Left Termination proof of /tmp/tmpNLGfjk/paper1.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 (⇒, 0 ms)

↳2 TRIPLES

↳3 TriplesToPiDPProof (⇒, 0 ms)

↳4 PiDP

↳5 DependencyGraphProof (⇔, 0 ms)

↳6 PiDP

↳7 PiDPToQDPProof (⇒, 0 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(Y))).

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="A: 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="6: ','(p(f(T7), f(X7)), p(X7, g(T9)))\n\nT7 is ground\nX7 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(T9))), SCOPE: 2, CLAUSE: 0 | ','(p(f(T7), f(X7)), p(X7, g(T9))), SCOPE: 2, CLAUSE: 1\n\nT7 is ground\nX7 is free", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(p(f(T7), f(X7)), p(X7, g(T9))), SCOPE: 2, CLAUSE: 0\n\nT7 is ground\nX7 is free", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(p(f(T7), f(X7)), p(X7, g(T9))), SCOPE: 2, CLAUSE: 1\n\nT7 is ground\nX7 is free", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: p(T22, g(T9))\n\nT22 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(p(f(T30), f(X25)), p(X25, g(T32)))\n\nT30 is ground\nX2 is free\nX25 is free\np(f(T30), T2) !=? p(X2, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: ','(p(f(T30), f(X25)), p(X25, g(T32))), SCOPE: 3, CLAUSE: 0 | ','(p(f(T30), f(X25)), p(X25, g(T32))), SCOPE: 3, CLAUSE: 1\n\nT30 is ground\nX2 is free\nX25 is free\np(f(T30), T2) !=? p(X2, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: ','(p(f(T30), f(X25)), p(X25, g(T32))), SCOPE: 3, CLAUSE: 0\n\nT30 is ground\nX2 is free\nX25 is free\np(f(T30), T2) !=? p(X2, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: ','(p(f(T30), f(X25)), p(X25, g(T32))), SCOPE: 3, CLAUSE: 1\n\nT30 is ground\nX2 is free\nX25 is free\np(f(T30), T2) !=? p(X2, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: p(T45, g(T32))\n\nT45 is ground\nX2 is free\np(f(T45), T2) !=? p(X2, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 20 [arrowhead = none ];
20 -> 2;

21 [label="EVAL with clause\np(X2, X2).\nand substitutionT1 -> T4,\nX2 -> T4,\nT2 -> T4", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 21 [arrowhead = none ];
21 -> 3;

22 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 22 [arrowhead = none ];
22 -> 4;

23 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 23 [arrowhead = none ];
23 -> 5;

24 [label="EVAL with clause\np(f(X23), g(X24)) :- ','(p(f(X23), f(X25)), p(X25, g(X24))).\nand substitutionX23 -> T30,\nT1 -> f(T30),\nX24 -> T32,\nT2 -> g(T32),\nT31 -> T32", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 24 [arrowhead = none ];
24 -> 13;

25 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 25 [arrowhead = none ];
25 -> 14;

26 [label="EVAL with clause\np(f(X5), g(X6)) :- ','(p(f(X5), f(X7)), p(X7, g(X6))).\nand substitutionX5 -> T7,\nT4 -> f(T7),\nX6 -> T9,\nT2 -> g(T9),\nT8 -> T9", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 26 [arrowhead = none ];
26 -> 6;

27 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 27 [arrowhead = none ];
27 -> 7;

28 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
6 -> 28 [arrowhead = none ];
28 -> 8;

29 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
8 -> 29 [arrowhead = none ];
29 -> 9;

30 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
8 -> 30 [arrowhead = none ];
30 -> 10;

31 [label="ONLY EVAL with clause\np(X16, X16).\nand substitutionT7 -> T22,\nX16 -> f(T22),\nX7 -> T22", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 31 [arrowhead = none ];
31 -> 11;

32 [label="BACKTRACK\nfor clause: p(f(X), g(Y)) :- ','(p(f(X), f(Z)), p(Z, g(Y)))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 32 [arrowhead = none ];
32 -> 12;

33 [label="INSTANCE with matching:\nT1 -> T22\nT2 -> g(T9)", fontsize=14, style = filled, fillcolor = lightgrey];
11 -> 33 [arrowhead = none , style=dashed];
33 -> 1[style=dashed];

34 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
13 -> 34 [arrowhead = none ];
34 -> 15;

35 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
15 -> 35 [arrowhead = none ];
35 -> 16;

36 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
15 -> 36 [arrowhead = none ];
36 -> 17;

37 [label="ONLY EVAL with clause\np(X34, X34).\nand substitutionT30 -> T45,\nX34 -> f(T45),\nX25 -> T45", fontsize=14, style = filled, fillcolor = lightblue];
16 -> 37 [arrowhead = none ];
37 -> 18;

38 [label="BACKTRACK\nfor clause: p(f(X), g(Y)) :- ','(p(f(X), f(Z)), p(Z, g(Y)))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 38 [arrowhead = none ];
38 -> 19;

39 [label="INSTANCE with matching:\nT1 -> T45\nT2 -> g(T32)", fontsize=14, style = filled, fillcolor = lightgrey];
18 -> 39 [arrowhead = none , style=dashed];
39 -> 1[style=dashed];

}

(2) Obligation:

Triples:

pA(f(X1), g(X2)) :- pA(X1, g(X2)).
pA(f(X1), g(X2)) :- pA(X1, g(X2)).

Clauses:

pcA(X1, X1).
pcA(f(X1), g(X2)) :- pcA(X1, g(X2)).
pcA(f(X1), g(X2)) :- pcA(X1, g(X2)).

Afs:

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

PA_IN_GA(f(X1), g(X2)) → U1_GA(X1, X2, pA_in_gg(X1, g(X2)))
PA_IN_GA(f(X1), g(X2)) → PA_IN_GG(X1, g(X2))
PA_IN_GG(f(X1), g(X2)) → U1_GG(X1, X2, pA_in_gg(X1, g(X2)))
PA_IN_GG(f(X1), g(X2)) → PA_IN_GG(X1, g(X2))

R is empty.
The argument filtering Pi contains the following mapping:
f(x1) = f(x1)
pA_in_gg(x1, x2) = pA_in_gg(x1, x2)
g(x1) = g
PA_IN_GA(x1, x2) = PA_IN_GA(x1)
U1_GA(x1, x2, x3) = U1_GA(x1, x3)
PA_IN_GG(x1, x2) = PA_IN_GG(x1, x2)
U1_GG(x1, x2, x3) = U1_GG(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:

PA_IN_GA(f(X1), g(X2)) → U1_GA(X1, X2, pA_in_gg(X1, g(X2)))
PA_IN_GA(f(X1), g(X2)) → PA_IN_GG(X1, g(X2))
PA_IN_GG(f(X1), g(X2)) → U1_GG(X1, X2, pA_in_gg(X1, g(X2)))
PA_IN_GG(f(X1), g(X2)) → PA_IN_GG(X1, g(X2))

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes.

(6) Obligation:

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

PA_IN_GG(f(X1), g(X2)) → PA_IN_GG(X1, g(X2))

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

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_GG(f(X1), g) → PA_IN_GG(X1, g)

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:

PA_IN_GG(f(X1), g) → PA_IN_GG(X1, g)
The graph contains the following edges 1 > 1, 2 >= 2