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

(0) Obligation:

Clauses:

p :- thief(john).
thief(john).
thief(X) :- ','(thief(X), !).

Queries:

p().

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: p\n\n", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: p, SCOPE: 1, CLAUSE: 0\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: thief(john)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: thief(john), SCOPE: 2, CLAUSE: 1 | thief(john), SCOPE: 2, CLAUSE: 2 | ?_2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: thief(john), SCOPE: 2, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: thief(john), SCOPE: 2, CLAUSE: 2 | ?_2\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: ','(thief(john), !_2) | ?_2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(thief(john), !_2), SCOPE: 3, CLAUSE: 1 | ','(thief(john), !_2), SCOPE: 3, CLAUSE: 2 | ?_3 | ?_2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: !_2 | ','(thief(john), !_2), SCOPE: 3, CLAUSE: 2 | ?_3 | ?_2\n\n", 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="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 14 [arrowhead = none ];
14 -> 2;

15 [label="EVAL with clause\np :- thief(john).\nand substitution", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 15 [arrowhead = none ];
15 -> 3;

16 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 16 [arrowhead = none ];
16 -> 4;

17 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
4 -> 17 [arrowhead = none ];
17 -> 5;

18 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
4 -> 18 [arrowhead = none ];
18 -> 6;

19 [label="EVAL with clause\nthief(john).\nand substitution", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 19 [arrowhead = none ];
19 -> 7;

20 [label="EVAL with clause\nthief(X2) :- ','(thief(X2), !).\nand substitution\nX2 -> john", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 20 [arrowhead = none ];
20 -> 9;

21 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
7 -> 21 [arrowhead = none ];
21 -> 8;

22 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
9 -> 22 [arrowhead = none ];
22 -> 10;

23 [label="EVAL with clause\nthief(john).\nand substitution", fontsize=14, style = filled, fillcolor = lightblue];
10 -> 23 [arrowhead = none ];
23 -> 11;

24 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 24 [arrowhead = none ];
24 -> 12;

25 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 25 [arrowhead = none ];
25 -> 13;

}

(2) Obligation:

Clauses:

p1.
p1.

Queries:

p1().

(3) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. 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:

p1_in_ → p1_out_

Pi is empty.

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

p1_in_ → p1_out_

Pi is empty.

(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:

p1_in_ → p1_out_

Pi is empty.
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:

p1_in_ → p1_out_

Pi is empty.
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) TRUE