, nil))).
pcA(cons(0, cons(s(s(X1)), cons(X2, X3)))) :- ','(pcA(cons(X1, cons(X2, X3))), pcA(cons(s(s(s(s(X2)))), X3))).
pcA(cons(0, cons(0, X1))) :- pcA(X1).

Afs:

pA(x1) = pA(x1)

(3) TriplesToPiDPProof (SOUND transformation)

We use the technique of [DT09]. With regard to the inferred argument filtering the predicates were used in the following modes:
pA_in: (b)
pcA_in: (b)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

PA_IN_G(cons(s(s(s(s(X1)))), cons(X2, X3))) → U1_G(X1, X2, X3, pA_in_g(cons(X1, cons(X2, X3))))
PA_IN_G(cons(s(s(s(s(X1)))), cons(X2, X3))) → PA_IN_G(cons(X1, cons(X2, X3)))
PA_IN_G(cons(s(s(s(s(X1)))), cons(X2, X3))) → U2_G(X1, X2, X3, pcA_in_g(cons(X1, cons(X2, X3))))
U2_G(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → U3_G(X1, X2, X3, pA_in_g(cons(s(s(s(s(X2)))), X3)))
U2_G(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → PA_IN_G(cons(s(s(s(s(X2)))), X3))
PA_IN_G(cons(s(s(0)), cons(X1, X2))) → U6_G(X1, X2, pA_in_g(cons(X1, X2)))
PA_IN_G(cons(s(s(0)), cons(X1, X2))) → PA_IN_G(cons(X1, X2))
PA_IN_G(cons(s(s(0)), cons(X1, X2))) → U7_G(X1, X2, pcA_in_g(cons(X1, X2)))
U7_G(X1, X2, pcA_out_g(cons(X1, X2))) → U8_G(X1, X2, pA_in_g(cons(s(s(s(s(X1)))), X2)))
U7_G(X1, X2, pcA_out_g(cons(X1, X2))) → PA_IN_G(cons(s(s(s(s(X1)))), X2))
PA_IN_G(cons(0, cons(s(s(X1)), cons(X2, X3)))) → U9_G(X1, X2, X3, pA_in_g(cons(X1, cons(X2, X3))))
PA_IN_G(cons(0, cons(s(s(X1)), cons(X2, X3)))) → PA_IN_G(cons(X1, cons(X2, X3)))
PA_IN_G(cons(0, cons(s(s(X1)), cons(X2, X3)))) → U10_G(X1, X2, X3, pcA_in_g(cons(X1, cons(X2, X3))))
U10_G(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → U11_G(X1, X2, X3, pA_in_g(cons(s(s(s(s(X2)))), X3)))
U10_G(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → PA_IN_G(cons(s(s(s(s(X2)))), X3))
PA_IN_G(cons(0, cons(0, X1))) → U12_G(X1, pA_in_g(X1))
PA_IN_G(cons(0, cons(0, X1))) → PA_IN_G(X1)
U2_G(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → U4_G(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U4_G(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → U5_G(X1, X2, X3, pA_in_g(cons(s(s(s(s(X2)))), X3)))
U4_G(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → PA_IN_G(cons(s(s(s(s(X2)))), X3))

The TRS R consists of the following rules:

