Left Termination of the query pattern p_in_1(a) w.r.t. the given Prolog program could successfully be proven:

(0) Obligation:

Clauses:

p(X) :- ','(q(X), r(X)).
p(X) :- =(X, 2).
q(X) :- ','(=(X, 1), !).
q(X) :- =(X, 3).
r(X) :- ','(=(X, 1), !).
r(X) :- =(X, 4).
=(X, X).

Queries:

p(a).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: p(T1)\n\n", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: p(T1), SCOPE: 1, CLAUSE: 0 | p(T1), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: p(T1), SCOPE: 1, CLAUSE: 0\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: p(T1), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: ','(q(T3), r(T3))\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: q(T3)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: r(T4)\n\nT4 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: q(T3), SCOPE: 2, CLAUSE: 2 | q(T3), SCOPE: 2, CLAUSE: 3 | ?_2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(=(T6, 1), !_2) | q(T3), SCOPE: 2, CLAUSE: 3 | ?_2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(=(T6, 1), !_2), SCOPE: 3, CLAUSE: 6 | q(T3), SCOPE: 2, CLAUSE: 3 | ?_2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: !_2 | q(T3), SCOPE: 2, CLAUSE: 3 | ?_2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: q(T3), SCOPE: 2, CLAUSE: 3\n\nX7 is free", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: =(T9, 3)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: =(T9, 3), SCOPE: 4, CLAUSE: 6\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: \n\nX11 is free", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: r(T4), SCOPE: 5, CLAUSE: 4 | r(T4), SCOPE: 5, CLAUSE: 5 | ?_5\n\nT4 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: ','(=(T11, 1), !_5) | r(T11), SCOPE: 5, CLAUSE: 5 | ?_5\n\nT11 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: ','(=(T11, 1), !_5), SCOPE: 6, CLAUSE: 6 | r(T11), SCOPE: 5, CLAUSE: 5 | ?_5\n\nT11 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: !_5 | r(1), SCOPE: 5, CLAUSE: 5 | ?_5\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: r(T11), SCOPE: 5, CLAUSE: 5\n\nT11 is ground\nX15 is free; \n=(T11, 1) !=? =(X15, X15)", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: =(T13, 4)\n\nT13 is ground\nX15 is free; \n=(T13, 1) !=? =(X15, X15)", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: =(T13, 4), SCOPE: 7, CLAUSE: 6\n\nT13 is ground\nX15 is free; \n=(T13, 1) !=? =(X15, X15)", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="29: true\n\nX15 is free", fontsize=16, style=filled, fillcolor=lightyellow];
30 [label="30: \n\nX15 is free\nX19 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
31 [label="31: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
32 [label="32: =(T16, 2)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
33 [label="33: =(T16, 2), SCOPE: 8, CLAUSE: 6\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
34 [label="34: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
35 [label="35: \n\nX23 is free", fontsize=16, style=filled, fillcolor=lightyellow];
36 [label="36: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
37 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 37 [arrowhead = none ];
37 -> 2;

38 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
2 -> 38 [arrowhead = none ];
38 -> 3;

39 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
2 -> 39 [arrowhead = none ];
39 -> 4;

40 [label="EVAL with clause\np(X3) :- ','(q(X3), r(X3)).\nand substitution\nT1 -> T3\nX3 -> T3\nT2 -> T3", fontsize=14, style = filled, fillcolor = lightblue];
3 -> 40 [arrowhead = none ];
40 -> 5;

41 [label="EVAL with clause\np(X21) :- =(X21, 2).\nand substitution\nT1 -> T16\nX21 -> T16\nT15 -> T16", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 41 [arrowhead = none ];
41 -> 32;

42 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
5 -> 42 [arrowhead = none ];
42 -> 6;

43 [label="SPLIT 2\nnew knowledge:\nT4 is ground\nreplacements:\nT3 -> T4", fontsize=14, style = filled, fillcolor = violet];
5 -> 43 [arrowhead = none ];
43 -> 7;

44 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
6 -> 44 [arrowhead = none ];
44 -> 8;

45 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 45 [arrowhead = none ];
45 -> 20;

46 [label="EVAL with clause\nq(X5) :- ','(=(X5, 1), !).\nand substitution\nT3 -> T6\nX5 -> T6\nT5 -> T6", fontsize=14, style = filled, fillcolor = lightblue];
8 -> 46 [arrowhead = none ];
46 -> 9;

47 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
9 -> 47 [arrowhead = none ];
47 -> 10;

48 [label="EVAL with clause\n=(X7, X7).\nand substitution\nT6 -> 1\nX7 -> 1\nT7 -> 1", fontsize=14, style = filled, fillcolor = lightblue];
10 -> 48 [arrowhead = none ];
48 -> 11;

