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

(0) Obligation:

Clauses:

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

Query: plus(g,a,a)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

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];
2 [label="2: plus(T1, T2, T3), SCOPE: 1, CLAUSE: 2 | plus(T1, T2, T3), SCOPE: 1, CLAUSE: 3\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: true | plus(0, T2, T3), SCOPE: 1, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: plus(T1, T2, T3), SCOPE: 1, CLAUSE: 3\n\nT1 is ground\nX2 is free\nplus(T1, T2, T3) !=? plus(0, X2, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: plus(0, T2, T3), SCOPE: 1, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: ','(p(s(T9), X12), plus(X12, T12, T13))\n\nT9 is ground\nX12 is free", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(p(s(T9), X12), plus(X12, T12, T13)), SCOPE: 2, CLAUSE: 0 | ','(p(s(T9), X12), plus(X12, T12, T13)), SCOPE: 2, CLAUSE: 1\n\nT9 is ground\nX12 is free", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(p(s(T9), X12), plus(X12, T12, T13)), SCOPE: 2, CLAUSE: 0\n\nT9 is ground\nX12 is free", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(p(s(T9), X12), plus(X12, T12, T13)), SCOPE: 2, CLAUSE: 1\n\nT9 is ground\nX12 is free", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: plus(0, T12, T13)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: plus(0, T12, T13), SCOPE: 3, CLAUSE: 2 | plus(0, T12, T13), SCOPE: 3, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: plus(0, T12, T13), SCOPE: 3, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: plus(0, T12, T13), SCOPE: 3, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: ','(p(s(T23), s(X31)), plus(s(s(X31)), T12, T13))\n\nT23 is ground\nX31 is free", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="B: p(s(T23), s(X31))\n\nT23 is ground\nX31 is free", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: plus(s(s(T24)), T12, T13)\n\nT24 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: p(s(T23), s(X31)), SCOPE: 4, CLAUSE: 0 | p(s(T23), s(X31)), SCOPE: 4, CLAUSE: 1\n\nT23 is ground\nX31 is free", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: p(s(T23), s(X31)), SCOPE: 4, CLAUSE: 1\n\nT23 is ground\nX31 is free", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: p(s(T27), s(X40))\n\nT27 is ground\nX40 is free", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 29 [arrowhead = none ];
29 -> 2;

30 [label="EVAL with clause\nplus(0, X2, X2).\nand substitutionT1 -> 0,\nT2 -> T5,\nX2 -> T5,\nT3 -> T5", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 30 [arrowhead = none ];
30 -> 3;

31 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 31 [arrowhead = none ];
31 -> 4;

32 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 32 [arrowhead = none ];
32 -> 5;

33 [label="EVAL with clause\nplus(s(X9), X10, s(X11)) :- ','(p(s(X9), X12), plus(X12, X10, X11)).\nand substitutionX9 -> T9,\nT1 -> s(T9),\nT2 -> T12,\nX10 -> T12,\nX11 -> T13,\nT3 -> s(T13),\nT10 -> T12,\nT11 -> T13", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 33 [arrowhead = none ];
33 -> 7;

34 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 34 [arrowhead = none ];
34 -> 8;

35 [label="BACKTRACK\nfor clause: plus(s(X), Y, s(Z)) :- ','(p(s(X), U), plus(U, Y, Z))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 35 [arrowhead = none ];
35 -> 6;

36 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 36 [arrowhead = none ];
36 -> 9;

37 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
9 -> 37 [arrowhead = none ];
37 -> 10;

38 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
9 -> 38 [arrowhead = none ];
38 -> 11;

39 [label="EVAL with clause\np(s(0), 0).\nand substitutionT9 -> 0,\nX12 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
10 -> 39 [arrowhead = none ];
39 -> 12;

40 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 40 [arrowhead = none ];
40 -> 13;

41 [label="EVAL with clause\np(s(s(X29)), s(s(X30))) :- p(s(X29), s(X30)).\nand substitutionX29 -> T23,\nT9 -> s(T23),\nX30 -> X31,\nX12 -> s(s(X31))", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 41 [arrowhead = none ];
41 -> 21;

