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

(0) Obligation:

Clauses:

preorder(T, Xs) :- preorder_dl(T, -(Xs, [])).
preorder_dl(nil, -(X, X)).
preorder_dl(tree(L, X, R), -(.(X, Xs), Zs)) :- ','(preorder_dl(L, -(Xs, Ys)), preorder_dl(R, -(Ys, Zs))).

Query: preorder(g,a)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="C: preorder(T1, T2)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: preorder(T1, T2), SCOPE: 1, CLAUSE: 0\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="A: preorder_dl(T5, -(T7, []))\n\nT5 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: preorder_dl(T5, -(T7, [])), SCOPE: 2, CLAUSE: 1 | preorder_dl(T5, -(T7, [])), SCOPE: 2, CLAUSE: 2\n\nT5 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: preorder_dl(T5, -(T7, [])), SCOPE: 2, CLAUSE: 1\n\nT5 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: preorder_dl(T5, -(T7, [])), SCOPE: 2, CLAUSE: 2\n\nT5 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(preorder_dl(T21, -(T25, X26)), preorder_dl(T23, -(X26, [])))\n\nT21 is ground\nT23 is ground\nX26 is free", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="B: preorder_dl(T21, -(T25, X26))\n\nT21 is ground\nX26 is free", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: preorder_dl(T23, -(T28, []))\n\nT23 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: preorder_dl(T21, -(T25, X26)), SCOPE: 3, CLAUSE: 1 | preorder_dl(T21, -(T25, X26)), SCOPE: 3, CLAUSE: 2\n\nT21 is ground\nX26 is free", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: preorder_dl(T21, -(T25, X26)), SCOPE: 3, CLAUSE: 1\n\nT21 is ground\nX26 is free", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: preorder_dl(T21, -(T25, X26)), SCOPE: 3, CLAUSE: 2\n\nT21 is ground\nX26 is free", 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: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: ','(preorder_dl(T44, -(T48, X54)), preorder_dl(T46, -(X54, X55)))\n\nT44 is ground\nT46 is ground\nX55 is free\nX54 is free", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: preorder_dl(T44, -(T48, X54))\n\nT44 is ground\nX54 is free", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: preorder_dl(T46, -(T51, X55))\n\nT46 is ground\nX55 is free", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 24 [arrowhead = none ];
24 -> 2;

25 [label="ONLY EVAL with clause\npreorder(X3, X4) :- preorder_dl(X3, -(X4, [])).\nand substitutionT1 -> T5,\nX3 -> T5,\nT2 -> T7,\nX4 -> T7,\nT6 -> T7", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 25 [arrowhead = none ];
25 -> 3;

26 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 26 [arrowhead = none ];
26 -> 4;

27 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
4 -> 27 [arrowhead = none ];
27 -> 5;

28 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
4 -> 28 [arrowhead = none ];
28 -> 6;

29 [label="EVAL with clause\npreorder_dl(nil, -(X9, X9)).\nand substitutionT5 -> nil,\nT7 -> [],\nX9 -> [],\nT12 -> []", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 29 [arrowhead = none ];
29 -> 7;

30 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 30 [arrowhead = none ];
30 -> 8;

31 [label="EVAL with clause\npreorder_dl(tree(X21, X22, X23), -(.(X22, X24), X25)) :- ','(preorder_dl(X21, -(X24, X26)), preorder_dl(X23, -(X26, X25))).\nand substitutionX21 -> T21,\nX22 -> T22,\nX23 -> T23,\nT5 -> tree(T21, T22, T23),\nX24 -> T25,\nT7 -> .(T22, T25),\nX25 -> [],\nT24 -> T25", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 31 [arrowhead = none ];
31 -> 10;

32 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 32 [arrowhead = none ];
32 -> 11;

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

34 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
10 -> 34 [arrowhead = none ];
34 -> 12;

35 [label="SPLIT 2\nnew knowledge:\nT21 is ground\nreplacements:X26 -> T28", fontsize=14, style = filled, fillcolor = violet];
10 -> 35 [arrowhead = none ];
35 -> 13;

36 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
12 -> 36 [arrowhead = none ];
36 -> 14;

37 [label="INSTANCE with matching:\nT5 -> T23\nT7 -> T28", fontsize=14, style = filled, fillcolor = lightgrey];
13 -> 37 [arrowhead = none , style=dashed];
37 -> 3[style=dashed];

