Left Termination proof of /tmp/tmpA8

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

0 Prolog

↳1 PrologToPrologProblemTransformerProof (⇒, 83 ms)

↳2 Prolog

↳3 PrologToPiTRSProof (⇒, 0 ms)

↳4 PiTRS

↳5 DependencyPairsProof (⇔, 0 ms)

↳6 PiDP

↳7 DependencyGraphProof (⇔, 3 ms)

↳8 AND

    ↳9 PiDP

        ↳10 UsableRulesProof (⇔, 0 ms)

        ↳11 PiDP

        ↳12 PiDPToQDPProof (⇒, 0 ms)

        ↳13 QDP

        ↳14 QDPSizeChangeProof (⇔, 0 ms)

        ↳15 YES

    ↳16 PiDP

        ↳17 UsableRulesProof (⇔, 0 ms)

        ↳18 PiDP

        ↳19 PiDPToQDPProof (⇒, 0 ms)

        ↳20 QDP

        ↳21 MRRProof (⇔, 12 ms)

        ↳22 QDP

        ↳23 PisEmptyProof (⇔, 0 ms)

        ↳24 YES

(0) Obligation:

Clauses:

half(0, 0).
half(s(0), 0).
half(s(s(X)), s(Y)) :- half(X, Y).
log(0, s(0)).
log(s(X), s(Y)) :- ','(half(s(X), Z), log(Z, Y)).

Query: log(g,a)

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: log(T1, T2)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: log(T1, T2), SCOPE: 1, CLAUSE: 3 | log(T1, T2), SCOPE: 1, CLAUSE: 4\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: true | log(0, T2), SCOPE: 1, CLAUSE: 4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: log(T1, T2), SCOPE: 1, CLAUSE: 4\n\nT1 is ground\nlog(T1, T2) !=? log(0, s(0))", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: log(0, T2), SCOPE: 1, CLAUSE: 4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: ','(half(s(T5), X7), log(X7, T7))\n\nT5 is ground\nX7 is free", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(half(s(T5), X7), log(X7, T7)), SCOPE: 2, CLAUSE: 0 | ','(half(s(T5), X7), log(X7, T7)), SCOPE: 2, CLAUSE: 1 | ','(half(s(T5), X7), log(X7, T7)), SCOPE: 2, CLAUSE: 2\n\nT5 is ground\nX7 is free", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(half(s(T5), X7), log(X7, T7)), SCOPE: 2, CLAUSE: 1 | ','(half(s(T5), X7), log(X7, T7)), SCOPE: 2, CLAUSE: 2\n\nT5 is ground\nX7 is free", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(half(s(T5), X7), log(X7, T7)), SCOPE: 2, CLAUSE: 1\n\nT5 is ground\nX7 is free", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(half(s(T5), X7), log(X7, T7)), SCOPE: 2, CLAUSE: 2\n\nT5 is ground\nX7 is free", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: log(0, T7)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: log(0, T7), SCOPE: 3, CLAUSE: 3 | log(0, T7), SCOPE: 3, CLAUSE: 4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: log(0, T7), SCOPE: 3, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: log(0, T7), SCOPE: 3, CLAUSE: 4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: ','(half(T10, X18), log(s(X18), T7))\n\nT10 is ground\nX18 is free", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: half(T10, X18)\n\nT10 is ground\nX18 is free", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: log(s(T11), T7)\n\nT11 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: half(T10, X18), SCOPE: 4, CLAUSE: 0 | half(T10, X18), SCOPE: 4, CLAUSE: 1 | half(T10, X18), SCOPE: 4, CLAUSE: 2\n\nT10 is ground\nX18 is free", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: half(T10, X18), SCOPE: 4, CLAUSE: 0\n\nT10 is ground\nX18 is free", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: half(T10, X18), SCOPE: 4, CLAUSE: 1 | half(T10, X18), SCOPE: 4, CLAUSE: 2\n\nT10 is ground\nX18 is free", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="29: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
30 [label="30: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
31 [label="31: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
32 [label="32: half(T10, X18), SCOPE: 4, CLAUSE: 1\n\nT10 is ground\nX18 is free", fontsize=16, style=filled, fillcolor=lightyellow];
33 [label="33: half(T10, X18), SCOPE: 4, CLAUSE: 2\n\nT10 is ground\nX18 is free", fontsize=16, style=filled, fillcolor=lightyellow];
34 [label="34: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
35 [label="35: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
36 [label="36: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
37 [label="37: half(T14, X27)\n\nT14 is ground\nX27 is free", fontsize=16, style=filled, fillcolor=lightyellow];
38 [label="38: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
39 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 39 [arrowhead = none ];
39 -> 2;

