Left Termination of the query pattern p_in_3(a, a, a) w.r.t. the given Prolog program could successfully be proven:

(0) Obligation:

Clauses:

p(X, X, 1) :- !.
p(X, Y, Z) :- ','(=(Z, 1), ','(=(Y, X), p(X, Y, Z))).
=(X, X).

Queries:

p(a,a,a).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: p(T1, T2, T3)\n\n", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: p(T1, T2, T3), SCOPE: 1, CLAUSE: 0 | p(T1, T2, T3), SCOPE: 1, CLAUSE: 1 | ?_1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: !_1 | p(T1, T2, T3), SCOPE: 1, CLAUSE: 1 | ?_1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: p(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nX2 is free\np(T1, T2, T3) !=? p(X2, X2, 1)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: ','(=(T8, 1), ','(=(T9, T10), p(T10, T9, T8)))\n\nX2 is free", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: ','(=(T8, 1), ','(=(T9, T10), p(T10, T9, T8))), SCOPE: 2, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(=(T12, T13), p(T13, T12, 1))\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: \n\nX10 is free", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(=(T12, T13), p(T13, T12, 1)), SCOPE: 3, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: p(T15, T15, 1)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: \n\nX12 is free", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: p(T15, T15, 1), SCOPE: 4, CLAUSE: 0 | p(T15, T15, 1), SCOPE: 4, CLAUSE: 1 | ?_4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: !_4 | p(T15, T15, 1), SCOPE: 4, CLAUSE: 1 | ?_4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 18 [arrowhead = none ];
18 -> 2;

19 [label="EVAL with clause\np(X2, X2, 1) :- !.\nand substitution\nT1 -> T4\nX2 -> T4\nT2 -> T4\nT3 -> 1", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 19 [arrowhead = none ];
19 -> 3;

20 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 20 [arrowhead = none ];
20 -> 4;

21 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 21 [arrowhead = none ];
21 -> 5;

22 [label="EVAL with clause\np(X6, X7, X8) :- ','(=(X8, 1), ','(=(X7, X6), p(X6, X7, X8))).\nand substitution\nT1 -> T10\nX6 -> T10\nT2 -> T9\nX7 -> T9\nT3 -> T8\nX8 -> T8\nT7 -> T8\nT6 -> T9\nT5 -> T10", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 22 [arrowhead = none ];
22 -> 7;

23 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 23 [arrowhead = none ];
23 -> 6;

24 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 24 [arrowhead = none ];
24 -> 8;

25 [label="EVAL with clause\n=(X10, X10).\nand substitution\nT8 -> 1\nX10 -> 1\nT11 -> 1\nT9 -> T12\nT10 -> T13", fontsize=14, style = filled, fillcolor = lightblue];
8 -> 25 [arrowhead = none ];
25 -> 9;

26 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
8 -> 26 [arrowhead = none ];
26 -> 10;

27 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
9 -> 27 [arrowhead = none ];
27 -> 11;

28 [label="EVAL with clause\n=(X12, X12).\nand substitution\nT12 -> T15\nX12 -> T15\nT13 -> T15\nT14 -> T15", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 28 [arrowhead = none ];
28 -> 12;

29 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 29 [arrowhead = none ];
29 -> 13;

30 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
12 -> 30 [arrowhead = none ];
30 -> 14;

31 [label="EVAL with clause\np(X14, X14, 1) :- !.\nand substitution\nT15 -> T16\nX14 -> T16", fontsize=14, style = filled, fillcolor = lightblue];
14 -> 31 [arrowhead = none ];
31 -> 15;

32 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 32 [arrowhead = none ];
32 -> 16;

33 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
16 -> 33 [arrowhead = none ];
33 -> 17;

}

(2) Obligation:

Clauses:

p1(T4, T4, 1).
p1(T16, T16, 1).

Queries:

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

p1_in_aaa(T4, T4, 1) → p1_out_aaa(T4, T4, 1)

The argument filtering Pi contains the following mapping:
p1_in_aaa(x1, x2, x3)  =  p1_in_aaa
p1_out_aaa(x1, x2, x3)  =  p1_out_aaa(x3)

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_aaa(T4, T4, 1) → p1_out_aaa(T4, T4, 1)

The argument filtering Pi contains the following mapping:
p1_in_aaa(x1, x2, x3)  =  p1_in_aaa
p1_out_aaa(x1, x2, x3)  =  p1_out_aaa(x3)

(5) DependencyPairsProof (EQUIVALENT transformation)

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

p1_in_aaa(T4, T4, 1) → p1_out_aaa(T4, T4, 1)

The argument filtering Pi contains the following mapping:
p1_in_aaa(x1, x2, x3)  =  p1_in_aaa
p1_out_aaa(x1, x2, x3)  =  p1_out_aaa(x3)

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

(6) Obligation:

Pi DP problem:
P is empty.
The TRS R consists of the following rules:

p1_in_aaa(T4, T4, 1) → p1_out_aaa(T4, T4, 1)

The argument filtering Pi contains the following mapping:
p1_in_aaa(x1, x2, x3)  =  p1_in_aaa
p1_out_aaa(x1, x2, x3)  =  p1_out_aaa(x3)

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

(7) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,R,Pi) chain.

(8) TRUE