Left Termination proof of /tmp/tmpTdnL6X/add2.pl

Left Termination of the query pattern add_in_3(a, g, a) w.r.t. the given Prolog program could not be shown:

0 Prolog

↳1 PrologToPrologProblemTransformerProof (⇐)

↳2 Prolog

    ↳3 PrologToPiTRSProof (⇐)

        ↳4 PiTRS

            ↳5 DependencyPairsProof (⇔)

                ↳6 PiDP

                    ↳7 DependencyGraphProof (⇔)

                        ↳8 AND

                            ↳9 PiDP

                                ↳10 UsableRulesProof (⇔)

                                    ↳11 PiDP

                                        ↳12 PiDPToQDPProof (⇐)

                                            ↳13 QDP

                                                ↳14 NonTerminationProof (⇔)

                                                    ↳15 FALSE

                            ↳16 PiDP

                                ↳17 UsableRulesProof (⇔)

                                    ↳18 PiDP

                                        ↳19 PiDPToQDPProof (⇐)

                                            ↳20 QDP

                                                ↳21 QDPSizeChangeProof (⇔)

                                                    ↳22 TRUE

    ↳23 PrologToPiTRSProof (⇐)

        ↳24 PiTRS

            ↳25 DependencyPairsProof (⇔)

                ↳26 PiDP

                    ↳27 DependencyGraphProof (⇔)

                        ↳28 AND

                            ↳29 PiDP

                                ↳30 UsableRulesProof (⇔)

                                    ↳31 PiDP

                                        ↳32 PiDPToQDPProof (⇐)

                                            ↳33 QDP

                                                ↳34 NonTerminationProof (⇔)

                                                    ↳35 FALSE

                            ↳36 PiDP

                                ↳37 UsableRulesProof (⇔)

                                    ↳38 PiDP

                                        ↳39 PiDPToQDPProof (⇐)

                                            ↳40 QDP

                                                ↳41 QDPSizeChangeProof (⇔)

                                                    ↳42 TRUE

(0) Obligation:

Clauses:

add(X, 0, X) :- !.
add(X, Y, s(Z)) :- ','(p(Y, P), add(X, P, Z)).
p(0, 0).
p(s(X), X).

Queries:

add(a,g,a).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: add(T1, T2, T3)\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: add(T1, T2, T3), SCOPE: 1, CLAUSE: 0 | add(T1, T2, T3), SCOPE: 1, CLAUSE: 1 | ?_1\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: !_1 | add(T1, 0, T3), SCOPE: 1, CLAUSE: 1 | ?_1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: add(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT2 is ground\nX2 is free; \nadd(T1, T2, T3) !=? add(X2, 0, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: ','(p(T6, X9), add(T8, X9, T9))\n\nT6 is ground\nX2 is free; \nX9 is free; \nadd(T1, T6, T3) !=? add(X2, 0, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\nX2 is free\nX6 is free; \nX7 is free; \nX8 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(p(T6, X9), add(T8, X9, T9)), SCOPE: 2, CLAUSE: 2 | ','(p(T6, X9), add(T8, X9, T9)), SCOPE: 2, CLAUSE: 3\n\nT6 is ground\nX2 is free; \nX9 is free; \nadd(T1, T6, T3) !=? add(X2, 0, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: add(T8, 0, T9) | ','(p(0, X9), add(T8, X9, T9)), SCOPE: 2, CLAUSE: 3\n\nX2 is free\nX9 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(p(T6, X9), add(T8, X9, T9)), SCOPE: 2, CLAUSE: 3\n\nT6 is ground\nX2 is free; \nX9 is free; \nadd(T1, T6, T3) !=? add(X2, 0, X2)\np(T6, X9) !=? p(0, 0)", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: add(T8, 0, T9)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(p(0, X9), add(T8, X9, T9)), SCOPE: 2, CLAUSE: 3\n\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: add(T8, 0, T9), SCOPE: 3, CLAUSE: 0 | add(T8, 0, T9), SCOPE: 3, CLAUSE: 1 | ?_3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: !_3 | add(T8, 0, T9), SCOPE: 3, CLAUSE: 1 | ?_3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: add(T8, 0, T9), SCOPE: 3, CLAUSE: 1\n\nX11 is free\nadd(T8, 0, T9) !=? add(X11, 0, X11)", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: ','(p(0, X18), add(T13, X18, T14))\n\nX11 is free\nX18 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: \n\nX11 is free\nX15 is free; \nX16 is free; \nX17 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: ','(p(0, X18), add(T13, X18, T14)), SCOPE: 4, CLAUSE: 2 | ','(p(0, X18), add(T13, X18, T14)), SCOPE: 4, CLAUSE: 3\n\nX18 is free", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: add(T13, 0, T14) | ','(p(0, X18), add(T13, X18, T14)), SCOPE: 4, CLAUSE: 3\n\nX18 is free", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: add(T8, T15, T9)\n\nT15 is ground\nX2 is free; \nX9 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: \n\nX2 is free\nX9 is free; \nX21 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 26 [arrowhead = none ];
26 -> 2;

