Left Termination proof of /tmp/tmpA6DsSh/sum-ioi.pl

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

0 Prolog

↳1 PrologToDTProblemTransformerProof (⇒, 0 ms)

↳2 TRIPLES

↳3 TriplesToPiDPProof (⇒, 0 ms)

↳4 PiDP

↳5 DependencyGraphProof (⇔, 0 ms)

↳6 PiDP

↳7 PiDPToQDPProof (⇒, 25 ms)

↳8 QDP

↳9 QDPSizeChangeProof (⇔, 0 ms)

↳10 YES

(0) Obligation:

Clauses:

sum([], [], []).
sum(.(X1, Y1), .(X2, Y2), .(X3, Y3)) :- ','(add(X1, X2, X3), sum(Y1, Y2, Y3)).
add(0, X, X).
add(s(X), Y, s(Z)) :- add(X, Y, Z).

Query: sum(g,a,g)

(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\nT1 is ground\nT3 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\nT1 is ground\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: true | sum([], T2, []), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: sum(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nT3 is ground\nsum(T1, T2, T3) !=? sum([], [], [])", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: sum([], T2, []), 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="B: ','(add(T10, T16, T14), sum(T11, T17, T15))\n\nT10 is ground\nT11 is ground\nT14 is ground\nT15 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(add(T10, T16, T14), sum(T11, T17, T15)), SCOPE: 2, CLAUSE: 2 | ','(add(T10, T16, T14), sum(T11, T17, T15)), SCOPE: 2, CLAUSE: 3\n\nT10 is ground\nT11 is ground\nT14 is ground\nT15 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(add(T10, T16, T14), sum(T11, T17, T15)), SCOPE: 2, CLAUSE: 2\n\nT10 is ground\nT11 is ground\nT14 is ground\nT15 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(add(T10, T16, T14), sum(T11, T17, T15)), SCOPE: 2, CLAUSE: 3\n\nT10 is ground\nT11 is ground\nT14 is ground\nT15 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: sum(T11, T23, T15)\n\nT11 is ground\nT15 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(add(T30, T33, T32), sum(T11, T34, T15))\n\nT11 is ground\nT15 is ground\nT30 is ground\nT32 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 16 [arrowhead = none ];
16 -> 2;

17 [label="EVAL with clause\nsum([], [], []).\nand substitutionT1 -> [],\nT2 -> [],\nT3 -> []", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 17 [arrowhead = none ];
17 -> 3;

18 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 18 [arrowhead = none ];
18 -> 4;

19 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 19 [arrowhead = none ];
19 -> 5;

20 [label="EVAL with clause\nsum(.(X16, X17), .(X18, X19), .(X20, X21)) :- ','(add(X16, X18, X20), sum(X17, X19, X21)).\nand substitutionX16 -> T10,\nX17 -> T11,\nT1 -> .(T10, T11),\nX18 -> T16,\nX19 -> T17,\nT2 -> .(T16, T17),\nX20 -> T14,\nX21 -> T15,\nT3 -> .(T14, T15),\nT12 -> T16,\nT13 -> T17", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 20 [arrowhead = none ];
20 -> 7;

21 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 21 [arrowhead = none ];
21 -> 8;

22 [label="BACKTRACK\nfor clause: sum(.(X1, Y1), .(X2, Y2), .(X3, Y3)) :- ','(add(X1, X2, X3), sum(Y1, Y2, Y3))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 22 [arrowhead = none ];
22 -> 6;

23 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 23 [arrowhead = none ];
23 -> 9;

24 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
9 -> 24 [arrowhead = none ];
24 -> 10;

25 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
9 -> 25 [arrowhead = none ];
25 -> 11;

26 [label="EVAL with clause\nadd(0, X26, X26).\nand substitutionT10 -> 0,\nT16 -> T22,\nX26 -> T22,\nT14 -> T22,\nT17 -> T23", fontsize=14, style = filled, fillcolor = lightblue];
10 -> 26 [arrowhead = none ];
26 -> 12;

27 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 27 [arrowhead = none ];
27 -> 13;

28 [label="EVAL with clause\nadd(s(X33), X34, s(X35)) :- add(X33, X34, X35).\nand substitutionX33 -> T30,\nT10 -> s(T30),\nT16 -> T33,\nX34 -> T33,\nX35 -> T32,\nT14 -> s(T32),\nT31 -> T33,\nT17 -> T34", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 28 [arrowhead = none ];
28 -> 14;

29 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 29 [arrowhead = none ];
29 -> 15;

30 [label="INSTANCE with matching:\nT1 -> T11\nT2 -> T23\nT3 -> T15", fontsize=14, style = filled, fillcolor = lightgrey];
12 -> 30 [arrowhead = none , style=dashed];
30 -> 1[style=dashed];

31 [label="INSTANCE with matching:\nT10 -> T30\nT16 -> T33\nT14 -> T32\nT17 -> T34", fontsize=14, style = filled, fillcolor = lightgrey];
14 -> 31 [arrowhead = none , style=dashed];
31 -> 7[style=dashed];

}

(2) Obligation:

Triples:

pB(0, X1, X1, X2, X3, X4) :- sumA(X2, X3, X4).
pB(s(X1), X2, s(X3), X4, X5, X6) :- pB(X1, X2, X3, X4, X5, X6).
sumA(.(X1, X2), .(X3, X4), .(X5, X6)) :- pB(X1, X3, X5, X2, X4, X6).

