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

(0) Obligation:

Clauses:

dis(or(B1, B2)) :- ','(con(B1), dis(B2)).
dis(B) :- con(B).
con(and(B1, B2)) :- ','(dis(B1), con(B2)).
con(B) :- bool(B).
bool(0).
bool(1).

Query: con(g)

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: con(T1)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: con(T1), SCOPE: 1, CLAUSE: 2 | con(T1), SCOPE: 1, CLAUSE: 3\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: ','(dis(T4), con(T5)) | con(and(T4, T5)), SCOPE: 1, CLAUSE: 3\n\nT4 is ground\nT5 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: con(T1), SCOPE: 1, CLAUSE: 3\n\nT1 is ground\nX3 is free\nX4 is free\ncon(T1) !=? con(and(X3, X4))", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: ','(dis(T4), con(T5)), SCOPE: 2, CLAUSE: 0 | ','(dis(T4), con(T5)), SCOPE: 2, CLAUSE: 1 | ?_2 | con(and(T4, T5)), SCOPE: 1, CLAUSE: 3\n\nT4 is ground\nT5 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: ','(dis(T4), con(T5)), SCOPE: 2, CLAUSE: 0\n\nT4 is ground\nT5 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: ','(dis(T4), con(T5)), SCOPE: 2, CLAUSE: 1 | ?_2 | con(and(T4, T5)), SCOPE: 1, CLAUSE: 3\n\nT4 is ground\nT5 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: ','(','(con(T14), dis(T15)), con(T5))\n\nT5 is ground\nT14 is ground\nT15 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: con(T14)\n\nT14 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(dis(T15), con(T5))\n\nT5 is ground\nT15 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: dis(T15)\n\nT15 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: con(T5)\n\nT5 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: dis(T15), SCOPE: 3, CLAUSE: 0 | dis(T15), SCOPE: 3, CLAUSE: 1\n\nT15 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: dis(T15), SCOPE: 3, CLAUSE: 0\n\nT15 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: dis(T15), SCOPE: 3, CLAUSE: 1\n\nT15 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: ','(con(T32), dis(T33))\n\nT32 is ground\nT33 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: con(T32)\n\nT32 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: dis(T33)\n\nT33 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: con(T44)\n\nT44 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: ','(dis(T4), con(T5)), SCOPE: 2, CLAUSE: 1\n\nT4 is ground\nT5 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: ?_2 | con(and(T4, T5)), SCOPE: 1, CLAUSE: 3\n\nT4 is ground\nT5 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: ','(con(T57), con(T5))\n\nT5 is ground\nT57 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: con(T57)\n\nT57 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: con(T5)\n\nT5 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: con(and(T4, T5)), SCOPE: 1, CLAUSE: 3\n\nT4 is ground\nT5 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: bool(and(T68, T69))\n\nT68 is ground\nT69 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="29: bool(and(T68, T69)), SCOPE: 4, CLAUSE: 4 | bool(and(T68, T69)), SCOPE: 4, CLAUSE: 5\n\nT68 is ground\nT69 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
30 [label="30: bool(and(T68, T69)), SCOPE: 4, CLAUSE: 5\n\nT68 is ground\nT69 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
31 [label="31: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
32 [label="32: bool(T71)\n\nT71 is ground\nX3 is free\nX4 is free\ncon(T71) !=? con(and(X3, X4))", fontsize=16, style=filled, fillcolor=lightyellow];
33 [label="33: bool(T71), SCOPE: 5, CLAUSE: 4 | bool(T71), SCOPE: 5, CLAUSE: 5\n\nT71 is ground\nX3 is free\nX4 is free\ncon(T71) !=? con(and(X3, X4))", fontsize=16, style=filled, fillcolor=lightyellow];
34 [label="34: bool(T71), SCOPE: 5, CLAUSE: 4\n\nT71 is ground\nX3 is free\nX4 is free\ncon(T71) !=? con(and(X3, X4))", fontsize=16, style=filled, fillcolor=lightyellow];
35 [label="35: bool(T71), SCOPE: 5, CLAUSE: 5\n\nT71 is ground\nX3 is free\nX4 is free\ncon(T71) !=? con(and(X3, X4))", fontsize=16, style=filled, fillcolor=lightyellow];
36 [label="36: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
37 [label="37: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
38 [label="38: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
39 [label="39: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
40 [label="40: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
41 [label="41: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
42 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 42 [arrowhead = none ];
42 -> 2;