pcA_in_g(cons(X1, nil)) → pcA_out_g(cons(X1, nil))
pcA_in_g(cons(s(s(s(s(X1)))), cons(X2, X3))) → U14_g(X1, X2, X3, pcA_in_g(cons(X1, cons(X2, X3))))
pcA_in_g(cons(s(s(0)), cons(X1, X2))) → U17_g(X1, X2, pcA_in_g(cons(X1, X2)))
pcA_in_g(cons(0, cons(X1, nil))) → pcA_out_g(cons(0, cons(X1, nil)))
pcA_in_g(cons(0, cons(s(s(X1)), cons(X2, X3)))) → U19_g(X1, X2, X3, pcA_in_g(cons(X1, cons(X2, X3))))
pcA_in_g(cons(0, cons(0, X1))) → U21_g(X1, pcA_in_g(X1))
U21_g(X1, pcA_out_g(X1)) → pcA_out_g(cons(0, cons(0, X1)))
U19_g(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → U20_g(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U20_g(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → pcA_out_g(cons(0, cons(s(s(X1)), cons(X2, X3))))
U17_g(X1, X2, pcA_out_g(cons(X1, X2))) → U18_g(X1, X2, pcA_in_g(cons(s(s(s(s(X1)))), X2)))
U18_g(X1, X2, pcA_out_g(cons(s(s(s(s(X1)))), X2))) → pcA_out_g(cons(s(s(0)), cons(X1, X2)))
U14_g(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → U15_g(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U15_g(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → U16_g(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U16_g(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → pcA_out_g(cons(s(s(s(s(X1)))), cons(X2, X3)))

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

Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES

(4) Obligation:

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

PA_IN_G(cons(s(s(s(s(X1)))), cons(X2, X3))) → U1_G(X1, X2, X3, pA_in_g(cons(X1, cons(X2, X3))))
PA_IN_G(cons(s(s(s(s(X1)))), cons(X2, X3))) → PA_IN_G(cons(X1, cons(X2, X3)))
PA_IN_G(cons(s(s(s(s(X1)))), cons(X2, X3))) → U2_G(X1, X2, X3, pcA_in_g(cons(X1, cons(X2, X3))))
U2_G(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → U3_G(X1, X2, X3, pA_in_g(cons(s(s(s(s(X2)))), X3)))
U2_G(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → PA_IN_G(cons(s(s(s(s(X2)))), X3))
PA_IN_G(cons(s(s(0)), cons(X1, X2))) → U6_G(X1, X2, pA_in_g(cons(X1, X2)))
PA_IN_G(cons(s(s(0)), cons(X1, X2))) → PA_IN_G(cons(X1, X2))
PA_IN_G(cons(s(s(0)), cons(X1, X2))) → U7_G(X1, X2, pcA_in_g(cons(X1, X2)))
U7_G(X1, X2, pcA_out_g(cons(X1, X2))) → U8_G(X1, X2, pA_in_g(cons(s(s(s(s(X1)))), X2)))
U7_G(X1, X2, pcA_out_g(cons(X1, X2))) → PA_IN_G(cons(s(s(s(s(X1)))), X2))
PA_IN_G(cons(0, cons(s(s(X1)), cons(X2, X3)))) → U9_G(X1, X2, X3, pA_in_g(cons(X1, cons(X2, X3))))
PA_IN_G(cons(0, cons(s(s(X1)), cons(X2, X3)))) → PA_IN_G(cons(X1, cons(X2, X3)))
PA_IN_G(cons(0, cons(s(s(X1)), cons(X2, X3)))) → U10_G(X1, X2, X3, pcA_in_g(cons(X1, cons(X2, X3))))
U10_G(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → U11_G(X1, X2, X3, pA_in_g(cons(s(s(s(s(X2)))), X3)))
U10_G(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → PA_IN_G(cons(s(s(s(s(X2)))), X3))
PA_IN_G(cons(0, cons(0, X1))) → U12_G(X1, pA_in_g(X1))
PA_IN_G(cons(0, cons(0, X1))) → PA_IN_G(X1)
U2_G(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → U4_G(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U4_G(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → U5_G(X1, X2, X3, pA_in_g(cons(s(s(s(s(X2)))), X3)))
U4_G(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → PA_IN_G(cons(s(s(s(s(X2)))), X3))

