Left Termination proof of /tmp/tmpRUWxtf/paper2.pl

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

0 Prolog

↳1 PrologToDTProblemTransformerProof (⇒, 89 ms)

↳2 TRIPLES

↳3 TriplesToPiDPProof (⇒, 0 ms)

↳4 PiDP

↳5 DependencyGraphProof (⇔, 0 ms)

↳6 PiDP

↳7 PiDPToQDPProof (⇒, 23 ms)

↳8 QDP

↳9 QDPSizeChangeProof (⇔, 0 ms)

↳10 YES

(0) Obligation:

Clauses:

p(X, X).
p(f(X), g(Y)) :- ','(p(f(X), f(Z)), p(Z, g(W))).

Query: p(g,a)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="C: p(T1, T2)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: p(T1, T2), SCOPE: 1, CLAUSE: 0 | p(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: true | p(T4, T2), SCOPE: 1, CLAUSE: 1\n\nT4 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: p(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nX2 is free\np(T1, T2) !=? p(X2, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: p(T4, T2), SCOPE: 1, CLAUSE: 1\n\nT4 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="A: ','(p(f(T7), f(X7)), p(X7, g(X8)))\n\nT7 is ground\nX7 is free\nX8 is free", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: ','(p(f(T7), f(X7)), p(X7, g(X8))), SCOPE: 2, CLAUSE: 0 | ','(p(f(T7), f(X7)), p(X7, g(X8))), SCOPE: 2, CLAUSE: 1\n\nT7 is ground\nX7 is free\nX8 is free", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(p(f(T7), f(X7)), p(X7, g(X8))), SCOPE: 2, CLAUSE: 0\n\nT7 is ground\nX7 is free\nX8 is free", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(p(f(T7), f(X7)), p(X7, g(X8))), SCOPE: 2, CLAUSE: 1\n\nT7 is ground\nX7 is free\nX8 is free", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="B: p(T21, g(X8))\n\nT21 is ground\nX8 is free", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: p(T21, g(X8)), SCOPE: 3, CLAUSE: 0 | p(T21, g(X8)), SCOPE: 3, CLAUSE: 1\n\nT21 is ground\nX8 is free", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: p(T21, g(X8)), SCOPE: 3, CLAUSE: 0\n\nT21 is ground\nX8 is free", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: p(T21, g(X8)), SCOPE: 3, CLAUSE: 1\n\nT21 is ground\nX8 is free", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: ','(p(f(T38), f(X35)), p(X35, g(X36)))\n\nT38 is ground\nX35 is free\nX36 is free", 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: ','(p(f(T43), f(X48)), p(X48, g(X49)))\n\nT43 is ground\nX2 is free\nX48 is free\nX49 is free\np(f(T43), T2) !=? p(X2, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: ','(p(f(T43), f(X48)), p(X48, g(X49))), SCOPE: 4, CLAUSE: 0 | ','(p(f(T43), f(X48)), p(X48, g(X49))), SCOPE: 4, CLAUSE: 1\n\nT43 is ground\nX2 is free\nX48 is free\nX49 is free\np(f(T43), T2) !=? p(X2, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: ','(p(f(T43), f(X48)), p(X48, g(X49))), SCOPE: 4, CLAUSE: 0\n\nT43 is ground\nX2 is free\nX48 is free\nX49 is free\np(f(T43), T2) !=? p(X2, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: ','(p(f(T43), f(X48)), p(X48, g(X49))), SCOPE: 4, CLAUSE: 1\n\nT43 is ground\nX2 is free\nX48 is free\nX49 is free\np(f(T43), T2) !=? p(X2, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: p(T57, g(X49))\n\nT57 is ground\nX2 is free\nX49 is free\np(f(T57), T2) !=? p(X2, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 28 [arrowhead = none ];
28 -> 2;

