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) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
3 [label="A: plus(T1, T2, T3)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
6 [label="6: 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];
11 [label="11: true | plus(0, T2, T3), SCOPE: 1, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: 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];
15 [label="15: plus(0, T2, T3), SCOPE: 1, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: ','(p(s(T9), X12), plus(X12, T12, T13))\n\nT9 is ground\nX12 is free", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: ','(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];
23 [label="23: ','(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];
24 [label="24: ','(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];
25 [label="25: plus(0, T12, T13)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
36 [label="36: plus(0, T12, T13), SCOPE: 3, CLAUSE: 2 | plus(0, T12, T13), SCOPE: 3, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
40 [label="40: plus(0, T12, T13), SCOPE: 3, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
42 [label="42: plus(0, T12, T13), SCOPE: 3, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
45 [label="45: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
47 [label="47: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
49 [label="49: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
52 [label="52: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
58 [label="B: ','(p(s(T23), s(X31)), plus(s(s(X31)), T12, T13))\n\nT23 is ground\nX31 is free", fontsize=16, style=filled, fillcolor=lightyellow];
60 [label="60: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
66 [label="C: p(s(T23), s(X31))\n\nT23 is ground\nX31 is free", fontsize=16, style=filled, fillcolor=lightyellow];
67 [label="67: plus(s(s(T24)), T12, T13)\n\nT24 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
68 [label="68: 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];
70 [label="70: p(s(T23), s(X31)), SCOPE: 4, CLAUSE: 1\n\nT23 is ground\nX31 is free", fontsize=16, style=filled, fillcolor=lightyellow];
76 [label="76: p(s(T27), s(X40))\n\nT27 is ground\nX40 is free", fontsize=16, style=filled, fillcolor=lightyellow];
77 [label="77: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
78 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 78 [arrowhead = none ];
78 -> 6;

79 [label="EVAL with clause\nplus(0, X2, X2).\nand substitutionT1 -> 0,\nT2 -> T5,\nX2 -> T5,\nT3 -> T5", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 79 [arrowhead = none ];
79 -> 11;

80 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 80 [arrowhead = none ];
80 -> 13;

81 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 81 [arrowhead = none ];
81 -> 15;

82 [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];
13 -> 82 [arrowhead = none ];
82 -> 17;

83 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
13 -> 83 [arrowhead = none ];
83 -> 19;

84 [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];
15 -> 84 [arrowhead = none ];
84 -> 16;

85 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
17 -> 85 [arrowhead = none ];
85 -> 22;

86 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
22 -> 86 [arrowhead = none ];
86 -> 23;

87 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
22 -> 87 [arrowhead = none ];
87 -> 24;

88 [label="EVAL with clause\np(s(0), 0).\nand substitutionT9 -> 0,\nX12 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
23 -> 88 [arrowhead = none ];
88 -> 25;

89 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
23 -> 89 [arrowhead = none ];
89 -> 26;

90 [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];
24 -> 90 [arrowhead = none ];
90 -> 58;

91 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
24 -> 91 [arrowhead = none ];
91 -> 60;

92 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
25 -> 92 [arrowhead = none ];
92 -> 36;

93 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
36 -> 93 [arrowhead = none ];
93 -> 40;

94 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
36 -> 94 [arrowhead = none ];
94 -> 42;

95 [label="EVAL with clause\nplus(0, X19, X19).\nand substitutionT12 -> T20,\nX19 -> T20,\nT13 -> T20", fontsize=14, style = filled, fillcolor = lightblue];
40 -> 95 [arrowhead = none ];
95 -> 45;

96 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
40 -> 96 [arrowhead = none ];
96 -> 47;

97 [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];
42 -> 97 [arrowhead = none ];
97 -> 52;

98 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
45 -> 98 [arrowhead = none ];
98 -> 49;

99 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
58 -> 99 [arrowhead = none ];
99 -> 66;

100 [label="SPLIT 2\nnew knowledge:\nT23 is ground\nT24 is ground\nreplacements:X31 -> T24", fontsize=14, style = filled, fillcolor = violet];
58 -> 100 [arrowhead = none ];
100 -> 67;

101 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
66 -> 101 [arrowhead = none ];
101 -> 68;

102 [label="INSTANCE with matching:\nT1 -> s(s(T24))\nT2 -> T12\nT3 -> T13", fontsize=14, style = filled, fillcolor = lightgrey];
67 -> 102 [arrowhead = none , style=dashed];
102 -> 3[style=dashed];

