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

(0) Obligation:

Clauses:

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

Query: p()

(1) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
3 [label="A: p\n\n", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
5 [label="5: p, SCOPE: 1, CLAUSE: 0\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: thief(john)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: thief(john), SCOPE: 2, CLAUSE: 1 | thief(john), SCOPE: 2, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: true | thief(john), SCOPE: 2, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: thief(john), SCOPE: 2, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: ','(thief(john), !_2)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: ','(thief(john), !_2), SCOPE: 3, CLAUSE: 1 | ','(thief(john), !_2), SCOPE: 3, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: !_2 | ','(thief(john), !_2), SCOPE: 3, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 28 [arrowhead = none ];
28 -> 5;

29 [label="ONLY EVAL with clause\np :- thief(john).\nand substitution", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 29 [arrowhead = none ];
29 -> 8;

30 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
8 -> 30 [arrowhead = none ];
30 -> 10;

31 [label="ONLY EVAL with clause\nthief(john).\nand substitution", fontsize=14, style = filled, fillcolor = lightblue];
10 -> 31 [arrowhead = none ];
31 -> 12;

32 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 32 [arrowhead = none ];
32 -> 13;

33 [label="ONLY EVAL with clause\nthief(X3) :- ','(thief(X3), !_2).\nand substitutionX3 -> john", fontsize=14, style = filled, fillcolor = lightblue];
13 -> 33 [arrowhead = none ];
33 -> 17;

34 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
17 -> 34 [arrowhead = none ];
34 -> 19;

35 [label="ONLY EVAL with clause\nthief(john).\nand substitution", fontsize=14, style = filled, fillcolor = lightblue];
19 -> 35 [arrowhead = none ];
35 -> 21;

36 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
21 -> 36 [arrowhead = none ];
36 -> 24;

37 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
24 -> 37 [arrowhead = none ];
37 -> 27;

}

(2) Obligation:

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

pA_in_ → pA_out_

Pi is empty.

(3) 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_ → pA_out_

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

(4) Obligation:

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

pA_in_ → pA_out_

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

(5) PisEmptyProof (EQUIVALENT transformation)

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

(6) YES