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) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: 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: true | p(b), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: p(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\np(T1) !=? p(b)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: p(b), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: p1(X1)\n\nX1 is free", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: p1(X1), SCOPE: 2, CLAUSE: 2 | p1(X1), SCOPE: 2, CLAUSE: 3\n\nX1 is free", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: !_2 | p1(X1), SCOPE: 2, CLAUSE: 3\n\nX1 is free", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 13 [arrowhead = none ];
13 -> 2;

14 [label="EVAL with clause\np(b).\nand substitutionT1 -> b", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 14 [arrowhead = none ];
14 -> 3;

15 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 15 [arrowhead = none ];
15 -> 4;

16 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 16 [arrowhead = none ];
16 -> 5;

17 [label="EVAL with clause\np(a) :- p1(X1).\nand substitutionT1 -> a", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 17 [arrowhead = none ];
17 -> 7;

18 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 18 [arrowhead = none ];
18 -> 8;

19 [label="BACKTRACK\nfor clause: p(a) :- p1(X)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 19 [arrowhead = none ];
19 -> 6;

20 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 20 [arrowhead = none ];
20 -> 9;

21 [label="ONLY EVAL with clause\np1(b) :- !_2.\nand substitutionX1 -> b", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 21 [arrowhead = none ];
21 -> 10;

22 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 22 [arrowhead = none ];
22 -> 11;

23 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 23 [arrowhead = none ];
23 -> 12;

}

(2) Obligation:

Triples:


Clauses:


Afs:

(3) TPisEmptyProof (EQUIVALENT transformation)

There are no more dependency triples. Hence, the dependency triple problem trivially terminates.

(4) YES