Left Termination proof of /tmp/tmpn3znrn/length1.pl

Left Termination of the query pattern
len1(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 (⇒, 31 ms)

↳8 QDP

↳9 QDPSizeChangeProof (⇔, 0 ms)

↳10 YES

(0) Obligation:

Clauses:

len1([], 0).
len1(.(X1, Ts), N) :- ','(len1(Ts, M), eq(N, s(M))).
eq(X, X).

Query: len1(g,a)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="B: len1(T1, T2)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: len1(T1, T2), SCOPE: 1, CLAUSE: 0 | len1(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: true | len1([], T2), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: len1(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nlen1(T1, T2) !=? len1([], 0)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: len1([], 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: ','(len1(T7, X11), eq(T9, s(X11)))\n\nT7 is ground\nX11 is free", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(len1(T7, X11), eq(T9, s(X11))), SCOPE: 2, CLAUSE: 0 | ','(len1(T7, X11), eq(T9, s(X11))), SCOPE: 2, CLAUSE: 1\n\nT7 is ground\nX11 is free", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(len1(T7, X11), eq(T9, s(X11))), SCOPE: 2, CLAUSE: 0\n\nT7 is ground\nX11 is free", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(len1(T7, X11), eq(T9, s(X11))), SCOPE: 2, CLAUSE: 1\n\nT7 is ground\nX11 is free", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: eq(T9, s(0))\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: eq(T9, s(0)), SCOPE: 3, CLAUSE: 2\n\n", 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: ','(','(len1(T18, X27), eq(X28, s(X27))), eq(T9, s(X28)))\n\nT18 is ground\nX28 is free\nX27 is free", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="A: len1(T18, X27)\n\nT18 is ground\nX27 is free", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: ','(eq(X28, s(T19)), eq(T9, s(X28)))\n\nT19 is ground\nX28 is free", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: len1(T18, X27), SCOPE: 4, CLAUSE: 0 | len1(T18, X27), SCOPE: 4, CLAUSE: 1\n\nT18 is ground\nX27 is free", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: len1(T18, X27), SCOPE: 4, CLAUSE: 0\n\nT18 is ground\nX27 is free", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: len1(T18, X27), SCOPE: 4, CLAUSE: 1\n\nT18 is ground\nX27 is free", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: ','(len1(T25, X41), eq(X42, s(X41)))\n\nT25 is ground\nX42 is free\nX41 is free", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="29: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
30 [label="30: len1(T25, X41)\n\nT25 is ground\nX41 is free", fontsize=16, style=filled, fillcolor=lightyellow];
31 [label="31: eq(X42, s(T26))\n\nT26 is ground\nX42 is free", fontsize=16, style=filled, fillcolor=lightyellow];
32 [label="32: eq(X42, s(T26)), SCOPE: 5, CLAUSE: 2\n\nT26 is ground\nX42 is free", fontsize=16, style=filled, fillcolor=lightyellow];
33 [label="33: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
34 [label="34: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
35 [label="35: ','(eq(X28, s(T19)), eq(T9, s(X28))), SCOPE: 6, CLAUSE: 2\n\nT19 is ground\nX28 is free", fontsize=16, style=filled, fillcolor=lightyellow];
36 [label="36: eq(T9, s(s(T32)))\n\nT32 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
37 [label="37: eq(T9, s(s(T32))), SCOPE: 7, CLAUSE: 2\n\nT32 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
38 [label="38: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
39 [label="39: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
40 [label="40: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
41 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 41 [arrowhead = none ];
41 -> 2;

42 [label="EVAL with clause\nlen1([], 0).\nand substitutionT1 -> [],\nT2 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 42 [arrowhead = none ];
42 -> 3;

43 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 43 [arrowhead = none ];
43 -> 4;

44 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 44 [arrowhead = none ];
44 -> 5;

45 [label="EVAL with clause\nlen1(.(X8, X9), X10) :- ','(len1(X9, X11), eq(X10, s(X11))).\nand substitutionX8 -> T6,\nX9 -> T7,\nT1 -> .(T6, T7),\nT2 -> T9,\nX10 -> T9,\nT8 -> T9", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 45 [arrowhead = none ];
45 -> 7;

46 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 46 [arrowhead = none ];
46 -> 8;

47 [label="BACKTRACK\nfor clause: len1(.(X1, Ts), N) :- ','(len1(Ts, M), eq(N, s(M)))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 47 [arrowhead = none ];
47 -> 6;

48 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 48 [arrowhead = none ];
48 -> 9;

49 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
9 -> 49 [arrowhead = none ];
49 -> 10;

50 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
9 -> 50 [arrowhead = none ];
50 -> 11;

51 [label="EVAL with clause\nlen1([], 0).\nand substitutionT7 -> [],\nX11 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
10 -> 51 [arrowhead = none ];
51 -> 12;

52 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 52 [arrowhead = none ];
52 -> 13;

53 [label="EVAL with clause\nlen1(.(X24, X25), X26) :- ','(len1(X25, X27), eq(X26, s(X27))).\nand substitutionX24 -> T17,\nX25 -> T18,\nT7 -> .(T17, T18),\nX11 -> X28,\nX26 -> X28", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 53 [arrowhead = none ];
53 -> 18;

54 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 54 [arrowhead = none ];
54 -> 19;

55 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
12 -> 55 [arrowhead = none ];
55 -> 14;

56 [label="EVAL with clause\neq(X14, X14).\nand substitutionT9 -> s(0),\nX14 -> s(0),\nT12 -> s(0)", fontsize=14, style = filled, fillcolor = lightblue];
14 -> 56 [arrowhead = none ];
56 -> 15;

57 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
14 -> 57 [arrowhead = none ];
57 -> 16;

58 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 58 [arrowhead = none ];
58 -> 17;

