Left Termination proof of /tmp/tmplLfIBI/add3.pl

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

0 Prolog

↳1 PrologToPrologProblemTransformerProof (⇒, 78 ms)

↳2 Prolog

↳3 PrologToPiTRSProof (⇒, 0 ms)

↳4 PiTRS

↳5 DependencyPairsProof (⇔, 0 ms)

↳6 PiDP

↳7 DependencyGraphProof (⇔, 0 ms)

↳8 PiDP

↳9 UsableRulesProof (⇔, 0 ms)

↳10 PiDP

↳11 PiDPToQDPProof (⇒, 7 ms)

↳12 QDP

↳13 QDPSizeChangeProof (⇔, 0 ms)

↳14 YES

(0) Obligation:

Clauses:

add(X, 0, X).
add(X, Y, s(Z)) :- ','(no(zero(Y)), ','(p(Y, P), add(X, P, Z))).
p(0, 0).
p(s(X), X).
zero(0).
no(X) :- ','(X, ','(!, failure(a))).
no(X).
failure(b).

Query: add(a,g,a)

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: add(T1, T2, T3)\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: add(T1, T2, T3), SCOPE: 1, CLAUSE: 0 | add(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: true | add(T1, 0, T3), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: add(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT2 is ground\nX2 is free\nadd(T1, T2, T3) !=? add(X2, 0, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: add(T1, 0, T3), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: ','(no(zero(0)), ','(p(0, X9), add(T10, X9, T11)))\n\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: ','(no(zero(0)), ','(p(0, X9), add(T10, X9, T11))), SCOPE: 2, CLAUSE: 5 | ','(no(zero(0)), ','(p(0, X9), add(T10, X9, T11))), SCOPE: 2, CLAUSE: 6\n\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(','(call(zero(0)), ','(!_2, failure(a))), ','(p(0, X9), add(T10, X9, T11))) | ','(no(zero(0)), ','(p(0, X9), add(T10, X9, T11))), SCOPE: 2, CLAUSE: 6\n\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(zero(0), ','(','(!_2, failure(a)), ','(p(0, X9), add(T10, X9, T11)))) | ?_3 | ','(no(zero(0)), ','(p(0, X9), add(T10, X9, T11))), SCOPE: 2, CLAUSE: 6\n\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(zero(0), ','(','(!_2, failure(a)), ','(p(0, X9), add(T10, X9, T11)))), SCOPE: 4, CLAUSE: 4 | ?_4 | ?_3 | ','(no(zero(0)), ','(p(0, X9), add(T10, X9, T11))), SCOPE: 2, CLAUSE: 6\n\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(','(!_2, failure(a)), ','(p(0, X9), add(T10, X9, T11))) | ?_4 | ?_3 | ','(no(zero(0)), ','(p(0, X9), add(T10, X9, T11))), SCOPE: 2, CLAUSE: 6\n\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(failure(a), ','(p(0, X9), add(T10, X9, T11)))\n\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(failure(a), ','(p(0, X9), add(T10, X9, T11))), SCOPE: 5, CLAUSE: 7\n\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: ','(no(zero(T18)), ','(p(T18, X21), add(T20, X21, T21)))\n\nT18 is ground\nX2 is free\nX21 is free\nadd(T1, T18, T3) !=? add(X2, 0, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: ','(no(zero(T18)), ','(p(T18, X21), add(T20, X21, T21))), SCOPE: 6, CLAUSE: 5 | ','(no(zero(T18)), ','(p(T18, X21), add(T20, X21, T21))), SCOPE: 6, CLAUSE: 6\n\nT18 is ground\nX2 is free\nX21 is free\nadd(T1, T18, T3) !=? add(X2, 0, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: ','(','(call(zero(T24)), ','(!