42 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 42 [arrowhead = none ];
42 -> 22;

43 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
12 -> 43 [arrowhead = none ];
43 -> 14;

44 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
14 -> 44 [arrowhead = none ];
44 -> 15;

45 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
14 -> 45 [arrowhead = none ];
45 -> 16;

46 [label="EVAL with clause\nplus(0, X19, X19).\nand substitutionT12 -> T20,\nX19 -> T20,\nT13 -> T20", fontsize=14, style = filled, fillcolor = lightblue];
15 -> 46 [arrowhead = none ];
46 -> 17;

47 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 47 [arrowhead = none ];
47 -> 18;

48 [label="BACKTRACK\nfor clause: plus(s(X), Y, s(Z)) :- ','(p(s(X), U), plus(U, Y, Z))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
16 -> 48 [arrowhead = none ];
48 -> 20;

49 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 49 [arrowhead = none ];
49 -> 19;

50 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
21 -> 50 [arrowhead = none ];
50 -> 23;

51 [label="SPLIT 2\nnew knowledge:\nT23 is ground\nT24 is ground\nreplacements:X31 -> T24", fontsize=14, style = filled, fillcolor = violet];
21 -> 51 [arrowhead = none ];
51 -> 24;

52 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
23 -> 52 [arrowhead = none ];
52 -> 25;

53 [label="INSTANCE with matching:\nT1 -> s(s(T24))\nT2 -> T12\nT3 -> T13", fontsize=14, style = filled, fillcolor = lightgrey];
24 -> 53 [arrowhead = none , style=dashed];
53 -> 1[style=dashed];

54 [label="BACKTRACK\nfor clause: p(s(0), 0)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
25 -> 54 [arrowhead = none ];
54 -> 26;

55 [label="EVAL with clause\np(s(s(X38)), s(s(X39))) :- p(s(X38), s(X39)).\nand substitutionX38 -> T27,\nT23 -> s(T27),\nX39 -> X40,\nX31 -> s(X40)", fontsize=14, style = filled, fillcolor = lightblue];
26 -> 55 [arrowhead = none ];
55 -> 27;

56 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
26 -> 56 [arrowhead = none ];
56 -> 28;

57 [label="INSTANCE with matching:\nT23 -> T27\nX31 -> X40", fontsize=14, style = filled, fillcolor = lightgrey];
27 -> 57 [arrowhead = none , style=dashed];
57 -> 23[style=dashed];

}

(2) Obligation:

Triples:

pB(s(X1), s(X2)) :- pB(X1, X2).
plusA(s(s(X1)), X2, s(X3)) :- pB(X1, X4).
plusA(s(s(X1)), X2, s(X3)) :- ','(pcB(X1, X4), plusA(s(s(X4)), X2, X3)).

Clauses:

pluscA(0, X1, X1).
pluscA(s(0), X1, s(X1)).
pluscA(s(s(X1)), X2, s(X3)) :- ','(pcB(X1, X4), pluscA(s(s(X4)), X2, X3)).
pcB(s(X1), s(X2)) :- pcB(X1, X2).

Afs:

plusA(x1, x2, x3)  =  plusA(x1)

(3) TriplesToPiDPProof (SOUND transformation)

We use the technique of [DT09]. With regard to the inferred argument filtering the predicates were used in the following modes:
plusA_in: (b,f,f)
pB_in: (b,f)
pcB_in: (b,f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

PLUSA_IN_GAA(s(s(X1)), X2, s(X3)) → U2_GAA(X1, X2, X3, pB_in_ga(X1, X4))
PLUSA_IN_GAA(s(s(X1)), X2, s(X3)) → PB_IN_GA(X1, X4)
PB_IN_GA(s(X1), s(X2)) → U1_GA(X1, X2, pB_in_ga(X1, X2))
PB_IN_GA(s(X1), s(X2)) → PB_IN_GA(X1, X2)
PLUSA_IN_GAA(s(s(X1)), X2, s(X3)) → U3_GAA(X1, X2, X3, pcB_in_ga(X1, X4))
U3_GAA(X1, X2, X3, pcB_out_ga(X1, X4)) → U4_GAA(X1, X2, X3, plusA_in_gaa(s(s(X4)), X2, X3))
U3_GAA(X1, X2, X3, pcB_out_ga(X1, X4)) → PLUSA_IN_GAA(s(s(X4)), X2, X3)