The TRS R consists of the following rules:

pcA_in_g(cons(X1, nil)) → pcA_out_g(cons(X1, nil))
pcA_in_g(cons(s(s(s(s(X1)))), cons(X2, X3))) → U14_g(X1, X2, X3, pcA_in_g(cons(X1, cons(X2, X3))))
pcA_in_g(cons(s(s(0)), cons(X1, X2))) → U17_g(X1, X2, pcA_in_g(cons(X1, X2)))
pcA_in_g(cons(0, cons(X1, nil))) → pcA_out_g(cons(0, cons(X1, nil)))
pcA_in_g(cons(0, cons(s(s(X1)), cons(X2, X3)))) → U19_g(X1, X2, X3, pcA_in_g(cons(X1, cons(X2, X3))))
pcA_in_g(cons(0, cons(0, X1))) → U21_g(X1, pcA_in_g(X1))
U21_g(X1, pcA_out_g(X1)) → pcA_out_g(cons(0, cons(0, X1)))
U19_g(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → U20_g(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U20_g(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → pcA_out_g(cons(0, cons(s(s(X1)), cons(X2, X3))))
U17_g(X1, X2, pcA_out_g(cons(X1, X2))) → U18_g(X1, X2, pcA_in_g(cons(s(s(s(s(X1)))), X2)))
U18_g(X1, X2, pcA_out_g(cons(s(s(s(s(X1)))), X2))) → pcA_out_g(cons(s(s(0)), cons(X1, X2)))
U14_g(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → U15_g(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U15_g(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → U16_g(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U16_g(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → pcA_out_g(cons(s(s(s(s(X1)))), cons(X2, X3)))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 8 less nodes.

(6) Obligation:

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

PA_IN_G(cons(s(s(s(s(X1)))), cons(X2, X3))) → U2_G(X1, X2, X3, pcA_in_g(cons(X1, cons(X2, X3))))
U2_G(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → PA_IN_G(cons(s(s(s(s(X2)))), X3))
PA_IN_G(cons(s(s(s(s(X1)))), cons(X2, X3))) → PA_IN_G(cons(X1, cons(X2, X3)))
PA_IN_G(cons(s(s(0)), cons(X1, X2))) → PA_IN_G(cons(X1, X2))
PA_IN_G(cons(s(s(0)), cons(X1, X2))) → U7_G(X1, X2, pcA_in_g(cons(X1, X2)))
U7_G(X1, X2, pcA_out_g(cons(X1, X2))) → PA_IN_G(cons(s(s(s(s(X1)))), X2))
PA_IN_G(cons(0, cons(s(s(X1)), cons(X2, X3)))) → PA_IN_G(cons(X1, cons(X2, X3)))
PA_IN_G(cons(0, cons(s(s(X1)), cons(X2, X3)))) → U10_G(X1, X2, X3, pcA_in_g(cons(X1, cons(X2, X3))))
U10_G(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → PA_IN_G(cons(s(s(s(s(X2)))), X3))
PA_IN_G(cons(0, cons(0, X1))) → PA_IN_G(X1)
U2_G(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → U4_G(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U4_G(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → PA_IN_G(cons(s(s(s(s(X2)))), X3))

The TRS R consists of the following rules:

pcA_in_g(cons(X1, nil)) → pcA_out_g(cons(X1, nil))
pcA_in_g(cons(s(s(s(s(X1)))), cons(X2, X3))) → U14_g(X1, X2, X3, pcA_in_g(cons(X1, cons(X2, X3))))
pcA_in_g(cons(s(s(0)), cons(X1, X2))) → U17_g(X1, X2, pcA_in_g(cons(X1, X2)))
pcA_in_g(cons(0, cons(X1, nil))) → pcA_out_g(cons(0, cons(X1, nil)))
pcA_in_g(cons(0, cons(s(s(X1)), cons(X2, X3)))) → U19_g(X1, X2, X3, pcA_in_g(cons(X1, cons(X2, X3))))
pcA_in_g(cons(0, cons(0, X1))) → U21_g(X1, pcA_in_g(X1))
U21_g(X1, pcA_out_g(X1)) → pcA_out_g(cons(0, cons(0, X1)))
U19_g(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → U20_g(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U20_g(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → pcA_out_g(cons(0, cons(s(s(X1)), cons(X2, X3))))
U17_g(X1, X2, pcA_out_g(cons(X1, X2))) → U18_g(X1, X2, pcA_in_g(cons(s(s(s(s(X1)))), X2)))
U18_g(X1, X2, pcA_out_g(cons(s(s(s(s(X1)))), X2))) → pcA_out_g(cons(s(s(0)), cons(X1, X2)))
U14_g(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → U15_g(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U15_g(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → U16_g(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U16_g(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → pcA_out_g(cons(s(s(s(s(X1)))), cons(X2, X3)))