43 [label="EVAL with clause\ncon(and(X3, X4)) :- ','(dis(X3), con(X4)).\nand substitutionX3 -> T4,\nX4 -> T5,\nT1 -> and(T4, T5)", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 43 [arrowhead = none ];
43 -> 3;

44 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 44 [arrowhead = none ];
44 -> 4;

45 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 45 [arrowhead = none ];
45 -> 5;

46 [label="ONLY EVAL with clause\ncon(X67) :- bool(X67).\nand substitutionT1 -> T71,\nX67 -> T71", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 46 [arrowhead = none ];
46 -> 32;

47 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
5 -> 47 [arrowhead = none ];
47 -> 6;

48 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
5 -> 48 [arrowhead = none ];
48 -> 7;

49 [label="EVAL with clause\ndis(or(X13, X14)) :- ','(con(X13), dis(X14)).\nand substitutionX13 -> T14,\nX14 -> T15,\nT4 -> or(T14, T15)", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 49 [arrowhead = none ];
49 -> 8;

50 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 50 [arrowhead = none ];
50 -> 9;

51 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
7 -> 51 [arrowhead = none ];
51 -> 22;

52 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
7 -> 52 [arrowhead = none ];
52 -> 23;

53 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
8 -> 53 [arrowhead = none ];
53 -> 10;

54 [label="SPLIT 2\nnew knowledge:\nT14 is ground", fontsize=14, style = filled, fillcolor = violet];
8 -> 54 [arrowhead = none ];
54 -> 11;

55 [label="INSTANCE with matching:\nT1 -> T14", fontsize=14, style = filled, fillcolor = lightgrey];
10 -> 55 [arrowhead = none , style=dashed];
55 -> 1[style=dashed];

56 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
11 -> 56 [arrowhead = none ];
56 -> 12;

57 [label="SPLIT 2\nnew knowledge:\nT15 is ground", fontsize=14, style = filled, fillcolor = violet];
11 -> 57 [arrowhead = none ];
57 -> 13;

58 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
12 -> 58 [arrowhead = none ];
58 -> 14;

59 [label="INSTANCE with matching:\nT1 -> T5", fontsize=14, style = filled, fillcolor = lightgrey];
13 -> 59 [arrowhead = none , style=dashed];
59 -> 1[style=dashed];

60 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
14 -> 60 [arrowhead = none ];
60 -> 15;

61 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
14 -> 61 [arrowhead = none ];
61 -> 16;

62 [label="EVAL with clause\ndis(or(X31, X32)) :- ','(con(X31), dis(X32)).\nand substitutionX31 -> T32,\nX32 -> T33,\nT15 -> or(T32, T33)", fontsize=14, style = filled, fillcolor = lightblue];
15 -> 62 [arrowhead = none ];
62 -> 17;

63 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 63 [arrowhead = none ];
63 -> 18;

64 [label="ONLY EVAL with clause\ndis(X43) :- con(X43).\nand substitutionT15 -> T44,\nX43 -> T44", fontsize=14, style = filled, fillcolor = lightblue];
16 -> 64 [arrowhead = none ];
64 -> 21;

65 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
17 -> 65 [arrowhead = none ];
65 -> 19;

66 [label="SPLIT 2\nnew knowledge:\nT32 is ground", fontsize=14, style = filled, fillcolor = violet];
17 -> 66 [arrowhead = none ];
66 -> 20;

67 [label="INSTANCE with matching:\nT1 -> T32", fontsize=14, style = filled, fillcolor = lightgrey];
19 -> 67 [arrowhead = none , style=dashed];
67 -> 1[style=dashed];

68 [label="INSTANCE with matching:\nT15 -> T33", fontsize=14, style = filled, fillcolor = lightgrey];
20 -> 68 [arrowhead = none , style=dashed];
68 -> 12[style=dashed];

69 [label="INSTANCE with matching:\nT1 -> T44", fontsize=14, style = filled, fillcolor = lightgrey];
21 -> 69 [arrowhead = none , style=dashed];
69 -> 1[style=dashed];

70 [label="ONLY EVAL with clause\ndis(X56) :- con(X56).\nand substitutionT4 -> T57,\nX56 -> T57", fontsize=14, style = filled, fillcolor = lightblue];
22 -> 70 [arrowhead = none ];
70 -> 24;

71 [label="FAILURE", fontsize=14, style = filled, fillcolor = lightgreen];
23 -> 71 [arrowhead = none ];
71 -> 27;

72 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
24 -> 72 [arrowhead = none ];
72 -> 25;

73 [label="SPLIT 2\nnew knowledge:\nT57 is ground", fontsize=14, style = filled, fillcolor = violet];
24 -> 73 [arrowhead = none ];
73 -> 26;