27 [label="EVAL with clause\nadd(X2, 0, X2) :- !.\nand substitution\nT1 -> T4\nX2 -> T4\nT2 -> 0\nT3 -> T4", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 27 [arrowhead = none ];
27 -> 3;

28 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 28 [arrowhead = none ];
28 -> 4;

29 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 29 [arrowhead = none ];
29 -> 5;

30 [label="EVAL with clause\nadd(X6, X7, s(X8)) :- ','(p(X7, X9), add(X6, X9, X8)).\nand substitution\nT1 -> T8\nX6 -> T8\nT2 -> T6\nX7 -> T6\nX8 -> T9\nT3 -> s(T9)\nT5 -> T8\nT7 -> T9", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 30 [arrowhead = none ];
30 -> 7;

31 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 31 [arrowhead = none ];
31 -> 8;

32 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 32 [arrowhead = none ];
32 -> 6;

33 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 33 [arrowhead = none ];
33 -> 9;

34 [label="EVAL with clause\np(0, 0).\nand substitution\nT6 -> 0\nX9 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 34 [arrowhead = none ];
34 -> 10;

35 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 35 [arrowhead = none ];
35 -> 11;

36 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
10 -> 36 [arrowhead = none ];
36 -> 12;

37 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
10 -> 37 [arrowhead = none ];
37 -> 13;

38 [label="EVAL with clause\np(s(X21), X21).\nand substitution\nX21 -> T15\nT6 -> s(T15)\nX9 -> T15", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 38 [arrowhead = none ];
38 -> 24;

39 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 39 [arrowhead = none ];
39 -> 25;

40 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
12 -> 40 [arrowhead = none ];
40 -> 14;

41 [label="BACKTRACK\nfor clause: p(s(X), X)because of non-unification", fontsize=14, style = filled, fillcolor = white];
13 -> 41 [arrowhead = none ];
41 -> 23;

42 [label="EVAL with clause\nadd(X11, 0, X11) :- !.\nand substitution\nT8 -> T10\nX11 -> T10\nT9 -> T10", fontsize=14, style = filled, fillcolor = lightblue];
14 -> 42 [arrowhead = none ];
42 -> 15;

43 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
14 -> 43 [arrowhead = none ];
43 -> 16;

44 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 44 [arrowhead = none ];
44 -> 17;

45 [label="EVAL with clause\nadd(X15, X16, s(X17)) :- ','(p(X16, X18), add(X15, X18, X17)).\nand substitution\nT8 -> T13\nX15 -> T13\nX16 -> 0\nX17 -> T14\nT9 -> s(T14)\nT11 -> T13\nT12 -> T14", fontsize=14, style = filled, fillcolor = lightblue];
16 -> 45 [arrowhead = none ];
45 -> 19;

46 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
16 -> 46 [arrowhead = none ];
46 -> 20;

47 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 47 [arrowhead = none ];
47 -> 18;

48 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
19 -> 48 [arrowhead = none ];
48 -> 21;

49 [label="EVAL with clause\np(0, 0).\nand substitution\nX18 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
21 -> 49 [arrowhead = none ];
49 -> 22;

50 [label="INSTANCE with matching:\nT8 -> T13\nT9 -> T14\nX9 -> X18\nX2 -> X18", fontsize=14, style = filled, fillcolor = lightgrey];
22 -> 50 [arrowhead = none , style=dashed];
50 -> 10[style=dashed];

51 [label="INSTANCE with matching:\nT1 -> T8\nT2 -> T15\nT3 -> T9", fontsize=14, style = filled, fillcolor = lightgrey];
24 -> 51 [arrowhead = none , style=dashed];
51 -> 1[style=dashed];

}

(2) Obligation:

Clauses:

add10(T10, T10, X9).
add10(T13, s(T14), X9) :- add10(T13, T14, X18).
add1(T4, 0, T4).
add1(T10, 0, s(T10)).
add1(T13, 0, s(s(T14))) :- add10(T13, T14, X18).
add1(T8, s(T15), s(T9)) :- add1(T8, T15, T9).