103 [label="BACKTRACK\nfor clause: p(s(0), 0)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
68 -> 103 [arrowhead = none ];
103 -> 70;

104 [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];
70 -> 104 [arrowhead = none ];
104 -> 76;

105 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
70 -> 105 [arrowhead = none ];
105 -> 77;

106 [label="INSTANCE with matching:\nT23 -> T27\nX31 -> X40", fontsize=14, style = filled, fillcolor = lightgrey];
76 -> 106 [arrowhead = none , style=dashed];
106 -> 66[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), T20, s(T20)) → plusA_out_gaa(s(0), T20, s(T20))
plusA_in_gaa(s(s(T23)), T12, s(T13)) → U1_gaa(T23, T12, T13, pB_in_gaaa(T23, X31, T12, T13))
pB_in_gaaa(T23, T24, T12, T13) → U3_gaaa(T23, T24, T12, T13, pC_in_ga(T23, T24))
pC_in_ga(s(T27), s(X40)) → U2_ga(T27, X40, pC_in_ga(T27, X40))
U2_ga(T27, X40, pC_out_ga(T27, X40)) → pC_out_ga(s(T27), s(X40))
U3_gaaa(T23, T24, T12, T13, pC_out_ga(T23, T24)) → U4_gaaa(T23, T24, T12, T13, plusA_in_gaa(s(s(T24)), T12, T13))
U4_gaaa(T23, T24, T12, T13, plusA_out_gaa(s(s(T24)), T12, T13)) → pB_out_gaaa(T23, T24, T12, T13)
U1_gaa(T23, T12, T13, pB_out_gaaa(T23, X31, T12, T13)) → plusA_out_gaa(s(s(T23)), T12, s(T13))

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)
pB_in_gaaa(x1, x2, x3, x4)  =  pB_in_gaaa(x1)
U3_gaaa(x1, x2, x3, x4, x5)  =  U3_gaaa(x1, x5)
pC_in_ga(x1, x2)  =  pC_in_ga(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
pC_out_ga(x1, x2)  =  pC_out_ga(x1, x2)
U4_gaaa(x1, x2, x3, x4, x5)  =  U4_gaaa(x1, x2, x5)
pB_out_gaaa(x1, x2, x3, x4)  =  pB_out_gaaa(x1, x2)

(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(T23)), T12, s(T13)) → U1_GAA(T23, T12, T13, pB_in_gaaa(T23, X31, T12, T13))
PLUSA_IN_GAA(s(s(T23)), T12, s(T13)) → PB_IN_GAAA(T23, X31, T12, T13)
PB_IN_GAAA(T23, T24, T12, T13) → U3_GAAA(T23, T24, T12, T13, pC_in_ga(T23, T24))
PB_IN_GAAA(T23, T24, T12, T13) → PC_IN_GA(T23, T24)
PC_IN_GA(s(T27), s(X40)) → U2_GA(T27, X40, pC_in_ga(T27, X40))
PC_IN_GA(s(T27), s(X40)) → PC_IN_GA(T27, X40)
U3_GAAA(T23, T24, T12, T13, pC_out_ga(T23, T24)) → U4_GAAA(T23, T24, T12, T13, plusA_in_gaa(s(s(T24)), T12, T13))
U3_GAAA(T23, T24, T12, T13, pC_out_ga(T23, T24)) → PLUSA_IN_GAA(s(s(T24)), T12, T13)

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), T20, s(T20)) → plusA_out_gaa(s(0), T20, s(T20))
plusA_in_gaa(s(s(T23)), T12, s(T13)) → U1_gaa(T23, T12, T13, pB_in_gaaa(T23, X31, T12, T13))
pB_in_gaaa(T23, T24, T12, T13) → U3_gaaa(T23, T24, T12, T13, pC_in_ga(T23, T24))
pC_in_ga(s(T27), s(X40)) → U2_ga(T27, X40, pC_in_ga(T27, X40))
U2_ga(T27, X40, pC_out_ga(T27, X40)) → pC_out_ga(s(T27), s(X40))
U3_gaaa(T23, T24, T12, T13, pC_out_ga(T23, T24)) → U4_gaaa(T23, T24, T12, T13, plusA_in_gaa(s(s(T24)), T12, T13))
U4_gaaa(T23, T24, T12, T13, plusA_out_gaa(s(s(T24)), T12, T13)) → pB_out_gaaa(T23, T24, T12, T13)
U1_gaa(T23, T12, T13, pB_out_gaaa(T23, X31, T12, T13)) → plusA_out_gaa(s(s(T23)), T12, s(T13))