40 [label="EVAL with clause\nlog(0, s(0)).\nand substitutionT1 -> 0,\nT2 -> s(0)", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 40 [arrowhead = none ];
40 -> 3;

41 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 41 [arrowhead = none ];
41 -> 4;

42 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 42 [arrowhead = none ];
42 -> 5;

43 [label="EVAL with clause\nlog(s(X5), s(X6)) :- ','(half(s(X5), X7), log(X7, X6)).\nand substitutionX5 -> T5,\nT1 -> s(T5),\nX6 -> T7,\nT2 -> s(T7),\nT6 -> T7", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 43 [arrowhead = none ];
43 -> 7;

44 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 44 [arrowhead = none ];
44 -> 8;

45 [label="BACKTRACK\nfor clause: log(s(X), s(Y)) :- ','(half(s(X), Z), log(Z, Y))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 45 [arrowhead = none ];
45 -> 6;

46 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 46 [arrowhead = none ];
46 -> 9;

47 [label="BACKTRACK\nfor clause: half(0, 0)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 47 [arrowhead = none ];
47 -> 10;

48 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
10 -> 48 [arrowhead = none ];
48 -> 11;

49 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
10 -> 49 [arrowhead = none ];
49 -> 12;

50 [label="EVAL with clause\nhalf(s(0), 0).\nand substitutionT5 -> 0,\nX7 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 50 [arrowhead = none ];
50 -> 13;

51 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 51 [arrowhead = none ];
51 -> 14;

52 [label="EVAL with clause\nhalf(s(s(X16)), s(X17)) :- half(X16, X17).\nand substitutionX16 -> T10,\nT5 -> s(T10),\nX17 -> X18,\nX7 -> s(X18)", fontsize=14, style = filled, fillcolor = lightblue];
12 -> 52 [arrowhead = none ];
52 -> 22;

53 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 53 [arrowhead = none ];
53 -> 23;

54 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
13 -> 54 [arrowhead = none ];
54 -> 15;

55 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
15 -> 55 [arrowhead = none ];
55 -> 16;

56 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
15 -> 56 [arrowhead = none ];
56 -> 17;

57 [label="EVAL with clause\nlog(0, s(0)).\nand substitutionT7 -> s(0)", fontsize=14, style = filled, fillcolor = lightblue];
16 -> 57 [arrowhead = none ];
57 -> 18;

58 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
16 -> 58 [arrowhead = none ];
58 -> 19;

59 [label="BACKTRACK\nfor clause: log(s(X), s(Y)) :- ','(half(s(X), Z), log(Z, Y))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 59 [arrowhead = none ];
59 -> 21;

60 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
18 -> 60 [arrowhead = none ];
60 -> 20;

61 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
22 -> 61 [arrowhead = none ];
61 -> 24;

62 [label="SPLIT 2\nnew knowledge:\nT10 is ground\nT11 is ground\nreplacements:X18 -> T11", fontsize=14, style = filled, fillcolor = violet];
22 -> 62 [arrowhead = none ];
62 -> 25;

63 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
24 -> 63 [arrowhead = none ];
63 -> 26;

64 [label="INSTANCE with matching:\nT1 -> s(T11)\nT2 -> T7", fontsize=14, style = filled, fillcolor = lightgrey];
25 -> 64 [arrowhead = none , style=dashed];
64 -> 1[style=dashed];

65 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
26 -> 65 [arrowhead = none ];
65 -> 27;

66 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
26 -> 66 [arrowhead = none ];
66 -> 28;

67 [label="EVAL with clause\nhalf(0, 0).\nand substitutionT10 -> 0,\nX18 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
27 -> 67 [arrowhead = none ];
67 -> 29;

68 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
27 -> 68 [arrowhead = none ];
68 -> 30;

69 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
28 -> 69 [arrowhead = none ];
69 -> 32;

70 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
28 -> 70 [arrowhead = none ];
70 -> 33;

71 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
29 -> 71 [arrowhead = none ];
71 -> 31;

72 [label="EVAL with clause\nhalf(s(0), 0).\nand substitutionT10 -> s(0),\nX18 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
32 -> 72 [arrowhead = none ];
72 -> 34;

73 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
32 -> 73 [arrowhead = none ];
73 -> 35;