The TRS R consists of the following rules:

pcB_in_ga(s(X1), s(X2)) → U8_ga(X1, X2, pcB_in_ga(X1, X2))
U8_ga(X1, X2, pcB_out_ga(X1, X2)) → pcB_out_ga(s(X1), s(X2))

The argument filtering Pi contains the following mapping:
plusA_in_gaa(x1, x2, x3)  =  plusA_in_gaa(x1)
s(x1)  =  s(x1)
pB_in_ga(x1, x2)  =  pB_in_ga(x1)
pcB_in_ga(x1, x2)  =  pcB_in_ga(x1)
U8_ga(x1, x2, x3)  =  U8_ga(x1, x3)
pcB_out_ga(x1, x2)  =  pcB_out_ga(x1, x2)
PLUSA_IN_GAA(x1, x2, x3)  =  PLUSA_IN_GAA(x1)
U2_GAA(x1, x2, x3, x4)  =  U2_GAA(x1, x4)
PB_IN_GA(x1, x2)  =  PB_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x1, x3)
U3_GAA(x1, x2, x3, x4)  =  U3_GAA(x1, x4)
U4_GAA(x1, x2, x3, x4)  =  U4_GAA(x1, x4)

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

Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES

(4) Obligation:

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

PLUSA_IN_GAA(s(s(X1)), X2, s(X3)) → U2_GAA(X1, X2, X3, pB_in_ga(X1, X4))
PLUSA_IN_GAA(s(s(X1)), X2, s(X3)) → PB_IN_GA(X1, X4)
PB_IN_GA(s(X1), s(X2)) → U1_GA(X1, X2, pB_in_ga(X1, X2))
PB_IN_GA(s(X1), s(X2)) → PB_IN_GA(X1, X2)
PLUSA_IN_GAA(s(s(X1)), X2, s(X3)) → U3_GAA(X1, X2, X3, pcB_in_ga(X1, X4))
U3_GAA(X1, X2, X3, pcB_out_ga(X1, X4)) → U4_GAA(X1, X2, X3, plusA_in_gaa(s(s(X4)), X2, X3))
U3_GAA(X1, X2, X3, pcB_out_ga(X1, X4)) → PLUSA_IN_GAA(s(s(X4)), X2, X3)

The TRS R consists of the following rules:

pcB_in_ga(s(X1), s(X2)) → U8_ga(X1, X2, pcB_in_ga(X1, X2))
U8_ga(X1, X2, pcB_out_ga(X1, X2)) → pcB_out_ga(s(X1), s(X2))

The argument filtering Pi contains the following mapping:
plusA_in_gaa(x1, x2, x3)  =  plusA_in_gaa(x1)
s(x1)  =  s(x1)
pB_in_ga(x1, x2)  =  pB_in_ga(x1)
pcB_in_ga(x1, x2)  =  pcB_in_ga(x1)
U8_ga(x1, x2, x3)  =  U8_ga(x1, x3)
pcB_out_ga(x1, x2)  =  pcB_out_ga(x1, x2)
PLUSA_IN_GAA(x1, x2, x3)  =  PLUSA_IN_GAA(x1)
U2_GAA(x1, x2, x3, x4)  =  U2_GAA(x1, x4)
PB_IN_GA(x1, x2)  =  PB_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x1, x3)
U3_GAA(x1, x2, x3, x4)  =  U3_GAA(x1, x4)
U4_GAA(x1, x2, x3, x4)  =  U4_GAA(x1, x4)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

PB_IN_GA(s(X1), s(X2)) → PB_IN_GA(X1, X2)

