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

(0) Obligation:

Clauses:

plus(0, Y, Y).
plus(s(X), Y, Z) :- plus(X, s(Y), Z).

Query: plus(g,a,a)

(1) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: plus(T1, T2, T3)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
5 [label="5: plus(T1, T2, T3), SCOPE: 1, CLAUSE: 0 | plus(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: true | plus(0, T2, T3), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: plus(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nX2 is free\nplus(T1, T2, T3) !=? plus(0, X2, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: plus(0, T2, T3), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: plus(T9, s(T13), T12)\n\nT9 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: plus(T9, s(T13), T12), SCOPE: 2, CLAUSE: 0 | plus(T9, s(T13), T12), SCOPE: 2, CLAUSE: 1\n\nT9 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
30 [label="30: plus(T9, s(T13), T12), SCOPE: 2, CLAUSE: 0\n\nT9 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
31 [label="31: plus(T9, s(T13), T12), SCOPE: 2, CLAUSE: 1\n\nT9 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
35 [label="35: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
36 [label="36: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
37 [label="37: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
40 [label="40: plus(T25, s(s(T29)), T28)\n\nT25 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
41 [label="41: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
42 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 42 [arrowhead = none ];
42 -> 5;

43 [label="EVAL with clause\nplus(0, X2, X2).\nand substitutionT1 -> 0,\nT2 -> T5,\nX2 -> T5,\nT3 -> T5", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 43 [arrowhead = none ];
43 -> 9;

44 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 44 [arrowhead = none ];
44 -> 11;

45 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 45 [arrowhead = none ];
45 -> 17;

46 [label="EVAL with clause\nplus(s(X9), X10, X11) :- plus(X9, s(X10), X11).\nand substitutionX9 -> T9,\nT1 -> s(T9),\nT2 -> T13,\nX10 -> T13,\nT3 -> T12,\nX11 -> T12,\nT11 -> T12,\nT10 -> T13", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 46 [arrowhead = none ];
46 -> 22;

47 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 47 [arrowhead = none ];
47 -> 26;

48 [label="BACKTRACK\nfor clause: plus(s(X), Y, Z) :- plus(X, s(Y), Z)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 48 [arrowhead = none ];
48 -> 18;

49 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
22 -> 49 [arrowhead = none ];
49 -> 27;

50 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
27 -> 50 [arrowhead = none ];
50 -> 30;

51 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
27 -> 51 [arrowhead = none ];
51 -> 31;

52 [label="EVAL with clause\nplus(0, X16, X16).\nand substitutionT9 -> 0,\nT13 -> T18,\nX16 -> s(T18),\nT12 -> s(T18)", fontsize=14, style = filled, fillcolor = lightblue];
30 -> 52 [arrowhead = none ];
52 -> 35;

53 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
30 -> 53 [arrowhead = none ];
53 -> 36;

54 [label="EVAL with clause\nplus(s(X23), X24, X25) :- plus(X23, s(X24), X25).\nand substitutionX23 -> T25,\nT9 -> s(T25),\nT13 -> T29,\nX24 -> s(T29),\nT12 -> T28,\nX25 -> T28,\nT27 -> T28,\nT26 -> T29", fontsize=14, style = filled, fillcolor = lightblue];
31 -> 54 [arrowhead = none ];
54 -> 40;

55 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
31 -> 55 [arrowhead = none ];
55 -> 41;

56 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
35 -> 56 [arrowhead = none ];
56 -> 37;

57 [label="INSTANCE with matching:\nT1 -> T25\nT2 -> s(s(T29))\nT3 -> T28", fontsize=14, style = filled, fillcolor = lightgrey];
40 -> 57 [arrowhead = none , style=dashed];
57 -> 1[style=dashed];

}

(2) Obligation:

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

plusA_in_gaa(0, T5, T5) → plusA_out_gaa(0, T5, T5)
plusA_in_gaa(s(0), T18, s(T18)) → plusA_out_gaa(s(0), T18, s(T18))
plusA_in_gaa(s(s(T25)), T29, T28) → U1_gaa(T25, T29, T28, plusA_in_gaa(T25, s(s(T29)), T28))
U1_gaa(T25, T29, T28, plusA_out_gaa(T25, s(s(T29)), T28)) → plusA_out_gaa(s(s(T25)), T29, T28)

The argument filtering Pi contains the following mapping:
plusA_in_gaa(x1, x2, x3)  =  plusA_in_gaa(x1)
0  =  0
plusA_out_gaa(x1, x2, x3)  =  plusA_out_gaa(x1)
s(x1)  =  s(x1)
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x1, x4)