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

(7) PiDPToQDPProof (EQUIVALENT transformation)

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

(8) Obligation:

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

PA_IN_G(cons(s(s(s(s(X1)))), cons(X2, X3))) → U2_G(X1, X2, X3, pcA_in_g(cons(X1, cons(X2, X3))))
U2_G(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → PA_IN_G(cons(s(s(s(s(X2)))), X3))
PA_IN_G(cons(s(s(s(s(X1)))), cons(X2, X3))) → PA_IN_G(cons(X1, cons(X2, X3)))
PA_IN_G(cons(s(s(0)), cons(X1, X2))) → PA_IN_G(cons(X1, X2))
PA_IN_G(cons(s(s(0)), cons(X1, X2))) → U7_G(X1, X2, pcA_in_g(cons(X1, X2)))
U7_G(X1, X2, pcA_out_g(cons(X1, X2))) → PA_IN_G(cons(s(s(s(s(X1)))), X2))
PA_IN_G(cons(0, cons(s(s(X1)), cons(X2, X3)))) → PA_IN_G(cons(X1, cons(X2, X3)))
PA_IN_G(cons(0, cons(s(s(X1)), cons(X2, X3)))) → U10_G(X1, X2, X3, pcA_in_g(cons(X1, cons(X2, X3))))
U10_G(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → PA_IN_G(cons(s(s(s(s(X2)))), X3))
PA_IN_G(cons(0, cons(0, X1))) → PA_IN_G(X1)
U2_G(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → U4_G(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U4_G(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → PA_IN_G(cons(s(s(s(s(X2)))), X3))

The TRS R consists of the following rules:

pcA_in_g(cons(X1, nil)) → pcA_out_g(cons(X1, nil))
pcA_in_g(cons(s(s(s(s(X1)))), cons(X2, X3))) → U14_g(X1, X2, X3, pcA_in_g(cons(X1, cons(X2, X3))))
pcA_in_g(cons(s(s(0)), cons(X1, X2))) → U17_g(X1, X2, pcA_in_g(cons(X1, X2)))
pcA_in_g(cons(0, cons(X1, nil))) → pcA_out_g(cons(0, cons(X1, nil)))
pcA_in_g(cons(0, cons(s(s(X1)), cons(X2, X3)))) → U19_g(X1, X2, X3, pcA_in_g(cons(X1, cons(X2, X3))))
pcA_in_g(cons(0, cons(0, X1))) → U21_g(X1, pcA_in_g(X1))
U21_g(X1, pcA_out_g(X1)) → pcA_out_g(cons(0, cons(0, X1)))
U19_g(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → U20_g(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U20_g(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → pcA_out_g(cons(0, cons(s(s(X1)), cons(X2, X3))))
U17_g(X1, X2, pcA_out_g(cons(X1, X2))) → U18_g(X1, X2, pcA_in_g(cons(s(s(s(s(X1)))), X2)))
U18_g(X1, X2, pcA_out_g(cons(s(s(s(s(X1)))), X2))) → pcA_out_g(cons(s(s(0)), cons(X1, X2)))
U14_g(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → U15_g(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U15_g(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → U16_g(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U16_g(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → pcA_out_g(cons(s(s(s(s(X1)))), cons(X2, X3)))

The set Q consists of the following terms:

pcA_in_g(x0)
U21_g(x0, x1)
U19_g(x0, x1, x2, x3)
U20_g(x0, x1, x2, x3)
U17_g(x0, x1, x2)
U18_g(x0, x1, x2)
U14_g(x0, x1, x2, x3)
U15_g(x0, x1, x2, x3)
U16_g(x0, x1, x2, x3)