38 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
14 -> 38 [arrowhead = none ];
38 -> 15;

39 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
14 -> 39 [arrowhead = none ];
39 -> 16;

40 [label="EVAL with clause\npreorder_dl(nil, -(X35, X35)).\nand substitutionT21 -> nil,\nT25 -> T35,\nX35 -> T35,\nX26 -> T35", fontsize=14, style = filled, fillcolor = lightblue];
15 -> 40 [arrowhead = none ];
40 -> 17;

41 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 41 [arrowhead = none ];
41 -> 18;

42 [label="EVAL with clause\npreorder_dl(tree(X49, X50, X51), -(.(X50, X52), X53)) :- ','(preorder_dl(X49, -(X52, X54)), preorder_dl(X51, -(X54, X53))).\nand substitutionX49 -> T44,\nX50 -> T45,\nX51 -> T46,\nT21 -> tree(T44, T45, T46),\nX52 -> T48,\nT25 -> .(T45, T48),\nX26 -> X55,\nX53 -> X55,\nT47 -> T48", fontsize=14, style = filled, fillcolor = lightblue];
16 -> 42 [arrowhead = none ];
42 -> 20;

43 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
16 -> 43 [arrowhead = none ];
43 -> 21;

44 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 44 [arrowhead = none ];
44 -> 19;

45 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
20 -> 45 [arrowhead = none ];
45 -> 22;

46 [label="SPLIT 2\nnew knowledge:\nT44 is ground\nreplacements:X54 -> T51", fontsize=14, style = filled, fillcolor = violet];
20 -> 46 [arrowhead = none ];
46 -> 23;

47 [label="INSTANCE with matching:\nT21 -> T44\nT25 -> T48\nX26 -> X54", fontsize=14, style = filled, fillcolor = lightgrey];
22 -> 47 [arrowhead = none , style=dashed];
47 -> 12[style=dashed];

48 [label="INSTANCE with matching:\nT21 -> T46\nT25 -> T51\nX26 -> X55", fontsize=14, style = filled, fillcolor = lightgrey];
23 -> 48 [arrowhead = none , style=dashed];
48 -> 12[style=dashed];

}

(2) Obligation:

Triples:

preorder_dlA(tree(X1, X2, X3), .(X2, X4)) :- preorder_dlB(X1, X4, X5).
preorder_dlA(tree(X1, X2, X3), .(X2, X4)) :- ','(preorder_dlcB(X1, X4, X5), preorder_dlA(X3, X5)).
preorder_dlB(tree(X1, X2, X3), .(X2, X4), X5) :- preorder_dlB(X1, X4, X6).
preorder_dlB(tree(X1, X2, X3), .(X2, X4), X5) :- ','(preorder_dlcB(X1, X4, X6), preorder_dlB(X3, X6, X5)).
preorderC(X1, X2) :- preorder_dlA(X1, X2).

Clauses:

preorder_dlcA(nil, []).
preorder_dlcA(tree(X1, X2, X3), .(X2, X4)) :- ','(preorder_dlcB(X1, X4, X5), preorder_dlcA(X3, X5)).
preorder_dlcB(nil, X1, X1).
preorder_dlcB(tree(X1, X2, X3), .(X2, X4), X5) :- ','(preorder_dlcB(X1, X4, X6), preorder_dlcB(X3, X6, X5)).

Afs:

preorderC(x1, x2)  =  preorderC(x1)

