Left Termination proof of /tmp/tmpsR2SSL/sum-fbf.pl

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

0 Prolog

↳1 PrologToDTProblemTransformerProof (⇒, 0 ms)

↳2 TRIPLES

↳3 TriplesToPiDPProof (⇒, 8 ms)

↳4 PiDP

↳5 DependencyGraphProof (⇔, 0 ms)

↳6 PiDP

↳7 PiDPToQDPProof (⇒, 0 ms)

↳8 QDP

↳9 QDPSizeChangeProof (⇔, 0 ms)

↳10 YES

(0) Obligation:

Clauses:

sum(X, 0, X).
sum(X, s(Y), s(Z)) :- sum(X, Y, Z).

Query: sum(a,g,a)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: sum(T1, T2, T3)\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: sum(T1, T2, T3), SCOPE: 1, CLAUSE: 0 | sum(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: true | sum(T1, 0, T3), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: sum(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT2 is ground\nX2 is free\nsum(T1, T2, T3) !=? sum(X2, 0, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: sum(T1, 0, T3), 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: sum(T13, T11, T14)\n\nT11 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: sum(T13, T11, T14), SCOPE: 2, CLAUSE: 0 | sum(T13, T11, T14), SCOPE: 2, CLAUSE: 1\n\nT11 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: sum(T13, T11, T14), SCOPE: 2, CLAUSE: 0\n\nT11 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: sum(T13, T11, T14), SCOPE: 2, CLAUSE: 1\n\nT11 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: sum(T29, T27, T30)\n\nT27 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 17 [arrowhead = none ];
17 -> 2;

18 [label="EVAL with clause\nsum(X2, 0, X2).\nand substitutionT1 -> T5,\nX2 -> T5,\nT2 -> 0,\nT3 -> T5", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 18 [arrowhead = none ];
18 -> 3;

19 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 19 [arrowhead = none ];
19 -> 4;

20 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 20 [arrowhead = none ];
20 -> 5;

21 [label="EVAL with clause\nsum(X9, s(X10), s(X11)) :- sum(X9, X10, X11).\nand substitutionT1 -> T13,\nX9 -> T13,\nX10 -> T11,\nT2 -> s(T11),\nX11 -> T14,\nT3 -> s(T14),\nT10 -> T13,\nT12 -> T14", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 21 [arrowhead = none ];
21 -> 7;

22 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 22 [arrowhead = none ];
22 -> 8;

23 [label="BACKTRACK\nfor clause: sum(X, s(Y), s(Z)) :- sum(X, Y, Z)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 23 [arrowhead = none ];
23 -> 6;

24 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 24 [arrowhead = none ];
24 -> 9;

25 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
9 -> 25 [arrowhead = none ];
25 -> 10;

26 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
9 -> 26 [arrowhead = none ];
26 -> 11;

27 [label="EVAL with clause\nsum(X16, 0, X16).\nand substitutionT13 -> T19,\nX16 -> T19,\nT11 -> 0,\nT14 -> T19", fontsize=14, style = filled, fillcolor = lightblue];
10 -> 27 [arrowhead = none ];
27 -> 12;

28 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 28 [arrowhead = none ];
28 -> 13;

29 [label="EVAL with clause\nsum(X23, s(X24), s(X25)) :- sum(X23, X24, X25).\nand substitutionT13 -> T29,\nX23 -> T29,\nX24 -> T27,\nT11 -> s(T27),\nX25 -> T30,\nT14 -> s(T30),\nT26 -> T29,\nT28 -> T30", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 29 [arrowhead = none ];
29 -> 15;

30 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 30 [arrowhead = none ];
30 -> 16;

31 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 31 [arrowhead = none ];
31 -> 14;

32 [label="INSTANCE with matching:\nT1 -> T29\nT2 -> T27\nT3 -> T30", fontsize=14, style = filled, fillcolor = lightgrey];
15 -> 32 [arrowhead = none , style=dashed];
32 -> 1[style=dashed];

}

(2) Obligation:

Triples:

sumA(X1, s(s(X2)), s(s(X3))) :- sumA(X1, X2, X3).

Clauses:

sumcA(X1, 0, X1).
sumcA(X1, s(0), s(X1)).
sumcA(X1, s(s(X2)), s(s(X3))) :- sumcA(X1, X2, X3).

Afs:

sumA(x1, x2, x3) = sumA(x2)

(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:
sumA_in: (f,b,f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

SUMA_IN_AGA(X1, s(s(X2)), s(s(X3))) → U1_AGA(X1, X2, X3, sumA_in_aga(X1, X2, X3))
SUMA_IN_AGA(X1, s(s(X2)), s(s(X3))) → SUMA_IN_AGA(X1, X2, X3)

R is empty.
The argument filtering Pi contains the following mapping:
sumA_in_aga(x1, x2, x3) = sumA_in_aga(x2)
s(x1) = s(x1)
SUMA_IN_AGA(x1, x2, x3) = SUMA_IN_AGA(x2)
U1_AGA(x1, x2, x3, x4) = U1_AGA(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:

SUMA_IN_AGA(X1, s(s(X2)), s(s(X3))) → U1_AGA(X1, X2, X3, sumA_in_aga(X1, X2, X3))
SUMA_IN_AGA(X1, s(s(X2)), s(s(X3))) → SUMA_IN_AGA(X1, X2, X3)

(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:

SUMA_IN_AGA(X1, s(s(X2)), s(s(X3))) → SUMA_IN_AGA(X1, X2, X3)

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

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:

SUMA_IN_AGA(s(s(X2))) → SUMA_IN_AGA(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:

SUMA_IN_AGA(s(s(X2))) → SUMA_IN_AGA(X2)
The graph contains the following edges 1 > 1