Left Termination proof of /tmp/tmpXJxKhD/evenodd.pl

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

0 Prolog

↳1 PrologToPiTRSViaGraphTransformerProof (⇒, 56 ms)

↳2 PiTRS

↳3 DependencyPairsProof (⇔, 0 ms)

↳4 PiDP

↳5 DependencyGraphProof (⇔, 0 ms)

↳6 PiDP

↳7 UsableRulesProof (⇔, 0 ms)

↳8 PiDP

↳9 PiDPToQDPProof (⇔, 4 ms)

↳10 QDP

↳11 QDPSizeChangeProof (⇔, 0 ms)

↳12 YES

(0) Obligation:

Clauses:

even(0).
even(s(s(0))).
even(s(s(s(X)))) :- odd(X).
odd(s(0)).
odd(s(X)) :- even(s(s(X))).

Query: even(g)

(1) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: even(T1)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
5 [label="5: even(T1), SCOPE: 1, CLAUSE: 0 | even(T1), SCOPE: 1, CLAUSE: 1 | even(T1), SCOPE: 1, CLAUSE: 2\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: true | even(0), SCOPE: 1, CLAUSE: 1 | even(0), SCOPE: 1, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: even(T1), SCOPE: 1, CLAUSE: 1 | even(T1), SCOPE: 1, CLAUSE: 2\n\nT1 is ground\neven(T1) !=? even(0)", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: even(0), SCOPE: 1, CLAUSE: 1 | even(0), SCOPE: 1, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: even(0), SCOPE: 1, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: true | even(s(s(0))), SCOPE: 1, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: even(T1), SCOPE: 1, CLAUSE: 2\n\nT1 is ground\neven(T1) !=? even(0)\neven(T1) !=? even(s(s(0)))", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: even(s(s(0))), SCOPE: 1, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="29: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
31 [label="31: odd(T3)\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
32 [label="32: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
33 [label="33: odd(T3), SCOPE: 2, CLAUSE: 3 | odd(T3), SCOPE: 2, CLAUSE: 4\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
34 [label="34: odd(T3), SCOPE: 2, CLAUSE: 3\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
35 [label="35: odd(T3), SCOPE: 2, CLAUSE: 4\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
55 [label="55: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
56 [label="56: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
57 [label="57: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
59 [label="59: even(s(s(T6)))\n\nT6 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
61 [label="61: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
62 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 62 [arrowhead = none ];
62 -> 5;

63 [label="EVAL with clause\neven(0).\nand substitutionT1 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 63 [arrowhead = none ];
63 -> 9;

64 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 64 [arrowhead = none ];
64 -> 10;

65 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 65 [arrowhead = none ];
65 -> 14;

66 [label="EVAL with clause\neven(s(s(0))).\nand substitutionT1 -> s(s(0))", fontsize=14, style = filled, fillcolor = lightblue];
10 -> 66 [arrowhead = none ];
66 -> 25;

67 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 67 [arrowhead = none ];
67 -> 26;

68 [label="BACKTRACK\nfor clause: even(s(s(0)))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
14 -> 68 [arrowhead = none ];
68 -> 18;

69 [label="BACKTRACK\nfor clause: even(s(s(s(X)))) :- odd(X)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
18 -> 69 [arrowhead = none ];
69 -> 21;

70 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
25 -> 70 [arrowhead = none ];
70 -> 27;

71 [label="EVAL with clause\neven(s(s(s(X4)))) :- odd(X4).\nand substitutionX4 -> T3,\nT1 -> s(s(s(T3)))", fontsize=14, style = filled, fillcolor = lightblue];
26 -> 71 [arrowhead = none ];
71 -> 31;

72 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
26 -> 72 [arrowhead = none ];
72 -> 32;

73 [label="BACKTRACK\nfor clause: even(s(s(s(X)))) :- odd(X)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
27 -> 73 [arrowhead = none ];
73 -> 29;

74 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
31 -> 74 [arrowhead = none ];
74 -> 33;

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

76 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
33 -> 76 [arrowhead = none ];
76 -> 35;

77 [label="EVAL with clause\nodd(s(0)).\nand substitutionT3 -> s(0)", fontsize=14, style = filled, fillcolor = lightblue];
34 -> 77 [arrowhead = none ];
77 -> 55;

78 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
34 -> 78 [arrowhead = none ];
78 -> 56;