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)
pB_in_gaaa(x1, x2, x3, x4)  =  pB_in_gaaa(x1)
U3_gaaa(x1, x2, x3, x4, x5)  =  U3_gaaa(x1, x5)
pC_in_ga(x1, x2)  =  pC_in_ga(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
pC_out_ga(x1, x2)  =  pC_out_ga(x1, x2)
U4_gaaa(x1, x2, x3, x4, x5)  =  U4_gaaa(x1, x2, x5)
pB_out_gaaa(x1, x2, x3, x4)  =  pB_out_gaaa(x1, x2)
PLUSA_IN_GAA(x1, x2, x3)  =  PLUSA_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4)  =  U1_GAA(x1, x4)
PB_IN_GAAA(x1, x2, x3, x4)  =  PB_IN_GAAA(x1)
U3_GAAA(x1, x2, x3, x4, x5)  =  U3_GAAA(x1, x5)
PC_IN_GA(x1, x2)  =  PC_IN_GA(x1)
U2_GA(x1, x2, x3)  =  U2_GA(x1, x3)
U4_GAAA(x1, x2, x3, x4, x5)  =  U4_GAAA(x1, x2, x5)

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(T23)), T12, s(T13)) → U1_GAA(T23, T12, T13, pB_in_gaaa(T23, X31, T12, T13))
PLUSA_IN_GAA(s(s(T23)), T12, s(T13)) → PB_IN_GAAA(T23, X31, T12, T13)
PB_IN_GAAA(T23, T24, T12, T13) → U3_GAAA(T23, T24, T12, T13, pC_in_ga(T23, T24))
PB_IN_GAAA(T23, T24, T12, T13) → PC_IN_GA(T23, T24)
PC_IN_GA(s(T27), s(X40)) → U2_GA(T27, X40, pC_in_ga(T27, X40))
PC_IN_GA(s(T27), s(X40)) → PC_IN_GA(T27, X40)
U3_GAAA(T23, T24, T12, T13, pC_out_ga(T23, T24)) → U4_GAAA(T23, T24, T12, T13, plusA_in_gaa(s(s(T24)), T12, T13))
U3_GAAA(T23, T24, T12, T13, pC_out_ga(T23, T24)) → PLUSA_IN_GAA(s(s(T24)), T12, T13)

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), T20, s(T20)) → plusA_out_gaa(s(0), T20, s(T20))
plusA_in_gaa(s(s(T23)), T12, s(T13)) → U1_gaa(T23, T12, T13, pB_in_gaaa(T23, X31, T12, T13))
pB_in_gaaa(T23, T24, T12, T13) → U3_gaaa(T23, T24, T12, T13, pC_in_ga(T23, T24))
pC_in_ga(s(T27), s(X40)) → U2_ga(T27, X40, pC_in_ga(T27, X40))
U2_ga(T27, X40, pC_out_ga(T27, X40)) → pC_out_ga(s(T27), s(X40))
U3_gaaa(T23, T24, T12, T13, pC_out_ga(T23, T24)) → U4_gaaa(T23, T24, T12, T13, plusA_in_gaa(s(s(T24)), T12, T13))
U4_gaaa(T23, T24, T12, T13, plusA_out_gaa(s(s(T24)), T12, T13)) → pB_out_gaaa(T23, T24, T12, T13)
U1_gaa(T23, T12, T13, pB_out_gaaa(T23, X31, T12, T13)) → plusA_out_gaa(s(s(T23)), T12, s(T13))