_6, failure(a))), ','(p(T24, X21), add(T20, X21, T21))) | ','(no(zero(T24)), ','(p(T24, X21), add(T20, X21, T21))), SCOPE: 6, CLAUSE: 6\n\nT24 is ground\nX2 is free\nX21 is free\nadd(T1, T24, T3) !=? add(X2, 0, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: ','(zero(T24), ','(','(!_6, failure(a)), ','(p(T24, X21), add(T20, X21, T21)))) | ?_7 | ','(no(zero(T24)), ','(p(T24, X21), add(T20, X21, T21))), SCOPE: 6, CLAUSE: 6\n\nT24 is ground\nX2 is free\nX21 is free\nadd(T1, T24, T3) !=? add(X2, 0, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: ','(zero(T24), ','(','(!_6, failure(a)), ','(p(T24, X21), add(T20, X21, T21)))), SCOPE: 8, CLAUSE: 4 | ?_8 | ?_7 | ','(no(zero(T24)), ','(p(T24, X21), add(T20, X21, T21))), SCOPE: 6, CLAUSE: 6\n\nT24 is ground\nX2 is free\nX21 is free\nadd(T1, T24, T3) !=? add(X2, 0, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: ','(','(!_6, failure(a)), ','(p(0, X21), add(T20, X21, T21))) | ?_8 | ?_7 | ','(no(zero(0)), ','(p(0, X21), add(T20, X21, T21))), SCOPE: 6, CLAUSE: 6\n\nX21 is free", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: ?_8 | ?_7 | ','(no(zero(T24)), ','(p(T24, X21), add(T20, X21, T21))), SCOPE: 6, CLAUSE: 6\n\nT24 is ground\nX2 is free\nX21 is free\nadd(T1, T24, T3) !=? add(X2, 0, X2)\nzero(T24) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: ','(failure(a), ','(p(0, X21), add(T20, X21, T21)))\n\nX21 is free", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: ','(failure(a), ','(p(0, X21), add(T20, X21, T21))), SCOPE: 9, CLAUSE: 7\n\nX21 is free", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: ?_7 | ','(no(zero(T24)), ','(p(T24, X21), add(T20, X21, T21))), SCOPE: 6, CLAUSE: 6\n\nT24 is ground\nX2 is free\nX21 is free\nadd(T1, T24, T3) !=? add(X2, 0, X2)\nzero(T24) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: ','(no(zero(T24)), ','(p(T24, X21), add(T20, X21, T21))), SCOPE: 6, CLAUSE: 6\n\nT24 is ground\nX2 is free\nX21 is free\nadd(T1, T24, T3) !=? add(X2, 0, X2)\nzero(T24) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="29: ','(p(T31, X21), add(T20, X21, T21))\n\nT31 is ground\nX2 is free\nX21 is free\nadd(T1, T31, T3) !=? add(X2, 0, X2)\nzero(T31) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
30 [label="30: ','(p(T31, X21), add(T20, X21, T21)), SCOPE: 10, CLAUSE: 2 | ','(p(T31, X21), add(T20, X21, T21)), SCOPE: 10, CLAUSE: 3\n\nT31 is ground\nX2 is free\nX21 is free\nadd(T1, T31, T3) !=? add(X2, 0, X2)\nzero(T31) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
31 [label="31: ','(p(T31, X21), add(T20, X21, T21)), SCOPE: 10, CLAUSE: 3\n\nT31 is ground\nX2 is free\nX21 is free\nadd(T1, T31, T3) !=? add(X2, 0, X2)\nzero(T31) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
32 [label="32: add(T20, T36, T21)\n\nT36 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
33 [label="33: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
34 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 34 [arrowhead = none ];
34 -> 2;