Queries:

add1(a,g,a).

(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:
add1_in: (f,b,f)
add10_in: (f,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

add1_in_aga(T4, 0, T4) → add1_out_aga(T4, 0, T4)
add1_in_aga(T10, 0, s(T10)) → add1_out_aga(T10, 0, s(T10))
add1_in_aga(T13, 0, s(s(T14))) → U2_aga(T13, T14, add10_in_aaa(T13, T14, X18))
add10_in_aaa(T10, T10, X9) → add10_out_aaa(T10, T10, X9)
add10_in_aaa(T13, s(T14), X9) → U1_aaa(T13, T14, X9, add10_in_aaa(T13, T14, X18))
U1_aaa(T13, T14, X9, add10_out_aaa(T13, T14, X18)) → add10_out_aaa(T13, s(T14), X9)
U2_aga(T13, T14, add10_out_aaa(T13, T14, X18)) → add1_out_aga(T13, 0, s(s(T14)))
add1_in_aga(T8, s(T15), s(T9)) → U3_aga(T8, T15, T9, add1_in_aga(T8, T15, T9))
U3_aga(T8, T15, T9, add1_out_aga(T8, T15, T9)) → add1_out_aga(T8, s(T15), s(T9))

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

add1_in_aga(T4, 0, T4) → add1_out_aga(T4, 0, T4)
add1_in_aga(T10, 0, s(T10)) → add1_out_aga(T10, 0, s(T10))
add1_in_aga(T13, 0, s(s(T14))) → U2_aga(T13, T14, add10_in_aaa(T13, T14, X18))
add10_in_aaa(T10, T10, X9) → add10_out_aaa(T10, T10, X9)
add10_in_aaa(T13, s(T14), X9) → U1_aaa(T13, T14, X9, add10_in_aaa(T13, T14, X18))
U1_aaa(T13, T14, X9, add10_out_aaa(T13, T14, X18)) → add10_out_aaa(T13, s(T14), X9)
U2_aga(T13, T14, add10_out_aaa(T13, T14, X18)) → add1_out_aga(T13, 0, s(s(T14)))
add1_in_aga(T8, s(T15), s(T9)) → U3_aga(T8, T15, T9, add1_in_aga(T8, T15, T9))
U3_aga(T8, T15, T9, add1_out_aga(T8, T15, T9)) → add1_out_aga(T8, s(T15), s(T9))

(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:

ADD1_IN_AGA(T13, 0, s(s(T14))) → U2_AGA(T13, T14, add10_in_aaa(T13, T14, X18))
ADD1_IN_AGA(T13, 0, s(s(T14))) → ADD10_IN_AAA(T13, T14, X18)
ADD10_IN_AAA(T13, s(T14), X9) → U1_AAA(T13, T14, X9, add10_in_aaa(T13, T14, X18))
ADD10_IN_AAA(T13, s(T14), X9) → ADD10_IN_AAA(T13, T14, X18)
ADD1_IN_AGA(T8, s(T15), s(T9)) → U3_AGA(T8, T15, T9, add1_in_aga(T8, T15, T9))
ADD1_IN_AGA(T8, s(T15), s(T9)) → ADD1_IN_AGA(T8, T15, T9)

The TRS R consists of the following rules:

add1_in_aga(T4, 0, T4) → add1_out_aga(T4, 0, T4)
add1_in_aga(T10, 0, s(T10)) → add1_out_aga(T10, 0, s(T10))
add1_in_aga(T13, 0, s(s(T14))) → U2_aga(T13, T14, add10_in_aaa(T13, T14, X18))
add10_in_aaa(T10, T10, X9) → add10_out_aaa(T10, T10, X9)
add10_in_aaa(T13, s(T14), X9) → U1_aaa(T13, T14, X9, add10_in_aaa(T13, T14, X18))
U1_aaa(T13, T14, X9, add10_out_aaa(T13, T14, X18)) → add10_out_aaa(T13, s(T14), X9)
U2_aga(T13, T14, add10_out_aaa(T13, T14, X18)) → add1_out_aga(T13, 0, s(s(T14)))
add1_in_aga(T8, s(T15), s(T9)) → U3_aga(T8, T15, T9, add1_in_aga(T8, T15, T9))
U3_aga(T8, T15, T9, add1_out_aga(T8, T15, T9)) → add1_out_aga(T8, s(T15), s(T9))