74 [label="INSTANCE with matching:\nT1 -> T57", fontsize=14, style = filled, fillcolor = lightgrey];
25 -> 74 [arrowhead = none , style=dashed];
74 -> 1[style=dashed];

75 [label="INSTANCE with matching:\nT1 -> T5", fontsize=14, style = filled, fillcolor = lightgrey];
26 -> 75 [arrowhead = none , style=dashed];
75 -> 1[style=dashed];

76 [label="ONLY EVAL with clause\ncon(X65) :- bool(X65).\nand substitutionT4 -> T68,\nT5 -> T69,\nX65 -> and(T68, T69)", fontsize=14, style = filled, fillcolor = lightblue];
27 -> 76 [arrowhead = none ];
76 -> 28;

77 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
28 -> 77 [arrowhead = none ];
77 -> 29;

78 [label="BACKTRACK\nfor clause: bool(0)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
29 -> 78 [arrowhead = none ];
78 -> 30;

79 [label="BACKTRACK\nfor clause: bool(1)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
30 -> 79 [arrowhead = none ];
79 -> 31;

80 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
32 -> 80 [arrowhead = none ];
80 -> 33;

81 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
33 -> 81 [arrowhead = none ];
81 -> 34;

82 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
33 -> 82 [arrowhead = none ];
82 -> 35;

83 [label="EVAL with clause\nbool(0).\nand substitutionT71 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
34 -> 83 [arrowhead = none ];
83 -> 36;

84 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
34 -> 84 [arrowhead = none ];
84 -> 37;

85 [label="EVAL with clause\nbool(1).\nand substitutionT71 -> 1", fontsize=14, style = filled, fillcolor = lightblue];
35 -> 85 [arrowhead = none ];
85 -> 39;

86 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
35 -> 86 [arrowhead = none ];
86 -> 40;

87 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
36 -> 87 [arrowhead = none ];
87 -> 38;

88 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
39 -> 88 [arrowhead = none ];
88 -> 41;

}

(2) Obligation:

Clauses:

disA(or(T32, T33)) :- conB(T32).
disA(or(T32, T33)) :- ','(conB(T32), disA(T33)).
disA(T44) :- conB(T44).
conB(and(or(T14, T15), T5)) :- conB(T14).
conB(and(or(T14, T15), T5)) :- ','(conB(T14), disA(T15)).
conB(and(or(T14, T15), T5)) :- ','(conB(T14), ','(disA(T15), conB(T5))).
conB(and(T57, T5)) :- conB(T57).
conB(and(T57, T5)) :- ','(conB(T57), conB(T5)).
conB(0).
conB(1).

Query: conB(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:
conB_in: (b)
disA_in: (b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

conB_in_g(and(or(T14, T15), T5)) → U4_g(T14, T15, T5, conB_in_g(T14))
conB_in_g(and(T57, T5)) → U7_g(T57, T5, conB_in_g(T57))
conB_in_g(0) → conB_out_g(0)
conB_in_g(1) → conB_out_g(1)
U7_g(T57, T5, conB_out_g(T57)) → conB_out_g(and(T57, T5))
U7_g(T57, T5, conB_out_g(T57)) → U8_g(T57, T5, conB_in_g(T5))
U8_g(T57, T5, conB_out_g(T5)) → conB_out_g(and(T57, T5))
U4_g(T14, T15, T5, conB_out_g(T14)) → conB_out_g(and(or(T14, T15), T5))
U4_g(T14, T15, T5, conB_out_g(T14)) → U5_g(T14, T15, T5, disA_in_g(T15))
disA_in_g(or(T32, T33)) → U1_g(T32, T33, conB_in_g(T32))
U1_g(T32, T33, conB_out_g(T32)) → disA_out_g(or(T32, T33))
U1_g(T32, T33, conB_out_g(T32)) → U2_g(T32, T33, disA_in_g(T33))
disA_in_g(T44) → U3_g(T44, conB_in_g(T44))
U3_g(T44, conB_out_g(T44)) → disA_out_g(T44)
U2_g(T32, T33, disA_out_g(T33)) → disA_out_g(or(T32, T33))
U5_g(T14, T15, T5, disA_out_g(T15)) → conB_out_g(and(or(T14, T15), T5))
U5_g(T14, T15, T5, disA_out_g(T15)) → U6_g(T14, T15, T5, conB_in_g(T5))
U6_g(T14, T15, T5, conB_out_g(T5)) → conB_out_g(and(or(T14, T15), T5))