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

PREORDERC_IN_GA(X1, X2) → U7_GA(X1, X2, preorder_dlA_in_ga(X1, X2))
PREORDERC_IN_GA(X1, X2) → PREORDER_DLA_IN_GA(X1, X2)
PREORDER_DLA_IN_GA(tree(X1, X2, X3), .(X2, X4)) → U1_GA(X1, X2, X3, X4, preorder_dlB_in_gaa(X1, X4, X5))
PREORDER_DLA_IN_GA(tree(X1, X2, X3), .(X2, X4)) → PREORDER_DLB_IN_GAA(X1, X4, X5)
PREORDER_DLB_IN_GAA(tree(X1, X2, X3), .(X2, X4), X5) → U4_GAA(X1, X2, X3, X4, X5, preorder_dlB_in_gaa(X1, X4, X6))
PREORDER_DLB_IN_GAA(tree(X1, X2, X3), .(X2, X4), X5) → PREORDER_DLB_IN_GAA(X1, X4, X6)
PREORDER_DLB_IN_GAA(tree(X1, X2, X3), .(X2, X4), X5) → U5_GAA(X1, X2, X3, X4, X5, preorder_dlcB_in_gaa(X1, X4, X6))
U5_GAA(X1, X2, X3, X4, X5, preorder_dlcB_out_gaa(X1, X4, X6)) → U6_GAA(X1, X2, X3, X4, X5, preorder_dlB_in_gaa(X3, X6, X5))
U5_GAA(X1, X2, X3, X4, X5, preorder_dlcB_out_gaa(X1, X4, X6)) → PREORDER_DLB_IN_GAA(X3, X6, X5)
PREORDER_DLA_IN_GA(tree(X1, X2, X3), .(X2, X4)) → U2_GA(X1, X2, X3, X4, preorder_dlcB_in_gaa(X1, X4, X5))
U2_GA(X1, X2, X3, X4, preorder_dlcB_out_gaa(X1, X4, X5)) → U3_GA(X1, X2, X3, X4, preorder_dlA_in_ga(X3, X5))
U2_GA(X1, X2, X3, X4, preorder_dlcB_out_gaa(X1, X4, X5)) → PREORDER_DLA_IN_GA(X3, X5)

The TRS R consists of the following rules:

preorder_dlcB_in_gaa(nil, X1, X1) → preorder_dlcB_out_gaa(nil, X1, X1)
preorder_dlcB_in_gaa(tree(X1, X2, X3), .(X2, X4), X5) → U11_gaa(X1, X2, X3, X4, X5, preorder_dlcB_in_gaa(X1, X4, X6))
U11_gaa(X1, X2, X3, X4, X5, preorder_dlcB_out_gaa(X1, X4, X6)) → U12_gaa(X1, X2, X3, X4, X5, X6, preorder_dlcB_in_gaa(X3, X6, X5))
U12_gaa(X1, X2, X3, X4, X5, X6, preorder_dlcB_out_gaa(X3, X6, X5)) → preorder_dlcB_out_gaa(tree(X1, X2, X3), .(X2, X4), X5)

The argument filtering Pi contains the following mapping:
preorder_dlA_in_ga(x1, x2)  =  preorder_dlA_in_ga(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
preorder_dlB_in_gaa(x1, x2, x3)  =  preorder_dlB_in_gaa(x1)
preorder_dlcB_in_gaa(x1, x2, x3)  =  preorder_dlcB_in_gaa(x1)
nil  =  nil
preorder_dlcB_out_gaa(x1, x2, x3)  =  preorder_dlcB_out_gaa(x1)
U11_gaa(x1, x2, x3, x4, x5, x6)  =  U11_gaa(x1, x2, x3, x6)
U12_gaa(x1, x2, x3, x4, x5, x6, x7)  =  U12_gaa(x1, x2, x3, x7)
PREORDERC_IN_GA(x1, x2)  =  PREORDERC_IN_GA(x1)
U7_GA(x1, x2, x3)  =  U7_GA(x1, x3)
PREORDER_DLA_IN_GA(x1, x2)  =  PREORDER_DLA_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5)  =  U1_GA(x1, x2, x3, x5)
PREORDER_DLB_IN_GAA(x1, x2, x3)  =  PREORDER_DLB_IN_GAA(x1)
U4_GAA(x1, x2, x3, x4, x5, x6)  =  U4_GAA(x1, x2, x3, x6)
U5_GAA(x1, x2, x3, x4, x5, x6)  =  U5_GAA(x1, x2, x3, x6)
U6_GAA(x1, x2, x3, x4, x5, x6)  =  U6_GAA(x1, x2, x3, x6)
U2_GA(x1, x2, x3, x4, x5)  =  U2_GA(x1, x2, x3, x5)
U3_GA(x1, x2, x3, x4, x5)  =  U3_GA(x1, x2, x3, x5)

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:

PREORDERC_IN_GA(X1, X2) → U7_GA(X1, X2, preorder_dlA_in_ga(X1, X2))
PREORDERC_IN_GA(X1, X2) → PREORDER_DLA_IN_GA(X1, X2)
PREORDER_DLA_IN_GA(tree(X1, X2, X3), .(X2, X4)) → U1_GA(X1, X2, X3, X4, preorder_dlB_in_gaa(X1, X4, X5))
PREORDER_DLA_IN_GA(tree(X1, X2, X3), .(X2, X4)) → PREORDER_DLB_IN_GAA(X1, X4, X5)
PREORDER_DLB_IN_GAA(tree(X1, X2, X3), .(X2, X4), X5) → U4_GAA(X1, X2, X3, X4, X5, preorder_dlB_in_gaa(X1, X4, X6))
PREORDER_DLB_IN_GAA(tree(X1, X2, X3), .(X2, X4), X5) → PREORDER_DLB_IN_GAA(X1, X4, X6)
PREORDER_DLB_IN_GAA(tree(X1, X2, X3), .(X2, X4), X5) → U5_GAA(X1, X2, X3, X4, X5, preorder_dlcB_in_gaa(X1, X4, X6))
U5_GAA(X1, X2, X3, X4, X5, preorder_dlcB_out_gaa(X1, X4, X6)) → U6_GAA(X1, X2, X3, X4, X5, preorder_dlB_in_gaa(X3, X6, X5))
U5_GAA(X1, X2, X3, X4, X5, preorder_dlcB_out_gaa(X1, X4, X6)) → PREORDER_DLB_IN_GAA(X3, X6, X5)
PREORDER_DLA_IN_GA(tree(X1, X2, X3), .(X2, X4)) → U2_GA(X1, X2, X3, X4, preorder_dlcB_in_gaa(X1, X4, X5))
U2_GA(X1, X2, X3, X4, preorder_dlcB_out_gaa(X1, X4, X5)) → U3_GA(X1, X2, X3, X4, preorder_dlA_in_ga(X3, X5))
U2_GA(X1, X2, X3, X4, preorder_dlcB_out_gaa(X1, X4, X5)) → PREORDER_DLA_IN_GA(X3, X5)

The TRS R consists of the following rules:

preorder_dlcB_in_gaa(nil, X1, X1) → preorder_dlcB_out_gaa(nil, X1, X1)
preorder_dlcB_in_gaa(tree(X1, X2, X3), .(X2, X4), X5) → U11_gaa(X1, X2, X3, X4, X5, preorder_dlcB_in_gaa(X1, X4, X6))
U11_gaa(X1, X2, X3, X4, X5, preorder_dlcB_out_gaa(X1, X4, X6)) → U12_gaa(X1, X2, X3, X4, X5, X6, preorder_dlcB_in_gaa(X3, X6, X5))
U12_gaa(X1, X2, X3, X4, X5, X6, preorder_dlcB_out_gaa(X3, X6, X5)) → preorder_dlcB_out_gaa(tree(X1, X2, X3), .(X2, X4), X5)

The argument filtering Pi contains the following mapping:
preorder_dlA_in_ga(x1, x2)  =  preorder_dlA_in_ga(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
preorder_dlB_in_gaa(x1, x2, x3)  =  preorder_dlB_in_gaa(x1)
preorder_dlcB_in_gaa(x1, x2, x3)  =  preorder_dlcB_in_gaa(x1)
nil  =  nil
preorder_dlcB_out_gaa(x1, x2, x3)  =  preorder_dlcB_out_gaa(x1)
U11_gaa(x1, x2, x3, x4, x5, x6)  =  U11_gaa(x1, x2, x3, x6)
U12_gaa(x1, x2, x3, x4, x5, x6, x7)  =  U12_gaa(x1, x2, x3, x7)
PREORDERC_IN_GA(x1, x2)  =  PREORDERC_IN_GA(x1)
U7_GA(x1, x2, x3)  =  U7_GA(x1, x3)
PREORDER_DLA_IN_GA(x1, x2)  =  PREORDER_DLA_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5)  =  U1_GA(x1, x2, x3, x5)
PREORDER_DLB_IN_GAA(x1, x2, x3)  =  PREORDER_DLB_IN_GAA(x1)
U4_GAA(x1, x2, x3, x4, x5, x6)  =  U4_GAA(x1, x2, x3, x6)
U5_GAA(x1, x2, x3, x4, x5, x6)  =  U5_GAA(x1, x2, x3, x6)
U6_GAA(x1, x2, x3, x4, x5, x6)  =  U6_GAA(x1, x2, x3, x6)
U2_GA(x1, x2, x3, x4, x5)  =  U2_GA(x1, x2, x3, x5)
U3_GA(x1, x2, x3, x4, x5)  =  U3_GA(x1, x2, x3, x5)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 7 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