The argument filtering Pi contains the following mapping:
add1_in_aga(x1, x2, x3) = add1_in_aga(x2)
0 = 0
add1_out_aga(x1, x2, x3) = add1_out_aga
U2_aga(x1, x2, x3) = U2_aga(x3)
add10_in_aaa(x1, x2, x3) = add10_in_aaa
add10_out_aaa(x1, x2, x3) = add10_out_aaa
U1_aaa(x1, x2, x3, x4) = U1_aaa(x4)
s(x1) = s(x1)
U3_aga(x1, x2, x3, x4) = U3_aga(x4)
ADD1_IN_AGA(x1, x2, x3) = ADD1_IN_AGA(x2)
U2_AGA(x1, x2, x3) = U2_AGA(x3)
ADD10_IN_AAA(x1, x2, x3) = ADD10_IN_AAA
U1_AAA(x1, x2, x3, x4) = U1_AAA(x4)
U3_AGA(x1, x2, x3, x4) = U3_AGA(x4)

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

(6) Obligation:

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

ADD1_IN_AGA(T13, 0, s(s(T14))) → U2_AGA(T13, T14, add10_in_aaa(T13, T14, X18))
ADD1_IN_AGA(T13, 0, s(s(T14))) → ADD10_IN_AAA(T13, T14, X18)
ADD10_IN_AAA(T13, s(T14), X9) → U1_AAA(T13, T14, X9, add10_in_aaa(T13, T14, X18))
ADD10_IN_AAA(T13, s(T14), X9) → ADD10_IN_AAA(T13, T14, X18)
ADD1_IN_AGA(T8, s(T15), s(T9)) → U3_AGA(T8, T15, T9, add1_in_aga(T8, T15, T9))
ADD1_IN_AGA(T8, s(T15), s(T9)) → ADD1_IN_AGA(T8, T15, T9)

The TRS R consists of the following rules:

add1_in_aga(T4, 0, T4) → add1_out_aga(T4, 0, T4)
add1_in_aga(T10, 0, s(T10)) → add1_out_aga(T10, 0, s(T10))
add1_in_aga(T13, 0, s(s(T14))) → U2_aga(T13, T14, add10_in_aaa(T13, T14, X18))
add10_in_aaa(T10, T10, X9) → add10_out_aaa(T10, T10, X9)
add10_in_aaa(T13, s(T14), X9) → U1_aaa(T13, T14, X9, add10_in_aaa(T13, T14, X18))
U1_aaa(T13, T14, X9, add10_out_aaa(T13, T14, X18)) → add10_out_aaa(T13, s(T14), X9)
U2_aga(T13, T14, add10_out_aaa(T13, T14, X18)) → add1_out_aga(T13, 0, s(s(T14)))
add1_in_aga(T8, s(T15), s(T9)) → U3_aga(T8, T15, T9, add1_in_aga(T8, T15, T9))
U3_aga(T8, T15, T9, add1_out_aga(T8, T15, T9)) → add1_out_aga(T8, s(T15), s(T9))

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 4 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

ADD10_IN_AAA(T13, s(T14), X9) → ADD10_IN_AAA(T13, T14, X18)

The TRS R consists of the following rules:

add1_in_aga(T4, 0, T4) → add1_out_aga(T4, 0, T4)
add1_in_aga(T10, 0, s(T10)) → add1_out_aga(T10, 0, s(T10))
add1_in_aga(T13, 0, s(s(T14))) → U2_aga(T13, T14, add10_in_aaa(T13, T14, X18))
add10_in_aaa(T10, T10, X9) → add10_out_aaa(T10, T10, X9)
add10_in_aaa(T13, s(T14), X9) → U1_aaa(T13, T14, X9, add10_in_aaa(T13, T14, X18))
U1_aaa(T13, T14, X9, add10_out_aaa(T13, T14, X18)) → add10_out_aaa(T13, s(T14), X9)
U2_aga(T13, T14, add10_out_aaa(T13, T14, X18)) → add1_out_aga(T13, 0, s(s(T14)))
add1_in_aga(T8, s(T15), s(T9)) → U3_aga(T8, T15, T9, add1_in_aga(T8, T15, T9))
U3_aga(T8, T15, T9, add1_out_aga(T8, T15, T9)) → add1_out_aga(T8, s(T15), s(T9))

