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

(0) Obligation:

Clauses:

duplicate([], L) :- ','(!, eq(L, [])).
duplicate(X, .(H, .(H, Z))) :- ','(head(X, H), ','(tail(X, T), duplicate(T, Z))).
head([], X1).
head(.(H, X2), H).
tail([], []).
tail(.(X3, T), T).
eq(X, X).

Query: duplicate(g,a)

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: duplicate(T1, T2)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: duplicate(T1, T2), SCOPE: 1, CLAUSE: 0 | duplicate(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: ','(!_1, eq(T5, [])) | duplicate([], T2), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: duplicate(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nX5 is free\nduplicate(T1, T2) !=? duplicate([], X5)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: eq(T5, [])\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: eq(T5, []), SCOPE: 2, CLAUSE: 6\n\n", 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: ','(head(T12, T15), ','(tail(T12, X15), duplicate(X15, T16)))\n\nT12 is ground\nX5 is free\nX15 is free\nduplicate(T12, T2) !=? duplicate([], X5)", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(head(T12, T15), ','(tail(T12, X15), duplicate(X15, T16))), SCOPE: 3, CLAUSE: 2 | ','(head(T12, T15), ','(tail(T12, X15), duplicate(X15, T16))), SCOPE: 3, CLAUSE: 3\n\nT12 is ground\nX5 is free\nX15 is free\nduplicate(T12, T2) !=? duplicate([], X5)", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(head(T12, T15), ','(tail(T12, X15), duplicate(X15, T16))), SCOPE: 3, CLAUSE: 3\n\nT12 is ground\nX5 is free\nX15 is free\nduplicate(T12, T2) !=? duplicate([], X5)", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(tail(.(T22, T23), X15), duplicate(X15, T24))\n\nT22 is ground\nT23 is ground\nX15 is free", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: ','(tail(.(T22, T23), X15), duplicate(X15, T24)), SCOPE: 4, CLAUSE: 4 | ','(tail(.(T22, T23), X15), duplicate(X15, T24)), SCOPE: 4, CLAUSE: 5\n\nT22 is ground\nT23 is ground\nX15 is free", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: ','(tail(.(T22, T23), X15), duplicate(X15, T24)), SCOPE: 4, CLAUSE: 5\n\nT22 is ground\nT23 is ground\nX15 is free", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: duplicate(T32, T24)\n\nT32 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 19 [arrowhead = none ];
19 -> 2;

20 [label="EVAL with clause\nduplicate([], X5) :- ','(!_1, eq(X5, [])).\nand substitutionT1 -> [],\nT2 -> T5,\nX5 -> T5,\nT4 -> T5", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 20 [arrowhead = none ];
20 -> 3;

21 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 21 [arrowhead = none ];
21 -> 4;

22 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 22 [arrowhead = none ];
22 -> 5;

23 [label="EVAL with clause\nduplicate(X12, .(X13, .(X13, X14))) :- ','(head(X12, X13), ','(tail(X12, X15), duplicate(X15, X14))).\nand substitutionT1 -> T12,\nX12 -> T12,\nX13 -> T15,\nX14 -> T16,\nT2 -> .(T15, .(T15, T16)),\nT13 -> T15,\nT14 -> T16", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 23 [arrowhead = none ];
23 -> 10;

24 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 24 [arrowhead = none ];
24 -> 11;

25 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
5 -> 25 [arrowhead = none ];
25 -> 6;

26 [label="EVAL with clause\neq(X8, X8).\nand substitutionT5 -> [],\nX8 -> [],\nT8 -> []", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 26 [arrowhead = none ];
26 -> 7;

27 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 27 [arrowhead = none ];
27 -> 8;

28 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
7 -> 28 [arrowhead = none ];
28 -> 9;

29 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
10 -> 29 [arrowhead = none ];
29 -> 12;

30 [label="BACKTRACK\nfor clause: head([], X1)\nwith clash: (duplicate(T12, T2), duplicate([], X5))", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 30 [arrowhead = none ];
30 -> 13;

31 [label="EVAL with clause\nhead(.(X21, X22), X21).\nand substitutionX21 -> T22,\nX22 -> T23,\nT12 -> .(T22, T23),\nT15 -> T22,\nT16 -> T24", fontsize=14, style = filled, fillcolor = lightblue];
13 -> 31 [arrowhead = none ];
31 -> 14;

32 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
13 -> 32 [arrowhead = none ];
32 -> 15;

33 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
14 -> 33 [arrowhead = none ];
33 -> 16;

34 [label="BACKTRACK\nfor clause: tail([], [])because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
16 -> 34 [arrowhead = none ];
34 -> 17;

35 [label="ONLY EVAL with clause\ntail(.(X29, X30), X30).\nand substitutionT22 -> T31,\nX29 -> T31,\nT23 -> T32,\nX30 -> T32,\nX15 -> T32", fontsize=14, style = filled, fillcolor = lightblue];
17 -> 35 [arrowhead = none ];
35 -> 18;

36 [label="INSTANCE with matching:\nT1 -> T32\nT2 -> T24", fontsize=14, style = filled, fillcolor = lightgrey];
18 -> 36 [arrowhead = none , style=dashed];
36 -> 1[style=dashed];

}

(2) Obligation:

Clauses:

duplicateA([], []).
duplicateA(.(T31, T32), .(T31, .(T31, T24))) :- duplicateA(T32, T24).

