Left Termination proof of /tmp/tmpBAXqaS/reverse.pl

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

0 Prolog

↳1 PrologToPrologProblemTransformerProof (⇒, 0 ms)

↳2 Prolog

↳3 PrologToPiTRSProof (⇒, 8 ms)

↳4 PiTRS

↳5 DependencyPairsProof (⇔, 0 ms)

↳6 PiDP

↳7 DependencyGraphProof (⇔, 0 ms)

↳8 PiDP

↳9 UsableRulesProof (⇔, 0 ms)

↳10 PiDP

↳11 PiDPToQDPProof (⇒, 0 ms)

↳12 QDP

↳13 QDPSizeChangeProof (⇔, 0 ms)

↳14 YES

(0) Obligation:

Clauses:

reverse([], X, X).
reverse(.(X, Y), Z, U) :- reverse(Y, Z, .(X, U)).

Query: reverse(g,a,g)

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: reverse(T1, T2, T3)\n\nT1 is ground\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: reverse(T1, T2, T3), SCOPE: 1, CLAUSE: 0 | reverse(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: true | reverse([], T2, T5), SCOPE: 1, CLAUSE: 1\n\nT5 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: reverse(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nT3 is ground\nX2 is free\nreverse(T1, T2, T3) !=? reverse([], X2, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: reverse([], T2, T5), SCOPE: 1, CLAUSE: 1\n\nT5 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: reverse(T11, T14, .(T10, T13))\n\nT10 is ground\nT11 is ground\nT13 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: reverse(T11, T14, .(T10, T13)), SCOPE: 2, CLAUSE: 0 | reverse(T11, T14, .(T10, T13)), SCOPE: 2, CLAUSE: 1\n\nT10 is ground\nT11 is ground\nT13 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: reverse(T11, T14, .(T10, T13)), SCOPE: 2, CLAUSE: 0\n\nT10 is ground\nT11 is ground\nT13 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: reverse(T11, T14, .(T10, T13)), SCOPE: 2, CLAUSE: 1\n\nT10 is ground\nT11 is ground\nT13 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: reverse(T41, T45, .(T40, .(T43, T44)))\n\nT40 is ground\nT41 is ground\nT43 is ground\nT44 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\nreverse([], X2, X2).\nand substitutionT1 -> [],\nT2 -> T5,\nX2 -> T5,\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\nreverse(.(X11, X12), X13, X14) :- reverse(X12, X13, .(X11, X14)).\nand substitutionX11 -> T10,\nX12 -> T11,\nT1 -> .(T10, T11),\nT2 -> T14,\nX13 -> T14,\nT3 -> T13,\nX14 -> 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: reverse(.(X, Y), Z, U) :- reverse(Y, Z, .(X, U))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\nreverse([], X19, X19).\nand substitutionT11 -> [],\nT14 -> .(T28, T29),\nX19 -> .(T28, T29),\nT10 -> T28,\nT13 -> T29,\nT27 -> .(T28, T29)", 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\nreverse(.(X28, X29), X30, X31) :- reverse(X29, X30, .(X28, X31)).\nand substitutionX28 -> T40,\nX29 -> T41,\nT11 -> .(T40, T41),\nT14 -> T45,\nX30 -> T45,\nT10 -> T43,\nT13 -> T44,\nX31 -> .(T43, T44),\nT42 -> T45", 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 -> T41\nT2 -> T45\nT3 -> .(T40, .(T43, T44))", fontsize=14, style = filled, fillcolor = lightgrey];
15 -> 32 [arrowhead = none , style=dashed];
32 -> 1[style=dashed];

}

(2) Obligation:

Clauses:

reverseA([], T5, T5).
reverseA(.(T28, []), .(T28, T29), T29).
reverseA(.(T43, .(T40, T41)), T45, T44) :- reverseA(T41, T45, .(T40, .(T43, T44))).

Query: reverseA(g,a,g)

(3) PrologToPiTRSProof (SOUND transformation)

We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
reverseA_in: (b,f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

reverseA_in_gag([], T5, T5) → reverseA_out_gag([], T5, T5)
reverseA_in_gag(.(T28, []), .(T28, T29), T29) → reverseA_out_gag(.(T28, []), .(T28, T29), T29)
reverseA_in_gag(.(T43, .(T40, T41)), T45, T44) → U1_gag(T43, T40, T41, T45, T44, reverseA_in_gag(T41, T45, .(T40, .(T43, T44))))
U1_gag(T43, T40, T41, T45, T44, reverseA_out_gag(T41, T45, .(T40, .(T43, T44)))) → reverseA_out_gag(.(T43, .(T40, T41)), T45, T44)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

reverseA_in_gag([], T5, T5) → reverseA_out_gag([], T5, T5)
reverseA_in_gag(.(T28, []), .(T28, T29), T29) → reverseA_out_gag(.(T28, []), .(T28, T29), T29)
reverseA_in_gag(.(T43, .(T40, T41)), T45, T44) → U1_gag(T43, T40, T41, T45, T44, reverseA_in_gag(T41, T45, .(T40, .(T43, T44))))
U1_gag(T43, T40, T41, T45, T44, reverseA_out_gag(T41, T45, .(T40, .(T43, T44)))) → reverseA_out_gag(.(T43, .(T40, T41)), T45, T44)