The argument filtering Pi contains the following mapping:
conB_in_g(x1)  =  conB_in_g(x1)
and(x1, x2)  =  and(x1, x2)
or(x1, x2)  =  or(x1, x2)
U4_g(x1, x2, x3, x4)  =  U4_g(x2, x3, x4)
U7_g(x1, x2, x3)  =  U7_g(x2, x3)
0  =  0
conB_out_g(x1)  =  conB_out_g
1  =  1
U8_g(x1, x2, x3)  =  U8_g(x3)
U5_g(x1, x2, x3, x4)  =  U5_g(x3, x4)
disA_in_g(x1)  =  disA_in_g(x1)
U1_g(x1, x2, x3)  =  U1_g(x2, x3)
disA_out_g(x1)  =  disA_out_g
U2_g(x1, x2, x3)  =  U2_g(x3)
U3_g(x1, x2)  =  U3_g(x2)
U6_g(x1, x2, x3, x4)  =  U6_g(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:

conB_in_g(and(or(T14, T15), T5)) → U4_g(T14, T15, T5, conB_in_g(T14))
conB_in_g(and(T57, T5)) → U7_g(T57, T5, conB_in_g(T57))
conB_in_g(0) → conB_out_g(0)
conB_in_g(1) → conB_out_g(1)
U7_g(T57, T5, conB_out_g(T57)) → conB_out_g(and(T57, T5))
U7_g(T57, T5, conB_out_g(T57)) → U8_g(T57, T5, conB_in_g(T5))
U8_g(T57, T5, conB_out_g(T5)) → conB_out_g(and(T57, T5))
U4_g(T14, T15, T5, conB_out_g(T14)) → conB_out_g(and(or(T14, T15), T5))
U4_g(T14, T15, T5, conB_out_g(T14)) → U5_g(T14, T15, T5, disA_in_g(T15))
disA_in_g(or(T32, T33)) → U1_g(T32, T33, conB_in_g(T32))
U1_g(T32, T33, conB_out_g(T32)) → disA_out_g(or(T32, T33))
U1_g(T32, T33, conB_out_g(T32)) → U2_g(T32, T33, disA_in_g(T33))
disA_in_g(T44) → U3_g(T44, conB_in_g(T44))
U3_g(T44, conB_out_g(T44)) → disA_out_g(T44)
U2_g(T32, T33, disA_out_g(T33)) → disA_out_g(or(T32, T33))
U5_g(T14, T15, T5, disA_out_g(T15)) → conB_out_g(and(or(T14, T15), T5))
U5_g(T14, T15, T5, disA_out_g(T15)) → U6_g(T14, T15, T5, conB_in_g(T5))
U6_g(T14, T15, T5, conB_out_g(T5)) → conB_out_g(and(or(T14, T15), T5))

The argument filtering Pi contains the following mapping:
conB_in_g(x1)  =  conB_in_g(x1)
and(x1, x2)  =  and(x1, x2)
or(x1, x2)  =  or(x1, x2)
U4_g(x1, x2, x3, x4)  =  U4_g(x2, x3, x4)
U7_g(x1, x2, x3)  =  U7_g(x2, x3)
0  =  0
conB_out_g(x1)  =  conB_out_g
1  =  1
U8_g(x1, x2, x3)  =  U8_g(x3)
U5_g(x1, x2, x3, x4)  =  U5_g(x3, x4)
disA_in_g(x1)  =  disA_in_g(x1)
U1_g(x1, x2, x3)  =  U1_g(x2, x3)
disA_out_g(x1)  =  disA_out_g
U2_g(x1, x2, x3)  =  U2_g(x3)
U3_g(x1, x2)  =  U3_g(x2)
U6_g(x1, x2, x3, x4)  =  U6_g(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:

CONB_IN_G(and(or(T14, T15), T5)) → U4_G(T14, T15, T5, conB_in_g(T14))
CONB_IN_G(and(or(T14, T15), T5)) → CONB_IN_G(T14)
CONB_IN_G(and(T57, T5)) → U7_G(T57, T5, conB_in_g(T57))
CONB_IN_G(and(T57, T5)) → CONB_IN_G(T57)
U7_G(T57, T5, conB_out_g(T57)) → U8_G(T57, T5, conB_in_g(T5))
U7_G(T57, T5, conB_out_g(T57)) → CONB_IN_G(T5)
U4_G(T14, T15, T5, conB_out_g(T14)) → U5_G(T14, T15, T5, disA_in_g(T15))
U4_G(T14, T15, T5, conB_out_g(T14)) → DISA_IN_G(T15)
DISA_IN_G(or(T32, T33)) → U1_G(T32, T33, conB_in_g(T32))
DISA_IN_G(or(T32, T33)) → CONB_IN_G(T32)
U1_G(T32, T33, conB_out_g(T32)) → U2_G(T32, T33, disA_in_g(T33))
U1_G(T32, T33, conB_out_g(T32)) → DISA_IN_G(T33)
DISA_IN_G(T44) → U3_G(T44, conB_in_g(T44))
DISA_IN_G(T44) → CONB_IN_G(T44)
U5_G(T14, T15, T5, disA_out_g(T15)) → U6_G(T14, T15, T5, conB_in_g(T5))
U5_G(T14, T15, T5, disA_out_g(T15)) → CONB_IN_G(T5)

The TRS R consists of the following rules:

conB_in_g(and(or(T14, T15), T5)) → U4_g(T14, T15, T5, conB_in_g(T14))
conB_in_g(and(T57, T5)) → U7_g(T57, T5, conB_in_g(T57))
conB_in_g(0) → conB_out_g(0)
conB_in_g(1) → conB_out_g(1)
U7_g(T57, T5, conB_out_g(T57)) → conB_out_g(and(T57, T5))
U7_g(T57, T5, conB_out_g(T57)) → U8_g(T57, T5, conB_in_g(T5))
U8_g(T57, T5, conB_out_g(T5)) → conB_out_g(and(T57, T5))
U4_g(T14, T15, T5, conB_out_g(T14)) → conB_out_g(and(or(T14, T15), T5))
U4_g(T14, T15, T5, conB_out_g(T14)) → U5_g(T14, T15, T5, disA_in_g(T15))
disA_in_g(or(T32, T33)) → U1_g(T32, T33, conB_in_g(T32))
U1_g(T32, T33, conB_out_g(T32)) → disA_out_g(or(T32, T33))
U1_g(T32, T33, conB_out_g(T32)) → U2_g(T32, T33, disA_in_g(T33))
disA_in_g(T44) → U3_g(T44, conB_in_g(T44))
U3_g(T44, conB_out_g(T44)) → disA_out_g(T44)
U2_g(T32, T33, disA_out_g(T33)) → disA_out_g(or(T32, T33))
U5_g(T14, T15, T5, disA_out_g(T15)) → conB_out_g(and(or(T14, T15), T5))
U5_g(T14, T15, T5, disA_out_g(T15)) → U6_g(T14, T15, T5, conB_in_g(T5))
U6_g(T14, T15, T5, conB_out_g(T5)) → conB_out_g(and(or(T14, T15), T5))

The argument filtering Pi contains the following mapping:
conB_in_g(x1)  =  conB_in_g(x1)
and(x1, x2)  =  and(x1, x2)
or(x1, x2)  =  or(x1, x2)
U4_g(x1, x2, x3, x4)  =  U4_g(x2, x3, x4)
U7_g(x1, x2, x3)  =  U7_g(x2, x3)
0  =  0
conB_out_g(x1)  =  conB_out_g
1  =  1
U8_g(x1, x2, x3)  =  U8_g(x3)
U5_g(x1, x2, x3, x4)  =  U5_g(x3, x4)
disA_in_g(x1)  =  disA_in_g(x1)
U1_g(x1, x2, x3)  =  U1_g(x2, x3)
disA_out_g(x1)  =  disA_out_g
U2_g(x1, x2, x3)  =  U2_g(x3)
U3_g(x1, x2)  =  U3_g(x2)
U6_g(x1, x2, x3, x4)  =  U6_g(x4)
CONB_IN_G(x1)  =  CONB_IN_G(x1)
U4_G(x1, x2, x3, x4)  =  U4_G(x2, x3, x4)
U7_G(x1, x2, x3)  =  U7_G(x2, x3)
U8_G(x1, x2, x3)  =  U8_G(x3)
U5_G(x1, x2, x3, x4)  =  U5_G(x3, x4)
DISA_IN_G(x1)  =  DISA_IN_G(x1)
U1_G(x1, x2, x3)  =  U1_G(x2, x3)
U2_G(x1, x2, x3)  =  U2_G(x3)
U3_G(x1, x2)  =  U3_G(x2)
U6_G(x1, x2, x3, x4)  =  U6_G(x4)

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

(6) Obligation:

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

CONB_IN_G(and(or(T14, T15), T5)) → U4_G(T14, T15, T5, conB_in_g(T14))
CONB_IN_G(and(or(T14, T15), T5)) → CONB_IN_G(T14)
CONB_IN_G(and(T57, T5)) → U7_G(T57, T5, conB_in_g(T57))
CONB_IN_G(and(T57, T5)) → CONB_IN_G(T57)
U7_G(T57, T5, conB_out_g(T57)) → U8_G(T57, T5, conB_in_g(T5))
U7_G(T57, T5, conB_out_g(T57)) → CONB_IN_G(T5)
U4_G(T14, T15, T5, conB_out_g(T14)) → U5_G(T14, T15, T5, disA_in_g(T15))
U4_G(T14, T15, T5, conB_out_g(T14)) → DISA_IN_G(T15)
DISA_IN_G(or(T32, T33)) → U1_G(T32, T33, conB_in_g(T32))
DISA_IN_G(or(T32, T33)) → CONB_IN_G(T32)
U1_G(T32, T33, conB_out_g(T32)) → U2_G(T32, T33, disA_in_g(T33))
U1_G(T32, T33, conB_out_g(T32)) → DISA_IN_G(T33)
DISA_IN_G(T44) → U3_G(T44, conB_in_g(T44))
DISA_IN_G(T44) → CONB_IN_G(T44)
U5_G(T14, T15, T5, disA_out_g(T15)) → U6_G(T14, T15, T5, conB_in_g(T5))
U5_G(T14, T15, T5, disA_out_g(T15)) → CONB_IN_G(T5)

The TRS R consists of the following rules:

conB_in_g(and(or(T14, T15), T5)) → U4_g(T14, T15, T5, conB_in_g(T14))
conB_in_g(and(T57, T5)) → U7_g(T57, T5, conB_in_g(T57))
conB_in_g(0) → conB_out_g(0)
conB_in_g(1) → conB_out_g(1)
U7_g(T57, T5, conB_out_g(T57)) → conB_out_g(and(T57, T5))
U7_g(T57, T5, conB_out_g(T57)) → U8_g(T57, T5, conB_in_g(T5))
U8_g(T57, T5, conB_out_g(T5)) → conB_out_g(and(T57, T5))
U4_g(T14, T15, T5, conB_out_g(T14)) → conB_out_g(and(or(T14, T15), T5))
U4_g(T14, T15, T5, conB_out_g(T14)) → U5_g(T14, T15, T5, disA_in_g(T15))
disA_in_g(or(T32, T33)) → U1_g(T32, T33, conB_in_g(T32))
U1_g(T32, T33, conB_out_g(T32)) → disA_out_g(or(T32, T33))
U1_g(T32, T33, conB_out_g(T32)) → U2_g(T32, T33, disA_in_g(T33))
disA_in_g(T44) → U3_g(T44, conB_in_g(T44))
U3_g(T44, conB_out_g(T44)) → disA_out_g(T44)
U2_g(T32, T33, disA_out_g(T33)) → disA_out_g(or(T32, T33))
U5_g(T14, T15, T5, disA_out_g(T15)) → conB_out_g(and(or(T14, T15), T5))
U5_g(T14, T15, T5, disA_out_g(T15)) → U6_g(T14, T15, T5, conB_in_g(T5))
U6_g(T14, T15, T5, conB_out_g(T5)) → conB_out_g(and(or(T14, T15), T5))

The argument filtering Pi contains the following mapping:
conB_in_g(x1)  =  conB_in_g(x1)
and(x1, x2)  =  and(x1, x2)
or(x1, x2)  =  or(x1, x2)
U4_g(x1, x2, x3, x4)  =  U4_g(x2, x3, x4)
U7_g(x1, x2, x3)  =  U7_g(x2, x3)
0  =  0
conB_out_g(x1)  =  conB_out_g
1  =  1
U8_g(x1, x2, x3)  =  U8_g(x3)
U5_g(x1, x2, x3, x4)  =  U5_g(x3, x4)
disA_in_g(x1)  =  disA_in_g(x1)
U1_g(x1, x2, x3)  =  U1_g(x2, x3)
disA_out_g(x1)  =  disA_out_g
U2_g(x1, x2, x3)  =  U2_g(x3)
U3_g(x1, x2)  =  U3_g(x2)
U6_g(x1, x2, x3, x4)  =  U6_g(x4)
CONB_IN_G(x1)  =  CONB_IN_G(x1)
U4_G(x1, x2, x3, x4)  =  U4_G(x2, x3, x4)
U7_G(x1, x2, x3)  =  U7_G(x2, x3)
U8_G(x1, x2, x3)  =  U8_G(x3)
U5_G(x1, x2, x3, x4)  =  U5_G(x3, x4)
DISA_IN_G(x1)  =  DISA_IN_G(x1)
U1_G(x1, x2, x3)  =  U1_G(x2, x3)
U2_G(x1, x2, x3)  =  U2_G(x3)
U3_G(x1, x2)  =  U3_G(x2)
U6_G(x1, x2, x3, x4)  =  U6_G(x4)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 4 less nodes.

