Left Termination of the query pattern p_in_1(a) w.r.t. the given Prolog program could successfully be proven:

(0) Obligation:

Clauses:

p(X) :- q(X).
p(X) :- ','(no(q(X)), p(X)).
q(X).
no(X) :- ','(X, ','(!, failure(a))).
no(X1).
failure(b).

Queries:

p(a).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

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 | p(T1), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: p(T1), SCOPE: 1, CLAUSE: 0\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: p(T1), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: q(T3)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: q(T3), SCOPE: 2, CLAUSE: 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: ','(no(q(T6)), p(T6))\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(no(q(T6)), p(T6)), SCOPE: 3, CLAUSE: 3 | ','(no(q(T6)), p(T6)), SCOPE: 3, CLAUSE: 4 | ?_3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(q(T8), ','(!_3, ','(failure(a), p(T8)))) | ','(no(q(T6)), p(T6)), SCOPE: 3, CLAUSE: 4 | ?_3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(q(T8), ','(!_3, ','(failure(a), p(T8)))), SCOPE: 4, CLAUSE: 2 | ','(no(q(T6)), p(T6)), SCOPE: 3, CLAUSE: 4 | ?_3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(!_3, ','(failure(a), p(T10))) | ','(no(q(T6)), p(T6)), SCOPE: 3, CLAUSE: 4 | ?_3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(failure(a), p(T10))\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: ','(failure(a), p(T10)), SCOPE: 5, CLAUSE: 5\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="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
2 -> 18 [arrowhead = none ];
18 -> 3;

19 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
2 -> 19 [arrowhead = none ];
19 -> 4;

20 [label="EVAL with clause\np(X4) :- q(X4).\nand substitution\nT1 -> T3\nX4 -> T3\nT2 -> T3", fontsize=14, style = filled, fillcolor = lightblue];
3 -> 20 [arrowhead = none ];
20 -> 5;

21 [label="EVAL with clause\np(X8) :- ','(no(q(X8)), p(X8)).\nand substitution\nT1 -> T6\nX8 -> T6\nT5 -> T6", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 21 [arrowhead = none ];
21 -> 9;

22 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
5 -> 22 [arrowhead = none ];
22 -> 6;

23 [label="EVAL with clause\nq(X6).\nand substitution\nT3 -> T4\nX6 -> T4", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 23 [arrowhead = none ];
23 -> 7;

24 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
7 -> 24 [arrowhead = none ];
24 -> 8;

25 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
9 -> 25 [arrowhead = none ];
25 -> 10;

26 [label="EVAL with clause\nno(X10) :- ','(X10, ','(!, failure(a))).\nand substitution\nT6 -> T8\nX10 -> q(T8)\nT7 -> T8", fontsize=14, style = filled, fillcolor = lightblue];
10 -> 26 [arrowhead = none ];
26 -> 11;

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

28 [label="EVAL with clause\nq(X12).\nand substitution\nT8 -> T10\nX12 -> T10\nT9 -> T10", fontsize=14, style = filled, fillcolor = lightblue];
12 -> 28 [arrowhead = none ];
28 -> 13;

29 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
13 -> 29 [arrowhead = none ];
29 -> 14;

30 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
14 -> 30 [arrowhead = none ];
30 -> 15;

31 [label="BACKTRACK\nfor clause: failure(b)because of non-unification", fontsize=14, style = filled, fillcolor = white];
15 -> 31 [arrowhead = none ];
31 -> 16;

}

(2) Obligation:

Clauses:

p1(T4).

Queries:

p1(a).

(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_a(T4) → p1_out_a(T4)

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a

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_a(T4) → p1_out_a(T4)

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a

(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_a(T4) → p1_out_a(T4)

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a

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_a(T4) → p1_out_a(T4)

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a

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