Left Termination proof of /tmp/tmpjiPBI4/log.pl

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

0 Prolog

↳1 PrologToDTProblemTransformerProof (⇒, 88 ms)

↳2 TRIPLES

↳3 TriplesToPiDPProof (⇒, 0 ms)

↳4 PiDP

↳5 DependencyGraphProof (⇔, 0 ms)

↳6 AND

    ↳7 PiDP

        ↳8 UsableRulesProof (⇔, 0 ms)

        ↳9 PiDP

        ↳10 PiDPToQDPProof (⇒, 14 ms)

        ↳11 QDP

        ↳12 QDPSizeChangeProof (⇔, 0 ms)

        ↳13 YES

    ↳14 PiDP

        ↳15 PiDPToQDPProof (⇒, 0 ms)

        ↳16 QDP

        ↳17 QDPOrderProof (⇔, 38 ms)

        ↳18 QDP

        ↳19 DependencyGraphProof (⇔, 0 ms)

        ↳20 TRUE

(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) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: 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="B: 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:

Triples:

halfB(s(s(X1)), s(X2)) :- halfB(X1, X2).
logA(s(s(X1)), s(X2)) :- halfB(X1, X3).
logA(s(s(X1)), s(X2)) :- ','(halfcB(X1, X3), logA(s(X3), X2)).

Clauses:

logcA(0, s(0)).
logcA(s(0), s(s(0))).
logcA(s(s(X1)), s(X2)) :- ','(halfcB(X1, X3), logcA(s(X3), X2)).
halfcB(0, 0).
halfcB(s(0), 0).
halfcB(s(s(X1)), s(X2)) :- halfcB(X1, X2).

Afs:

logA(x1, x2) = logA(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:
logA_in: (b,f)
halfB_in: (b,f)
halfcB_in: (b,f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

LOGA_IN_GA(s(s(X1)), s(X2)) → U2_GA(X1, X2, halfB_in_ga(X1, X3))
LOGA_IN_GA(s(s(X1)), s(X2)) → HALFB_IN_GA(X1, X3)
HALFB_IN_GA(s(s(X1)), s(X2)) → U1_GA(X1, X2, halfB_in_ga(X1, X2))
HALFB_IN_GA(s(s(X1)), s(X2)) → HALFB_IN_GA(X1, X2)
LOGA_IN_GA(s(s(X1)), s(X2)) → U3_GA(X1, X2, halfcB_in_ga(X1, X3))
U3_GA(X1, X2, halfcB_out_ga(X1, X3)) → U4_GA(X1, X2, logA_in_ga(s(X3), X2))
U3_GA(X1, X2, halfcB_out_ga(X1, X3)) → LOGA_IN_GA(s(X3), X2)

The TRS R consists of the following rules:

halfcB_in_ga(0, 0) → halfcB_out_ga(0, 0)
halfcB_in_ga(s(0), 0) → halfcB_out_ga(s(0), 0)
halfcB_in_ga(s(s(X1)), s(X2)) → U8_ga(X1, X2, halfcB_in_ga(X1, X2))
U8_ga(X1, X2, halfcB_out_ga(X1, X2)) → halfcB_out_ga(s(s(X1)), s(X2))

The argument filtering Pi contains the following mapping:
logA_in_ga(x1, x2) = logA_in_ga(x1)
s(x1) = s(x1)
halfB_in_ga(x1, x2) = halfB_in_ga(x1)
halfcB_in_ga(x1, x2) = halfcB_in_ga(x1)
0 = 0
halfcB_out_ga(x1, x2) = halfcB_out_ga(x1, x2)
U8_ga(x1, x2, x3) = U8_ga(x1, x3)
LOGA_IN_GA(x1, x2) = LOGA_IN_GA(x1)
U2_GA(x1, x2, x3) = U2_GA(x1, x3)
HALFB_IN_GA(x1, x2) = HALFB_IN_GA(x1)
U1_GA(x1, x2, x3) = U1_GA(x1, x3)
U3_GA(x1, x2, x3) = U3_GA(x1, x3)
U4_GA(x1, x2, x3) = U4_GA(x1, x3)

