Left Termination of the query pattern
minus(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).

Query: minus(g,g,a)

(1) BuiltinConflictTransformerProof (EQUIVALENT transformation)

Renamed defined predicates conflicting with built-in predicates [PROLOG].

(2) Obligation:

Clauses:

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

Query: minus(g,g,a)

(3) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
2 [label="A: minus(T1, T2, T3)\n\nT1 is ground\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
6 [label="6: minus(T1, T2, T3), SCOPE: 1, CLAUSE: 0\n\nT1 is ground\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: ','(f(T7, 0), ','(!_1, user_defined_=(T10, 0)))\n\nT7 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(f(T7, 0), ','(!_1, user_defined_=(T10, 0))), SCOPE: 2, CLAUSE: 1 | ','(f(T7, 0), ','(!_1, user_defined_=(T10, 0))), SCOPE: 2, CLAUSE: 2\n\nT7 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: ','(','(!_2, user_defined_=(T13, 0)), ','(!_1, user_defined_=(T10, 0))) | ','(f(T13, 0), ','(!_1, user_defined_=(T10, 0))), SCOPE: 2, CLAUSE: 2\n\nT13 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: ','(user_defined_=(T13, 0), ','(!_1, user_defined_=(T10, 0)))\n\nT13 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: ','(user_defined_=(T13, 0), ','(!_1, user_defined_=(T10, 0))), SCOPE: 3, CLAUSE: 3\n\nT13 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: ','(!_1, user_defined_=(T10, 0))\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: user_defined_=(T10, 0)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
30 [label="30: user_defined_=(T10, 0), SCOPE: 4, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
32 [label="32: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
33 [label="33: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
34 [label="34: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
35 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
2 -> 35 [arrowhead = none ];
35 -> 6;

36 [label="ONLY EVAL with clause\nminus(X4, X5, X6) :- ','(f(X4, 0), ','(!_1, user_defined_=(X6, 0))).\nand substitutionT1 -> T7,\nX4 -> T7,\nT2 -> T8,\nX5 -> T8,\nT3 -> T10,\nX6 -> T10,\nT9 -> T10", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 36 [arrowhead = none ];
36 -> 8;

37 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
8 -> 37 [arrowhead = none ];
37 -> 11;

38 [label="ONLY EVAL with clause\nf(X11, X12) :- ','(!_2, user_defined_=(X11, X12)).\nand substitutionT7 -> T13,\nX11 -> T13,\nX12 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 38 [arrowhead = none ];
38 -> 17;

39 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 39 [arrowhead = none ];
39 -> 18;

40 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
18 -> 40 [arrowhead = none ];
40 -> 20;

41 [label="EVAL with clause\nuser_defined_=(X15, X15).\nand substitutionT13 -> 0,\nX15 -> 0,\nT16 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
20 -> 41 [arrowhead = none ];
41 -> 24;

42 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
20 -> 42 [arrowhead = none ];
42 -> 25;

43 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
24 -> 43 [arrowhead = none ];
43 -> 28;

44 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
28 -> 44 [arrowhead = none ];
44 -> 30;

45 [label="EVAL with clause\nuser_defined_=(X18, X18).\nand substitutionT10 -> 0,\nX18 -> 0,\nT19 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
30 -> 45 [arrowhead = none ];
45 -> 32;

46 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
30 -> 46 [arrowhead = none ];
46 -> 33;

47 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
32 -> 47 [arrowhead = none ];
47 -> 34;

}

(4) Obligation:

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

minusA_in_gga(0, T8, 0) → minusA_out_gga(0, T8, 0)

The argument filtering Pi contains the following mapping:
minusA_in_gga(x1, x2, x3)  =  minusA_in_gga(x1, x2)
0  =  0
minusA_out_gga(x1, x2, x3)  =  minusA_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:

minusA_in_gga(0, T8, 0) → minusA_out_gga(0, T8, 0)

The argument filtering Pi contains the following mapping:
minusA_in_gga(x1, x2, x3)  =  minusA_in_gga(x1, x2)
0  =  0
minusA_out_gga(x1, x2, x3)  =  minusA_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:

minusA_in_gga(0, T8, 0) → minusA_out_gga(0, T8, 0)

The argument filtering Pi contains the following mapping:
minusA_in_gga(x1, x2, x3)  =  minusA_in_gga(x1, x2)
0  =  0
minusA_out_gga(x1, x2, x3)  =  minusA_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) YES