The argument filtering Pi contains the following mapping:
add1_in_aga(x1, x2, x3) = add1_in_aga(x2)
0 = 0
add1_out_aga(x1, x2, x3) = add1_out_aga
U2_aga(x1, x2, x3) = U2_aga(x3)
add10_in_aaa(x1, x2, x3) = add10_in_aaa
add10_out_aaa(x1, x2, x3) = add10_out_aaa
U1_aaa(x1, x2, x3, x4) = U1_aaa(x4)
s(x1) = s(x1)
U3_aga(x1, x2, x3, x4) = U3_aga(x4)
ADD10_IN_AAA(x1, x2, x3) = ADD10_IN_AAA

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

(10) UsableRulesProof (EQUIVALENT transformation)

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

(11) Obligation:

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

ADD10_IN_AAA(T13, s(T14), X9) → ADD10_IN_AAA(T13, T14, X18)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1) = s(x1)
ADD10_IN_AAA(x1, x2, x3) = ADD10_IN_AAA

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

(12) PiDPToQDPProof (SOUND transformation)

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

(13) Obligation:

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

ADD10_IN_AAA → ADD10_IN_AAA

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

(14) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = ADD10_IN_AAA evaluates to t =ADD10_IN_AAA

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from ADD10_IN_AAA to ADD10_IN_AAA.

(15) FALSE

(16) Obligation:

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

ADD1_IN_AGA(T8, s(T15), s(T9)) → ADD1_IN_AGA(T8, T15, T9)

The TRS R consists of the following rules:

add1_in_aga(T4, 0, T4) → add1_out_aga(T4, 0, T4)
add1_in_aga(T10, 0, s(T10)) → add1_out_aga(T10, 0, s(T10))
add1_in_aga(T13, 0, s(s(T14))) → U2_aga(T13, T14, add10_in_aaa(T13, T14, X18))
add10_in_aaa(T10, T10, X9) → add10_out_aaa(T10, T10, X9)
add10_in_aaa(T13, s(T14), X9) → U1_aaa(T13, T14, X9, add10_in_aaa(T13, T14, X18))
U1_aaa(T13, T14, X9, add10_out_aaa(T13, T14, X18)) → add10_out_aaa(T13, s(T14), X9)
U2_aga(T13, T14, add10_out_aaa(T13, T14, X18)) → add1_out_aga(T13, 0, s(s(T14)))
add1_in_aga(T8, s(T15), s(T9)) → U3_aga(T8, T15, T9, add1_in_aga(T8, T15, T9))
U3_aga(T8, T15, T9, add1_out_aga(T8, T15, T9)) → add1_out_aga(T8, s(T15), s(T9))

(17) UsableRulesProof (EQUIVALENT transformation)

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

(18) Obligation:

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

ADD1_IN_AGA(T8, s(T15), s(T9)) → ADD1_IN_AGA(T8, T15, T9)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1) = s(x1)
ADD1_IN_AGA(x1, x2, x3) = ADD1_IN_AGA(x2)

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

(19) PiDPToQDPProof (SOUND transformation)

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

(20) Obligation:

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

ADD1_IN_AGA(s(T15)) → ADD1_IN_AGA(T15)

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

(21) 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:

ADD1_IN_AGA(s(T15)) → ADD1_IN_AGA(T15)
The graph contains the following edges 1 > 1