(8) Obligation:

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

U4_G(T14, T15, T5, conB_out_g(T14)) → U5_G(T14, T15, T5, disA_in_g(T15))
U5_G(T14, T15, T5, disA_out_g(T15)) → CONB_IN_G(T5)
CONB_IN_G(and(or(T14, T15), T5)) → U4_G(T14, T15, T5, conB_in_g(T14))
U4_G(T14, T15, T5, conB_out_g(T14)) → DISA_IN_G(T15)
DISA_IN_G(or(T32, T33)) → U1_G(T32, T33, conB_in_g(T32))
U1_G(T32, T33, conB_out_g(T32)) → DISA_IN_G(T33)
DISA_IN_G(or(T32, T33)) → CONB_IN_G(T32)
CONB_IN_G(and(or(T14, T15), T5)) → CONB_IN_G(T14)
CONB_IN_G(and(T57, T5)) → U7_G(T57, T5, conB_in_g(T57))
U7_G(T57, T5, conB_out_g(T57)) → CONB_IN_G(T5)
CONB_IN_G(and(T57, T5)) → CONB_IN_G(T57)
DISA_IN_G(T44) → CONB_IN_G(T44)

The TRS R consists of the following rules:

conB_in_g(and(or(T14, T15), T5)) → U4_g(T14, T15, T5, conB_in_g(T14))
conB_in_g(and(T57, T5)) → U7_g(T57, T5, conB_in_g(T57))
conB_in_g(0) → conB_out_g(0)
conB_in_g(1) → conB_out_g(1)
U7_g(T57, T5, conB_out_g(T57)) → conB_out_g(and(T57, T5))
U7_g(T57, T5, conB_out_g(T57)) → U8_g(T57, T5, conB_in_g(T5))
U8_g(T57, T5, conB_out_g(T5)) → conB_out_g(and(T57, T5))
U4_g(T14, T15, T5, conB_out_g(T14)) → conB_out_g(and(or(T14, T15), T5))
U4_g(T14, T15, T5, conB_out_g(T14)) → U5_g(T14, T15, T5, disA_in_g(T15))
disA_in_g(or(T32, T33)) → U1_g(T32, T33, conB_in_g(T32))
U1_g(T32, T33, conB_out_g(T32)) → disA_out_g(or(T32, T33))
U1_g(T32, T33, conB_out_g(T32)) → U2_g(T32, T33, disA_in_g(T33))
disA_in_g(T44) → U3_g(T44, conB_in_g(T44))
U3_g(T44, conB_out_g(T44)) → disA_out_g(T44)
U2_g(T32, T33, disA_out_g(T33)) → disA_out_g(or(T32, T33))
U5_g(T14, T15, T5, disA_out_g(T15)) → conB_out_g(and(or(T14, T15), T5))
U5_g(T14, T15, T5, disA_out_g(T15)) → U6_g(T14, T15, T5, conB_in_g(T5))
U6_g(T14, T15, T5, conB_out_g(T5)) → conB_out_g(and(or(T14, T15), T5))

The argument filtering Pi contains the following mapping:
conB_in_g(x1)  =  conB_in_g(x1)
and(x1, x2)  =  and(x1, x2)
or(x1, x2)  =  or(x1, x2)
U4_g(x1, x2, x3, x4)  =  U4_g(x2, x3, x4)
U7_g(x1, x2, x3)  =  U7_g(x2, x3)
0  =  0
conB_out_g(x1)  =  conB_out_g
1  =  1
U8_g(x1, x2, x3)  =  U8_g(x3)
U5_g(x1, x2, x3, x4)  =  U5_g(x3, x4)
disA_in_g(x1)  =  disA_in_g(x1)
U1_g(x1, x2, x3)  =  U1_g(x2, x3)
disA_out_g(x1)  =  disA_out_g
U2_g(x1, x2, x3)  =  U2_g(x3)
U3_g(x1, x2)  =  U3_g(x2)
U6_g(x1, x2, x3, x4)  =  U6_g(x4)
CONB_IN_G(x1)  =  CONB_IN_G(x1)
U4_G(x1, x2, x3, x4)  =  U4_G(x2, x3, x4)
U7_G(x1, x2, x3)  =  U7_G(x2, x3)
U5_G(x1, x2, x3, x4)  =  U5_G(x3, x4)
DISA_IN_G(x1)  =  DISA_IN_G(x1)
U1_G(x1, x2, x3)  =  U1_G(x2, x3)