74 [label="EVAL with clause\nhalf(s(s(X25)), s(X26)) :- half(X25, X26).\nand substitutionX25 -> T14,\nT10 -> s(s(T14)),\nX26 -> X27,\nX18 -> s(X27)", fontsize=14, style = filled, fillcolor = lightblue];
33 -> 74 [arrowhead = none ];
74 -> 37;

75 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
33 -> 75 [arrowhead = none ];
75 -> 38;

76 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
34 -> 76 [arrowhead = none ];
76 -> 36;

77 [label="INSTANCE with matching:\nT10 -> T14\nX18 -> X27", fontsize=14, style = filled, fillcolor = lightgrey];
37 -> 77 [arrowhead = none , style=dashed];
77 -> 24[style=dashed];

}

(2) Obligation:

Clauses:

halfA(0, 0).
halfA(s(0), 0).
halfA(s(s(T14)), s(X27)) :- halfA(T14, X27).
logB(0, s(0)).
logB(s(0), s(s(0))).
logB(s(s(T10)), s(T7)) :- halfA(T10, X18).
logB(s(s(T10)), s(T7)) :- ','(halfA(T10, T11), logB(s(T11), T7)).

Query: logB(g,a)

(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:
logB_in: (b,f)
halfA_in: (b,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

logB_in_ga(0, s(0)) → logB_out_ga(0, s(0))
logB_in_ga(s(0), s(s(0))) → logB_out_ga(s(0), s(s(0)))
logB_in_ga(s(s(T10)), s(T7)) → U2_ga(T10, T7, halfA_in_ga(T10, X18))
halfA_in_ga(0, 0) → halfA_out_ga(0, 0)
halfA_in_ga(s(0), 0) → halfA_out_ga(s(0), 0)
halfA_in_ga(s(s(T14)), s(X27)) → U1_ga(T14, X27, halfA_in_ga(T14, X27))
U1_ga(T14, X27, halfA_out_ga(T14, X27)) → halfA_out_ga(s(s(T14)), s(X27))
U2_ga(T10, T7, halfA_out_ga(T10, X18)) → logB_out_ga(s(s(T10)), s(T7))
logB_in_ga(s(s(T10)), s(T7)) → U3_ga(T10, T7, halfA_in_ga(T10, T11))
U3_ga(T10, T7, halfA_out_ga(T10, T11)) → U4_ga(T10, T7, logB_in_ga(s(T11), T7))
U4_ga(T10, T7, logB_out_ga(s(T11), T7)) → logB_out_ga(s(s(T10)), s(T7))

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

logB_in_ga(0, s(0)) → logB_out_ga(0, s(0))
logB_in_ga(s(0), s(s(0))) → logB_out_ga(s(0), s(s(0)))
logB_in_ga(s(s(T10)), s(T7)) → U2_ga(T10, T7, halfA_in_ga(T10, X18))
halfA_in_ga(0, 0) → halfA_out_ga(0, 0)
halfA_in_ga(s(0), 0) → halfA_out_ga(s(0), 0)
halfA_in_ga(s(s(T14)), s(X27)) → U1_ga(T14, X27, halfA_in_ga(T14, X27))
U1_ga(T14, X27, halfA_out_ga(T14, X27)) → halfA_out_ga(s(s(T14)), s(X27))
U2_ga(T10, T7, halfA_out_ga(T10, X18)) → logB_out_ga(s(s(T10)), s(T7))
logB_in_ga(s(s(T10)), s(T7)) → U3_ga(T10, T7, halfA_in_ga(T10, T11))
U3_ga(T10, T7, halfA_out_ga(T10, T11)) → U4_ga(T10, T7, logB_in_ga(s(T11), T7))
U4_ga(T10, T7, logB_out_ga(s(T11), T7)) → logB_out_ga(s(s(T10)), s(T7))

(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:

LOGB_IN_GA(s(s(T10)), s(T7)) → U2_GA(T10, T7, halfA_in_ga(T10, X18))
LOGB_IN_GA(s(s(T10)), s(T7)) → HALFA_IN_GA(T10, X18)
HALFA_IN_GA(s(s(T14)), s(X27)) → U1_GA(T14, X27, halfA_in_ga(T14, X27))
HALFA_IN_GA(s(s(T14)), s(X27)) → HALFA_IN_GA(T14, X27)
LOGB_IN_GA(s(s(T10)), s(T7)) → U3_GA(T10, T7, halfA_in_ga(T10, T11))
U3_GA(T10, T7, halfA_out_ga(T10, T11)) → U4_GA(T10, T7, logB_in_ga(s(T11), T7))
U3_GA(T10, T7, halfA_out_ga(T10, T11)) → LOGB_IN_GA(s(T11), T7)

