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

(0) Obligation:

Clauses:

p(b).
p(a) :- p1(X).
p1(b) :- !.
p1(a) :- p1(X).

Queries:

p(g).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: p(T1)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: p(T1), SCOPE: 1, CLAUSE: 0 | p(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: p(T1), SCOPE: 1, CLAUSE: 0\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: p(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: p1(X1)\n\nX1 is free", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: p1(X1), SCOPE: 2, CLAUSE: 2 | p1(X1), SCOPE: 2, CLAUSE: 3 | ?_2\n\nX1 is free", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: !_2 | p1(X1), SCOPE: 2, CLAUSE: 3 | ?_2\n\nX1 is free", 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="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
2 -> 15 [arrowhead = none ];
15 -> 3;

16 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
2 -> 16 [arrowhead = none ];
16 -> 4;

17 [label="EVAL with clause\np(b).\nand substitution\nT1 -> b", fontsize=14, style = filled, fillcolor = lightblue];
3 -> 17 [arrowhead = none ];
17 -> 5;

18 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
3 -> 18 [arrowhead = none ];
18 -> 6;

19 [label="EVAL with clause\np(a) :- p1(X1).\nand substitution\nT1 -> a", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 19 [arrowhead = none ];
19 -> 8;

20 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 20 [arrowhead = none ];
20 -> 9;

21 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 21 [arrowhead = none ];
21 -> 7;

22 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
8 -> 22 [arrowhead = none ];
22 -> 10;

23 [label="EVAL with clause\np1(b) :- !.\nand substitution\nX1 -> b", 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:

p11(b).
p11(a).

Queries:

p11(g).

(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:

p11_in_g(b) → p11_out_g(b)
p11_in_g(a) → p11_out_g(a)

The argument filtering Pi contains the following mapping:
p11_in_g(x1)  =  p11_in_g(x1)
b  =  b
p11_out_g(x1)  =  p11_out_g
a  =  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:

p11_in_g(b) → p11_out_g(b)
p11_in_g(a) → p11_out_g(a)

The argument filtering Pi contains the following mapping:
p11_in_g(x1)  =  p11_in_g(x1)
b  =  b
p11_out_g(x1)  =  p11_out_g
a  =  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:

p11_in_g(b) → p11_out_g(b)
p11_in_g(a) → p11_out_g(a)

The argument filtering Pi contains the following mapping:
p11_in_g(x1)  =  p11_in_g(x1)
b  =  b
p11_out_g(x1)  =  p11_out_g
a  =  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:

p11_in_g(b) → p11_out_g(b)
p11_in_g(a) → p11_out_g(a)

The argument filtering Pi contains the following mapping:
p11_in_g(x1)  =  p11_in_g(x1)
b  =  b
p11_out_g(x1)  =  p11_out_g
a  =  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