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

(0) Obligation:

Clauses:

goal :- ','(lte(X, s(s(s(s(0))))), even(X)).
lte(0, Y).
lte(X, s(Y)) :- ','(no(zero(X)), ','(p(X, P), lte(P, Y))).
even(0).
even(s(s(X))) :- even(X).
p(0, 0).
p(s(X), X).
zero(0).
no(X) :- ','(X, ','(!, failure(a))).
no(X1).
failure(b).

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(X2, s(s(s(s(0))))), even(X2))\n\nX2 is free", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: lte(X2, s(s(s(s(0)))))\n\nX2 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(X2, s(s(s(s(0))))), SCOPE: 2, CLAUSE: 1 | lte(X2, s(s(s(s(0))))), SCOPE: 2, CLAUSE: 2\n\nX2 is free", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: lte(X2, s(s(s(s(0))))), SCOPE: 2, CLAUSE: 1\n\nX2 is free", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: lte(X2, s(s(s(s(0))))), SCOPE: 2, CLAUSE: 2\n\nX2 is free", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(no(zero(X11)), ','(p(X11, X10), lte(X10, s(s(s(0))))))\n\nX11 is free\nX10 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(no(zero(X11)), ','(p(X11, X10), lte(X10, s(s(s(0)))))), SCOPE: 3, CLAUSE: 8 | ','(no(zero(X11)), ','(p(X11, X10), lte(X10, s(s(s(0)))))), SCOPE: 3, CLAUSE: 9 | ?_3\n\nX11 is free\nX10 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(zero(X14), ','(!_3, ','(failure(a), ','(p(X14, X10), lte(X10, s(s(s(0)))))))) | ','(no(zero(X11)), ','(p(X11, X10), lte(X10, s(s(s(0)))))), SCOPE: 3, CLAUSE: 9 | ?_3\n\nX11 is free\nX10 is free; \nX14 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(zero(X14), ','(!_3, ','(failure(a), ','(p(X14, X10), lte(X10, s(s(s(0)))))))), SCOPE: 4, CLAUSE: 7 | ','(no(zero(X11)), ','(p(X11, X10), lte(X10, s(s(s(0)))))), SCOPE: 3, CLAUSE: 9 | ?_3\n\nX11 is free\nX10 is free; \nX14 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: ','(!_3, ','(failure(a), ','(p(0, X10), lte(X10, s(s(s(0))))))) | ','(no(zero(X11)), ','(p(X11, X10), lte(X10, s(s(s(0)))))), SCOPE: 3, CLAUSE: 9 | ?_3\n\nX11 is free\nX10 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: ','(failure(a), ','(p(0, X10), lte(X10, s(s(s(0))))))\n\nX10 is free", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: ','(failure(a), ','(p(0, X10), lte(X10, s(s(s(0)))))), SCOPE: 5, CLAUSE: 10\n\nX10 is free", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: even(T1), SCOPE: 6, CLAUSE: 3 | even(T1), SCOPE: 6, CLAUSE: 4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: even(T1), SCOPE: 6, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: even(T1), SCOPE: 6, CLAUSE: 4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: even(T3)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: \n\nX16 is free", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 27 [arrowhead = none ];
27 -> 2;

28 [label="EVAL with clause\ngoal :- ','(lte(X2, s(s(s(s(0))))), even(X2)).\nand substitution", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 28 [arrowhead = none ];
28 -> 3;

29 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
3 -> 29 [arrowhead = none ];
29 -> 4;

30 [label="SPLIT 2\nreplacements:\nX2 -> T1", fontsize=14, style = filled, fillcolor = violet];
3 -> 30 [arrowhead = none ];
30 -> 5;

31 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
4 -> 31 [arrowhead = none ];
31 -> 6;

32 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
5 -> 32 [arrowhead = none ];
32 -> 19;

33 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
6 -> 33 [arrowhead = none ];
33 -> 7;

34 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
6 -> 34 [arrowhead = none ];
34 -> 8;