Clauses:

sumcA([], [], []).
sumcA(.(X1, X2), .(X3, X4), .(X5, X6)) :- qcB(X1, X3, X5, X2, X4, X6).
qcB(0, X1, X1, X2, X3, X4) :- sumcA(X2, X3, X4).
qcB(s(X1), X2, s(X3), X4, X5, X6) :- qcB(X1, X2, X3, X4, X5, X6).

Afs:

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

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

SUMA_IN_GAG(.(X1, X2), .(X3, X4), .(X5, X6)) → U3_GAG(X1, X2, X3, X4, X5, X6, pB_in_gaggag(X1, X3, X5, X2, X4, X6))
SUMA_IN_GAG(.(X1, X2), .(X3, X4), .(X5, X6)) → PB_IN_GAGGAG(X1, X3, X5, X2, X4, X6)
PB_IN_GAGGAG(0, X1, X1, X2, X3, X4) → U1_GAGGAG(X1, X2, X3, X4, sumA_in_gag(X2, X3, X4))
PB_IN_GAGGAG(0, X1, X1, X2, X3, X4) → SUMA_IN_GAG(X2, X3, X4)
PB_IN_GAGGAG(s(X1), X2, s(X3), X4, X5, X6) → U2_GAGGAG(X1, X2, X3, X4, X5, X6, pB_in_gaggag(X1, X2, X3, X4, X5, X6))
PB_IN_GAGGAG(s(X1), X2, s(X3), X4, X5, X6) → PB_IN_GAGGAG(X1, X2, X3, X4, X5, X6)

R is empty.
The argument filtering Pi contains the following mapping:
sumA_in_gag(x1, x2, x3) = sumA_in_gag(x1, x3)
.(x1, x2) = .(x1, x2)
pB_in_gaggag(x1, x2, x3, x4, x5, x6) = pB_in_gaggag(x1, x3, x4, x6)
0 = 0
s(x1) = s(x1)
SUMA_IN_GAG(x1, x2, x3) = SUMA_IN_GAG(x1, x3)
U3_GAG(x1, x2, x3, x4, x5, x6, x7) = U3_GAG(x1, x2, x5, x6, x7)
PB_IN_GAGGAG(x1, x2, x3, x4, x5, x6) = PB_IN_GAGGAG(x1, x3, x4, x6)
U1_GAGGAG(x1, x2, x3, x4, x5) = U1_GAGGAG(x1, x2, x4, x5)
U2_GAGGAG(x1, x2, x3, x4, x5, x6, x7) = U2_GAGGAG(x1, x3, x4, x6, x7)

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_GAG(.(X1, X2), .(X3, X4), .(X5, X6)) → U3_GAG(X1, X2, X3, X4, X5, X6, pB_in_gaggag(X1, X3, X5, X2, X4, X6))
SUMA_IN_GAG(.(X1, X2), .(X3, X4), .(X5, X6)) → PB_IN_GAGGAG(X1, X3, X5, X2, X4, X6)
PB_IN_GAGGAG(0, X1, X1, X2, X3, X4) → U1_GAGGAG(X1, X2, X3, X4, sumA_in_gag(X2, X3, X4))
PB_IN_GAGGAG(0, X1, X1, X2, X3, X4) → SUMA_IN_GAG(X2, X3, X4)
PB_IN_GAGGAG(s(X1), X2, s(X3), X4, X5, X6) → U2_GAGGAG(X1, X2, X3, X4, X5, X6, pB_in_gaggag(X1, X2, X3, X4, X5, X6))
PB_IN_GAGGAG(s(X1), X2, s(X3), X4, X5, X6) → PB_IN_GAGGAG(X1, X2, X3, X4, X5, X6)

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes.

(6) Obligation:

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

SUMA_IN_GAG(.(X1, X2), .(X3, X4), .(X5, X6)) → PB_IN_GAGGAG(X1, X3, X5, X2, X4, X6)
PB_IN_GAGGAG(0, X1, X1, X2, X3, X4) → SUMA_IN_GAG(X2, X3, X4)
PB_IN_GAGGAG(s(X1), X2, s(X3), X4, X5, X6) → PB_IN_GAGGAG(X1, X2, X3, X4, X5, X6)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
0 = 0
s(x1) = s(x1)
SUMA_IN_GAG(x1, x2, x3) = SUMA_IN_GAG(x1, x3)
PB_IN_GAGGAG(x1, x2, x3, x4, x5, x6) = PB_IN_GAGGAG(x1, x3, x4, x6)

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_GAG(.(X1, X2), .(X5, X6)) → PB_IN_GAGGAG(X1, X5, X2, X6)
PB_IN_GAGGAG(0, X1, X2, X4) → SUMA_IN_GAG(X2, X4)
PB_IN_GAGGAG(s(X1), s(X3), X4, X6) → PB_IN_GAGGAG(X1, X3, X4, X6)

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:

PB_IN_GAGGAG(0, X1, X2, X4) → SUMA_IN_GAG(X2, X4)
The graph contains the following edges 3 >= 1, 4 >= 2
PB_IN_GAGGAG(s(X1), s(X3), X4, X6) → PB_IN_GAGGAG(X1, X3, X4, X6)
The graph contains the following edges 1 > 1, 2 > 2, 3 >= 3, 4 >= 4
SUMA_IN_GAG(.(X1, X2), .(X5, X6)) → PB_IN_GAGGAG(X1, X5, X2, X6)
The graph contains the following edges 1 > 1, 2 > 2, 1 > 3, 2 > 4