Left Termination proof of /tmp/tmpgXjGcm/lte1.pl

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

0 Prolog

↳1 PrologToPrologProblemTransformerProof (⇐)

↳2 Prolog

↳3 PrologToPiTRSProof (⇐)

↳4 PiTRS

↳5 DependencyPairsProof (⇔)

↳6 PiDP

↳7 DependencyGraphProof (⇔)

↳8 PiDP

↳9 UsableRulesProof (⇔)

↳10 PiDP

↳11 PiDPToQDPProof (⇔)

↳12 QDP

↳13 QDPSizeChangeProof (⇔)

↳14 TRUE

(0) Obligation:

Clauses:

goal :- ','(lte(X, s(s(s(s(0))))), even(X)).
lte(0, Y) :- !.
lte(X, s(Y)) :- ','(p(X, P), lte(P, Y)).
even(0).
even(s(s(X))) :- even(X).
p(0, 0).
p(s(X), X).

Queries:

goal().

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: goal\n\n", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: goal, SCOPE: 1, CLAUSE: 0\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: ','(lte(X1, s(s(s(s(0))))), even(X1))\n\nX1 is free", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: lte(X1, s(s(s(s(0)))))\n\nX1 is free", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: even(T1)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: lte(X1, s(s(s(s(0))))), SCOPE: 2, CLAUSE: 1 | lte(X1, s(s(s(s(0))))), SCOPE: 2, CLAUSE: 2 | ?_2\n\nX1 is free", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: !_2 | lte(X1, s(s(s(s(0))))), SCOPE: 2, CLAUSE: 2 | ?_2\n\nX1 is free", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: even(T1), SCOPE: 3, CLAUSE: 3 | even(T1), SCOPE: 3, CLAUSE: 4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: even(T1), SCOPE: 3, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: even(T1), SCOPE: 3, CLAUSE: 4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: even(T3)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: \n\nX5 is free", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 18 [arrowhead = none ];
18 -> 2;

19 [label="EVAL with clause\ngoal :- ','(lte(X1, s(s(s(s(0))))), even(X1)).\nand substitution", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 19 [arrowhead = none ];
19 -> 3;

20 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
3 -> 20 [arrowhead = none ];
20 -> 4;

21 [label="SPLIT 2\nreplacements:\nX1 -> T1", fontsize=14, style = filled, fillcolor = violet];
3 -> 21 [arrowhead = none ];
21 -> 5;

22 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
4 -> 22 [arrowhead = none ];
22 -> 6;

23 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
5 -> 23 [arrowhead = none ];
23 -> 10;

24 [label="EVAL with clause\nlte(0, X3) :- !.\nand substitution\nX1 -> 0\nX3 -> s(s(s(s(0))))", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 24 [arrowhead = none ];
24 -> 7;

25 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
7 -> 25 [arrowhead = none ];
25 -> 8;

26 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
8 -> 26 [arrowhead = none ];
26 -> 9;

27 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
10 -> 27 [arrowhead = none ];
27 -> 11;

28 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
10 -> 28 [arrowhead = none ];
28 -> 12;

29 [label="EVAL with clause\neven(0).\nand substitution\nT1 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 29 [arrowhead = none ];
29 -> 13;

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

31 [label="EVAL with clause\neven(s(s(X5))) :- even(X5).\nand substitution\nX5 -> T3\nT1 -> s(s(T3))\nT2 -> T3", fontsize=14, style = filled, fillcolor = lightblue];
12 -> 31 [arrowhead = none ];
31 -> 16;

32 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
12 -> 32 [arrowhead = none ];
32 -> 17;

33 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
13 -> 33 [arrowhead = none ];
33 -> 15;

34 [label="INSTANCE with matching:\nT1 -> T3", fontsize=14, style = filled, fillcolor = lightgrey];
16 -> 34 [arrowhead = none , style=dashed];
34 -> 5[style=dashed];

}

(2) Obligation:

Clauses:

even5(0).
even5(s(s(T3))) :- even5(T3).
lte4(0).
goal1 :- lte4(X1).
goal1 :- lte4(0).
goal1 :- ','(lte4(s(s(T3))), even5(T3)).

Queries:

goal1().