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)
pB_in_gaaa(x1, x2, x3, x4)  =  pB_in_gaaa(x1)
U3_gaaa(x1, x2, x3, x4, x5)  =  U3_gaaa(x1, x5)
pC_in_ga(x1, x2)  =  pC_in_ga(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
pC_out_ga(x1, x2)  =  pC_out_ga(x1, x2)
U4_gaaa(x1, x2, x3, x4, x5)  =  U4_gaaa(x1, x2, x5)
pB_out_gaaa(x1, x2, x3, x4)  =  pB_out_gaaa(x1, x2)
PLUSA_IN_GAA(x1, x2, x3)  =  PLUSA_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4)  =  U1_GAA(x1, x4)
PB_IN_GAAA(x1, x2, x3, x4)  =  PB_IN_GAAA(x1)
U3_GAAA(x1, x2, x3, x4, x5)  =  U3_GAAA(x1, x5)
PC_IN_GA(x1, x2)  =  PC_IN_GA(x1)
U2_GA(x1, x2, x3)  =  U2_GA(x1, x3)
U4_GAAA(x1, x2, x3, x4, x5)  =  U4_GAAA(x1, x2, x5)

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:

PC_IN_GA(s(T27), s(X40)) → PC_IN_GA(T27, X40)

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), T20, s(T20)) → plusA_out_gaa(s(0), T20, s(T20))
plusA_in_gaa(s(s(T23)), T12, s(T13)) → U1_gaa(T23, T12, T13, pB_in_gaaa(T23, X31, T12, T13))
pB_in_gaaa(T23, T24, T12, T13) → U3_gaaa(T23, T24, T12, T13, pC_in_ga(T23, T24))
pC_in_ga(s(T27), s(X40)) → U2_ga(T27, X40, pC_in_ga(T27, X40))
U2_ga(T27, X40, pC_out_ga(T27, X40)) → pC_out_ga(s(T27), s(X40))
U3_gaaa(T23, T24, T12, T13, pC_out_ga(T23, T24)) → U4_gaaa(T23, T24, T12, T13, plusA_in_gaa(s(s(T24)), T12, T13))
U4_gaaa(T23, T24, T12, T13, plusA_out_gaa(s(s(T24)), T12, T13)) → pB_out_gaaa(T23, T24, T12, T13)
U1_gaa(T23, T12, T13, pB_out_gaaa(T23, X31, T12, T13)) → plusA_out_gaa(s(s(T23)), T12, s(T13))

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)
pB_in_gaaa(x1, x2, x3, x4)  =  pB_in_gaaa(x1)
U3_gaaa(x1, x2, x3, x4, x5)  =  U3_gaaa(x1, x5)
pC_in_ga(x1, x2)  =  pC_in_ga(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
pC_out_ga(x1, x2)  =  pC_out_ga(x1, x2)
U4_gaaa(x1, x2, x3, x4, x5)  =  U4_gaaa(x1, x2, x5)
pB_out_gaaa(x1, x2, x3, x4)  =  pB_out_gaaa(x1, x2)
PC_IN_GA(x1, x2)  =  PC_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:

PC_IN_GA(s(T27), s(X40)) → PC_IN_GA(T27, X40)

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

PC_IN_GA(s(T27)) → PC_IN_GA(T27)

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:

• PC_IN_GA(s(T27)) → PC_IN_GA(T27)
  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(T23)), T12, s(T13)) → PB_IN_GAAA(T23, X31, T12, T13)
PB_IN_GAAA(T23, T24, T12, T13) → U3_GAAA(T23, T24, T12, T13, pC_in_ga(T23, T24))
U3_GAAA(T23, T24, T12, T13, pC_out_ga(T23, T24)) → PLUSA_IN_GAA(s(s(T24)), T12, T13)

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), T20, s(T20)) → plusA_out_gaa(s(0), T20, s(T20))
plusA_in_gaa(s(s(T23)), T12, s(T13)) → U1_gaa(T23, T12, T13, pB_in_gaaa(T23, X31, T12, T13))
pB_in_gaaa(T23, T24, T12, T13) → U3_gaaa(T23, T24, T12, T13, pC_in_ga(T23, T24))
pC_in_ga(s(T27), s(X40)) → U2_ga(T27, X40, pC_in_ga(T27, X40))
U2_ga(T27, X40, pC_out_ga(T27, X40)) → pC_out_ga(s(T27), s(X40))
U3_gaaa(T23, T24, T12, T13, pC_out_ga(T23, T24)) → U4_gaaa(T23, T24, T12, T13, plusA_in_gaa(s(s(T24)), T12, T13))
U4_gaaa(T23, T24, T12, T13, plusA_out_gaa(s(s(T24)), T12, T13)) → pB_out_gaaa(T23, T24, T12, T13)
U1_gaa(T23, T12, T13, pB_out_gaaa(T23, X31, T12, T13)) → plusA_out_gaa(s(s(T23)), T12, s(T13))