59 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
18 -> 59 [arrowhead = none ];
59 -> 20;

60 [label="SPLIT 2\nnew knowledge:\nT18 is ground\nT19 is ground\nreplacements:X27 -> T19", fontsize=14, style = filled, fillcolor = violet];
18 -> 60 [arrowhead = none ];
60 -> 21;

61 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
20 -> 61 [arrowhead = none ];
61 -> 22;

62 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
21 -> 62 [arrowhead = none ];
62 -> 35;

63 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
22 -> 63 [arrowhead = none ];
63 -> 23;

64 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
22 -> 64 [arrowhead = none ];
64 -> 24;

65 [label="EVAL with clause\nlen1([], 0).\nand substitutionT18 -> [],\nX27 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
23 -> 65 [arrowhead = none ];
65 -> 25;

66 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
23 -> 66 [arrowhead = none ];
66 -> 26;

67 [label="EVAL with clause\nlen1(.(X38, X39), X40) :- ','(len1(X39, X41), eq(X40, s(X41))).\nand substitutionX38 -> T24,\nX39 -> T25,\nT18 -> .(T24, T25),\nX27 -> X42,\nX40 -> X42", fontsize=14, style = filled, fillcolor = lightblue];
24 -> 67 [arrowhead = none ];
67 -> 28;

68 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
24 -> 68 [arrowhead = none ];
68 -> 29;

69 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
25 -> 69 [arrowhead = none ];
69 -> 27;

70 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
28 -> 70 [arrowhead = none ];
70 -> 30;

71 [label="SPLIT 2\nnew knowledge:\nT25 is ground\nT26 is ground\nreplacements:X41 -> T26", fontsize=14, style = filled, fillcolor = violet];
28 -> 71 [arrowhead = none ];
71 -> 31;

72 [label="INSTANCE with matching:\nT18 -> T25\nX27 -> X41", fontsize=14, style = filled, fillcolor = lightgrey];
30 -> 72 [arrowhead = none , style=dashed];
72 -> 20[style=dashed];

73 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
31 -> 73 [arrowhead = none ];
73 -> 32;

74 [label="ONLY EVAL with clause\neq(X47, X47).\nand substitutionX42 -> s(T29),\nX47 -> s(T29),\nT26 -> T29,\nX48 -> s(T29)", fontsize=14, style = filled, fillcolor = lightblue];
32 -> 74 [arrowhead = none ];
74 -> 33;

75 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
33 -> 75 [arrowhead = none ];
75 -> 34;

76 [label="ONLY EVAL with clause\neq(X53, X53).\nand substitutionX28 -> s(T32),\nX53 -> s(T32),\nT19 -> T32,\nX54 -> s(T32)", fontsize=14, style = filled, fillcolor = lightblue];
35 -> 76 [arrowhead = none ];
76 -> 36;

77 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
36 -> 77 [arrowhead = none ];
77 -> 37;

78 [label="EVAL with clause\neq(X57, X57).\nand substitutionT9 -> s(s(T38)),\nX57 -> s(s(T38)),\nT32 -> T38,\nT37 -> s(s(T38))", fontsize=14, style = filled, fillcolor = lightblue];
37 -> 78 [arrowhead = none ];
78 -> 38;

79 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
37 -> 79 [arrowhead = none ];
79 -> 39;

80 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
38 -> 80 [arrowhead = none ];
80 -> 40;

}

(2) Obligation:

Triples:

len1A(.(X1, X2), X3) :- len1A(X2, X4).
len1B(.(X1, .(X2, X3)), X4) :- len1A(X3, X5).

Clauses:

len1cA([], 0).
len1cA(.(X1, X2), s(X3)) :- len1cA(X2, X3).

Afs:

len1B(x1, x2) = len1B(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:
len1B_in: (b,f)
len1A_in: (b,f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

LEN1B_IN_GA(.(X1, .(X2, X3)), X4) → U2_GA(X1, X2, X3, X4, len1A_in_ga(X3, X5))
LEN1B_IN_GA(.(X1, .(X2, X3)), X4) → LEN1A_IN_GA(X3, X5)
LEN1A_IN_GA(.(X1, X2), X3) → U1_GA(X1, X2, X3, len1A_in_ga(X2, X4))
LEN1A_IN_GA(.(X1, X2), X3) → LEN1A_IN_GA(X2, X4)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
len1A_in_ga(x1, x2) = len1A_in_ga(x1)
LEN1B_IN_GA(x1, x2) = LEN1B_IN_GA(x1)
U2_GA(x1, x2, x3, x4, x5) = U2_GA(x1, x2, x3, x5)
LEN1A_IN_GA(x1, x2) = LEN1A_IN_GA(x1)
U1_GA(x1, x2, x3, x4) = U1_GA(x1, x2, x4)

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:

LEN1B_IN_GA(.(X1, .(X2, X3)), X4) → U2_GA(X1, X2, X3, X4, len1A_in_ga(X3, X5))
LEN1B_IN_GA(.(X1, .(X2, X3)), X4) → LEN1A_IN_GA(X3, X5)
LEN1A_IN_GA(.(X1, X2), X3) → U1_GA(X1, X2, X3, len1A_in_ga(X2, X4))
LEN1A_IN_GA(.(X1, X2), X3) → LEN1A_IN_GA(X2, X4)

(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:

LEN1A_IN_GA(.(X1, X2), X3) → LEN1A_IN_GA(X2, X4)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
LEN1A_IN_GA(x1, x2) = LEN1A_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:

LEN1A_IN_GA(.(X1, X2)) → LEN1A_IN_GA(X2)

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:

LEN1A_IN_GA(.(X1, X2)) → LEN1A_IN_GA(X2)
The graph contains the following edges 1 > 1