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

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

PA_IN_G(cons(s(s(0)), cons(X1, X2))) → PA_IN_G(cons(X1, X2))
PA_IN_G(cons(s(s(0)), cons(X1, X2))) → U7_G(X1, X2, pcA_in_g(cons(X1, X2)))
PA_IN_G(cons(0, cons(s(s(X1)), cons(X2, X3)))) → PA_IN_G(cons(X1, cons(X2, X3)))
PA_IN_G(cons(0, cons(s(s(X1)), cons(X2, X3)))) → U10_G(X1, X2, X3, pcA_in_g(cons(X1, cons(X2, X3))))
PA_IN_G(cons(0, cons(0, X1))) → PA_IN_G(X1)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 1
POL(PA_IN_G(x₁)) = x₁
POL(U10_G(x₁, x₂, x₃, x₄)) = x₂ + x₃
POL(U14_g(x₁, x₂, x₃, x₄)) = 0
POL(U15_g(x₁, x₂, x₃, x₄)) = 0
POL(U16_g(x₁, x₂, x₃, x₄)) = 0
POL(U17_g(x₁, x₂, x₃)) = 0
POL(U18_g(x₁, x₂, x₃)) = 0
POL(U19_g(x₁, x₂, x₃, x₄)) = 0
POL(U20_g(x₁, x₂, x₃, x₄)) = 0
POL(U21_g(x₁, x₂)) = 0
POL(U2_G(x₁, x₂, x₃, x₄)) = x₂ + x₃
POL(U4_G(x₁, x₂, x₃, x₄)) = x₂ + x₃
POL(U7_G(x₁, x₂, x₃)) = x₁ + x₂
POL(cons(x₁, x₂)) = x₁ + x₂
POL(nil) = 1
POL(pcA_in_g(x₁)) = 1 + x₁
POL(pcA_out_g(x₁)) = 1
POL(s(x₁)) = x₁

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none

(10) Obligation:

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

PA_IN_G(cons(s(s(s(s(X1)))), cons(X2, X3))) → U2_G(X1, X2, X3, pcA_in_g(cons(X1, cons(X2, X3))))
U2_G(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → PA_IN_G(cons(s(s(s(s(X2)))), X3))
PA_IN_G(cons(s(s(s(s(X1)))), cons(X2, X3))) → PA_IN_G(cons(X1, cons(X2, X3)))
U7_G(X1, X2, pcA_out_g(cons(X1, X2))) → PA_IN_G(cons(s(s(s(s(X1)))), X2))
U10_G(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → PA_IN_G(cons(s(s(s(s(X2)))), X3))
U2_G(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → U4_G(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U4_G(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → PA_IN_G(cons(s(s(s(s(X2)))), X3))

The TRS R consists of the following rules:

pcA_in_g(cons(X1, nil)) → pcA_out_g(cons(X1, nil))
pcA_in_g(cons(s(s(s(s(X1)))), cons(X2, X3))) → U14_g(X1, X2, X3, pcA_in_g(cons(X1, cons(X2, X3))))
pcA_in_g(cons(s(s(0)), cons(X1, X2))) → U17_g(X1, X2, pcA_in_g(cons(X1, X2)))
pcA_in_g(cons(0, cons(X1, nil))) → pcA_out_g(cons(0, cons(X1, nil)))
pcA_in_g(cons(0, cons(s(s(X1)), cons(X2, X3)))) → U19_g(X1, X2, X3, pcA_in_g(cons(X1, cons(X2, X3))))
pcA_in_g(cons(0, cons(0, X1))) → U21_g(X1, pcA_in_g(X1))
U21_g(X1, pcA_out_g(X1)) → pcA_out_g(cons(0, cons(0, X1)))
U19_g(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → U20_g(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U20_g(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → pcA_out_g(cons(0, cons(s(s(X1)), cons(X2, X3))))
U17_g(X1, X2, pcA_out_g(cons(X1, X2))) → U18_g(X1, X2, pcA_in_g(cons(s(s(s(s(X1)))), X2)))
U18_g(X1, X2, pcA_out_g(cons(s(s(s(s(X1)))), X2))) → pcA_out_g(cons(s(s(0)), cons(X1, X2)))
U14_g(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → U15_g(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U15_g(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → U16_g(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U16_g(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → pcA_out_g(cons(s(s(s(s(X1)))), cons(X2, X3)))

