Left Termination proof of /tmp/tmpnzoVBZ/length.pl

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

0 Prolog

↳1 PrologToDTProblemTransformerProof (⇒, 0 ms)

↳2 TRIPLES

↳3 TriplesToPiDPProof (⇒, 16 ms)

↳4 PiDP

↳5 DependencyGraphProof (⇔, 0 ms)

↳6 PiDP

↳7 PiDPToQDPProof (⇒, 0 ms)

↳8 QDP

↳9 QDPSizeChangeProof (⇔, 0 ms)

↳10 YES

(0) Obligation:

Clauses:

len([], 0).
len(.(X1, Ts), s(N)) :- len(Ts, N).

Query: len(g,a)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: len(T1, T2)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: len(T1, T2), SCOPE: 1, CLAUSE: 0 | len(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: true | len([], T2), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: len(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nlen(T1, T2) !=? len([], 0)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: len([], T2), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: len(T7, T9)\n\nT7 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: len(T7, T9), SCOPE: 2, CLAUSE: 0 | len(T7, T9), SCOPE: 2, CLAUSE: 1\n\nT7 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: len(T7, T9), SCOPE: 2, CLAUSE: 0\n\nT7 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: len(T7, T9), SCOPE: 2, CLAUSE: 1\n\nT7 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: len(T17, T19)\n\nT17 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 17 [arrowhead = none ];
17 -> 2;

18 [label="EVAL with clause\nlen([], 0).\nand substitutionT1 -> [],\nT2 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 18 [arrowhead = none ];
18 -> 3;

19 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 19 [arrowhead = none ];
19 -> 4;

20 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 20 [arrowhead = none ];
20 -> 5;

21 [label="EVAL with clause\nlen(.(X8, X9), s(X10)) :- len(X9, X10).\nand substitutionX8 -> T6,\nX9 -> T7,\nT1 -> .(T6, T7),\nX10 -> T9,\nT2 -> s(T9),\nT8 -> T9", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 21 [arrowhead = none ];
21 -> 7;

22 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 22 [arrowhead = none ];
22 -> 8;

23 [label="BACKTRACK\nfor clause: len(.(X1, Ts), s(N)) :- len(Ts, N)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 23 [arrowhead = none ];
23 -> 6;

24 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 24 [arrowhead = none ];
24 -> 9;

25 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
9 -> 25 [arrowhead = none ];
25 -> 10;

26 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
9 -> 26 [arrowhead = none ];
26 -> 11;

27 [label="EVAL with clause\nlen([], 0).\nand substitutionT7 -> [],\nT9 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
10 -> 27 [arrowhead = none ];
27 -> 12;

28 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 28 [arrowhead = none ];
28 -> 13;

29 [label="EVAL with clause\nlen(.(X17, X18), s(X19)) :- len(X18, X19).\nand substitutionX17 -> T16,\nX18 -> T17,\nT7 -> .(T16, T17),\nX19 -> T19,\nT9 -> s(T19),\nT18 -> T19", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 29 [arrowhead = none ];
29 -> 15;

30 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 30 [arrowhead = none ];
30 -> 16;

31 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 31 [arrowhead = none ];
31 -> 14;

32 [label="INSTANCE with matching:\nT1 -> T17\nT2 -> T19", fontsize=14, style = filled, fillcolor = lightgrey];
15 -> 32 [arrowhead = none , style=dashed];
32 -> 1[style=dashed];

}

(2) Obligation:

Triples:

lenA(.(X1, .(X2, X3)), s(s(X4))) :- lenA(X3, X4).

Clauses:

lencA([], 0).
lencA(.(X1, []), s(0)).
lencA(.(X1, .(X2, X3)), s(s(X4))) :- lencA(X3, X4).

Afs:

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

LENA_IN_GA(.(X1, .(X2, X3)), s(s(X4))) → U1_GA(X1, X2, X3, X4, lenA_in_ga(X3, X4))
LENA_IN_GA(.(X1, .(X2, X3)), s(s(X4))) → LENA_IN_GA(X3, X4)

R is empty.
The argument filtering Pi contains the following mapping:
lenA_in_ga(x1, x2) = lenA_in_ga(x1)
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
LENA_IN_GA(x1, x2) = LENA_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5) = U1_GA(x1, x2, x3, x5)

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:

LENA_IN_GA(.(X1, .(X2, X3)), s(s(X4))) → U1_GA(X1, X2, X3, X4, lenA_in_ga(X3, X4))
LENA_IN_GA(.(X1, .(X2, X3)), s(s(X4))) → LENA_IN_GA(X3, X4)

(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:

LENA_IN_GA(.(X1, .(X2, X3)), s(s(X4))) → LENA_IN_GA(X3, X4)

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

LENA_IN_GA(.(X1, .(X2, X3))) → LENA_IN_GA(X3)

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:

LENA_IN_GA(.(X1, .(X2, X3))) → LENA_IN_GA(X3)
The graph contains the following edges 1 > 1