Left Termination of the query pattern
p1(g)
w.r.t. the given Prolog program could successfully be proven:

(0) Obligation:

Clauses:

p1(f(X)) :- p1(X).
p2(f(X)) :- p2(X).

Query: p1(g)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: p1(T1)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: p1(T1), SCOPE: 1, CLAUSE: 0\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: p1(T3)\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: p1(T3), SCOPE: 2, CLAUSE: 0\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: p1(T6)\n\nT6 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 8 [arrowhead = none ];
8 -> 2;

9 [label="EVAL with clause\np1(f(X2)) :- p1(X2).\nand substitutionX2 -> T3,\nT1 -> f(T3)", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 9 [arrowhead = none ];
9 -> 3;

10 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 10 [arrowhead = none ];
10 -> 4;

11 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 11 [arrowhead = none ];
11 -> 5;

12 [label="EVAL with clause\np1(f(X5)) :- p1(X5).\nand substitutionX5 -> T6,\nT3 -> f(T6)", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 12 [arrowhead = none ];
12 -> 6;

13 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 13 [arrowhead = none ];
13 -> 7;

14 [label="INSTANCE with matching:\nT1 -> T6", fontsize=14, style = filled, fillcolor = lightgrey];
6 -> 14 [arrowhead = none , style=dashed];
14 -> 1[style=dashed];

}

(2) Obligation:

Triples:

p1A(f(f(X1))) :- p1A(X1).

Clauses:

p1cA(f(f(X1))) :- p1cA(X1).

Afs:

p1A(x1)  =  p1A(x1)

(3) TriplesToPiDPProof (SOUND transformation)

We use the technique of [DT09]. With regard to the inferred argument filtering the predicates were used in the following modes:
p1A_in: (b)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

P1A_IN_G(f(f(X1))) → U1_G(X1, p1A_in_g(X1))
P1A_IN_G(f(f(X1))) → P1A_IN_G(X1)

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

Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES

(4) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

P1A_IN_G(f(f(X1))) → U1_G(X1, p1A_in_g(X1))
P1A_IN_G(f(f(X1))) → P1A_IN_G(X1)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.

(6) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

P1A_IN_G(f(f(X1))) → P1A_IN_G(X1)

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

(7) PiDPToQDPProof (EQUIVALENT transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(8) Obligation:

Q DP problem:
The TRS P consists of the following rules:

P1A_IN_G(f(f(X1))) → P1A_IN_G(X1)

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

(9) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• P1A_IN_G(f(f(X1))) → P1A_IN_G(X1)
  The graph contains the following edges 1 > 1

(10) YES