Left Termination of the query pattern
p(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).

Query: p(g)

(1) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
2 [label="A: p(T1)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
4 [label="4: p(T1), SCOPE: 1, CLAUSE: 0 | p(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: true | p(b), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: p(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\np(T1) !=? p(b)", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: p(b), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: p1(X1)\n\nX1 is free", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: p1(X1), SCOPE: 2, CLAUSE: 2 | p1(X1), SCOPE: 2, CLAUSE: 3\n\nX1 is free", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: !_2 | p1(X1), SCOPE: 2, CLAUSE: 3\n\nX1 is free", fontsize=16, style=filled, fillcolor=lightyellow];
31 [label="31: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
32 [label="32: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
33 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
2 -> 33 [arrowhead = none ];
33 -> 4;

34 [label="EVAL with clause\np(b).\nand substitutionT1 -> b", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 34 [arrowhead = none ];
34 -> 10;

35 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 35 [arrowhead = none ];
35 -> 11;

36 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 36 [arrowhead = none ];
36 -> 17;

37 [label="EVAL with clause\np(a) :- p1(X1).\nand substitutionT1 -> a", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 37 [arrowhead = none ];
37 -> 23;

38 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 38 [arrowhead = none ];
38 -> 25;

39 [label="BACKTRACK\nfor clause: p(a) :- p1(X)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 39 [arrowhead = none ];
39 -> 19;

40 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
23 -> 40 [arrowhead = none ];
40 -> 27;

41 [label="ONLY EVAL with clause\np1(b) :- !_2.\nand substitutionX1 -> b", fontsize=14, style = filled, fillcolor = lightblue];
27 -> 41 [arrowhead = none ];
41 -> 28;

42 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
28 -> 42 [arrowhead = none ];
42 -> 31;

43 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
31 -> 43 [arrowhead = none ];
43 -> 32;

}

(2) Obligation:

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

pA_in_g(b) → pA_out_g(b)
pA_in_g(a) → pA_out_g(a)

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_g(b) → pA_out_g(b)
pA_in_g(a) → pA_out_g(a)

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_g(b) → pA_out_g(b)
pA_in_g(a) → pA_out_g(a)

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