The TRS R consists of the following rules:

pcB_in_ga(s(X1), s(X2)) → U8_ga(X1, X2, pcB_in_ga(X1, X2))
U8_ga(X1, X2, pcB_out_ga(X1, X2)) → pcB_out_ga(s(X1), s(X2))

The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
pcB_in_ga(x1, x2)  =  pcB_in_ga(x1)
U8_ga(x1, x2, x3)  =  U8_ga(x1, x3)
pcB_out_ga(x1, x2)  =  pcB_out_ga(x1, x2)
PB_IN_GA(x1, x2)  =  PB_IN_GA(x1)

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

(8) UsableRulesProof (EQUIVALENT transformation)

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

(9) Obligation:

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

PB_IN_GA(s(X1), s(X2)) → PB_IN_GA(X1, X2)

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

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

(10) PiDPToQDPProof (SOUND transformation)

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

(11) Obligation:

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

PB_IN_GA(s(X1)) → PB_IN_GA(X1)

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

(12) 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:

• PB_IN_GA(s(X1)) → PB_IN_GA(X1)
  The graph contains the following edges 1 > 1

(13) YES

(14) Obligation:

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

PLUSA_IN_GAA(s(s(X1)), X2, s(X3)) → U3_GAA(X1, X2, X3, pcB_in_ga(X1, X4))
U3_GAA(X1, X2, X3, pcB_out_ga(X1, X4)) → PLUSA_IN_GAA(s(s(X4)), X2, X3)

The TRS R consists of the following rules:

pcB_in_ga(s(X1), s(X2)) → U8_ga(X1, X2, pcB_in_ga(X1, X2))
U8_ga(X1, X2, pcB_out_ga(X1, X2)) → pcB_out_ga(s(X1), s(X2))

The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
pcB_in_ga(x1, x2)  =  pcB_in_ga(x1)
U8_ga(x1, x2, x3)  =  U8_ga(x1, x3)
pcB_out_ga(x1, x2)  =  pcB_out_ga(x1, x2)
PLUSA_IN_GAA(x1, x2, x3)  =  PLUSA_IN_GAA(x1)
U3_GAA(x1, x2, x3, x4)  =  U3_GAA(x1, x4)

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

(15) PiDPToQDPProof (SOUND transformation)

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

(16) Obligation:

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

PLUSA_IN_GAA(s(s(X1))) → U3_GAA(X1, pcB_in_ga(X1))
U3_GAA(X1, pcB_out_ga(X1, X4)) → PLUSA_IN_GAA(s(s(X4)))

The TRS R consists of the following rules:

pcB_in_ga(s(X1)) → U8_ga(X1, pcB_in_ga(X1))
U8_ga(X1, pcB_out_ga(X1, X2)) → pcB_out_ga(s(X1), s(X2))

The set Q consists of the following terms:

pcB_in_ga(x0)
U8_ga(x0, x1)

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

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

PLUSA_IN_GAA(s(s(X1))) → U3_GAA(X1, pcB_in_ga(X1))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(PLUSA_IN_GAA(x₁)) = 1 + x₁   
POL(U3_GAA(x₁, x₂)) = x₂   
POL(U8_ga(x₁, x₂)) = x₂   
POL(pcB_in_ga(x₁)) = x₁   
POL(pcB_out_ga(x₁, x₂)) = 1 + x₂   
POL(s(x₁)) = x₁   

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

pcB_in_ga(s(X1)) → U8_ga(X1, pcB_in_ga(X1))
U8_ga(X1, pcB_out_ga(X1, X2)) → pcB_out_ga(s(X1), s(X2))

(18) Obligation:

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

U3_GAA(X1, pcB_out_ga(X1, X4)) → PLUSA_IN_GAA(s(s(X4)))

The TRS R consists of the following rules:

pcB_in_ga(s(X1)) → U8_ga(X1, pcB_in_ga(X1))
U8_ga(X1, pcB_out_ga(X1, X2)) → pcB_out_ga(s(X1), s(X2))

The set Q consists of the following terms:

pcB_in_ga(x0)
U8_ga(x0, x1)

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

(19) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(20) TRUE