(3) 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:

PLUSA_IN_GAA(s(s(T25)), T29, T28) → U1_GAA(T25, T29, T28, plusA_in_gaa(T25, s(s(T29)), T28))
PLUSA_IN_GAA(s(s(T25)), T29, T28) → PLUSA_IN_GAA(T25, s(s(T29)), T28)

The TRS R consists of the following rules:

plusA_in_gaa(0, T5, T5) → plusA_out_gaa(0, T5, T5)
plusA_in_gaa(s(0), T18, s(T18)) → plusA_out_gaa(s(0), T18, s(T18))
plusA_in_gaa(s(s(T25)), T29, T28) → U1_gaa(T25, T29, T28, plusA_in_gaa(T25, s(s(T29)), T28))
U1_gaa(T25, T29, T28, plusA_out_gaa(T25, s(s(T29)), T28)) → plusA_out_gaa(s(s(T25)), T29, T28)

The argument filtering Pi contains the following mapping:
plusA_in_gaa(x1, x2, x3)  =  plusA_in_gaa(x1)
0  =  0
plusA_out_gaa(x1, x2, x3)  =  plusA_out_gaa(x1)
s(x1)  =  s(x1)
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x1, x4)
PLUSA_IN_GAA(x1, x2, x3)  =  PLUSA_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4)  =  U1_GAA(x1, x4)

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

(4) Obligation:

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

PLUSA_IN_GAA(s(s(T25)), T29, T28) → U1_GAA(T25, T29, T28, plusA_in_gaa(T25, s(s(T29)), T28))
PLUSA_IN_GAA(s(s(T25)), T29, T28) → PLUSA_IN_GAA(T25, s(s(T29)), T28)

The TRS R consists of the following rules:

plusA_in_gaa(0, T5, T5) → plusA_out_gaa(0, T5, T5)
plusA_in_gaa(s(0), T18, s(T18)) → plusA_out_gaa(s(0), T18, s(T18))
plusA_in_gaa(s(s(T25)), T29, T28) → U1_gaa(T25, T29, T28, plusA_in_gaa(T25, s(s(T29)), T28))
U1_gaa(T25, T29, T28, plusA_out_gaa(T25, s(s(T29)), T28)) → plusA_out_gaa(s(s(T25)), T29, T28)

The argument filtering Pi contains the following mapping:
plusA_in_gaa(x1, x2, x3)  =  plusA_in_gaa(x1)
0  =  0
plusA_out_gaa(x1, x2, x3)  =  plusA_out_gaa(x1)
s(x1)  =  s(x1)
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x1, x4)
PLUSA_IN_GAA(x1, x2, x3)  =  PLUSA_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4)  =  U1_GAA(x1, x4)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.

(6) Obligation:

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

PLUSA_IN_GAA(s(s(T25)), T29, T28) → PLUSA_IN_GAA(T25, s(s(T29)), T28)

The TRS R consists of the following rules:

plusA_in_gaa(0, T5, T5) → plusA_out_gaa(0, T5, T5)
plusA_in_gaa(s(0), T18, s(T18)) → plusA_out_gaa(s(0), T18, s(T18))
plusA_in_gaa(s(s(T25)), T29, T28) → U1_gaa(T25, T29, T28, plusA_in_gaa(T25, s(s(T29)), T28))
U1_gaa(T25, T29, T28, plusA_out_gaa(T25, s(s(T29)), T28)) → plusA_out_gaa(s(s(T25)), T29, T28)

The argument filtering Pi contains the following mapping:
plusA_in_gaa(x1, x2, x3)  =  plusA_in_gaa(x1)
0  =  0
plusA_out_gaa(x1, x2, x3)  =  plusA_out_gaa(x1)
s(x1)  =  s(x1)
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x1, x4)
PLUSA_IN_GAA(x1, x2, x3)  =  PLUSA_IN_GAA(x1)

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

(7) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(8) Obligation:

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

PLUSA_IN_GAA(s(s(T25)), T29, T28) → PLUSA_IN_GAA(T25, s(s(T29)), T28)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
PLUSA_IN_GAA(x1, x2, x3)  =  PLUSA_IN_GAA(x1)

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

(9) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(10) Obligation:

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

PLUSA_IN_GAA(s(s(T25))) → PLUSA_IN_GAA(T25)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(11) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• PLUSA_IN_GAA(s(s(T25))) → PLUSA_IN_GAA(T25)
  The graph contains the following edges 1 > 1

(12) YES