Query: duplicateA(g,a)

(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:
duplicateA_in: (b,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

duplicateA_in_ga([], []) → duplicateA_out_ga([], [])
duplicateA_in_ga(.(T31, T32), .(T31, .(T31, T24))) → U1_ga(T31, T32, T24, duplicateA_in_ga(T32, T24))
U1_ga(T31, T32, T24, duplicateA_out_ga(T32, T24)) → duplicateA_out_ga(.(T31, T32), .(T31, .(T31, T24)))

The argument filtering Pi contains the following mapping:
duplicateA_in_ga(x1, x2)  =  duplicateA_in_ga(x1)
[]  =  []
duplicateA_out_ga(x1, x2)  =  duplicateA_out_ga(x2)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4)  =  U1_ga(x1, x4)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

duplicateA_in_ga([], []) → duplicateA_out_ga([], [])
duplicateA_in_ga(.(T31, T32), .(T31, .(T31, T24))) → U1_ga(T31, T32, T24, duplicateA_in_ga(T32, T24))
U1_ga(T31, T32, T24, duplicateA_out_ga(T32, T24)) → duplicateA_out_ga(.(T31, T32), .(T31, .(T31, T24)))

The argument filtering Pi contains the following mapping:
duplicateA_in_ga(x1, x2)  =  duplicateA_in_ga(x1)
[]  =  []
duplicateA_out_ga(x1, x2)  =  duplicateA_out_ga(x2)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4)  =  U1_ga(x1, x4)

(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:

DUPLICATEA_IN_GA(.(T31, T32), .(T31, .(T31, T24))) → U1_GA(T31, T32, T24, duplicateA_in_ga(T32, T24))
DUPLICATEA_IN_GA(.(T31, T32), .(T31, .(T31, T24))) → DUPLICATEA_IN_GA(T32, T24)

The TRS R consists of the following rules:

duplicateA_in_ga([], []) → duplicateA_out_ga([], [])
duplicateA_in_ga(.(T31, T32), .(T31, .(T31, T24))) → U1_ga(T31, T32, T24, duplicateA_in_ga(T32, T24))
U1_ga(T31, T32, T24, duplicateA_out_ga(T32, T24)) → duplicateA_out_ga(.(T31, T32), .(T31, .(T31, T24)))

The argument filtering Pi contains the following mapping:
duplicateA_in_ga(x1, x2)  =  duplicateA_in_ga(x1)
[]  =  []
duplicateA_out_ga(x1, x2)  =  duplicateA_out_ga(x2)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4)  =  U1_ga(x1, x4)
DUPLICATEA_IN_GA(x1, x2)  =  DUPLICATEA_IN_GA(x1)
U1_GA(x1, x2, x3, x4)  =  U1_GA(x1, x4)

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

(6) Obligation:

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

DUPLICATEA_IN_GA(.(T31, T32), .(T31, .(T31, T24))) → U1_GA(T31, T32, T24, duplicateA_in_ga(T32, T24))
DUPLICATEA_IN_GA(.(T31, T32), .(T31, .(T31, T24))) → DUPLICATEA_IN_GA(T32, T24)

The TRS R consists of the following rules:

duplicateA_in_ga([], []) → duplicateA_out_ga([], [])
duplicateA_in_ga(.(T31, T32), .(T31, .(T31, T24))) → U1_ga(T31, T32, T24, duplicateA_in_ga(T32, T24))
U1_ga(T31, T32, T24, duplicateA_out_ga(T32, T24)) → duplicateA_out_ga(.(T31, T32), .(T31, .(T31, T24)))

The argument filtering Pi contains the following mapping:
duplicateA_in_ga(x1, x2)  =  duplicateA_in_ga(x1)
[]  =  []
duplicateA_out_ga(x1, x2)  =  duplicateA_out_ga(x2)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4)  =  U1_ga(x1, x4)
DUPLICATEA_IN_GA(x1, x2)  =  DUPLICATEA_IN_GA(x1)
U1_GA(x1, x2, x3, x4)  =  U1_GA(x1, x4)

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

(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:

DUPLICATEA_IN_GA(.(T31, T32), .(T31, .(T31, T24))) → DUPLICATEA_IN_GA(T32, T24)

The TRS R consists of the following rules:

duplicateA_in_ga([], []) → duplicateA_out_ga([], [])
duplicateA_in_ga(.(T31, T32), .(T31, .(T31, T24))) → U1_ga(T31, T32, T24, duplicateA_in_ga(T32, T24))
U1_ga(T31, T32, T24, duplicateA_out_ga(T32, T24)) → duplicateA_out_ga(.(T31, T32), .(T31, .(T31, T24)))

The argument filtering Pi contains the following mapping:
duplicateA_in_ga(x1, x2)  =  duplicateA_in_ga(x1)
[]  =  []
duplicateA_out_ga(x1, x2)  =  duplicateA_out_ga(x2)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4)  =  U1_ga(x1, x4)
DUPLICATEA_IN_GA(x1, x2)  =  DUPLICATEA_IN_GA(x1)

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

(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:

DUPLICATEA_IN_GA(.(T31, T32), .(T31, .(T31, T24))) → DUPLICATEA_IN_GA(T32, T24)

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

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:

DUPLICATEA_IN_GA(.(T31, T32)) → DUPLICATEA_IN_GA(T32)

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:

• DUPLICATEA_IN_GA(.(T31, T32)) → DUPLICATEA_IN_GA(T32)
  The graph contains the following edges 1 > 1

(14) YES