(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:

REVERSEA_IN_GAG(.(T43, .(T40, T41)), T45, T44) → U1_GAG(T43, T40, T41, T45, T44, reverseA_in_gag(T41, T45, .(T40, .(T43, T44))))
REVERSEA_IN_GAG(.(T43, .(T40, T41)), T45, T44) → REVERSEA_IN_GAG(T41, T45, .(T40, .(T43, T44)))

The TRS R consists of the following rules:

reverseA_in_gag([], T5, T5) → reverseA_out_gag([], T5, T5)
reverseA_in_gag(.(T28, []), .(T28, T29), T29) → reverseA_out_gag(.(T28, []), .(T28, T29), T29)
reverseA_in_gag(.(T43, .(T40, T41)), T45, T44) → U1_gag(T43, T40, T41, T45, T44, reverseA_in_gag(T41, T45, .(T40, .(T43, T44))))
U1_gag(T43, T40, T41, T45, T44, reverseA_out_gag(T41, T45, .(T40, .(T43, T44)))) → reverseA_out_gag(.(T43, .(T40, T41)), T45, T44)

The argument filtering Pi contains the following mapping:
reverseA_in_gag(x1, x2, x3) = reverseA_in_gag(x1, x3)
[] = []
reverseA_out_gag(x1, x2, x3) = reverseA_out_gag(x2)
.(x1, x2) = .(x1, x2)
U1_gag(x1, x2, x3, x4, x5, x6) = U1_gag(x6)
REVERSEA_IN_GAG(x1, x2, x3) = REVERSEA_IN_GAG(x1, x3)
U1_GAG(x1, x2, x3, x4, x5, x6) = U1_GAG(x6)

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

(6) Obligation:

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

REVERSEA_IN_GAG(.(T43, .(T40, T41)), T45, T44) → U1_GAG(T43, T40, T41, T45, T44, reverseA_in_gag(T41, T45, .(T40, .(T43, T44))))
REVERSEA_IN_GAG(.(T43, .(T40, T41)), T45, T44) → REVERSEA_IN_GAG(T41, T45, .(T40, .(T43, T44)))

The TRS R consists of the following rules:

reverseA_in_gag([], T5, T5) → reverseA_out_gag([], T5, T5)
reverseA_in_gag(.(T28, []), .(T28, T29), T29) → reverseA_out_gag(.(T28, []), .(T28, T29), T29)
reverseA_in_gag(.(T43, .(T40, T41)), T45, T44) → U1_gag(T43, T40, T41, T45, T44, reverseA_in_gag(T41, T45, .(T40, .(T43, T44))))
U1_gag(T43, T40, T41, T45, T44, reverseA_out_gag(T41, T45, .(T40, .(T43, T44)))) → reverseA_out_gag(.(T43, .(T40, T41)), T45, T44)

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(8) Obligation:

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

REVERSEA_IN_GAG(.(T43, .(T40, T41)), T45, T44) → REVERSEA_IN_GAG(T41, T45, .(T40, .(T43, T44)))

The TRS R consists of the following rules:

reverseA_in_gag([], T5, T5) → reverseA_out_gag([], T5, T5)
reverseA_in_gag(.(T28, []), .(T28, T29), T29) → reverseA_out_gag(.(T28, []), .(T28, T29), T29)
reverseA_in_gag(.(T43, .(T40, T41)), T45, T44) → U1_gag(T43, T40, T41, T45, T44, reverseA_in_gag(T41, T45, .(T40, .(T43, T44))))
U1_gag(T43, T40, T41, T45, T44, reverseA_out_gag(T41, T45, .(T40, .(T43, T44)))) → reverseA_out_gag(.(T43, .(T40, T41)), T45, T44)

(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:

REVERSEA_IN_GAG(.(T43, .(T40, T41)), T45, T44) → REVERSEA_IN_GAG(T41, T45, .(T40, .(T43, T44)))

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

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

(11) PiDPToQDPProof (SOUND 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:

REVERSEA_IN_GAG(.(T43, .(T40, T41)), T44) → REVERSEA_IN_GAG(T41, .(T40, .(T43, T44)))

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:

REVERSEA_IN_GAG(.(T43, .(T40, T41)), T44) → REVERSEA_IN_GAG(T41, .(T40, .(T43, T44)))
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 (SOUND transformation)

(12) Obligation:

(13) QDPSizeChangeProof (EQUIVALENT transformation)

(14) YES