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

(0) Obligation:

Clauses:

f(X) :- ','(p(X), q(X)).
p(a) :- !.
p(X) :- p(Y).
q(b).

Query: f(a)

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: f(T1)\n\n", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: f(T1), SCOPE: 1, CLAUSE: 0\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: ','(p(T4), q(T4))\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: ','(p(T4), q(T4)), SCOPE: 2, CLAUSE: 1 | ','(p(T4), q(T4)), SCOPE: 2, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: ','(!_2, q(a)) | ','(p(T4), q(T4)), SCOPE: 2, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: ','(p(T4), q(T4)), SCOPE: 2, CLAUSE: 2\n\np(T4) !=? p(a)", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: q(a)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: q(a), SCOPE: 3, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(p(X7), q(T9))\n\nX7 is free", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: p(X7)\n\nX7 is free", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: q(T9)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: p(X7), SCOPE: 4, CLAUSE: 1 | p(X7), SCOPE: 4, CLAUSE: 2\n\nX7 is free", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: !_4 | p(X7), SCOPE: 4, CLAUSE: 2\n\nX7 is free", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: q(T9), SCOPE: 5, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 21 [arrowhead = none ];
21 -> 2;

22 [label="ONLY EVAL with clause\nf(X2) :- ','(p(X2), q(X2)).\nand substitutionT1 -> T4,\nX2 -> T4,\nT3 -> T4", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 22 [arrowhead = none ];
22 -> 3;

23 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 23 [arrowhead = none ];
23 -> 4;

24 [label="EVAL with clause\np(a) :- !_2.\nand substitutionT4 -> a", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 24 [arrowhead = none ];
24 -> 5;

25 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 25 [arrowhead = none ];
25 -> 6;

26 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 26 [arrowhead = none ];
26 -> 7;

27 [label="ONLY EVAL with clause\np(X6) :- p(X7).\nand substitutionT4 -> T9,\nX6 -> T9,\nT8 -> T9", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 27 [arrowhead = none ];
27 -> 10;

28 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 28 [arrowhead = none ];
28 -> 8;

29 [label="BACKTRACK\nfor clause: q(b)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
8 -> 29 [arrowhead = none ];
29 -> 9;

30 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
10 -> 30 [arrowhead = none ];
30 -> 11;

31 [label="SPLIT 2\nreplacements:X7 -> T10", fontsize=14, style = filled, fillcolor = violet];
10 -> 31 [arrowhead = none ];
31 -> 12;

32 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
11 -> 32 [arrowhead = none ];
32 -> 13;

33 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
12 -> 33 [arrowhead = none ];
33 -> 17;

34 [label="ONLY EVAL with clause\np(a) :- !_4.\nand substitutionX7 -> a", fontsize=14, style = filled, fillcolor = lightblue];
13 -> 34 [arrowhead = none ];
34 -> 14;

35 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
14 -> 35 [arrowhead = none ];
35 -> 15;

36 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 36 [arrowhead = none ];
36 -> 16;

37 [label="EVAL with clause\nq(b).\nand substitutionT9 -> b", fontsize=14, style = filled, fillcolor = lightblue];
17 -> 37 [arrowhead = none ];
37 -> 18;

38 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 38 [arrowhead = none ];
38 -> 19;

39 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
18 -> 39 [arrowhead = none ];
39 -> 20;

}

(2) Obligation:

Clauses:

pA(a).
fB(T9) :- pA(X7).
fB(b) :- pA(T10).

Query: fB(a)

(3) PrologToPiTRSProof (SOUND transformation)

We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
fB_in: (f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

fB_in_a(T9) → U1_a(T9, pA_in_a(X7))
pA_in_a(a) → pA_out_a(a)
U1_a(T9, pA_out_a(X7)) → fB_out_a(T9)
fB_in_a(b) → U2_a(pA_in_a(T10))
U2_a(pA_out_a(T10)) → fB_out_a(b)

The argument filtering Pi contains the following mapping:
fB_in_a(x1)  =  fB_in_a
U1_a(x1, x2)  =  U1_a(x2)
pA_in_a(x1)  =  pA_in_a
pA_out_a(x1)  =  pA_out_a(x1)
fB_out_a(x1)  =  fB_out_a
U2_a(x1)  =  U2_a(x1)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

fB_in_a(T9) → U1_a(T9, pA_in_a(X7))
pA_in_a(a) → pA_out_a(a)
U1_a(T9, pA_out_a(X7)) → fB_out_a(T9)
fB_in_a(b) → U2_a(pA_in_a(T10))
U2_a(pA_out_a(T10)) → fB_out_a(b)

The argument filtering Pi contains the following mapping:
fB_in_a(x1)  =  fB_in_a
U1_a(x1, x2)  =  U1_a(x2)
pA_in_a(x1)  =  pA_in_a
pA_out_a(x1)  =  pA_out_a(x1)
fB_out_a(x1)  =  fB_out_a
U2_a(x1)  =  U2_a(x1)

(5) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

FB_IN_A(T9) → U1_A(T9, pA_in_a(X7))
FB_IN_A(T9) → PA_IN_A(X7)
FB_IN_A(b) → U2_A(pA_in_a(T10))
FB_IN_A(b) → PA_IN_A(T10)

The TRS R consists of the following rules:

fB_in_a(T9) → U1_a(T9, pA_in_a(X7))
pA_in_a(a) → pA_out_a(a)
U1_a(T9, pA_out_a(X7)) → fB_out_a(T9)
fB_in_a(b) → U2_a(pA_in_a(T10))
U2_a(pA_out_a(T10)) → fB_out_a(b)

The argument filtering Pi contains the following mapping:
fB_in_a(x1)  =  fB_in_a
U1_a(x1, x2)  =  U1_a(x2)
pA_in_a(x1)  =  pA_in_a
pA_out_a(x1)  =  pA_out_a(x1)
fB_out_a(x1)  =  fB_out_a
U2_a(x1)  =  U2_a(x1)
FB_IN_A(x1)  =  FB_IN_A
U1_A(x1, x2)  =  U1_A(x2)
PA_IN_A(x1)  =  PA_IN_A
U2_A(x1)  =  U2_A(x1)

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

(6) Obligation:

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

FB_IN_A(T9) → U1_A(T9, pA_in_a(X7))
FB_IN_A(T9) → PA_IN_A(X7)
FB_IN_A(b) → U2_A(pA_in_a(T10))
FB_IN_A(b) → PA_IN_A(T10)

The TRS R consists of the following rules:

fB_in_a(T9) → U1_a(T9, pA_in_a(X7))
pA_in_a(a) → pA_out_a(a)
U1_a(T9, pA_out_a(X7)) → fB_out_a(T9)
fB_in_a(b) → U2_a(pA_in_a(T10))
U2_a(pA_out_a(T10)) → fB_out_a(b)

The argument filtering Pi contains the following mapping:
fB_in_a(x1)  =  fB_in_a
U1_a(x1, x2)  =  U1_a(x2)
pA_in_a(x1)  =  pA_in_a
pA_out_a(x1)  =  pA_out_a(x1)
fB_out_a(x1)  =  fB_out_a
U2_a(x1)  =  U2_a(x1)
FB_IN_A(x1)  =  FB_IN_A
U1_A(x1, x2)  =  U1_A(x2)
PA_IN_A(x1)  =  PA_IN_A
U2_A(x1)  =  U2_A(x1)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 0 SCCs with 4 less nodes.

(8) TRUE