Left Termination proof of /tmp/tmp1W1xIU/gopher.pl

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

0 Prolog

↳1 PrologToDTProblemTransformerProof (⇒, 0 ms)

↳2 TRIPLES

↳3 TriplesToPiDPProof (⇒, 2 ms)

↳4 PiDP

↳5 DependencyGraphProof (⇔, 0 ms)

↳6 PiDP

↳7 PiDPToQDPProof (⇒, 0 ms)

↳8 QDP

↳9 MRRProof (⇔, 9 ms)

↳10 QDP

↳11 PisEmptyProof (⇔, 0 ms)

↳12 YES

(0) Obligation:

Clauses:

gopher(nil, nil).
gopher(cons(nil, Y), cons(nil, Y)).
gopher(cons(cons(U, V), W), X) :- gopher(cons(U, cons(V, W)), X).

Query: gopher(g,a)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: gopher(T1, T2)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: gopher(T1, T2), SCOPE: 1, CLAUSE: 0 | gopher(T1, T2), SCOPE: 1, CLAUSE: 1 | gopher(T1, T2), SCOPE: 1, CLAUSE: 2\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: true | gopher(nil, T2), SCOPE: 1, CLAUSE: 1 | gopher(nil, T2), SCOPE: 1, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: gopher(T1, T2), SCOPE: 1, CLAUSE: 1 | gopher(T1, T2), SCOPE: 1, CLAUSE: 2\n\nT1 is ground\ngopher(T1, T2) !=? gopher(nil, nil)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: gopher(nil, T2), SCOPE: 1, CLAUSE: 1 | gopher(nil, T2), SCOPE: 1, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: gopher(nil, T2), SCOPE: 1, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: true | gopher(cons(nil, T4), T2), SCOPE: 1, CLAUSE: 2\n\nT4 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: gopher(T1, T2), SCOPE: 1, CLAUSE: 2\n\nT1 is ground\nX7 is free\ngopher(T1, T2) !=? gopher(nil, nil)\ngopher(T1, T2) !=? gopher(cons(nil, X7), cons(nil, X7))", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: gopher(cons(nil, T4), T2), SCOPE: 1, CLAUSE: 2\n\nT4 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: gopher(cons(T9, cons(T10, T11)), T13)\n\nT9 is ground\nT10 is ground\nT11 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: gopher(cons(T9, cons(T10, T11)), T13), SCOPE: 2, CLAUSE: 0 | gopher(cons(T9, cons(T10, T11)), T13), SCOPE: 2, CLAUSE: 1 | gopher(cons(T9, cons(T10, T11)), T13), SCOPE: 2, CLAUSE: 2\n\nT9 is ground\nT10 is ground\nT11 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: gopher(cons(T9, cons(T10, T11)), T13), SCOPE: 2, CLAUSE: 1 | gopher(cons(T9, cons(T10, T11)), T13), SCOPE: 2, CLAUSE: 2\n\nT9 is ground\nT10 is ground\nT11 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: gopher(cons(T9, cons(T10, T11)), T13), SCOPE: 2, CLAUSE: 1\n\nT9 is ground\nT10 is ground\nT11 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: gopher(cons(T9, cons(T10, T11)), T13), SCOPE: 2, CLAUSE: 2\n\nT9 is ground\nT10 is ground\nT11 is ground", 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: gopher(cons(T34, cons(T35, cons(T36, T37))), T39)\n\nT34 is ground\nT35 is ground\nT36 is ground\nT37 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 23 [arrowhead = none ];
23 -> 2;

24 [label="EVAL with clause\ngopher(nil, nil).\nand substitutionT1 -> nil,\nT2 -> nil", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 24 [arrowhead = none ];
24 -> 3;

25 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 25 [arrowhead = none ];
25 -> 4;

26 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 26 [arrowhead = none ];
26 -> 5;

27 [label="EVAL with clause\ngopher(cons(nil, X7), cons(nil, X7)).\nand substitutionX7 -> T4,\nT1 -> cons(nil, T4),\nT2 -> cons(nil, T4)", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 27 [arrowhead = none ];
27 -> 8;

28 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 28 [arrowhead = none ];
28 -> 9;

29 [label="BACKTRACK\nfor clause: gopher(cons(nil, Y), cons(nil, Y))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 29 [arrowhead = none ];
29 -> 6;

30 [label="BACKTRACK\nfor clause: gopher(cons(cons(U, V), W), X) :- gopher(cons(U, cons(V, W)), X)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 30 [arrowhead = none ];
30 -> 7;

31 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
8 -> 31 [arrowhead = none ];
31 -> 10;

32 [label="EVAL with clause\ngopher(cons(cons(X16, X17), X18), X19) :- gopher(cons(X16, cons(X17, X18)), X19).\nand substitutionX16 -> T9,\nX17 -> T10,\nX18 -> T11,\nT1 -> cons(cons(T9, T10), T11),\nT2 -> T13,\nX19 -> T13,\nT12 -> T13", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 32 [arrowhead = none ];
32 -> 12;