(3) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
even5_in: (b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

goal1_in_ → U2_(lte4_in_a(X1))
lte4_in_a(0) → lte4_out_a(0)
U2_(lte4_out_a(X1)) → goal1_out_
goal1_in_ → U3_(lte4_in_g(0))
lte4_in_g(0) → lte4_out_g(0)
U3_(lte4_out_g(0)) → goal1_out_
goal1_in_ → U4_(lte4_in_a(s(s(T3))))
U4_(lte4_out_a(s(s(T3)))) → U5_(even5_in_g(T3))
even5_in_g(0) → even5_out_g(0)
even5_in_g(s(s(T3))) → U1_g(T3, even5_in_g(T3))
U1_g(T3, even5_out_g(T3)) → even5_out_g(s(s(T3)))
U5_(even5_out_g(T3)) → goal1_out_

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

goal1_in_ → U2_(lte4_in_a(X1))
lte4_in_a(0) → lte4_out_a(0)
U2_(lte4_out_a(X1)) → goal1_out_
goal1_in_ → U3_(lte4_in_g(0))
lte4_in_g(0) → lte4_out_g(0)
U3_(lte4_out_g(0)) → goal1_out_
goal1_in_ → U4_(lte4_in_a(s(s(T3))))
U4_(lte4_out_a(s(s(T3)))) → U5_(even5_in_g(T3))
even5_in_g(0) → even5_out_g(0)
even5_in_g(s(s(T3))) → U1_g(T3, even5_in_g(T3))
U1_g(T3, even5_out_g(T3)) → even5_out_g(s(s(T3)))
U5_(even5_out_g(T3)) → goal1_out_

(5) 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:

GOAL1_IN_ → U2_¹(lte4_in_a(X1))
GOAL1_IN_ → LTE4_IN_A(X1)
GOAL1_IN_ → U3_¹(lte4_in_g(0))
GOAL1_IN_ → LTE4_IN_G(0)
GOAL1_IN_ → U4_¹(lte4_in_a(s(s(T3))))
GOAL1_IN_ → LTE4_IN_A(s(s(T3)))
U4_¹(lte4_out_a(s(s(T3)))) → U5_¹(even5_in_g(T3))
U4_¹(lte4_out_a(s(s(T3)))) → EVEN5_IN_G(T3)
EVEN5_IN_G(s(s(T3))) → U1_G(T3, even5_in_g(T3))
EVEN5_IN_G(s(s(T3))) → EVEN5_IN_G(T3)

The TRS R consists of the following rules:

goal1_in_ → U2_(lte4_in_a(X1))
lte4_in_a(0) → lte4_out_a(0)
U2_(lte4_out_a(X1)) → goal1_out_
goal1_in_ → U3_(lte4_in_g(0))
lte4_in_g(0) → lte4_out_g(0)
U3_(lte4_out_g(0)) → goal1_out_
goal1_in_ → U4_(lte4_in_a(s(s(T3))))
U4_(lte4_out_a(s(s(T3)))) → U5_(even5_in_g(T3))
even5_in_g(0) → even5_out_g(0)
even5_in_g(s(s(T3))) → U1_g(T3, even5_in_g(T3))
U1_g(T3, even5_out_g(T3)) → even5_out_g(s(s(T3)))
U5_(even5_out_g(T3)) → goal1_out_

The argument filtering Pi contains the following mapping:
goal1_in_ = goal1_in_
U2_(x1) = U2_(x1)
lte4_in_a(x1) = lte4_in_a
lte4_out_a(x1) = lte4_out_a(x1)
goal1_out_ = goal1_out_
U3_(x1) = U3_(x1)
lte4_in_g(x1) = lte4_in_g(x1)
0 = 0
lte4_out_g(x1) = lte4_out_g(x1)
U4_(x1) = U4_(x1)
s(x1) = s(x1)
U5_(x1) = U5_(x1)
even5_in_g(x1) = even5_in_g(x1)
even5_out_g(x1) = even5_out_g(x1)
U1_g(x1, x2) = U1_g(x1, x2)
GOAL1_IN_ = GOAL1_IN_
U2_¹(x1) = U2_¹(x1)
LTE4_IN_A(x1) = LTE4_IN_A
U3_¹(x1) = U3_¹(x1)
LTE4_IN_G(x1) = LTE4_IN_G(x1)
U4_¹(x1) = U4_¹(x1)
U5_¹(x1) = U5_¹(x1)
EVEN5_IN_G(x1) = EVEN5_IN_G(x1)
U1_G(x1, x2) = U1_G(x1, x2)

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

(6) Obligation:

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

GOAL1_IN_ → U2_¹(lte4_in_a(X1))
GOAL1_IN_ → LTE4_IN_A(X1)
GOAL1_IN_ → U3_¹(lte4_in_g(0))
GOAL1_IN_ → LTE4_IN_G(0)
GOAL1_IN_ → U4_¹(lte4_in_a(s(s(T3))))
GOAL1_IN_ → LTE4_IN_A(s(s(T3)))
U4_¹(lte4_out_a(s(s(T3)))) → U5_¹(even5_in_g(T3))
U4_¹(lte4_out_a(s(s(T3)))) → EVEN5_IN_G(T3)
EVEN5_IN_G(s(s(T3))) → U1_G(T3, even5_in_g(T3))
EVEN5_IN_G(s(s(T3))) → EVEN5_IN_G(T3)

The TRS R consists of the following rules:

goal1_in_ → U2_(lte4_in_a(X1))
lte4_in_a(0) → lte4_out_a(0)
U2_(lte4_out_a(X1)) → goal1_out_
goal1_in_ → U3_(lte4_in_g(0))
lte4_in_g(0) → lte4_out_g(0)
U3_(lte4_out_g(0)) → goal1_out_
goal1_in_ → U4_(lte4_in_a(s(s(T3))))
U4_(lte4_out_a(s(s(T3)))) → U5_(even5_in_g(T3))
even5_in_g(0) → even5_out_g(0)
even5_in_g(s(s(T3))) → U1_g(T3, even5_in_g(T3))
U1_g(T3, even5_out_g(T3)) → even5_out_g(s(s(T3)))
U5_(even5_out_g(T3)) → goal1_out_

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 9 less nodes.

(8) Obligation:

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

EVEN5_IN_G(s(s(T3))) → EVEN5_IN_G(T3)

The TRS R consists of the following rules:

goal1_in_ → U2_(lte4_in_a(X1))
lte4_in_a(0) → lte4_out_a(0)
U2_(lte4_out_a(X1)) → goal1_out_
goal1_in_ → U3_(lte4_in_g(0))
lte4_in_g(0) → lte4_out_g(0)
U3_(lte4_out_g(0)) → goal1_out_
goal1_in_ → U4_(lte4_in_a(s(s(T3))))
U4_(lte4_out_a(s(s(T3)))) → U5_(even5_in_g(T3))
even5_in_g(0) → even5_out_g(0)
even5_in_g(s(s(T3))) → U1_g(T3, even5_in_g(T3))
U1_g(T3, even5_out_g(T3)) → even5_out_g(s(s(T3)))
U5_(even5_out_g(T3)) → goal1_out_

The argument filtering Pi contains the following mapping:
goal1_in_ = goal1_in_
U2_(x1) = U2_(x1)
lte4_in_a(x1) = lte4_in_a
lte4_out_a(x1) = lte4_out_a(x1)
goal1_out_ = goal1_out_
U3_(x1) = U3_(x1)
lte4_in_g(x1) = lte4_in_g(x1)
0 = 0
lte4_out_g(x1) = lte4_out_g(x1)
U4_(x1) = U4_(x1)
s(x1) = s(x1)
U5_(x1) = U5_(x1)
even5_in_g(x1) = even5_in_g(x1)
even5_out_g(x1) = even5_out_g(x1)
U1_g(x1, x2) = U1_g(x1, x2)
EVEN5_IN_G(x1) = EVEN5_IN_G(x1)

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

(9) UsableRulesProof (EQUIVALENT transformation)

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

(10) Obligation:

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

EVEN5_IN_G(s(s(T3))) → EVEN5_IN_G(T3)

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

(11) PiDPToQDPProof (EQUIVALENT transformation)

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

(12) Obligation:

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

EVEN5_IN_G(s(s(T3))) → EVEN5_IN_G(T3)

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

(13) 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:

EVEN5_IN_G(s(s(T3))) → EVEN5_IN_G(T3)
The graph contains the following edges 1 > 1

(0) Obligation:

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

(2) Obligation:

(3) PrologToPiTRSProof (SOUND transformation)

(4) Obligation:

(5) DependencyPairsProof (EQUIVALENT transformation)

(6) Obligation:

(7) DependencyGraphProof (EQUIVALENT transformation)

(8) Obligation:

(9) UsableRulesProof (EQUIVALENT transformation)

(10) Obligation:

(11) PiDPToQDPProof (EQUIVALENT transformation)

(12) Obligation:

(13) QDPSizeChangeProof (EQUIVALENT transformation)

(14) TRUE