Left Termination of the query pattern f_in_1(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(a), !).
q(b).

Queries:

f(a).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

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(T3), q(T3))\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: p(T3)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: q(T4)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: p(T3), SCOPE: 2, CLAUSE: 1 | p(T3), SCOPE: 2, CLAUSE: 2 | ?_2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: p(T3), SCOPE: 2, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: p(T3), SCOPE: 2, CLAUSE: 2 | ?_2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(p(a), !_2) | ?_2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(p(a), !_2), SCOPE: 3, CLAUSE: 1 | ','(p(a), !_2), SCOPE: 3, CLAUSE: 2 | ?_3 | ?_2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: !_2 | ','(p(a), !_2), SCOPE: 3, CLAUSE: 2 | ?_3 | ?_2\n\n", 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(T4), SCOPE: 4, 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="EVAL with clause\nf(X2) :- ','(p(X2), q(X2)).\nand substitution\nT1 -> T3\nX2 -> T3\nT2 -> T3", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 22 [arrowhead = none ];
22 -> 3;

23 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
3 -> 23 [arrowhead = none ];
23 -> 4;

24 [label="SPLIT 2\nreplacements:\nT3 -> T4", fontsize=14, style = filled, fillcolor = violet];
3 -> 24 [arrowhead = none ];
24 -> 5;

25 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
4 -> 25 [arrowhead = none ];
25 -> 6;

26 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
5 -> 26 [arrowhead = none ];
26 -> 17;

27 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
6 -> 27 [arrowhead = none ];
27 -> 7;

28 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
6 -> 28 [arrowhead = none ];
28 -> 8;

29 [label="EVAL with clause\np(a).\nand substitution\nT3 -> a", fontsize=14, style = filled, fillcolor = lightblue];
7 -> 29 [arrowhead = none ];
29 -> 9;

30 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
7 -> 30 [arrowhead = none ];
30 -> 10;

31 [label="EVAL with clause\np(X4) :- ','(p(a), !).\nand substitution\nT3 -> T5\nX4 -> T5", fontsize=14, style = filled, fillcolor = lightblue];
8 -> 31 [arrowhead = none ];
31 -> 12;

32 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 32 [arrowhead = none ];
32 -> 11;

33 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
12 -> 33 [arrowhead = none ];
33 -> 13;

34 [label="EVAL with clause\np(a).\nand substitution", 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 substitution\nT4 -> b", fontsize=14, style = filled, fillcolor = lightblue];
17 -> 37 [arrowhead = none ];
37 -> 18;

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

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

}

(2) Obligation:

Clauses:

p4(a).
p4(T5).
f1(T3) :- p4(T3).
f1(b) :- p4(b).

Queries:

f1(a).

(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:
f1_in: (f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

f1_in_a(T3) → U1_a(T3, p4_in_a(T3))
p4_in_a(a) → p4_out_a(a)
p4_in_a(T5) → p4_out_a(T5)
U1_a(T3, p4_out_a(T3)) → f1_out_a(T3)
f1_in_a(b) → U2_a(p4_in_g(b))
p4_in_g(a) → p4_out_g(a)
p4_in_g(T5) → p4_out_g(T5)
U2_a(p4_out_g(b)) → f1_out_a(b)

The argument filtering Pi contains the following mapping:
f1_in_a(x1)  =  f1_in_a
U1_a(x1, x2)  =  U1_a(x2)
p4_in_a(x1)  =  p4_in_a
p4_out_a(x1)  =  p4_out_a
f1_out_a(x1)  =  f1_out_a
U2_a(x1)  =  U2_a(x1)
p4_in_g(x1)  =  p4_in_g(x1)
a  =  a
p4_out_g(x1)  =  p4_out_g
b  =  b