35 [label="EVAL with clause\nlte(0, X5).\nand substitution\nX2 -> 0\nX5 -> s(s(s(s(0))))", fontsize=14, style = filled, fillcolor = lightblue];
7 -> 35 [arrowhead = none ];
35 -> 9;

36 [label="EVAL with clause\nlte(X8, s(X9)) :- ','(no(zero(X8)), ','(p(X8, X10), lte(X10, X9))).\nand substitution\nX2 -> X11\nX8 -> X11\nX9 -> s(s(s(0)))", fontsize=14, style = filled, fillcolor = lightblue];
8 -> 36 [arrowhead = none ];
36 -> 11;

37 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 37 [arrowhead = none ];
37 -> 10;

38 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
11 -> 38 [arrowhead = none ];
38 -> 12;

39 [label="EVAL with clause\nno(X13) :- ','(X13, ','(!, failure(a))).\nand substitution\nX11 -> X14\nX13 -> zero(X14)", fontsize=14, style = filled, fillcolor = lightblue];
12 -> 39 [arrowhead = none ];
39 -> 13;

40 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
13 -> 40 [arrowhead = none ];
40 -> 14;

41 [label="EVAL with clause\nzero(0).\nand substitution\nX14 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
14 -> 41 [arrowhead = none ];
41 -> 15;

42 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 42 [arrowhead = none ];
42 -> 16;

43 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
16 -> 43 [arrowhead = none ];
43 -> 17;

44 [label="BACKTRACK\nfor clause: failure(b)because of non-unification", fontsize=14, style = filled, fillcolor = white];
17 -> 44 [arrowhead = none ];
44 -> 18;

45 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
19 -> 45 [arrowhead = none ];
45 -> 20;

46 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
19 -> 46 [arrowhead = none ];
46 -> 21;

47 [label="EVAL with clause\neven(0).\nand substitution\nT1 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
20 -> 47 [arrowhead = none ];
47 -> 22;

48 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
20 -> 48 [arrowhead = none ];
48 -> 23;

49 [label="EVAL with clause\neven(s(s(X16))) :- even(X16).\nand substitution\nX16 -> T3\nT1 -> s(s(T3))\nT2 -> T3", fontsize=14, style = filled, fillcolor = lightblue];
21 -> 49 [arrowhead = none ];
49 -> 25;

50 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
21 -> 50 [arrowhead = none ];
50 -> 26;

51 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
22 -> 51 [arrowhead = none ];
51 -> 24;

52 [label="INSTANCE with matching:\nT1 -> T3", fontsize=14, style = filled, fillcolor = lightgrey];
25 -> 52 [arrowhead = none , style=dashed];
52 -> 5[style=dashed];

}

(2) Obligation:

Clauses:

even5(0).
even5(s(s(T3))) :- even5(T3).
lte4(0).
goal1 :- lte4(X2).
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(X2))
lte4_in_a(0) → lte4_out_a(0)
U2_(lte4_out_a(X2)) → 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
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
U1_g(x1, x2)  =  U1_g(x2)

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(X2))
lte4_in_a(0) → lte4_out_a(0)
U2_(lte4_out_a(X2)) → 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
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
U1_g(x1, x2)  =  U1_g(x2)

(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(X2))
GOAL1_IN_ → LTE4_IN_A(X2)
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(X2))
lte4_in_a(0) → lte4_out_a(0)
U2_(lte4_out_a(X2)) → 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
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
U1_g(x1, x2)  =  U1_g(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(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(X2))
GOAL1_IN_ → LTE4_IN_A(X2)
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(X2))
lte4_in_a(0) → lte4_out_a(0)
U2_(lte4_out_a(X2)) → 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
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
U1_g(x1, x2)  =  U1_g(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(x2)

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

(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(X2))
lte4_in_a(0) → lte4_out_a(0)
U2_(lte4_out_a(X2)) → 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
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
U1_g(x1, x2)  =  U1_g(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

(14) TRUE