The TRS R consists of the following rules:

logB_in_ga(0, s(0)) → logB_out_ga(0, s(0))
logB_in_ga(s(0), s(s(0))) → logB_out_ga(s(0), s(s(0)))
logB_in_ga(s(s(T10)), s(T7)) → U2_ga(T10, T7, halfA_in_ga(T10, X18))
halfA_in_ga(0, 0) → halfA_out_ga(0, 0)
halfA_in_ga(s(0), 0) → halfA_out_ga(s(0), 0)
halfA_in_ga(s(s(T14)), s(X27)) → U1_ga(T14, X27, halfA_in_ga(T14, X27))
U1_ga(T14, X27, halfA_out_ga(T14, X27)) → halfA_out_ga(s(s(T14)), s(X27))
U2_ga(T10, T7, halfA_out_ga(T10, X18)) → logB_out_ga(s(s(T10)), s(T7))
logB_in_ga(s(s(T10)), s(T7)) → U3_ga(T10, T7, halfA_in_ga(T10, T11))
U3_ga(T10, T7, halfA_out_ga(T10, T11)) → U4_ga(T10, T7, logB_in_ga(s(T11), T7))
U4_ga(T10, T7, logB_out_ga(s(T11), T7)) → logB_out_ga(s(s(T10)), s(T7))

The argument filtering Pi contains the following mapping:
logB_in_ga(x1, x2) = logB_in_ga(x1)
0 = 0
logB_out_ga(x1, x2) = logB_out_ga
s(x1) = s(x1)
U2_ga(x1, x2, x3) = U2_ga(x3)
halfA_in_ga(x1, x2) = halfA_in_ga(x1)
halfA_out_ga(x1, x2) = halfA_out_ga(x2)
U1_ga(x1, x2, x3) = U1_ga(x3)
U3_ga(x1, x2, x3) = U3_ga(x3)
U4_ga(x1, x2, x3) = U4_ga(x3)
LOGB_IN_GA(x1, x2) = LOGB_IN_GA(x1)
U2_GA(x1, x2, x3) = U2_GA(x3)
HALFA_IN_GA(x1, x2) = HALFA_IN_GA(x1)
U1_GA(x1, x2, x3) = U1_GA(x3)
U3_GA(x1, x2, x3) = U3_GA(x3)
U4_GA(x1, x2, x3) = U4_GA(x3)

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

(6) Obligation:

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

LOGB_IN_GA(s(s(T10)), s(T7)) → U2_GA(T10, T7, halfA_in_ga(T10, X18))
LOGB_IN_GA(s(s(T10)), s(T7)) → HALFA_IN_GA(T10, X18)
HALFA_IN_GA(s(s(T14)), s(X27)) → U1_GA(T14, X27, halfA_in_ga(T14, X27))
HALFA_IN_GA(s(s(T14)), s(X27)) → HALFA_IN_GA(T14, X27)
LOGB_IN_GA(s(s(T10)), s(T7)) → U3_GA(T10, T7, halfA_in_ga(T10, T11))
U3_GA(T10, T7, halfA_out_ga(T10, T11)) → U4_GA(T10, T7, logB_in_ga(s(T11), T7))
U3_GA(T10, T7, halfA_out_ga(T10, T11)) → LOGB_IN_GA(s(T11), T7)

The TRS R consists of the following rules:

logB_in_ga(0, s(0)) → logB_out_ga(0, s(0))
logB_in_ga(s(0), s(s(0))) → logB_out_ga(s(0), s(s(0)))
logB_in_ga(s(s(T10)), s(T7)) → U2_ga(T10, T7, halfA_in_ga(T10, X18))
halfA_in_ga(0, 0) → halfA_out_ga(0, 0)
halfA_in_ga(s(0), 0) → halfA_out_ga(s(0), 0)
halfA_in_ga(s(s(T14)), s(X27)) → U1_ga(T14, X27, halfA_in_ga(T14, X27))
U1_ga(T14, X27, halfA_out_ga(T14, X27)) → halfA_out_ga(s(s(T14)), s(X27))
U2_ga(T10, T7, halfA_out_ga(T10, X18)) → logB_out_ga(s(s(T10)), s(T7))
logB_in_ga(s(s(T10)), s(T7)) → U3_ga(T10, T7, halfA_in_ga(T10, T11))
U3_ga(T10, T7, halfA_out_ga(T10, T11)) → U4_ga(T10, T7, logB_in_ga(s(T11), T7))
U4_ga(T10, T7, logB_out_ga(s(T11), T7)) → logB_out_ga(s(s(T10)), s(T7))