33 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 33 [arrowhead = none ];
33 -> 13;

34 [label="BACKTRACK\nfor clause: gopher(cons(cons(U, V), W), X) :- gopher(cons(U, cons(V, W)), X)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 34 [arrowhead = none ];
34 -> 11;

35 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
12 -> 35 [arrowhead = none ];
35 -> 14;

36 [label="BACKTRACK\nfor clause: gopher(nil, nil)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
14 -> 36 [arrowhead = none ];
36 -> 15;

37 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
15 -> 37 [arrowhead = none ];
37 -> 16;

38 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
15 -> 38 [arrowhead = none ];
38 -> 17;

39 [label="EVAL with clause\ngopher(cons(nil, X24), cons(nil, X24)).\nand substitutionT9 -> nil,\nT10 -> T22,\nT11 -> T23,\nX24 -> cons(T22, T23),\nT13 -> cons(nil, cons(T22, T23))", fontsize=14, style = filled, fillcolor = lightblue];
16 -> 39 [arrowhead = none ];
39 -> 18;

40 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
16 -> 40 [arrowhead = none ];
40 -> 19;

41 [label="EVAL with clause\ngopher(cons(cons(X33, X34), X35), X36) :- gopher(cons(X33, cons(X34, X35)), X36).\nand substitutionX33 -> T34,\nX34 -> T35,\nT9 -> cons(T34, T35),\nT10 -> T36,\nT11 -> T37,\nX35 -> cons(T36, T37),\nT13 -> T39,\nX36 -> T39,\nT38 -> T39", fontsize=14, style = filled, fillcolor = lightblue];
17 -> 41 [arrowhead = none ];
41 -> 21;

42 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 42 [arrowhead = none ];
42 -> 22;

43 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
18 -> 43 [arrowhead = none ];
43 -> 20;

44 [label="INSTANCE with matching:\nT1 -> cons(T34, cons(T35, cons(T36, T37)))\nT2 -> T39", fontsize=14, style = filled, fillcolor = lightgrey];
21 -> 44 [arrowhead = none , style=dashed];
44 -> 1[style=dashed];

}

(2) Obligation:

Triples:

gopherA(cons(cons(cons(X1, X2), X3), X4), X5) :- gopherA(cons(X1, cons(X2, cons(X3, X4))), X5).

Clauses:

gophercA(nil, nil).
gophercA(cons(nil, X1), cons(nil, X1)).
gophercA(cons(cons(nil, X1), X2), cons(nil, cons(X1, X2))).
gophercA(cons(cons(cons(X1, X2), X3), X4), X5) :- gophercA(cons(X1, cons(X2, cons(X3, X4))), X5).

Afs:

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

GOPHERA_IN_GA(cons(cons(cons(X1, X2), X3), X4), X5) → U1_GA(X1, X2, X3, X4, X5, gopherA_in_ga(cons(X1, cons(X2, cons(X3, X4))), X5))
GOPHERA_IN_GA(cons(cons(cons(X1, X2), X3), X4), X5) → GOPHERA_IN_GA(cons(X1, cons(X2, cons(X3, X4))), X5)

R is empty.
The argument filtering Pi contains the following mapping:
gopherA_in_ga(x1, x2) = gopherA_in_ga(x1)
cons(x1, x2) = cons(x1, x2)
GOPHERA_IN_GA(x1, x2) = GOPHERA_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5, x6) = U1_GA(x1, x2, x3, x4, x6)

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:

GOPHERA_IN_GA(cons(cons(cons(X1, X2), X3), X4), X5) → U1_GA(X1, X2, X3, X4, X5, gopherA_in_ga(cons(X1, cons(X2, cons(X3, X4))), X5))
GOPHERA_IN_GA(cons(cons(cons(X1, X2), X3), X4), X5) → GOPHERA_IN_GA(cons(X1, cons(X2, cons(X3, X4))), X5)

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Obligation:

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

GOPHERA_IN_GA(cons(cons(cons(X1, X2), X3), X4), X5) → GOPHERA_IN_GA(cons(X1, cons(X2, cons(X3, X4))), X5)

R is empty.
The argument filtering Pi contains the following mapping:
cons(x1, x2) = cons(x1, x2)
GOPHERA_IN_GA(x1, x2) = GOPHERA_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:

GOPHERA_IN_GA(cons(cons(cons(X1, X2), X3), X4)) → GOPHERA_IN_GA(cons(X1, cons(X2, cons(X3, X4))))

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

(9) 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:

GOPHERA_IN_GA(cons(cons(cons(X1, X2), X3), X4)) → GOPHERA_IN_GA(cons(X1, cons(X2, cons(X3, X4))))

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

cons₂ > GOPHERA_{IN_GA₁}

and weight map:

GOPHERA_IN_GA_1=1
cons_2=0

The variable weight is 1

(10) Obligation:

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

(11) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.