The set Q consists of the following terms:

pcA_in_g(x0)
U21_g(x0, x1)
U19_g(x0, x1, x2, x3)
U20_g(x0, x1, x2, x3)
U17_g(x0, x1, x2)
U18_g(x0, x1, x2)
U14_g(x0, x1, x2, x3)
U15_g(x0, x1, x2, x3)
U16_g(x0, x1, x2, x3)

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

(11) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(12) Obligation:

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

U2_G(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → PA_IN_G(cons(s(s(s(s(X2)))), X3))
PA_IN_G(cons(s(s(s(s(X1)))), cons(X2, X3))) → U2_G(X1, X2, X3, pcA_in_g(cons(X1, cons(X2, X3))))
U2_G(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → U4_G(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U4_G(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → PA_IN_G(cons(s(s(s(s(X2)))), X3))
PA_IN_G(cons(s(s(s(s(X1)))), cons(X2, X3))) → PA_IN_G(cons(X1, cons(X2, X3)))

The TRS R consists of the following rules:

pcA_in_g(cons(X1, nil)) → pcA_out_g(cons(X1, nil))
pcA_in_g(cons(s(s(s(s(X1)))), cons(X2, X3))) → U14_g(X1, X2, X3, pcA_in_g(cons(X1, cons(X2, X3))))
pcA_in_g(cons(s(s(0)), cons(X1, X2))) → U17_g(X1, X2, pcA_in_g(cons(X1, X2)))
pcA_in_g(cons(0, cons(X1, nil))) → pcA_out_g(cons(0, cons(X1, nil)))
pcA_in_g(cons(0, cons(s(s(X1)), cons(X2, X3)))) → U19_g(X1, X2, X3, pcA_in_g(cons(X1, cons(X2, X3))))
pcA_in_g(cons(0, cons(0, X1))) → U21_g(X1, pcA_in_g(X1))
U21_g(X1, pcA_out_g(X1)) → pcA_out_g(cons(0, cons(0, X1)))
U19_g(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → U20_g(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U20_g(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → pcA_out_g(cons(0, cons(s(s(X1)), cons(X2, X3))))
U17_g(X1, X2, pcA_out_g(cons(X1, X2))) → U18_g(X1, X2, pcA_in_g(cons(s(s(s(s(X1)))), X2)))
U18_g(X1, X2, pcA_out_g(cons(s(s(s(s(X1)))), X2))) → pcA_out_g(cons(s(s(0)), cons(X1, X2)))
U14_g(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → U15_g(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U15_g(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → U16_g(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U16_g(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → pcA_out_g(cons(s(s(s(s(X1)))), cons(X2, X3)))

The set Q consists of the following terms:

pcA_in_g(x0)
U21_g(x0, x1)
U19_g(x0, x1, x2, x3)
U20_g(x0, x1, x2, x3)
U17_g(x0, x1, x2)
U18_g(x0, x1, x2)
U14_g(x0, x1, x2, x3)
U15_g(x0, x1, x2, x3)
U16_g(x0, x1, x2, x3)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

PA_IN_G(cons(s(s(s(s(X1)))), cons(X2, X3))) → U2_G(X1, X2, X3, pcA_in_g(cons(X1, cons(X2, X3))))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0
POL(PA_IN_G(x₁)) = x₁
POL(U14_g(x₁, x₂, x₃, x₄)) = 0
POL(U15_g(x₁, x₂, x₃, x₄)) = 0
POL(U16_g(x₁, x₂, x₃, x₄)) = 0
POL(U17_g(x₁, x₂, x₃)) = 0
POL(U18_g(x₁, x₂, x₃)) = 0
POL(U19_g(x₁, x₂, x₃, x₄)) = 0
POL(U20_g(x₁, x₂, x₃, x₄)) = 0
POL(U21_g(x₁, x₂)) = 0
POL(U2_G(x₁, x₂, x₃, x₄)) = 1 + x₃
POL(U4_G(x₁, x₂, x₃, x₄)) = 1 + x₃
POL(cons(x₁, x₂)) = 1 + x₂
POL(nil) = 0
POL(pcA_in_g(x₁)) = x₁
POL(pcA_out_g(x₁)) = 0
POL(s(x₁)) = 0