(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:

HALFA_IN_GA(s(s(T14)), s(X27)) → HALFA_IN_GA(T14, X27)

The TRS R consists of the following rules:

logB_in_ga(0, s(0)) → logB_out_ga(0, s(0))
logB_in_ga(s(0), s(s(0))) → logB_out_ga(s(0), s(s(0)))
logB_in_ga(s(s(T10)), s(T7)) → U2_ga(T10, T7, halfA_in_ga(T10, X18))
halfA_in_ga(0, 0) → halfA_out_ga(0, 0)
halfA_in_ga(s(0), 0) → halfA_out_ga(s(0), 0)
halfA_in_ga(s(s(T14)), s(X27)) → U1_ga(T14, X27, halfA_in_ga(T14, X27))
U1_ga(T14, X27, halfA_out_ga(T14, X27)) → halfA_out_ga(s(s(T14)), s(X27))
U2_ga(T10, T7, halfA_out_ga(T10, X18)) → logB_out_ga(s(s(T10)), s(T7))
logB_in_ga(s(s(T10)), s(T7)) → U3_ga(T10, T7, halfA_in_ga(T10, T11))
U3_ga(T10, T7, halfA_out_ga(T10, T11)) → U4_ga(T10, T7, logB_in_ga(s(T11), T7))
U4_ga(T10, T7, logB_out_ga(s(T11), T7)) → logB_out_ga(s(s(T10)), s(T7))

The argument filtering Pi contains the following mapping:
logB_in_ga(x1, x2) = logB_in_ga(x1)
0 = 0
logB_out_ga(x1, x2) = logB_out_ga
s(x1) = s(x1)
U2_ga(x1, x2, x3) = U2_ga(x3)
halfA_in_ga(x1, x2) = halfA_in_ga(x1)
halfA_out_ga(x1, x2) = halfA_out_ga(x2)
U1_ga(x1, x2, x3) = U1_ga(x3)
U3_ga(x1, x2, x3) = U3_ga(x3)
U4_ga(x1, x2, x3) = U4_ga(x3)
HALFA_IN_GA(x1, x2) = HALFA_IN_GA(x1)

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:

HALFA_IN_GA(s(s(T14)), s(X27)) → HALFA_IN_GA(T14, X27)

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

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

(12) PiDPToQDPProof (SOUND 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:

HALFA_IN_GA(s(s(T14))) → HALFA_IN_GA(T14)

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:

HALFA_IN_GA(s(s(T14))) → HALFA_IN_GA(T14)
The graph contains the following edges 1 > 1

(15) YES

(16) Obligation:

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

LOGB_IN_GA(s(s(T10)), s(T7)) → U3_GA(T10, T7, halfA_in_ga(T10, T11))
U3_GA(T10, T7, halfA_out_ga(T10, T11)) → LOGB_IN_GA(s(T11), T7)

The TRS R consists of the following rules:

logB_in_ga(0, s(0)) → logB_out_ga(0, s(0))
logB_in_ga(s(0), s(s(0))) → logB_out_ga(s(0), s(s(0)))
logB_in_ga(s(s(T10)), s(T7)) → U2_ga(T10, T7, halfA_in_ga(T10, X18))
halfA_in_ga(0, 0) → halfA_out_ga(0, 0)
halfA_in_ga(s(0), 0) → halfA_out_ga(s(0), 0)
halfA_in_ga(s(s(T14)), s(X27)) → U1_ga(T14, X27, halfA_in_ga(T14, X27))
U1_ga(T14, X27, halfA_out_ga(T14, X27)) → halfA_out_ga(s(s(T14)), s(X27))
U2_ga(T10, T7, halfA_out_ga(T10, X18)) → logB_out_ga(s(s(T10)), s(T7))
logB_in_ga(s(s(T10)), s(T7)) → U3_ga(T10, T7, halfA_in_ga(T10, T11))
U3_ga(T10, T7, halfA_out_ga(T10, T11)) → U4_ga(T10, T7, logB_in_ga(s(T11), T7))
U4_ga(T10, T7, logB_out_ga(s(T11), T7)) → logB_out_ga(s(s(T10)), s(T7))