PREORDER_DLB_IN_GAA(tree(X1, X2, X3), .(X2, X4), X5) → U5_GAA(X1, X2, X3, X4, X5, preorder_dlcB_in_gaa(X1, X4, X6))
U5_GAA(X1, X2, X3, X4, X5, preorder_dlcB_out_gaa(X1, X4, X6)) → PREORDER_DLB_IN_GAA(X3, X6, X5)
PREORDER_DLB_IN_GAA(tree(X1, X2, X3), .(X2, X4), X5) → PREORDER_DLB_IN_GAA(X1, X4, X6)

The TRS R consists of the following rules:

preorder_dlcB_in_gaa(nil, X1, X1) → preorder_dlcB_out_gaa(nil, X1, X1)
preorder_dlcB_in_gaa(tree(X1, X2, X3), .(X2, X4), X5) → U11_gaa(X1, X2, X3, X4, X5, preorder_dlcB_in_gaa(X1, X4, X6))
U11_gaa(X1, X2, X3, X4, X5, preorder_dlcB_out_gaa(X1, X4, X6)) → U12_gaa(X1, X2, X3, X4, X5, X6, preorder_dlcB_in_gaa(X3, X6, X5))
U12_gaa(X1, X2, X3, X4, X5, X6, preorder_dlcB_out_gaa(X3, X6, X5)) → preorder_dlcB_out_gaa(tree(X1, X2, X3), .(X2, X4), X5)

The argument filtering Pi contains the following mapping:
tree(x1, x2, x3)  =  tree(x1, x2, x3)
preorder_dlcB_in_gaa(x1, x2, x3)  =  preorder_dlcB_in_gaa(x1)
nil  =  nil
preorder_dlcB_out_gaa(x1, x2, x3)  =  preorder_dlcB_out_gaa(x1)
U11_gaa(x1, x2, x3, x4, x5, x6)  =  U11_gaa(x1, x2, x3, x6)
U12_gaa(x1, x2, x3, x4, x5, x6, x7)  =  U12_gaa(x1, x2, x3, x7)
PREORDER_DLB_IN_GAA(x1, x2, x3)  =  PREORDER_DLB_IN_GAA(x1)
U5_GAA(x1, x2, x3, x4, x5, x6)  =  U5_GAA(x1, x2, x3, x6)

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

(8) PiDPToQDPProof (SOUND transformation)

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

(9) Obligation:

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

PREORDER_DLB_IN_GAA(tree(X1, X2, X3)) → U5_GAA(X1, X2, X3, preorder_dlcB_in_gaa(X1))
U5_GAA(X1, X2, X3, preorder_dlcB_out_gaa(X1)) → PREORDER_DLB_IN_GAA(X3)
PREORDER_DLB_IN_GAA(tree(X1, X2, X3)) → PREORDER_DLB_IN_GAA(X1)

The TRS R consists of the following rules:

preorder_dlcB_in_gaa(nil) → preorder_dlcB_out_gaa(nil)
preorder_dlcB_in_gaa(tree(X1, X2, X3)) → U11_gaa(X1, X2, X3, preorder_dlcB_in_gaa(X1))
U11_gaa(X1, X2, X3, preorder_dlcB_out_gaa(X1)) → U12_gaa(X1, X2, X3, preorder_dlcB_in_gaa(X3))
U12_gaa(X1, X2, X3, preorder_dlcB_out_gaa(X3)) → preorder_dlcB_out_gaa(tree(X1, X2, X3))

The set Q consists of the following terms:

preorder_dlcB_in_gaa(x0)
U11_gaa(x0, x1, x2, x3)
U12_gaa(x0, x1, x2, x3)

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

(10) 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:

• U5_GAA(X1, X2, X3, preorder_dlcB_out_gaa(X1)) → PREORDER_DLB_IN_GAA(X3)
  The graph contains the following edges 3 >= 1