(22) TRUE

(23) PrologToPiTRSProof (SOUND transformation)

add1_in_aga(T4, 0, T4) → add1_out_aga(T4, 0, T4)
add1_in_aga(T10, 0, s(T10)) → add1_out_aga(T10, 0, s(T10))
add1_in_aga(T13, 0, s(s(T14))) → U2_aga(T13, T14, add10_in_aaa(T13, T14, X18))
add10_in_aaa(T10, T10, X9) → add10_out_aaa(T10, T10, X9)
add10_in_aaa(T13, s(T14), X9) → U1_aaa(T13, T14, X9, add10_in_aaa(T13, T14, X18))
U1_aaa(T13, T14, X9, add10_out_aaa(T13, T14, X18)) → add10_out_aaa(T13, s(T14), X9)
U2_aga(T13, T14, add10_out_aaa(T13, T14, X18)) → add1_out_aga(T13, 0, s(s(T14)))
add1_in_aga(T8, s(T15), s(T9)) → U3_aga(T8, T15, T9, add1_in_aga(T8, T15, T9))
U3_aga(T8, T15, T9, add1_out_aga(T8, T15, T9)) → add1_out_aga(T8, s(T15), s(T9))

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(24) Obligation:

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

add1_in_aga(T4, 0, T4) → add1_out_aga(T4, 0, T4)
add1_in_aga(T10, 0, s(T10)) → add1_out_aga(T10, 0, s(T10))
add1_in_aga(T13, 0, s(s(T14))) → U2_aga(T13, T14, add10_in_aaa(T13, T14, X18))
add10_in_aaa(T10, T10, X9) → add10_out_aaa(T10, T10, X9)
add10_in_aaa(T13, s(T14), X9) → U1_aaa(T13, T14, X9, add10_in_aaa(T13, T14, X18))
U1_aaa(T13, T14, X9, add10_out_aaa(T13, T14, X18)) → add10_out_aaa(T13, s(T14), X9)
U2_aga(T13, T14, add10_out_aaa(T13, T14, X18)) → add1_out_aga(T13, 0, s(s(T14)))
add1_in_aga(T8, s(T15), s(T9)) → U3_aga(T8, T15, T9, add1_in_aga(T8, T15, T9))
U3_aga(T8, T15, T9, add1_out_aga(T8, T15, T9)) → add1_out_aga(T8, s(T15), s(T9))

(25) 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:

ADD1_IN_AGA(T13, 0, s(s(T14))) → U2_AGA(T13, T14, add10_in_aaa(T13, T14, X18))
ADD1_IN_AGA(T13, 0, s(s(T14))) → ADD10_IN_AAA(T13, T14, X18)
ADD10_IN_AAA(T13, s(T14), X9) → U1_AAA(T13, T14, X9, add10_in_aaa(T13, T14, X18))
ADD10_IN_AAA(T13, s(T14), X9) → ADD10_IN_AAA(T13, T14, X18)
ADD1_IN_AGA(T8, s(T15), s(T9)) → U3_AGA(T8, T15, T9, add1_in_aga(T8, T15, T9))
ADD1_IN_AGA(T8, s(T15), s(T9)) → ADD1_IN_AGA(T8, T15, T9)

The TRS R consists of the following rules:

add1_in_aga(T4, 0, T4) → add1_out_aga(T4, 0, T4)
add1_in_aga(T10, 0, s(T10)) → add1_out_aga(T10, 0, s(T10))
add1_in_aga(T13, 0, s(s(T14))) → U2_aga(T13, T14, add10_in_aaa(T13, T14, X18))
add10_in_aaa(T10, T10, X9) → add10_out_aaa(T10, T10, X9)
add10_in_aaa(T13, s(T14), X9) → U1_aaa(T13, T14, X9, add10_in_aaa(T13, T14, X18))
U1_aaa(T13, T14, X9, add10_out_aaa(T13, T14, X18)) → add10_out_aaa(T13, s(T14), X9)
U2_aga(T13, T14, add10_out_aaa(T13, T14, X18)) → add1_out_aga(T13, 0, s(s(T14)))
add1_in_aga(T8, s(T15), s(T9)) → U3_aga(T8, T15, T9, add1_in_aga(T8, T15, T9))
U3_aga(T8, T15, T9, add1_out_aga(T8, T15, T9)) → add1_out_aga(T8, s(T15), s(T9))

