Left Termination of the query pattern minus_in_3(g, g, a) w.r.t. the given Prolog program could successfully be proven:

(0) Obligation:

Clauses:

minus(X, Y, Z) :- ','(f(X, 0), ','(!, =(Z, 0))).
f(X, Y) :- ','(!, =(X, Y)).
f(X, Y) :- f(X, Y).
=(X, X).

Queries:

minus(g,g,a).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: minus(T1, T2, T3)\n\nT1 is ground\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: minus(T1, T2, T3), SCOPE: 1, CLAUSE: 0 | ?_1\n\nT1 is ground\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: ','(f(T4, 0), ','(!_1, =(T7, 0))) | ?_1\n\nT4 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: ','(f(T4, 0), ','(!_1, =(T7, 0))), SCOPE: 2, CLAUSE: 1 | ','(f(T4, 0), ','(!_1, =(T7, 0))), SCOPE: 2, CLAUSE: 2 | ?_2 | ?_1\n\nT4 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: ','(!_2, ','(=(T8, 0), ','(!_1, =(T7, 0)))) | ','(f(T8, 0), ','(!_1, =(T7, 0))), SCOPE: 2, CLAUSE: 2 | ?_2 | ?_1\n\nT8 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: ','(=(T8, 0), ','(!_1, =(T7, 0))) | ?_1\n\nT8 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: ','(=(T8, 0), ','(!_1, =(T7, 0))), SCOPE: 3, CLAUSE: 3 | ?_1\n\nT8 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: ','(!_1, =(T7, 0)) | ?_1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: \n\nX12 is free", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: =(T7, 0)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: =(T7, 0), SCOPE: 4, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: \n\nX14 is free", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 15 [arrowhead = none ];
15 -> 2;

16 [label="EVAL with clause\nminus(X4, X5, X6) :- ','(f(X4, 0), ','(!, =(X6, 0))).\nand substitution\nT1 -> T4\nX4 -> T4\nT2 -> T5\nX5 -> T5\nT3 -> T7\nX6 -> T7\nT6 -> T7", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 16 [arrowhead = none ];
16 -> 3;

17 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 17 [arrowhead = none ];
17 -> 4;

18 [label="EVAL with clause\nf(X9, X10) :- ','(!, =(X9, X10)).\nand substitution\nT4 -> T8\nX9 -> T8\nX10 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 18 [arrowhead = none ];
18 -> 5;

19 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 19 [arrowhead = none ];
19 -> 6;

20 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
6 -> 20 [arrowhead = none ];
20 -> 7;

21 [label="EVAL with clause\n=(X12, X12).\nand substitution\nT8 -> 0\nX12 -> 0\nT9 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
7 -> 21 [arrowhead = none ];
21 -> 8;

22 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
7 -> 22 [arrowhead = none ];
22 -> 9;

23 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
8 -> 23 [arrowhead = none ];
23 -> 10;

24 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
10 -> 24 [arrowhead = none ];
24 -> 11;

25 [label="EVAL with clause\n=(X14, X14).\nand substitution\nT7 -> 0\nX14 -> 0\nT10 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 25 [arrowhead = none ];
25 -> 12;

26 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 26 [arrowhead = none ];
26 -> 13;

27 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 27 [arrowhead = none ];
27 -> 14;

}

(2) Obligation:

Clauses:

minus1(0, T5, 0).

Queries:

minus1(g,g,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:

minus1_in_gga(0, T5, 0) → minus1_out_gga(0, T5, 0)

The argument filtering Pi contains the following mapping:
minus1_in_gga(x1, x2, x3)  =  minus1_in_gga(x1, x2)
0  =  0
minus1_out_gga(x1, x2, x3)  =  minus1_out_gga(x1, x2, 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:

minus1_in_gga(0, T5, 0) → minus1_out_gga(0, T5, 0)

The argument filtering Pi contains the following mapping:
minus1_in_gga(x1, x2, x3)  =  minus1_in_gga(x1, x2)
0  =  0
minus1_out_gga(x1, x2, x3)  =  minus1_out_gga(x1, x2, 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:

minus1_in_gga(0, T5, 0) → minus1_out_gga(0, T5, 0)

The argument filtering Pi contains the following mapping:
minus1_in_gga(x1, x2, x3)  =  minus1_in_gga(x1, x2)
0  =  0
minus1_out_gga(x1, x2, x3)  =  minus1_out_gga(x1, x2, 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:

minus1_in_gga(0, T5, 0) → minus1_out_gga(0, T5, 0)

The argument filtering Pi contains the following mapping:
minus1_in_gga(x1, x2, x3)  =  minus1_in_gga(x1, x2)
0  =  0
minus1_out_gga(x1, x2, x3)  =  minus1_out_gga(x1, x2, 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