49 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
10 -> 49 [arrowhead = none ];
49 -> 12;

50 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 50 [arrowhead = none ];
50 -> 13;

51 [label="EVAL with clause\nq(X9) :- =(X9, 3).\nand substitution\nT3 -> T9\nX9 -> T9\nT8 -> T9", fontsize=14, style = filled, fillcolor = lightblue];
12 -> 51 [arrowhead = none ];
51 -> 15;

52 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
13 -> 52 [arrowhead = none ];
52 -> 14;

53 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
15 -> 53 [arrowhead = none ];
53 -> 16;

54 [label="EVAL with clause\n=(X11, X11).\nand substitution\nT9 -> 3\nX11 -> 3\nT10 -> 3", fontsize=14, style = filled, fillcolor = lightblue];
16 -> 54 [arrowhead = none ];
54 -> 17;

55 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
16 -> 55 [arrowhead = none ];
55 -> 18;

56 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 56 [arrowhead = none ];
56 -> 19;

57 [label="EVAL with clause\nr(X13) :- ','(=(X13, 1), !).\nand substitution\nT4 -> T11\nX13 -> T11", fontsize=14, style = filled, fillcolor = lightblue];
20 -> 57 [arrowhead = none ];
57 -> 21;

58 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
21 -> 58 [arrowhead = none ];
58 -> 22;

59 [label="EVAL with clause\n=(X15, X15).\nand substitution\nT11 -> 1\nX15 -> 1\nT12 -> 1", fontsize=14, style = filled, fillcolor = lightblue];
22 -> 59 [arrowhead = none ];
59 -> 23;

60 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
22 -> 60 [arrowhead = none ];
60 -> 24;

61 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
23 -> 61 [arrowhead = none ];
61 -> 25;

62 [label="EVAL with clause\nr(X17) :- =(X17, 4).\nand substitution\nT11 -> T13\nX17 -> T13", fontsize=14, style = filled, fillcolor = lightblue];
24 -> 62 [arrowhead = none ];
62 -> 27;

63 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
25 -> 63 [arrowhead = none ];
63 -> 26;

64 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
27 -> 64 [arrowhead = none ];
64 -> 28;

65 [label="EVAL with clause\n=(X19, X19).\nand substitution\nT13 -> 4\nX19 -> 4\nT14 -> 4", fontsize=14, style = filled, fillcolor = lightblue];
28 -> 65 [arrowhead = none ];
65 -> 29;

66 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
28 -> 66 [arrowhead = none ];
66 -> 30;

67 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
29 -> 67 [arrowhead = none ];
67 -> 31;

68 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
32 -> 68 [arrowhead = none ];
68 -> 33;

69 [label="EVAL with clause\n=(X23, X23).\nand substitution\nT16 -> 2\nX23 -> 2\nT17 -> 2", fontsize=14, style = filled, fillcolor = lightblue];
33 -> 69 [arrowhead = none ];
69 -> 34;

70 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
33 -> 70 [arrowhead = none ];
70 -> 35;

71 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
34 -> 71 [arrowhead = none ];
71 -> 36;

}

(2) Obligation:

Clauses:

q6(1).
q6(3).
p1(T3) :- q6(T3).
p1(1) :- q6(1).
p1(4) :- q6(4).
p1(2).

Queries:

p1(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:
p1_in: (f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

p1_in_a(T3) → U1_a(T3, q6_in_a(T3))
q6_in_a(1) → q6_out_a(1)
q6_in_a(3) → q6_out_a(3)
U1_a(T3, q6_out_a(T3)) → p1_out_a(T3)
p1_in_a(1) → U2_a(q6_in_g(1))
q6_in_g(1) → q6_out_g(1)
q6_in_g(3) → q6_out_g(3)
U2_a(q6_out_g(1)) → p1_out_a(1)
p1_in_a(4) → U3_a(q6_in_g(4))
U3_a(q6_out_g(4)) → p1_out_a(4)
p1_in_a(2) → p1_out_a(2)

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
U1_a(x1, x2)  =  U1_a(x2)
q6_in_a(x1)  =  q6_in_a
q6_out_a(x1)  =  q6_out_a(x1)
p1_out_a(x1)  =  p1_out_a(x1)
U2_a(x1)  =  U2_a(x1)
q6_in_g(x1)  =  q6_in_g(x1)
1  =  1
q6_out_g(x1)  =  q6_out_g
3  =  3
U3_a(x1)  =  U3_a(x1)
4  =  4

