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

(0) Obligation:

Clauses:

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

Query: reverse(g,a,g)

(1) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
2 [label="A: reverse(T1, T2, T3)\n\nT1 is ground\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
5 [label="5: 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];
10 [label="10: true | reverse([], T2, T5), SCOPE: 1, CLAUSE: 1\n\nT5 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: 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];
17 [label="17: reverse([], T2, T5), SCOPE: 1, CLAUSE: 1\n\nT5 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: reverse(T11, T14, .(T10, T13))\n\nT10 is ground\nT11 is ground\nT13 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: 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];
30 [label="30: 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];
31 [label="31: 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];
37 [label="37: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
38 [label="38: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
39 [label="39: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
40 [label="40: 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];
41 [label="41: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
42 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
2 -> 42 [arrowhead = none ];
42 -> 5;

43 [label="EVAL with clause\nreverse([], X2, X2).\nand substitutionT1 -> [],\nT2 -> T5,\nX2 -> T5,\nT3 -> T5", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 43 [arrowhead = none ];
43 -> 10;

44 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 44 [arrowhead = none ];
44 -> 14;

45 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 45 [arrowhead = none ];
45 -> 17;

46 [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];
14 -> 46 [arrowhead = none ];
46 -> 25;

47 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
14 -> 47 [arrowhead = none ];
47 -> 26;

48 [label="BACKTRACK\nfor clause: reverse(.(X, Y), Z, U) :- reverse(Y, Z, .(X, U))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 48 [arrowhead = none ];
48 -> 19;

49 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
25 -> 49 [arrowhead = none ];
49 -> 27;

50 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
27 -> 50 [arrowhead = none ];
50 -> 30;

51 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
27 -> 51 [arrowhead = none ];
51 -> 31;

52 [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];
30 -> 52 [arrowhead = none ];
52 -> 37;

53 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
30 -> 53 [arrowhead = none ];
53 -> 38;

54 [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];
31 -> 54 [arrowhead = none ];
54 -> 40;

55 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
31 -> 55 [arrowhead = none ];
55 -> 41;

56 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
37 -> 56 [arrowhead = none ];
56 -> 39;

57 [label="INSTANCE with matching:\nT1 -> T41\nT2 -> T45\nT3 -> .(T40, .(T43, T44))", fontsize=14, style = filled, fillcolor = lightgrey];
40 -> 57 [arrowhead = none , style=dashed];
57 -> 2[style=dashed];

}

(2) 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)

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(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_gag(x1, x2, x3, x4, x5, x6)  =  U1_gag(x1, x2, x3, x5, x6)

(3) 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(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_gag(x1, x2, x3, x4, x5, x6)  =  U1_gag(x1, x2, x3, x5, x6)
REVERSEA_IN_GAG(x1, x2, x3)  =  REVERSEA_IN_GAG(x1, x3)
U1_GAG(x1, x2, x3, x4, x5, x6)  =  U1_GAG(x1, x2, x3, x5, x6)

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

(4) 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)

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(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_gag(x1, x2, x3, x4, x5, x6)  =  U1_gag(x1, x2, x3, x5, x6)
REVERSEA_IN_GAG(x1, x2, x3)  =  REVERSEA_IN_GAG(x1, x3)
U1_GAG(x1, x2, x3, x4, x5, x6)  =  U1_GAG(x1, x2, x3, x5, x6)

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

(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:

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(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_gag(x1, x2, x3, x4, x5, x6)  =  U1_gag(x1, x2, x3, x5, x6)
REVERSEA_IN_GAG(x1, x2, x3)  =  REVERSEA_IN_GAG(x1, x3)

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

(7) UsableRulesProof (EQUIVALENT transformation)

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

(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)))

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

(9) PiDPToQDPProof (SOUND transformation)

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

(10) 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.

(11) 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

(12) YES