• PREORDER_DLB_IN_GAA(tree(X1, X2, X3)) → PREORDER_DLB_IN_GAA(X1)
  The graph contains the following edges 1 > 1

• PREORDER_DLB_IN_GAA(tree(X1, X2, X3)) → U5_GAA(X1, X2, X3, preorder_dlcB_in_gaa(X1))
  The graph contains the following edges 1 > 1, 1 > 2, 1 > 3

(11) YES

(12) Obligation:

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

PREORDER_DLA_IN_GA(tree(X1, X2, X3), .(X2, X4)) → U2_GA(X1, X2, X3, X4, preorder_dlcB_in_gaa(X1, X4, X5))
U2_GA(X1, X2, X3, X4, preorder_dlcB_out_gaa(X1, X4, X5)) → PREORDER_DLA_IN_GA(X3, X5)

The TRS R consists of the following rules:

preorder_dlcB_in_gaa(nil, X1, X1) → preorder_dlcB_out_gaa(nil, X1, X1)
preorder_dlcB_in_gaa(tree(X1, X2, X3), .(X2, X4), X5) → U11_gaa(X1, X2, X3, X4, X5, preorder_dlcB_in_gaa(X1, X4, X6))
U11_gaa(X1, X2, X3, X4, X5, preorder_dlcB_out_gaa(X1, X4, X6)) → U12_gaa(X1, X2, X3, X4, X5, X6, preorder_dlcB_in_gaa(X3, X6, X5))
U12_gaa(X1, X2, X3, X4, X5, X6, preorder_dlcB_out_gaa(X3, X6, X5)) → preorder_dlcB_out_gaa(tree(X1, X2, X3), .(X2, X4), X5)

The argument filtering Pi contains the following mapping:
tree(x1, x2, x3)  =  tree(x1, x2, x3)
preorder_dlcB_in_gaa(x1, x2, x3)  =  preorder_dlcB_in_gaa(x1)
nil  =  nil
preorder_dlcB_out_gaa(x1, x2, x3)  =  preorder_dlcB_out_gaa(x1)
U11_gaa(x1, x2, x3, x4, x5, x6)  =  U11_gaa(x1, x2, x3, x6)
U12_gaa(x1, x2, x3, x4, x5, x6, x7)  =  U12_gaa(x1, x2, x3, x7)
PREORDER_DLA_IN_GA(x1, x2)  =  PREORDER_DLA_IN_GA(x1)
U2_GA(x1, x2, x3, x4, x5)  =  U2_GA(x1, x2, x3, x5)

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

(13) PiDPToQDPProof (SOUND transformation)

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

(14) Obligation:

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

PREORDER_DLA_IN_GA(tree(X1, X2, X3)) → U2_GA(X1, X2, X3, preorder_dlcB_in_gaa(X1))
U2_GA(X1, X2, X3, preorder_dlcB_out_gaa(X1)) → PREORDER_DLA_IN_GA(X3)

The TRS R consists of the following rules:

preorder_dlcB_in_gaa(nil) → preorder_dlcB_out_gaa(nil)
preorder_dlcB_in_gaa(tree(X1, X2, X3)) → U11_gaa(X1, X2, X3, preorder_dlcB_in_gaa(X1))
U11_gaa(X1, X2, X3, preorder_dlcB_out_gaa(X1)) → U12_gaa(X1, X2, X3, preorder_dlcB_in_gaa(X3))
U12_gaa(X1, X2, X3, preorder_dlcB_out_gaa(X3)) → preorder_dlcB_out_gaa(tree(X1, X2, X3))

The set Q consists of the following terms:

preorder_dlcB_in_gaa(x0)
U11_gaa(x0, x1, x2, x3)
U12_gaa(x0, x1, x2, x3)

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

(15) 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:

• U2_GA(X1, X2, X3, preorder_dlcB_out_gaa(X1)) → PREORDER_DLA_IN_GA(X3)
  The graph contains the following edges 3 >= 1

• PREORDER_DLA_IN_GA(tree(X1, X2, X3)) → U2_GA(X1, X2, X3, preorder_dlcB_in_gaa(X1))
  The graph contains the following edges 1 > 1, 1 > 2, 1 > 3

(16) YES