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

p1_in_a(T3) → U1_a(T3, q6_in_a(T3))
q6_in_a(1) → q6_out_a(1)
q6_in_a(3) → q6_out_a(3)
U1_a(T3, q6_out_a(T3)) → p1_out_a(T3)
p1_in_a(1) → U2_a(q6_in_g(1))
q6_in_g(1) → q6_out_g(1)
q6_in_g(3) → q6_out_g(3)
U2_a(q6_out_g(1)) → p1_out_a(1)
p1_in_a(4) → U3_a(q6_in_g(4))
U3_a(q6_out_g(4)) → p1_out_a(4)
p1_in_a(2) → p1_out_a(2)

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
U1_a(x1, x2)  =  U1_a(x2)
q6_in_a(x1)  =  q6_in_a
q6_out_a(x1)  =  q6_out_a(x1)
p1_out_a(x1)  =  p1_out_a(x1)
U2_a(x1)  =  U2_a(x1)
q6_in_g(x1)  =  q6_in_g(x1)
1  =  1
q6_out_g(x1)  =  q6_out_g
3  =  3
U3_a(x1)  =  U3_a(x1)
4  =  4

(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:

P1_IN_A(T3) → U1_A(T3, q6_in_a(T3))
P1_IN_A(T3) → Q6_IN_A(T3)
P1_IN_A(1) → U2_A(q6_in_g(1))
P1_IN_A(1) → Q6_IN_G(1)
P1_IN_A(4) → U3_A(q6_in_g(4))
P1_IN_A(4) → Q6_IN_G(4)

The TRS R consists of the following rules:

p1_in_a(T3) → U1_a(T3, q6_in_a(T3))
q6_in_a(1) → q6_out_a(1)
q6_in_a(3) → q6_out_a(3)
U1_a(T3, q6_out_a(T3)) → p1_out_a(T3)
p1_in_a(1) → U2_a(q6_in_g(1))
q6_in_g(1) → q6_out_g(1)
q6_in_g(3) → q6_out_g(3)
U2_a(q6_out_g(1)) → p1_out_a(1)
p1_in_a(4) → U3_a(q6_in_g(4))
U3_a(q6_out_g(4)) → p1_out_a(4)
p1_in_a(2) → p1_out_a(2)

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
U1_a(x1, x2)  =  U1_a(x2)
q6_in_a(x1)  =  q6_in_a
q6_out_a(x1)  =  q6_out_a(x1)
p1_out_a(x1)  =  p1_out_a(x1)
U2_a(x1)  =  U2_a(x1)
q6_in_g(x1)  =  q6_in_g(x1)
1  =  1
q6_out_g(x1)  =  q6_out_g
3  =  3
U3_a(x1)  =  U3_a(x1)
4  =  4
P1_IN_A(x1)  =  P1_IN_A
U1_A(x1, x2)  =  U1_A(x2)
Q6_IN_A(x1)  =  Q6_IN_A
U2_A(x1)  =  U2_A(x1)
Q6_IN_G(x1)  =  Q6_IN_G(x1)
U3_A(x1)  =  U3_A(x1)

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

(6) Obligation:

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

P1_IN_A(T3) → U1_A(T3, q6_in_a(T3))
P1_IN_A(T3) → Q6_IN_A(T3)
P1_IN_A(1) → U2_A(q6_in_g(1))
P1_IN_A(1) → Q6_IN_G(1)
P1_IN_A(4) → U3_A(q6_in_g(4))
P1_IN_A(4) → Q6_IN_G(4)

The TRS R consists of the following rules:

p1_in_a(T3) → U1_a(T3, q6_in_a(T3))
q6_in_a(1) → q6_out_a(1)
q6_in_a(3) → q6_out_a(3)
U1_a(T3, q6_out_a(T3)) → p1_out_a(T3)
p1_in_a(1) → U2_a(q6_in_g(1))
q6_in_g(1) → q6_out_g(1)
q6_in_g(3) → q6_out_g(3)
U2_a(q6_out_g(1)) → p1_out_a(1)
p1_in_a(4) → U3_a(q6_in_g(4))
U3_a(q6_out_g(4)) → p1_out_a(4)
p1_in_a(2) → p1_out_a(2)

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
U1_a(x1, x2)  =  U1_a(x2)
q6_in_a(x1)  =  q6_in_a
q6_out_a(x1)  =  q6_out_a(x1)
p1_out_a(x1)  =  p1_out_a(x1)
U2_a(x1)  =  U2_a(x1)
q6_in_g(x1)  =  q6_in_g(x1)
1  =  1
q6_out_g(x1)  =  q6_out_g
3  =  3
U3_a(x1)  =  U3_a(x1)
4  =  4
P1_IN_A(x1)  =  P1_IN_A
U1_A(x1, x2)  =  U1_A(x2)
Q6_IN_A(x1)  =  Q6_IN_A
U2_A(x1)  =  U2_A(x1)
Q6_IN_G(x1)  =  Q6_IN_G(x1)
U3_A(x1)  =  U3_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 6 less nodes.

(8) TRUE