79 [label="EVAL with clause\nodd(s(X7)) :- even(s(s(X7))).\nand substitutionX7 -> T6,\nT3 -> s(T6)", fontsize=14, style = filled, fillcolor = lightblue];
35 -> 79 [arrowhead = none ];
79 -> 59;

80 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
35 -> 80 [arrowhead = none ];
80 -> 61;

81 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
55 -> 81 [arrowhead = none ];
81 -> 57;

82 [label="INSTANCE with matching:\nT1 -> s(s(T6))", fontsize=14, style = filled, fillcolor = lightgrey];
59 -> 82 [arrowhead = none , style=dashed];
82 -> 1[style=dashed];

}

(2) Obligation:

Pi-finite rewrite system:
The TRS R consists of the following rules:

evenA_in_g(0) → evenA_out_g(0)
evenA_in_g(s(s(0))) → evenA_out_g(s(s(0)))
evenA_in_g(s(s(s(s(0))))) → evenA_out_g(s(s(s(s(0)))))
evenA_in_g(s(s(s(s(T6))))) → U1_g(T6, evenA_in_g(s(s(T6))))
U1_g(T6, evenA_out_g(s(s(T6)))) → evenA_out_g(s(s(s(s(T6)))))

Pi is empty.

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

EVENA_IN_G(s(s(s(s(T6))))) → U1_G(T6, evenA_in_g(s(s(T6))))
EVENA_IN_G(s(s(s(s(T6))))) → EVENA_IN_G(s(s(T6)))

The TRS R consists of the following rules:

evenA_in_g(0) → evenA_out_g(0)
evenA_in_g(s(s(0))) → evenA_out_g(s(s(0)))
evenA_in_g(s(s(s(s(0))))) → evenA_out_g(s(s(s(s(0)))))
evenA_in_g(s(s(s(s(T6))))) → U1_g(T6, evenA_in_g(s(s(T6))))
U1_g(T6, evenA_out_g(s(s(T6)))) → evenA_out_g(s(s(s(s(T6)))))

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

(4) Obligation:

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

EVENA_IN_G(s(s(s(s(T6))))) → U1_G(T6, evenA_in_g(s(s(T6))))
EVENA_IN_G(s(s(s(s(T6))))) → EVENA_IN_G(s(s(T6)))

The TRS R consists of the following rules:

evenA_in_g(0) → evenA_out_g(0)
evenA_in_g(s(s(0))) → evenA_out_g(s(s(0)))
evenA_in_g(s(s(s(s(0))))) → evenA_out_g(s(s(s(s(0)))))
evenA_in_g(s(s(s(s(T6))))) → U1_g(T6, evenA_in_g(s(s(T6))))
U1_g(T6, evenA_out_g(s(s(T6)))) → evenA_out_g(s(s(s(s(T6)))))

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

(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:

EVENA_IN_G(s(s(s(s(T6))))) → EVENA_IN_G(s(s(T6)))

The TRS R consists of the following rules:

evenA_in_g(0) → evenA_out_g(0)
evenA_in_g(s(s(0))) → evenA_out_g(s(s(0)))
evenA_in_g(s(s(s(s(0))))) → evenA_out_g(s(s(s(s(0)))))
evenA_in_g(s(s(s(s(T6))))) → U1_g(T6, evenA_in_g(s(s(T6))))
U1_g(T6, evenA_out_g(s(s(T6)))) → evenA_out_g(s(s(s(s(T6)))))

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

(7) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(8) Obligation:

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

EVENA_IN_G(s(s(s(s(T6))))) → EVENA_IN_G(s(s(T6)))

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

(9) PiDPToQDPProof (EQUIVALENT transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(10) Obligation:

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

EVENA_IN_G(s(s(s(s(T6))))) → EVENA_IN_G(s(s(T6)))

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

(11) 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:

EVENA_IN_G(s(s(s(s(T6))))) → EVENA_IN_G(s(s(T6)))
The graph contains the following edges 1 > 1

(0) Obligation:

(1) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

(2) Obligation:

(3) DependencyPairsProof (EQUIVALENT transformation)

(4) Obligation:

(5) DependencyGraphProof (EQUIVALENT transformation)

(6) Obligation:

(7) UsableRulesProof (EQUIVALENT transformation)

(8) Obligation:

(9) PiDPToQDPProof (EQUIVALENT transformation)

(10) Obligation:

(11) QDPSizeChangeProof (EQUIVALENT transformation)

(12) YES