29 [label="EVAL with clause\np(X2, X2).\nand substitutionT1 -> T4,\nX2 -> T4,\nT2 -> T4", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 29 [arrowhead = none ];
29 -> 3;

30 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 30 [arrowhead = none ];
30 -> 4;

31 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 31 [arrowhead = none ];
31 -> 5;

32 [label="EVAL with clause\np(f(X46), g(X47)) :- ','(p(f(X46), f(X48)), p(X48, g(X49))).\nand substitutionX46 -> T43,\nT1 -> f(T43),\nX47 -> T44,\nT2 -> g(T44)", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 32 [arrowhead = none ];
32 -> 21;

33 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 33 [arrowhead = none ];
33 -> 22;

34 [label="EVAL with clause\np(f(X5), g(X6)) :- ','(p(f(X5), f(X7)), p(X7, g(X8))).\nand substitutionX5 -> T7,\nT4 -> f(T7),\nX6 -> T8,\nT2 -> g(T8)", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 34 [arrowhead = none ];
34 -> 6;

35 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 35 [arrowhead = none ];
35 -> 7;

36 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
6 -> 36 [arrowhead = none ];
36 -> 8;

37 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
8 -> 37 [arrowhead = none ];
37 -> 9;

38 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
8 -> 38 [arrowhead = none ];
38 -> 10;

39 [label="ONLY EVAL with clause\np(X17, X17).\nand substitutionT7 -> T21,\nX17 -> f(T21),\nX7 -> T21", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 39 [arrowhead = none ];
39 -> 11;

40 [label="BACKTRACK\nfor clause: p(f(X), g(Y)) :- ','(p(f(X), f(Z)), p(Z, g(W)))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 40 [arrowhead = none ];
40 -> 20;

41 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
11 -> 41 [arrowhead = none ];
41 -> 12;

42 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
12 -> 42 [arrowhead = none ];
42 -> 13;

43 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
12 -> 43 [arrowhead = none ];
43 -> 14;

44 [label="EVAL with clause\np(X24, X24).\nand substitutionT21 -> g(T35),\nX24 -> g(T35),\nX8 -> T35,\nT34 -> g(T35)", fontsize=14, style = filled, fillcolor = lightblue];
13 -> 44 [arrowhead = none ];
44 -> 15;

45 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
13 -> 45 [arrowhead = none ];
45 -> 16;

46 [label="EVAL with clause\np(f(X33), g(X34)) :- ','(p(f(X33), f(X35)), p(X35, g(X36))).\nand substitutionX33 -> T38,\nT21 -> f(T38),\nX8 -> X37,\nX34 -> X37", fontsize=14, style = filled, fillcolor = lightblue];
14 -> 46 [arrowhead = none ];
46 -> 18;

47 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
14 -> 47 [arrowhead = none ];
47 -> 19;

48 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 48 [arrowhead = none ];
48 -> 17;

49 [label="INSTANCE with matching:\nT7 -> T38\nX7 -> X35\nX8 -> X36", fontsize=14, style = filled, fillcolor = lightgrey];
18 -> 49 [arrowhead = none , style=dashed];
49 -> 6[style=dashed];

50 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
21 -> 50 [arrowhead = none ];
50 -> 23;

51 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
23 -> 51 [arrowhead = none ];
51 -> 24;

52 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
23 -> 52 [arrowhead = none ];
52 -> 25;

53 [label="ONLY EVAL with clause\np(X58, X58).\nand substitutionT43 -> T57,\nX58 -> f(T57),\nX48 -> T57", fontsize=14, style = filled, fillcolor = lightblue];
24 -> 53 [arrowhead = none ];
53 -> 26;

54 [label="BACKTRACK\nfor clause: p(f(X), g(Y)) :- ','(p(f(X), f(Z)), p(Z, g(W)))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
25 -> 54 [arrowhead = none ];
54 -> 27;

55 [label="INSTANCE with matching:\nT21 -> T57\nX8 -> X49", fontsize=14, style = filled, fillcolor = lightgrey];
26 -> 55 [arrowhead = none , style=dashed];
55 -> 11[style=dashed];

}

(2) Obligation:

Triples:

pA(X1, X1, X2) :- pB(X1, X2).
pB(f(X1), X2) :- pA(X1, X3, X4).
pC(f(X1), g(X2)) :- pA(X1, X3, X4).
pC(f(X1), g(X2)) :- pB(X1, X3).

