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

(0) Obligation:

Clauses:

interleave([], Xs, Xs).
interleave(.(X, Xs), Ys, .(X, Zs)) :- interleave(Ys, Xs, Zs).

Query: interleave(g,g,a)

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: interleave(T1, T2, T3)\n\nT1 is ground\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: interleave(T1, T2, T3), SCOPE: 1, CLAUSE: 0 | interleave(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: true | interleave([], T5, T3), SCOPE: 1, CLAUSE: 1\n\nT5 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: interleave(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nT2 is ground\nX2 is free\ninterleave(T1, T2, T3) !=? interleave([], X2, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: interleave([], T5, T3), 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: interleave(T12, T11, T14)\n\nT11 is ground\nT12 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: interleave(T12, T11, T14), SCOPE: 2, CLAUSE: 0 | interleave(T12, T11, T14), SCOPE: 2, CLAUSE: 1\n\nT11 is ground\nT12 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: interleave(T12, T11, T14), SCOPE: 2, CLAUSE: 0\n\nT11 is ground\nT12 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: interleave(T12, T11, T14), SCOPE: 2, CLAUSE: 1\n\nT11 is ground\nT12 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: interleave(T30, T29, T32)\n\nT29 is ground\nT30 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\ninterleave([], 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\ninterleave(.(X11, X12), X13, .(X11, X14)) :- interleave(X13, X12, X14).\nand substitutionX11 -> T10,\nX12 -> T11,\nT1 -> .(T10, T11),\nT2 -> T12,\nX13 -> T12,\nX14 -> T14,\nT3 -> .(T10, T14),\nT13 -> 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: interleave(.(X, Xs), Ys, .(X, Zs)) :- interleave(Ys, Xs, Zs)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\ninterleave([], X19, X19).\nand substitutionT12 -> [],\nT11 -> T19,\nX19 -> T19,\nT14 -> T19", 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\ninterleave(.(X28, X29), X30, .(X28, X31)) :- interleave(X30, X29, X31).\nand substitutionX28 -> T28,\nX29 -> T29,\nT12 -> .(T28, T29),\nT11 -> T30,\nX30 -> T30,\nX31 -> T32,\nT14 -> .(T28, T32),\nT31 -> T32", 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 -> T30\nT2 -> T29\nT3 -> T32", fontsize=14, style = filled, fillcolor = lightgrey];
15 -> 32 [arrowhead = none , style=dashed];
32 -> 1[style=dashed];

}

(2) Obligation:

Clauses:

interleaveA([], T5, T5).
interleaveA(.(T10, T19), [], .(T10, T19)).
interleaveA(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) :- interleaveA(T30, T29, T32).

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

interleaveA_in_gga([], T5, T5) → interleaveA_out_gga([], T5, T5)
interleaveA_in_gga(.(T10, T19), [], .(T10, T19)) → interleaveA_out_gga(.(T10, T19), [], .(T10, T19))
interleaveA_in_gga(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) → U1_gga(T10, T30, T28, T29, T32, interleaveA_in_gga(T30, T29, T32))
U1_gga(T10, T30, T28, T29, T32, interleaveA_out_gga(T30, T29, T32)) → interleaveA_out_gga(.(T10, T30), .(T28, T29), .(T10, .(T28, T32)))

The argument filtering Pi contains the following mapping:
interleaveA_in_gga(x1, x2, x3)  =  interleaveA_in_gga(x1, x2)
[]  =  []
interleaveA_out_gga(x1, x2, x3)  =  interleaveA_out_gga(x3)
.(x1, x2)  =  .(x1, x2)
U1_gga(x1, x2, x3, x4, x5, x6)  =  U1_gga(x1, x3, x6)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

interleaveA_in_gga([], T5, T5) → interleaveA_out_gga([], T5, T5)
interleaveA_in_gga(.(T10, T19), [], .(T10, T19)) → interleaveA_out_gga(.(T10, T19), [], .(T10, T19))
interleaveA_in_gga(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) → U1_gga(T10, T30, T28, T29, T32, interleaveA_in_gga(T30, T29, T32))
U1_gga(T10, T30, T28, T29, T32, interleaveA_out_gga(T30, T29, T32)) → interleaveA_out_gga(.(T10, T30), .(T28, T29), .(T10, .(T28, T32)))

The argument filtering Pi contains the following mapping:
interleaveA_in_gga(x1, x2, x3)  =  interleaveA_in_gga(x1, x2)
[]  =  []
interleaveA_out_gga(x1, x2, x3)  =  interleaveA_out_gga(x3)
.(x1, x2)  =  .(x1, x2)
U1_gga(x1, x2, x3, x4, x5, x6)  =  U1_gga(x1, x3, x6)

(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:

INTERLEAVEA_IN_GGA(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) → U1_GGA(T10, T30, T28, T29, T32, interleaveA_in_gga(T30, T29, T32))
INTERLEAVEA_IN_GGA(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) → INTERLEAVEA_IN_GGA(T30, T29, T32)