35 [label="EVAL with clause\nadd(X2, 0, X2).\nand substitutionT1 -> T5,\nX2 -> T5,\nT2 -> 0,\nT3 -> T5", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 35 [arrowhead = none ];
35 -> 3;

36 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 36 [arrowhead = none ];
36 -> 4;

37 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 37 [arrowhead = none ];
37 -> 5;

38 [label="EVAL with clause\nadd(X18, X19, s(X20)) :- ','(no(zero(X19)), ','(p(X19, X21), add(X18, X21, X20))).\nand substitutionT1 -> T20,\nX18 -> T20,\nT2 -> T18,\nX19 -> T18,\nX20 -> T21,\nT3 -> s(T21),\nT17 -> T20,\nT19 -> T21", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 38 [arrowhead = none ];
38 -> 16;

39 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 39 [arrowhead = none ];
39 -> 17;

40 [label="EVAL with clause\nadd(X6, X7, s(X8)) :- ','(no(zero(X7)), ','(p(X7, X9), add(X6, X9, X8))).\nand substitutionT1 -> T10,\nX6 -> T10,\nX7 -> 0,\nX8 -> T11,\nT3 -> s(T11),\nT8 -> T10,\nT9 -> T11", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 40 [arrowhead = none ];
40 -> 6;

41 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 41 [arrowhead = none ];
41 -> 7;

42 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
6 -> 42 [arrowhead = none ];
42 -> 8;

43 [label="ONLY EVAL with clause\nno(X12) :- ','(call(X12), ','(!_2, failure(a))).\nand substitutionX12 -> zero(0)", fontsize=14, style = filled, fillcolor = lightblue];
8 -> 43 [arrowhead = none ];
43 -> 9;

44 [label="CALL", fontsize=14, style = filled, fillcolor = lightcyan];
9 -> 44 [arrowhead = none ];
44 -> 10;

45 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
10 -> 45 [arrowhead = none ];
45 -> 11;

46 [label="ONLY EVAL with clause\nzero(0).\nand substitution", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 46 [arrowhead = none ];
46 -> 12;

47 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 47 [arrowhead = none ];
47 -> 13;

48 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
13 -> 48 [arrowhead = none ];
48 -> 14;

49 [label="BACKTRACK\nfor clause: failure(b)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
14 -> 49 [arrowhead = none ];
49 -> 15;

50 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
16 -> 50 [arrowhead = none ];
50 -> 18;

51 [label="ONLY EVAL with clause\nno(X24) :- ','(call(X24), ','(!_6, failure(a))).\nand substitutionT18 -> T24,\nX24 -> zero(T24)", fontsize=14, style = filled, fillcolor = lightblue];
18 -> 51 [arrowhead = none ];
51 -> 19;

52 [label="CALL", fontsize=14, style = filled, fillcolor = lightcyan];
19 -> 52 [arrowhead = none ];
52 -> 20;

53 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
20 -> 53 [arrowhead = none ];
53 -> 21;

54 [label="EVAL with clause\nzero(0).\nand substitutionT24 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
21 -> 54 [arrowhead = none ];
54 -> 22;

55 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
21 -> 55 [arrowhead = none ];
55 -> 23;

56 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
22 -> 56 [arrowhead = none ];
56 -> 24;

57 [label="FAILURE", fontsize=14, style = filled, fillcolor = lightgreen];
23 -> 57 [arrowhead = none ];
57 -> 27;

58 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
24 -> 58 [arrowhead = none ];
58 -> 25;

59 [label="BACKTRACK\nfor clause: failure(b)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
25 -> 59 [arrowhead = none ];
59 -> 26;

60 [label="FAILURE", fontsize=14, style = filled, fillcolor = lightgreen];
27 -> 60 [arrowhead = none ];
60 -> 28;

61 [label="ONLY EVAL with clause\nno(X31).\nand substitutionT24 -> T31,\nX31 -> zero(T31)", fontsize=14, style = filled, fillcolor = lightblue];
28 -> 61 [arrowhead = none ];
61 -> 29;

62 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
29 -> 62 [arrowhead = none ];
62 -> 30;

63 [label="BACKTRACK\nfor clause: p(0, 0)\nwith clash: (zero(T31), zero(0))", fontsize=14, style = filled, fillcolor = lightgreen];
30 -> 63 [arrowhead = none ];
63 -> 31;

64 [label="EVAL with clause\np(s(X36), X36).\nand substitutionX36 -> T36,\nT31 -> s(T36),\nX21 -> T36", fontsize=14, style = filled, fillcolor = lightblue];
31 -> 64 [arrowhead = none ];
64 -> 32;

65 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
31 -> 65 [arrowhead = none ];
65 -> 33;

66 [label="INSTANCE with matching:\nT1 -> T20\nT2 -> T36\nT3 -> T21", fontsize=14, style = filled, fillcolor = lightgrey];
32 -> 66 [arrowhead = none , style=dashed];
66 -> 1[style=dashed];

}

(2) Obligation:

Clauses:

addA(T5, 0, T5).
addA(T20, s(T36), s(T21)) :- addA(T20, T36, T21).

