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

(0) Obligation:

Clauses:

p(.(A, [])) :- l(.(A, [])).
r(1).
l([]).
l(.(H, T)) :- ','(r(H), l(T)).

Query: p(a)

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: p(T1)\n\n", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: p(T1), SCOPE: 1, CLAUSE: 0\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: l(.(T4, []))\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: l(.(T4, [])), SCOPE: 2, CLAUSE: 2 | l(.(T4, [])), SCOPE: 2, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: l(.(T4, [])), SCOPE: 2, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: ','(r(T9), l([]))\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: ','(r(T9), l([])), SCOPE: 3, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: l([])\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: l([]), SCOPE: 4, CLAUSE: 2 | l([]), SCOPE: 4, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: l([]), SCOPE: 4, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: l([]), SCOPE: 4, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: \n\n", 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\np(.(X2, [])) :- l(.(X2, [])).\nand substitutionX2 -> T4,\nT1 -> .(T4, []),\nT3 -> T4", 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="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 20 [arrowhead = none ];
20 -> 5;

21 [label="BACKTRACK\nfor clause: l([])because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 21 [arrowhead = none ];
21 -> 6;

22 [label="ONLY EVAL with clause\nl(.(X7, X8)) :- ','(r(X7), l(X8)).\nand substitutionT4 -> T9,\nX7 -> T9,\nX8 -> [],\nT8 -> T9", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 22 [arrowhead = none ];
22 -> 7;

23 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 23 [arrowhead = none ];
23 -> 8;

24 [label="EVAL with clause\nr(1).\nand substitutionT9 -> 1", fontsize=14, style = filled, fillcolor = lightblue];
8 -> 24 [arrowhead = none ];
24 -> 9;

25 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
8 -> 25 [arrowhead = none ];
25 -> 10;

26 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
9 -> 26 [arrowhead = none ];
26 -> 11;

27 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
11 -> 27 [arrowhead = none ];
27 -> 12;

28 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
11 -> 28 [arrowhead = none ];
28 -> 13;

29 [label="ONLY EVAL with clause\nl([]).\nand substitution", fontsize=14, style = filled, fillcolor = lightblue];
12 -> 29 [arrowhead = none ];
29 -> 14;

30 [label="BACKTRACK\nfor clause: l(.(H, T)) :- ','(r(H), l(T))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
13 -> 30 [arrowhead = none ];
30 -> 16;

31 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
14 -> 31 [arrowhead = none ];
31 -> 15;

}

(2) Obligation:

Clauses:

pA(.(1, [])).

Query: pA(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:
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

pA_in_a(.(1, [])) → pA_out_a(.(1, []))

The argument filtering Pi contains the following mapping:
pA_in_a(x1)  =  pA_in_a
pA_out_a(x1)  =  pA_out_a(x1)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

pA_in_a(.(1, [])) → pA_out_a(.(1, []))

The argument filtering Pi contains the following mapping:
pA_in_a(x1)  =  pA_in_a
pA_out_a(x1)  =  pA_out_a(x1)

(5) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
P is empty.
The TRS R consists of the following rules:

pA_in_a(.(1, [])) → pA_out_a(.(1, []))

The argument filtering Pi contains the following mapping:
pA_in_a(x1)  =  pA_in_a
pA_out_a(x1)  =  pA_out_a(x1)

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

(6) Obligation:

Pi DP problem:
P is empty.
The TRS R consists of the following rules:

pA_in_a(.(1, [])) → pA_out_a(.(1, []))

The argument filtering Pi contains the following mapping:
pA_in_a(x1)  =  pA_in_a
pA_out_a(x1)  =  pA_out_a(x1)

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

(7) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,R,Pi) chain.

(8) YES