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

f1_in_a(T3) → U1_a(T3, p4_in_a(T3))
p4_in_a(a) → p4_out_a(a)
p4_in_a(T5) → p4_out_a(T5)
U1_a(T3, p4_out_a(T3)) → f1_out_a(T3)
f1_in_a(b) → U2_a(p4_in_g(b))
p4_in_g(a) → p4_out_g(a)
p4_in_g(T5) → p4_out_g(T5)
U2_a(p4_out_g(b)) → f1_out_a(b)

The argument filtering Pi contains the following mapping:
f1_in_a(x1)  =  f1_in_a
U1_a(x1, x2)  =  U1_a(x2)
p4_in_a(x1)  =  p4_in_a
p4_out_a(x1)  =  p4_out_a
f1_out_a(x1)  =  f1_out_a
U2_a(x1)  =  U2_a(x1)
p4_in_g(x1)  =  p4_in_g(x1)
a  =  a
p4_out_g(x1)  =  p4_out_g
b  =  b

(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:

F1_IN_A(T3) → U1_A(T3, p4_in_a(T3))
F1_IN_A(T3) → P4_IN_A(T3)
F1_IN_A(b) → U2_A(p4_in_g(b))
F1_IN_A(b) → P4_IN_G(b)

The TRS R consists of the following rules:

f1_in_a(T3) → U1_a(T3, p4_in_a(T3))
p4_in_a(a) → p4_out_a(a)
p4_in_a(T5) → p4_out_a(T5)
U1_a(T3, p4_out_a(T3)) → f1_out_a(T3)
f1_in_a(b) → U2_a(p4_in_g(b))
p4_in_g(a) → p4_out_g(a)
p4_in_g(T5) → p4_out_g(T5)
U2_a(p4_out_g(b)) → f1_out_a(b)

The argument filtering Pi contains the following mapping:
f1_in_a(x1)  =  f1_in_a
U1_a(x1, x2)  =  U1_a(x2)
p4_in_a(x1)  =  p4_in_a
p4_out_a(x1)  =  p4_out_a
f1_out_a(x1)  =  f1_out_a
U2_a(x1)  =  U2_a(x1)
p4_in_g(x1)  =  p4_in_g(x1)
a  =  a
p4_out_g(x1)  =  p4_out_g
b  =  b
F1_IN_A(x1)  =  F1_IN_A
U1_A(x1, x2)  =  U1_A(x2)
P4_IN_A(x1)  =  P4_IN_A
U2_A(x1)  =  U2_A(x1)
P4_IN_G(x1)  =  P4_IN_G(x1)

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

(6) Obligation:

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

F1_IN_A(T3) → U1_A(T3, p4_in_a(T3))
F1_IN_A(T3) → P4_IN_A(T3)
F1_IN_A(b) → U2_A(p4_in_g(b))
F1_IN_A(b) → P4_IN_G(b)

The TRS R consists of the following rules:

f1_in_a(T3) → U1_a(T3, p4_in_a(T3))
p4_in_a(a) → p4_out_a(a)
p4_in_a(T5) → p4_out_a(T5)
U1_a(T3, p4_out_a(T3)) → f1_out_a(T3)
f1_in_a(b) → U2_a(p4_in_g(b))
p4_in_g(a) → p4_out_g(a)
p4_in_g(T5) → p4_out_g(T5)
U2_a(p4_out_g(b)) → f1_out_a(b)

The argument filtering Pi contains the following mapping:
f1_in_a(x1)  =  f1_in_a
U1_a(x1, x2)  =  U1_a(x2)
p4_in_a(x1)  =  p4_in_a
p4_out_a(x1)  =  p4_out_a
f1_out_a(x1)  =  f1_out_a
U2_a(x1)  =  U2_a(x1)
p4_in_g(x1)  =  p4_in_g(x1)
a  =  a
p4_out_g(x1)  =  p4_out_g
b  =  b
F1_IN_A(x1)  =  F1_IN_A
U1_A(x1, x2)  =  U1_A(x2)
P4_IN_A(x1)  =  P4_IN_A
U2_A(x1)  =  U2_A(x1)
P4_IN_G(x1)  =  P4_IN_G(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