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

addA_in_aga(T5, 0, T5) → addA_out_aga(T5, 0, T5)
addA_in_aga(T20, s(T36), s(T21)) → U1_aga(T20, T36, T21, addA_in_aga(T20, T36, T21))
U1_aga(T20, T36, T21, addA_out_aga(T20, T36, T21)) → addA_out_aga(T20, s(T36), s(T21))

The argument filtering Pi contains the following mapping:
addA_in_aga(x1, x2, x3) = addA_in_aga(x2)
0 = 0
addA_out_aga(x1, x2, x3) = addA_out_aga
s(x1) = s(x1)
U1_aga(x1, x2, x3, x4) = U1_aga(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:

addA_in_aga(T5, 0, T5) → addA_out_aga(T5, 0, T5)
addA_in_aga(T20, s(T36), s(T21)) → U1_aga(T20, T36, T21, addA_in_aga(T20, T36, T21))
U1_aga(T20, T36, T21, addA_out_aga(T20, T36, T21)) → addA_out_aga(T20, s(T36), s(T21))

The argument filtering Pi contains the following mapping:
addA_in_aga(x1, x2, x3) = addA_in_aga(x2)
0 = 0
addA_out_aga(x1, x2, x3) = addA_out_aga
s(x1) = s(x1)
U1_aga(x1, x2, x3, x4) = U1_aga(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:

ADDA_IN_AGA(T20, s(T36), s(T21)) → U1_AGA(T20, T36, T21, addA_in_aga(T20, T36, T21))
ADDA_IN_AGA(T20, s(T36), s(T21)) → ADDA_IN_AGA(T20, T36, T21)

The TRS R consists of the following rules:

addA_in_aga(T5, 0, T5) → addA_out_aga(T5, 0, T5)
addA_in_aga(T20, s(T36), s(T21)) → U1_aga(T20, T36, T21, addA_in_aga(T20, T36, T21))
U1_aga(T20, T36, T21, addA_out_aga(T20, T36, T21)) → addA_out_aga(T20, s(T36), s(T21))

The argument filtering Pi contains the following mapping:
addA_in_aga(x1, x2, x3) = addA_in_aga(x2)
0 = 0
addA_out_aga(x1, x2, x3) = addA_out_aga
s(x1) = s(x1)
U1_aga(x1, x2, x3, x4) = U1_aga(x4)
ADDA_IN_AGA(x1, x2, x3) = ADDA_IN_AGA(x2)
U1_AGA(x1, x2, x3, x4) = U1_AGA(x4)

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

(6) Obligation:

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

ADDA_IN_AGA(T20, s(T36), s(T21)) → U1_AGA(T20, T36, T21, addA_in_aga(T20, T36, T21))
ADDA_IN_AGA(T20, s(T36), s(T21)) → ADDA_IN_AGA(T20, T36, T21)

The TRS R consists of the following rules:

addA_in_aga(T5, 0, T5) → addA_out_aga(T5, 0, T5)
addA_in_aga(T20, s(T36), s(T21)) → U1_aga(T20, T36, T21, addA_in_aga(T20, T36, T21))
U1_aga(T20, T36, T21, addA_out_aga(T20, T36, T21)) → addA_out_aga(T20, s(T36), s(T21))

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(8) Obligation:

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

ADDA_IN_AGA(T20, s(T36), s(T21)) → ADDA_IN_AGA(T20, T36, T21)

The TRS R consists of the following rules:

addA_in_aga(T5, 0, T5) → addA_out_aga(T5, 0, T5)
addA_in_aga(T20, s(T36), s(T21)) → U1_aga(T20, T36, T21, addA_in_aga(T20, T36, T21))
U1_aga(T20, T36, T21, addA_out_aga(T20, T36, T21)) → addA_out_aga(T20, s(T36), s(T21))

(9) UsableRulesProof (EQUIVALENT transformation)

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

(10) Obligation:

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

ADDA_IN_AGA(T20, s(T36), s(T21)) → ADDA_IN_AGA(T20, T36, T21)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1) = s(x1)
ADDA_IN_AGA(x1, x2, x3) = ADDA_IN_AGA(x2)

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

(11) PiDPToQDPProof (SOUND transformation)

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

(12) Obligation:

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

ADDA_IN_AGA(s(T36)) → ADDA_IN_AGA(T36)

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

(13) 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:

ADDA_IN_AGA(s(T36)) → ADDA_IN_AGA(T36)
The graph contains the following edges 1 > 1

(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) Obligation:

(9) UsableRulesProof (EQUIVALENT transformation)

(10) Obligation:

(11) PiDPToQDPProof (SOUND transformation)

(12) Obligation:

(13) QDPSizeChangeProof (EQUIVALENT transformation)

(14) YES