Clauses:

qcA(X1, X1, X2) :- pcB(X1, X2).
pcB(g(X1), X1).
pcB(f(X1), X2) :- qcA(X1, X3, X4).

Afs:

pC(x1, x2) = pC(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:
pC_in: (b,f)
pA_in: (b,f,f)
pB_in: (b,f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

PC_IN_GA(f(X1), g(X2)) → U3_GA(X1, X2, pA_in_gaa(X1, X3, X4))
PC_IN_GA(f(X1), g(X2)) → PA_IN_GAA(X1, X3, X4)
PA_IN_GAA(X1, X1, X2) → U1_GAA(X1, X2, pB_in_ga(X1, X2))
PA_IN_GAA(X1, X1, X2) → PB_IN_GA(X1, X2)
PB_IN_GA(f(X1), X2) → U2_GA(X1, X2, pA_in_gaa(X1, X3, X4))
PB_IN_GA(f(X1), X2) → PA_IN_GAA(X1, X3, X4)
PC_IN_GA(f(X1), g(X2)) → U4_GA(X1, X2, pB_in_ga(X1, X3))
PC_IN_GA(f(X1), g(X2)) → PB_IN_GA(X1, X3)

R is empty.
The argument filtering Pi contains the following mapping:
f(x1) = f(x1)
pA_in_gaa(x1, x2, x3) = pA_in_gaa(x1)
pB_in_ga(x1, x2) = pB_in_ga(x1)
g(x1) = g
PC_IN_GA(x1, x2) = PC_IN_GA(x1)
U3_GA(x1, x2, x3) = U3_GA(x1, x3)
PA_IN_GAA(x1, x2, x3) = PA_IN_GAA(x1)
U1_GAA(x1, x2, x3) = U1_GAA(x1, x3)
PB_IN_GA(x1, x2) = PB_IN_GA(x1)
U2_GA(x1, x2, x3) = U2_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:

PC_IN_GA(f(X1), g(X2)) → U3_GA(X1, X2, pA_in_gaa(X1, X3, X4))
PC_IN_GA(f(X1), g(X2)) → PA_IN_GAA(X1, X3, X4)
PA_IN_GAA(X1, X1, X2) → U1_GAA(X1, X2, pB_in_ga(X1, X2))
PA_IN_GAA(X1, X1, X2) → PB_IN_GA(X1, X2)
PB_IN_GA(f(X1), X2) → U2_GA(X1, X2, pA_in_gaa(X1, X3, X4))
PB_IN_GA(f(X1), X2) → PA_IN_GAA(X1, X3, X4)
PC_IN_GA(f(X1), g(X2)) → U4_GA(X1, X2, pB_in_ga(X1, X3))
PC_IN_GA(f(X1), g(X2)) → PB_IN_GA(X1, X3)

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 6 less nodes.

(6) Obligation:

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

PA_IN_GAA(X1, X1, X2) → PB_IN_GA(X1, X2)
PB_IN_GA(f(X1), X2) → PA_IN_GAA(X1, X3, X4)

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

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

(7) PiDPToQDPProof (SOUND 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_GAA(X1) → PB_IN_GA(X1)
PB_IN_GA(f(X1)) → PA_IN_GAA(X1)

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

(9) 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:

PB_IN_GA(f(X1)) → PA_IN_GAA(X1)
The graph contains the following edges 1 > 1
PA_IN_GAA(X1) → PB_IN_GA(X1)
The graph contains the following edges 1 >= 1