The argument filtering Pi contains the following mapping:
logB_in_ga(x1, x2) = logB_in_ga(x1)
0 = 0
logB_out_ga(x1, x2) = logB_out_ga
s(x1) = s(x1)
U2_ga(x1, x2, x3) = U2_ga(x3)
halfA_in_ga(x1, x2) = halfA_in_ga(x1)
halfA_out_ga(x1, x2) = halfA_out_ga(x2)
U1_ga(x1, x2, x3) = U1_ga(x3)
U3_ga(x1, x2, x3) = U3_ga(x3)
U4_ga(x1, x2, x3) = U4_ga(x3)
LOGB_IN_GA(x1, x2) = LOGB_IN_GA(x1)
U3_GA(x1, x2, x3) = U3_GA(x3)

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:

LOGB_IN_GA(s(s(T10)), s(T7)) → U3_GA(T10, T7, halfA_in_ga(T10, T11))
U3_GA(T10, T7, halfA_out_ga(T10, T11)) → LOGB_IN_GA(s(T11), T7)

The TRS R consists of the following rules:

halfA_in_ga(0, 0) → halfA_out_ga(0, 0)
halfA_in_ga(s(0), 0) → halfA_out_ga(s(0), 0)
halfA_in_ga(s(s(T14)), s(X27)) → U1_ga(T14, X27, halfA_in_ga(T14, X27))
U1_ga(T14, X27, halfA_out_ga(T14, X27)) → halfA_out_ga(s(s(T14)), s(X27))

The argument filtering Pi contains the following mapping:
0 = 0
s(x1) = s(x1)
halfA_in_ga(x1, x2) = halfA_in_ga(x1)
halfA_out_ga(x1, x2) = halfA_out_ga(x2)
U1_ga(x1, x2, x3) = U1_ga(x3)
LOGB_IN_GA(x1, x2) = LOGB_IN_GA(x1)
U3_GA(x1, x2, x3) = U3_GA(x3)

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:

LOGB_IN_GA(s(s(T10))) → U3_GA(halfA_in_ga(T10))
U3_GA(halfA_out_ga(T11)) → LOGB_IN_GA(s(T11))

The TRS R consists of the following rules:

halfA_in_ga(0) → halfA_out_ga(0)
halfA_in_ga(s(0)) → halfA_out_ga(0)
halfA_in_ga(s(s(T14))) → U1_ga(halfA_in_ga(T14))
U1_ga(halfA_out_ga(X27)) → halfA_out_ga(s(X27))

The set Q consists of the following terms:

halfA_in_ga(x0)
U1_ga(x0)

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

(21) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

LOGB_IN_GA(s(s(T10))) → U3_GA(halfA_in_ga(T10))
U3_GA(halfA_out_ga(T11)) → LOGB_IN_GA(s(T11))

Strictly oriented rules of the TRS R:

halfA_in_ga(0) → halfA_out_ga(0)
halfA_in_ga(s(0)) → halfA_out_ga(0)
halfA_in_ga(s(s(T14))) → U1_ga(halfA_in_ga(T14))
U1_ga(halfA_out_ga(X27)) → halfA_out_ga(s(X27))

Used ordering: Knuth-Bendix order [KBO] with precedence:

U3_GA₁ > LOGB_{IN_GA₁} > s₁ > halfA_{in_ga₁} > halfA_{out_ga₁} > 0 > U1_ga₁

and weight map:

0=1
halfA_in_ga_1=7
halfA_out_ga_1=6
s_1=4
U1_ga_1=5
LOGB_IN_GA_1=5
U3_GA_1=3

The variable weight is 1

(22) Obligation:

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

halfA_in_ga(x0)
U1_ga(x0)

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) PrologToPrologProblemTransformerProof (SOUND transformation)

(2) Obligation:

(3) PrologToPiTRSProof (SOUND transformation)

(4) Obligation:

(5) DependencyPairsProof (EQUIVALENT transformation)

(6) Obligation:

(7) DependencyGraphProof (EQUIVALENT transformation)

(8) Complex Obligation (AND)

(9) Obligation:

(10) UsableRulesProof (EQUIVALENT transformation)

(11) Obligation:

(12) PiDPToQDPProof (SOUND transformation)

(13) Obligation:

(14) QDPSizeChangeProof (EQUIVALENT transformation)

(15) YES

(16) Obligation:

(17) UsableRulesProof (EQUIVALENT transformation)

(18) Obligation:

(19) PiDPToQDPProof (SOUND transformation)

(20) Obligation:

(21) MRRProof (EQUIVALENT transformation)

(22) Obligation:

(23) PisEmptyProof (EQUIVALENT transformation)

(24) YES