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)
pB_in_gaaa(x1, x2, x3, x4)  =  pB_in_gaaa(x1)
U3_gaaa(x1, x2, x3, x4, x5)  =  U3_gaaa(x1, x5)
pC_in_ga(x1, x2)  =  pC_in_ga(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
pC_out_ga(x1, x2)  =  pC_out_ga(x1, x2)
U4_gaaa(x1, x2, x3, x4, x5)  =  U4_gaaa(x1, x2, x5)
pB_out_gaaa(x1, x2, x3, x4)  =  pB_out_gaaa(x1, x2)
PLUSA_IN_GAA(x1, x2, x3)  =  PLUSA_IN_GAA(x1)
PB_IN_GAAA(x1, x2, x3, x4)  =  PB_IN_GAAA(x1)
U3_GAAA(x1, x2, x3, x4, x5)  =  U3_GAAA(x1, x5)

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

(15) UsableRulesProof (EQUIVALENT transformation)

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

(16) Obligation:

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

PLUSA_IN_GAA(s(s(T23)), T12, s(T13)) → PB_IN_GAAA(T23, X31, T12, T13)
PB_IN_GAAA(T23, T24, T12, T13) → U3_GAAA(T23, T24, T12, T13, pC_in_ga(T23, T24))
U3_GAAA(T23, T24, T12, T13, pC_out_ga(T23, T24)) → PLUSA_IN_GAA(s(s(T24)), T12, T13)

The TRS R consists of the following rules:

pC_in_ga(s(T27), s(X40)) → U2_ga(T27, X40, pC_in_ga(T27, X40))
U2_ga(T27, X40, pC_out_ga(T27, X40)) → pC_out_ga(s(T27), s(X40))

The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
pC_in_ga(x1, x2)  =  pC_in_ga(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
pC_out_ga(x1, x2)  =  pC_out_ga(x1, x2)
PLUSA_IN_GAA(x1, x2, x3)  =  PLUSA_IN_GAA(x1)
PB_IN_GAAA(x1, x2, x3, x4)  =  PB_IN_GAAA(x1)
U3_GAAA(x1, x2, x3, x4, x5)  =  U3_GAAA(x1, x5)

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

(17) PiDPToQDPProof (SOUND transformation)

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

(18) Obligation:

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

PLUSA_IN_GAA(s(s(T23))) → PB_IN_GAAA(T23)
PB_IN_GAAA(T23) → U3_GAAA(T23, pC_in_ga(T23))
U3_GAAA(T23, pC_out_ga(T23, T24)) → PLUSA_IN_GAA(s(s(T24)))

The TRS R consists of the following rules:

pC_in_ga(s(T27)) → U2_ga(T27, pC_in_ga(T27))
U2_ga(T27, pC_out_ga(T27, X40)) → pC_out_ga(s(T27), s(X40))

The set Q consists of the following terms:

pC_in_ga(x0)
U2_ga(x0, x1)

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

(19) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

PB_IN_GAAA(T23) → U3_GAAA(T23, pC_in_ga(T23))

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

POL(PB_IN_GAAA(x₁)) = 1   
POL(PLUSA_IN_GAA(x₁)) = x₁   
POL(U2_ga(x₁, x₂)) = x₂   
POL(U3_GAAA(x₁, x₂)) = x₂   
POL(pC_in_ga(x₁)) = 0   
POL(pC_out_ga(x₁, x₂)) = 1   
POL(s(x₁)) = 1   

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

pC_in_ga(s(T27)) → U2_ga(T27, pC_in_ga(T27))
U2_ga(T27, pC_out_ga(T27, X40)) → pC_out_ga(s(T27), s(X40))

(20) Obligation:

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

PLUSA_IN_GAA(s(s(T23))) → PB_IN_GAAA(T23)
U3_GAAA(T23, pC_out_ga(T23, T24)) → PLUSA_IN_GAA(s(s(T24)))

The TRS R consists of the following rules:

pC_in_ga(s(T27)) → U2_ga(T27, pC_in_ga(T27))
U2_ga(T27, pC_out_ga(T27, X40)) → pC_out_ga(s(T27), s(X40))

The set Q consists of the following terms:

pC_in_ga(x0)
U2_ga(x0, x1)

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

(21) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(22) TRUE