The argument filtering Pi contains the following mapping:
add1_in_aga(x1, x2, x3) = add1_in_aga(x2)
0 = 0
add1_out_aga(x1, x2, x3) = add1_out_aga(x2)
U2_aga(x1, x2, x3) = U2_aga(x3)
add10_in_aaa(x1, x2, x3) = add10_in_aaa
add10_out_aaa(x1, x2, x3) = add10_out_aaa
U1_aaa(x1, x2, x3, x4) = U1_aaa(x4)
s(x1) = s(x1)
U3_aga(x1, x2, x3, x4) = U3_aga(x2, x4)
ADD1_IN_AGA(x1, x2, x3) = ADD1_IN_AGA(x2)
U2_AGA(x1, x2, x3) = U2_AGA(x3)
ADD10_IN_AAA(x1, x2, x3) = ADD10_IN_AAA
U1_AAA(x1, x2, x3, x4) = U1_AAA(x4)
U3_AGA(x1, x2, x3, x4) = U3_AGA(x2, x4)

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

(26) Obligation:

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

ADD1_IN_AGA(T13, 0, s(s(T14))) → U2_AGA(T13, T14, add10_in_aaa(T13, T14, X18))
ADD1_IN_AGA(T13, 0, s(s(T14))) → ADD10_IN_AAA(T13, T14, X18)
ADD10_IN_AAA(T13, s(T14), X9) → U1_AAA(T13, T14, X9, add10_in_aaa(T13, T14, X18))
ADD10_IN_AAA(T13, s(T14), X9) → ADD10_IN_AAA(T13, T14, X18)
ADD1_IN_AGA(T8, s(T15), s(T9)) → U3_AGA(T8, T15, T9, add1_in_aga(T8, T15, T9))
ADD1_IN_AGA(T8, s(T15), s(T9)) → ADD1_IN_AGA(T8, T15, T9)

The TRS R consists of the following rules:

add1_in_aga(T4, 0, T4) → add1_out_aga(T4, 0, T4)
add1_in_aga(T10, 0, s(T10)) → add1_out_aga(T10, 0, s(T10))
add1_in_aga(T13, 0, s(s(T14))) → U2_aga(T13, T14, add10_in_aaa(T13, T14, X18))
add10_in_aaa(T10, T10, X9) → add10_out_aaa(T10, T10, X9)
add10_in_aaa(T13, s(T14), X9) → U1_aaa(T13, T14, X9, add10_in_aaa(T13, T14, X18))
U1_aaa(T13, T14, X9, add10_out_aaa(T13, T14, X18)) → add10_out_aaa(T13, s(T14), X9)
U2_aga(T13, T14, add10_out_aaa(T13, T14, X18)) → add1_out_aga(T13, 0, s(s(T14)))
add1_in_aga(T8, s(T15), s(T9)) → U3_aga(T8, T15, T9, add1_in_aga(T8, T15, T9))
U3_aga(T8, T15, T9, add1_out_aga(T8, T15, T9)) → add1_out_aga(T8, s(T15), s(T9))

(27) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 4 less nodes.

(28) Complex Obligation (AND)

(29) Obligation:

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

ADD10_IN_AAA(T13, s(T14), X9) → ADD10_IN_AAA(T13, T14, X18)

The TRS R consists of the following rules:

add1_in_aga(T4, 0, T4) → add1_out_aga(T4, 0, T4)
add1_in_aga(T10, 0, s(T10)) → add1_out_aga(T10, 0, s(T10))
add1_in_aga(T13, 0, s(s(T14))) → U2_aga(T13, T14, add10_in_aaa(T13, T14, X18))
add10_in_aaa(T10, T10, X9) → add10_out_aaa(T10, T10, X9)
add10_in_aaa(T13, s(T14), X9) → U1_aaa(T13, T14, X9, add10_in_aaa(T13, T14, X18))
U1_aaa(T13, T14, X9, add10_out_aaa(T13, T14, X18)) → add10_out_aaa(T13, s(T14), X9)
U2_aga(T13, T14, add10_out_aaa(T13, T14, X18)) → add1_out_aga(T13, 0, s(s(T14)))
add1_in_aga(T8, s(T15), s(T9)) → U3_aga(T8, T15, T9, add1_in_aga(T8, T15, T9))
U3_aga(T8, T15, T9, add1_out_aga(T8, T15, T9)) → add1_out_aga(T8, s(T15), s(T9))

The argument filtering Pi contains the following mapping:
add1_in_aga(x1, x2, x3) = add1_in_aga(x2)
0 = 0
add1_out_aga(x1, x2, x3) = add1_out_aga(x2)
U2_aga(x1, x2, x3) = U2_aga(x3)
add10_in_aaa(x1, x2, x3) = add10_in_aaa
add10_out_aaa(x1, x2, x3) = add10_out_aaa
U1_aaa(x1, x2, x3, x4) = U1_aaa(x4)
s(x1) = s(x1)
U3_aga(x1, x2, x3, x4) = U3_aga(x2, x4)
ADD10_IN_AAA(x1, x2, x3) = ADD10_IN_AAA