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:

LOGA_IN_GA(s(s(X1)), s(X2)) → U2_GA(X1, X2, halfB_in_ga(X1, X3))
LOGA_IN_GA(s(s(X1)), s(X2)) → HALFB_IN_GA(X1, X3)
HALFB_IN_GA(s(s(X1)), s(X2)) → U1_GA(X1, X2, halfB_in_ga(X1, X2))
HALFB_IN_GA(s(s(X1)), s(X2)) → HALFB_IN_GA(X1, X2)
LOGA_IN_GA(s(s(X1)), s(X2)) → U3_GA(X1, X2, halfcB_in_ga(X1, X3))
U3_GA(X1, X2, halfcB_out_ga(X1, X3)) → U4_GA(X1, X2, logA_in_ga(s(X3), X2))
U3_GA(X1, X2, halfcB_out_ga(X1, X3)) → LOGA_IN_GA(s(X3), X2)

The TRS R consists of the following rules:

halfcB_in_ga(0, 0) → halfcB_out_ga(0, 0)
halfcB_in_ga(s(0), 0) → halfcB_out_ga(s(0), 0)
halfcB_in_ga(s(s(X1)), s(X2)) → U8_ga(X1, X2, halfcB_in_ga(X1, X2))
U8_ga(X1, X2, halfcB_out_ga(X1, X2)) → halfcB_out_ga(s(s(X1)), s(X2))

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) Obligation:

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

HALFB_IN_GA(s(s(X1)), s(X2)) → HALFB_IN_GA(X1, X2)

The TRS R consists of the following rules:

halfcB_in_ga(0, 0) → halfcB_out_ga(0, 0)
halfcB_in_ga(s(0), 0) → halfcB_out_ga(s(0), 0)
halfcB_in_ga(s(s(X1)), s(X2)) → U8_ga(X1, X2, halfcB_in_ga(X1, X2))
U8_ga(X1, X2, halfcB_out_ga(X1, X2)) → halfcB_out_ga(s(s(X1)), s(X2))

The argument filtering Pi contains the following mapping:
s(x1) = s(x1)
halfcB_in_ga(x1, x2) = halfcB_in_ga(x1)
0 = 0
halfcB_out_ga(x1, x2) = halfcB_out_ga(x1, x2)
U8_ga(x1, x2, x3) = U8_ga(x1, x3)
HALFB_IN_GA(x1, x2) = HALFB_IN_GA(x1)

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

(8) UsableRulesProof (EQUIVALENT transformation)

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

(9) Obligation:

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

HALFB_IN_GA(s(s(X1)), s(X2)) → HALFB_IN_GA(X1, X2)

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

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

(10) PiDPToQDPProof (SOUND transformation)

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

(11) Obligation:

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

HALFB_IN_GA(s(s(X1))) → HALFB_IN_GA(X1)

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

(12) 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:

HALFB_IN_GA(s(s(X1))) → HALFB_IN_GA(X1)
The graph contains the following edges 1 > 1

(13) YES

(14) Obligation:

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

LOGA_IN_GA(s(s(X1)), s(X2)) → U3_GA(X1, X2, halfcB_in_ga(X1, X3))
U3_GA(X1, X2, halfcB_out_ga(X1, X3)) → LOGA_IN_GA(s(X3), X2)

The TRS R consists of the following rules:

halfcB_in_ga(0, 0) → halfcB_out_ga(0, 0)
halfcB_in_ga(s(0), 0) → halfcB_out_ga(s(0), 0)
halfcB_in_ga(s(s(X1)), s(X2)) → U8_ga(X1, X2, halfcB_in_ga(X1, X2))
U8_ga(X1, X2, halfcB_out_ga(X1, X2)) → halfcB_out_ga(s(s(X1)), s(X2))