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:

U4_G(T15, T5, conB_out_g) → U5_G(T5, disA_in_g(T15))
U5_G(T5, disA_out_g) → CONB_IN_G(T5)
CONB_IN_G(and(or(T14, T15), T5)) → U4_G(T15, T5, conB_in_g(T14))
U4_G(T15, T5, conB_out_g) → DISA_IN_G(T15)
DISA_IN_G(or(T32, T33)) → U1_G(T33, conB_in_g(T32))
U1_G(T33, conB_out_g) → DISA_IN_G(T33)
DISA_IN_G(or(T32, T33)) → CONB_IN_G(T32)
CONB_IN_G(and(or(T14, T15), T5)) → CONB_IN_G(T14)
CONB_IN_G(and(T57, T5)) → U7_G(T5, conB_in_g(T57))
U7_G(T5, conB_out_g) → CONB_IN_G(T5)
CONB_IN_G(and(T57, T5)) → CONB_IN_G(T57)
DISA_IN_G(T44) → CONB_IN_G(T44)

The TRS R consists of the following rules:

conB_in_g(and(or(T14, T15), T5)) → U4_g(T15, T5, conB_in_g(T14))
conB_in_g(and(T57, T5)) → U7_g(T5, conB_in_g(T57))
conB_in_g(0) → conB_out_g
conB_in_g(1) → conB_out_g
U7_g(T5, conB_out_g) → conB_out_g
U7_g(T5, conB_out_g) → U8_g(conB_in_g(T5))
U8_g(conB_out_g) → conB_out_g
U4_g(T15, T5, conB_out_g) → conB_out_g
U4_g(T15, T5, conB_out_g) → U5_g(T5, disA_in_g(T15))
disA_in_g(or(T32, T33)) → U1_g(T33, conB_in_g(T32))
U1_g(T33, conB_out_g) → disA_out_g
U1_g(T33, conB_out_g) → U2_g(disA_in_g(T33))
disA_in_g(T44) → U3_g(conB_in_g(T44))
U3_g(conB_out_g) → disA_out_g
U2_g(disA_out_g) → disA_out_g
U5_g(T5, disA_out_g) → conB_out_g
U5_g(T5, disA_out_g) → U6_g(conB_in_g(T5))
U6_g(conB_out_g) → conB_out_g

The set Q consists of the following terms:

conB_in_g(x0)
U7_g(x0, x1)
U8_g(x0)
U4_g(x0, x1, x2)
disA_in_g(x0)
U1_g(x0, x1)
U3_g(x0)
U2_g(x0)
U5_g(x0, x1)
U6_g(x0)

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:

• U5_G(T5, disA_out_g) → CONB_IN_G(T5)
  The graph contains the following edges 1 >= 1

• CONB_IN_G(and(or(T14, T15), T5)) → U4_G(T15, T5, conB_in_g(T14))
  The graph contains the following edges 1 > 1, 1 > 2

• U4_G(T15, T5, conB_out_g) → U5_G(T5, disA_in_g(T15))
  The graph contains the following edges 2 >= 1

• CONB_IN_G(and(T57, T5)) → U7_G(T5, conB_in_g(T57))
  The graph contains the following edges 1 > 1

• U4_G(T15, T5, conB_out_g) → DISA_IN_G(T15)
  The graph contains the following edges 1 >= 1

• U7_G(T5, conB_out_g) → CONB_IN_G(T5)
  The graph contains the following edges 1 >= 1

• DISA_IN_G(or(T32, T33)) → U1_G(T33, conB_in_g(T32))
  The graph contains the following edges 1 > 1

• U1_G(T33, conB_out_g) → DISA_IN_G(T33)
  The graph contains the following edges 1 >= 1

• CONB_IN_G(and(or(T14, T15), T5)) → CONB_IN_G(T14)
  The graph contains the following edges 1 > 1

• CONB_IN_G(and(T57, T5)) → CONB_IN_G(T57)
  The graph contains the following edges 1 > 1

• DISA_IN_G(or(T32, T33)) → CONB_IN_G(T32)
  The graph contains the following edges 1 > 1

• DISA_IN_G(T44) → CONB_IN_G(T44)
  The graph contains the following edges 1 >= 1

(12) YES