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none

(14) Obligation:

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

U2_G(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → PA_IN_G(cons(s(s(s(s(X2)))), X3))
U2_G(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → U4_G(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U4_G(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → PA_IN_G(cons(s(s(s(s(X2)))), X3))
PA_IN_G(cons(s(s(s(s(X1)))), cons(X2, X3))) → PA_IN_G(cons(X1, cons(X2, X3)))

The TRS R consists of the following rules:

pcA_in_g(cons(X1, nil)) → pcA_out_g(cons(X1, nil))
pcA_in_g(cons(s(s(s(s(X1)))), cons(X2, X3))) → U14_g(X1, X2, X3, pcA_in_g(cons(X1, cons(X2, X3))))
pcA_in_g(cons(s(s(0)), cons(X1, X2))) → U17_g(X1, X2, pcA_in_g(cons(X1, X2)))
pcA_in_g(cons(0, cons(X1, nil))) → pcA_out_g(cons(0, cons(X1, nil)))
pcA_in_g(cons(0, cons(s(s(X1)), cons(X2, X3)))) → U19_g(X1, X2, X3, pcA_in_g(cons(X1, cons(X2, X3))))
pcA_in_g(cons(0, cons(0, X1))) → U21_g(X1, pcA_in_g(X1))
U21_g(X1, pcA_out_g(X1)) → pcA_out_g(cons(0, cons(0, X1)))
U19_g(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → U20_g(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U20_g(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → pcA_out_g(cons(0, cons(s(s(X1)), cons(X2, X3))))
U17_g(X1, X2, pcA_out_g(cons(X1, X2))) → U18_g(X1, X2, pcA_in_g(cons(s(s(s(s(X1)))), X2)))
U18_g(X1, X2, pcA_out_g(cons(s(s(s(s(X1)))), X2))) → pcA_out_g(cons(s(s(0)), cons(X1, X2)))
U14_g(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → U15_g(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U15_g(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → U16_g(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U16_g(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → pcA_out_g(cons(s(s(s(s(X1)))), cons(X2, X3)))

The set Q consists of the following terms:

pcA_in_g(x0)
U21_g(x0, x1)
U19_g(x0, x1, x2, x3)
U20_g(x0, x1, x2, x3)
U17_g(x0, x1, x2)
U18_g(x0, x1, x2)
U14_g(x0, x1, x2, x3)
U15_g(x0, x1, x2, x3)
U16_g(x0, x1, x2, x3)

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

(15) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(16) Obligation:

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

PA_IN_G(cons(s(s(s(s(X1)))), cons(X2, X3))) → PA_IN_G(cons(X1, cons(X2, X3)))

The TRS R consists of the following rules:

pcA_in_g(cons(X1, nil)) → pcA_out_g(cons(X1, nil))
pcA_in_g(cons(s(s(s(s(X1)))), cons(X2, X3))) → U14_g(X1, X2, X3, pcA_in_g(cons(X1, cons(X2, X3))))
pcA_in_g(cons(s(s(0)), cons(X1, X2))) → U17_g(X1, X2, pcA_in_g(cons(X1, X2)))
pcA_in_g(cons(0, cons(X1, nil))) → pcA_out_g(cons(0, cons(X1, nil)))
pcA_in_g(cons(0, cons(s(s(X1)), cons(X2, X3)))) → U19_g(X1, X2, X3, pcA_in_g(cons(X1, cons(X2, X3))))
pcA_in_g(cons(0, cons(0, X1))) → U21_g(X1, pcA_in_g(X1))
U21_g(X1, pcA_out_g(X1)) → pcA_out_g(cons(0, cons(0, X1)))
U19_g(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → U20_g(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U20_g(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → pcA_out_g(cons(0, cons(s(s(X1)), cons(X2, X3))))
U17_g(X1, X2, pcA_out_g(cons(X1, X2))) → U18_g(X1, X2, pcA_in_g(cons(s(s(s(s(X1)))), X2)))
U18_g(X1, X2, pcA_out_g(cons(s(s(s(s(X1)))), X2))) → pcA_out_g(cons(s(s(0)), cons(X1, X2)))
U14_g(X1, X2, X3, pcA_out_g(cons(X1, cons(X2, X3)))) → U15_g(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U15_g(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → U16_g(X1, X2, X3, pcA_in_g(cons(s(s(s(s(X2)))), X3)))
U16_g(X1, X2, X3, pcA_out_g(cons(s(s(s(s(X2)))), X3))) → pcA_out_g(cons(s(s(s(s(X1)))), cons(X2, X3)))

The set Q consists of the following terms:

pcA_in_g(x0)
U21_g(x0, x1)
U19_g(x0, x1, x2, x3)
U20_g(x0, x1, x2, x3)
U17_g(x0, x1, x2)
U18_g(x0, x1, x2)
U14_g(x0, x1, x2, x3)
U15_g(x0, x1, x2, x3)
U16_g(x0, x1, x2, x3)

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

(17) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(18) Obligation:

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

PA_IN_G(cons(s(s(s(s(X1)))), cons(X2, X3))) → PA_IN_G(cons(X1, cons(X2, X3)))

R is empty.
The set Q consists of the following terms:

pcA_in_g(x0)
U21_g(x0, x1)
U19_g(x0, x1, x2, x3)
U20_g(x0, x1, x2, x3)
U17_g(x0, x1, x2)
U18_g(x0, x1, x2)
U14_g(x0, x1, x2, x3)
U15_g(x0, x1, x2, x3)
U16_g(x0, x1, x2, x3)

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

(19) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

pcA_in_g(x0)
U21_g(x0, x1)
U19_g(x0, x1, x2, x3)
U20_g(x0, x1, x2, x3)
U17_g(x0, x1, x2)
U18_g(x0, x1, x2)
U14_g(x0, x1, x2, x3)
U15_g(x0, x1, x2, x3)
U16_g(x0, x1, x2, x3)

(20) Obligation:

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

PA_IN_G(cons(s(s(s(s(X1)))), cons(X2, X3))) → PA_IN_G(cons(X1, cons(X2, X3)))

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

(21) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

PA_IN_G(cons(s(s(s(s(X1)))), cons(X2, X3))) → PA_IN_G(cons(X1, cons(X2, X3)))

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(PA_IN_G(x₁)) = x₁
POL(cons(x₁, x₂)) = x₁ + x₂
POL(s(x₁)) = x₁

(22) Obligation:

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

(23) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(0) Obligation:

(1) PrologToDTProblemTransformerProof (SOUND transformation)

(2) Obligation:

(3) TriplesToPiDPProof (SOUND transformation)

(4) Obligation:

(5) DependencyGraphProof (EQUIVALENT transformation)

(6) Obligation:

(7) PiDPToQDPProof (EQUIVALENT transformation)

(8) Obligation:

(9) QDPOrderProof (EQUIVALENT transformation)

(10) Obligation:

(11) DependencyGraphProof (EQUIVALENT transformation)

(12) Obligation:

(13) QDPOrderProof (EQUIVALENT transformation)

(14) Obligation:

(15) DependencyGraphProof (EQUIVALENT transformation)

(16) Obligation:

(17) UsableRulesProof (EQUIVALENT transformation)

(18) Obligation:

(19) QReductionProof (EQUIVALENT transformation)

(20) Obligation:

(21) UsableRulesReductionPairsProof (EQUIVALENT transformation)

(22) Obligation:

(23) PisEmptyProof (EQUIVALENT transformation)

(24) YES