The argument filtering Pi contains the following mapping:
s(x1) = s(x1)
halfcB_in_ga(x1, x2) = halfcB_in_ga(x1)
0 = 0
halfcB_out_ga(x1, x2) = halfcB_out_ga(x1, x2)
U8_ga(x1, x2, x3) = U8_ga(x1, x3)
LOGA_IN_GA(x1, x2) = LOGA_IN_GA(x1)
U3_GA(x1, x2, x3) = U3_GA(x1, x3)

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

(15) PiDPToQDPProof (SOUND transformation)

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

(16) Obligation:

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

LOGA_IN_GA(s(s(X1))) → U3_GA(X1, halfcB_in_ga(X1))
U3_GA(X1, halfcB_out_ga(X1, X3)) → LOGA_IN_GA(s(X3))

The TRS R consists of the following rules:

halfcB_in_ga(0) → halfcB_out_ga(0, 0)
halfcB_in_ga(s(0)) → halfcB_out_ga(s(0), 0)
halfcB_in_ga(s(s(X1))) → U8_ga(X1, halfcB_in_ga(X1))
U8_ga(X1, halfcB_out_ga(X1, X2)) → halfcB_out_ga(s(s(X1)), s(X2))

The set Q consists of the following terms:

halfcB_in_ga(x0)
U8_ga(x0, x1)

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

(17) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

LOGA_IN_GA(s(s(X1))) → U3_GA(X1, halfcB_in_ga(X1))

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

POL(0) = 0
POL(LOGA_IN_GA(x₁)) = x₁
POL(U3_GA(x₁, x₂)) = 1 + x₂
POL(U8_ga(x₁, x₂)) = 1 + x₂
POL(halfcB_in_ga(x₁)) = x₁
POL(halfcB_out_ga(x₁, x₂)) = x₂
POL(s(x₁)) = 1 + x₁

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

halfcB_in_ga(0) → halfcB_out_ga(0, 0)
halfcB_in_ga(s(0)) → halfcB_out_ga(s(0), 0)
halfcB_in_ga(s(s(X1))) → U8_ga(X1, halfcB_in_ga(X1))
U8_ga(X1, halfcB_out_ga(X1, X2)) → halfcB_out_ga(s(s(X1)), s(X2))

(18) Obligation:

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

U3_GA(X1, halfcB_out_ga(X1, X3)) → LOGA_IN_GA(s(X3))

The TRS R consists of the following rules:

halfcB_in_ga(0) → halfcB_out_ga(0, 0)
halfcB_in_ga(s(0)) → halfcB_out_ga(s(0), 0)
halfcB_in_ga(s(s(X1))) → U8_ga(X1, halfcB_in_ga(X1))
U8_ga(X1, halfcB_out_ga(X1, X2)) → halfcB_out_ga(s(s(X1)), s(X2))

The set Q consists of the following terms:

halfcB_in_ga(x0)
U8_ga(x0, x1)

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

(19) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(0) Obligation:

(1) PrologToDTProblemTransformerProof (SOUND transformation)

(2) Obligation:

(3) TriplesToPiDPProof (SOUND transformation)

(4) Obligation:

(5) DependencyGraphProof (EQUIVALENT transformation)

(6) Complex Obligation (AND)

(7) Obligation:

(8) UsableRulesProof (EQUIVALENT transformation)

(9) Obligation:

(10) PiDPToQDPProof (SOUND transformation)

(11) Obligation:

(12) QDPSizeChangeProof (EQUIVALENT transformation)

(13) YES

(14) Obligation:

(15) PiDPToQDPProof (SOUND transformation)

(16) Obligation:

(17) QDPOrderProof (EQUIVALENT transformation)

(18) Obligation:

(19) DependencyGraphProof (EQUIVALENT transformation)

(20) TRUE