The TRS R consists of the following rules:

interleaveA_in_gga([], T5, T5) → interleaveA_out_gga([], T5, T5)
interleaveA_in_gga(.(T10, T19), [], .(T10, T19)) → interleaveA_out_gga(.(T10, T19), [], .(T10, T19))
interleaveA_in_gga(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) → U1_gga(T10, T30, T28, T29, T32, interleaveA_in_gga(T30, T29, T32))
U1_gga(T10, T30, T28, T29, T32, interleaveA_out_gga(T30, T29, T32)) → interleaveA_out_gga(.(T10, T30), .(T28, T29), .(T10, .(T28, T32)))

The argument filtering Pi contains the following mapping:
interleaveA_in_gga(x1, x2, x3)  =  interleaveA_in_gga(x1, x2)
[]  =  []
interleaveA_out_gga(x1, x2, x3)  =  interleaveA_out_gga(x3)
.(x1, x2)  =  .(x1, x2)
U1_gga(x1, x2, x3, x4, x5, x6)  =  U1_gga(x1, x3, x6)
INTERLEAVEA_IN_GGA(x1, x2, x3)  =  INTERLEAVEA_IN_GGA(x1, x2)
U1_GGA(x1, x2, x3, x4, x5, x6)  =  U1_GGA(x1, x3, x6)

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

(6) Obligation:

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

INTERLEAVEA_IN_GGA(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) → U1_GGA(T10, T30, T28, T29, T32, interleaveA_in_gga(T30, T29, T32))
INTERLEAVEA_IN_GGA(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) → INTERLEAVEA_IN_GGA(T30, T29, T32)

The TRS R consists of the following rules:

interleaveA_in_gga([], T5, T5) → interleaveA_out_gga([], T5, T5)
interleaveA_in_gga(.(T10, T19), [], .(T10, T19)) → interleaveA_out_gga(.(T10, T19), [], .(T10, T19))
interleaveA_in_gga(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) → U1_gga(T10, T30, T28, T29, T32, interleaveA_in_gga(T30, T29, T32))
U1_gga(T10, T30, T28, T29, T32, interleaveA_out_gga(T30, T29, T32)) → interleaveA_out_gga(.(T10, T30), .(T28, T29), .(T10, .(T28, T32)))

The argument filtering Pi contains the following mapping:
interleaveA_in_gga(x1, x2, x3)  =  interleaveA_in_gga(x1, x2)
[]  =  []
interleaveA_out_gga(x1, x2, x3)  =  interleaveA_out_gga(x3)
.(x1, x2)  =  .(x1, x2)
U1_gga(x1, x2, x3, x4, x5, x6)  =  U1_gga(x1, x3, x6)
INTERLEAVEA_IN_GGA(x1, x2, x3)  =  INTERLEAVEA_IN_GGA(x1, x2)
U1_GGA(x1, x2, x3, x4, x5, x6)  =  U1_GGA(x1, x3, x6)

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:

INTERLEAVEA_IN_GGA(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) → INTERLEAVEA_IN_GGA(T30, T29, T32)

The TRS R consists of the following rules:

interleaveA_in_gga([], T5, T5) → interleaveA_out_gga([], T5, T5)
interleaveA_in_gga(.(T10, T19), [], .(T10, T19)) → interleaveA_out_gga(.(T10, T19), [], .(T10, T19))
interleaveA_in_gga(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) → U1_gga(T10, T30, T28, T29, T32, interleaveA_in_gga(T30, T29, T32))
U1_gga(T10, T30, T28, T29, T32, interleaveA_out_gga(T30, T29, T32)) → interleaveA_out_gga(.(T10, T30), .(T28, T29), .(T10, .(T28, T32)))

The argument filtering Pi contains the following mapping:
interleaveA_in_gga(x1, x2, x3)  =  interleaveA_in_gga(x1, x2)
[]  =  []
interleaveA_out_gga(x1, x2, x3)  =  interleaveA_out_gga(x3)
.(x1, x2)  =  .(x1, x2)
U1_gga(x1, x2, x3, x4, x5, x6)  =  U1_gga(x1, x3, x6)
INTERLEAVEA_IN_GGA(x1, x2, x3)  =  INTERLEAVEA_IN_GGA(x1, x2)

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:

INTERLEAVEA_IN_GGA(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) → INTERLEAVEA_IN_GGA(T30, T29, T32)

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

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:

INTERLEAVEA_IN_GGA(.(T10, T30), .(T28, T29)) → INTERLEAVEA_IN_GGA(T30, T29)

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:

• INTERLEAVEA_IN_GGA(.(T10, T30), .(T28, T29)) → INTERLEAVEA_IN_GGA(T30, T29)
  The graph contains the following edges 1 > 1, 2 > 2

(14) YES