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

(30) UsableRulesProof (EQUIVALENT transformation)

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

(31) Obligation:

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

ADD10_IN_AAA(T13, s(T14), X9) → ADD10_IN_AAA(T13, T14, X18)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1) = s(x1)
ADD10_IN_AAA(x1, x2, x3) = ADD10_IN_AAA

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

(32) PiDPToQDPProof (SOUND transformation)

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

(33) Obligation:

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

ADD10_IN_AAA → ADD10_IN_AAA

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

(34) NonTerminationProof (EQUIVALENT transformation)

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from ADD10_IN_AAA to ADD10_IN_AAA.

(35) FALSE

(36) Obligation:

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

ADD1_IN_AGA(T8, s(T15), s(T9)) → ADD1_IN_AGA(T8, T15, T9)

The TRS R consists of the following rules:

add1_in_aga(T4, 0, T4) → add1_out_aga(T4, 0, T4)
add1_in_aga(T10, 0, s(T10)) → add1_out_aga(T10, 0, s(T10))
add1_in_aga(T13, 0, s(s(T14))) → U2_aga(T13, T14, add10_in_aaa(T13, T14, X18))
add10_in_aaa(T10, T10, X9) → add10_out_aaa(T10, T10, X9)
add10_in_aaa(T13, s(T14), X9) → U1_aaa(T13, T14, X9, add10_in_aaa(T13, T14, X18))
U1_aaa(T13, T14, X9, add10_out_aaa(T13, T14, X18)) → add10_out_aaa(T13, s(T14), X9)
U2_aga(T13, T14, add10_out_aaa(T13, T14, X18)) → add1_out_aga(T13, 0, s(s(T14)))
add1_in_aga(T8, s(T15), s(T9)) → U3_aga(T8, T15, T9, add1_in_aga(T8, T15, T9))
U3_aga(T8, T15, T9, add1_out_aga(T8, T15, T9)) → add1_out_aga(T8, s(T15), s(T9))

(37) UsableRulesProof (EQUIVALENT transformation)

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

(38) Obligation:

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

ADD1_IN_AGA(T8, s(T15), s(T9)) → ADD1_IN_AGA(T8, T15, T9)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1) = s(x1)
ADD1_IN_AGA(x1, x2, x3) = ADD1_IN_AGA(x2)

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

(39) PiDPToQDPProof (SOUND transformation)

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

(40) Obligation:

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

ADD1_IN_AGA(s(T15)) → ADD1_IN_AGA(T15)

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

(41) 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:

ADD1_IN_AGA(s(T15)) → ADD1_IN_AGA(T15)
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) Complex Obligation (AND)

(9) Obligation:

(10) UsableRulesProof (EQUIVALENT transformation)

(11) Obligation:

(12) PiDPToQDPProof (SOUND transformation)

(13) Obligation:

(14) NonTerminationProof (EQUIVALENT transformation)

(15) FALSE

(16) Obligation:

(17) UsableRulesProof (EQUIVALENT transformation)

(18) Obligation:

(19) PiDPToQDPProof (SOUND transformation)

(20) Obligation:

(21) QDPSizeChangeProof (EQUIVALENT transformation)

(22) TRUE

(23) PrologToPiTRSProof (SOUND transformation)

(24) Obligation:

(25) DependencyPairsProof (EQUIVALENT transformation)

(26) Obligation:

(27) DependencyGraphProof (EQUIVALENT transformation)

(28) Complex Obligation (AND)

(29) Obligation:

(30) UsableRulesProof (EQUIVALENT transformation)

(31) Obligation:

(32) PiDPToQDPProof (SOUND transformation)

(33) Obligation:

(34) NonTerminationProof (EQUIVALENT transformation)

(35) FALSE

(36) Obligation:

(37) UsableRulesProof (EQUIVALENT transformation)

(38) Obligation:

(39) PiDPToQDPProof (SOUND transformation)

(40) Obligation:

(41) QDPSizeChangeProof (EQUIVALENT transformation)

(42) TRUE