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

(0) Obligation:

Clauses:

half(0, 0).
half(s(0), 0).
half(s(s(X)), s(Y)) :- half(X, Y).
log(0, s(0)).
log(s(X), s(Y)) :- ','(half(s(X), Z), log(Z, Y)).

Query: log(g,a)

(1) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
2 [label="A: log(T1, T2)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
5 [label="5: log(T1, T2), SCOPE: 1, CLAUSE: 3 | log(T1, T2), SCOPE: 1, CLAUSE: 4\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: true | log(0, T2), SCOPE: 1, CLAUSE: 4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: log(T1, T2), SCOPE: 1, CLAUSE: 4\n\nT1 is ground\nlog(T1, T2) !=? log(0, s(0))", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: log(0, T2), SCOPE: 1, CLAUSE: 4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: ','(half(s(T5), X7), log(X7, T7))\n\nT5 is ground\nX7 is free", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: ','(half(s(T5), X7), log(X7, T7)), SCOPE: 2, CLAUSE: 0 | ','(half(s(T5), X7), log(X7, T7)), SCOPE: 2, CLAUSE: 1 | ','(half(s(T5), X7), log(X7, T7)), SCOPE: 2, CLAUSE: 2\n\nT5 is ground\nX7 is free", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: ','(half(s(T5), X7), log(X7, T7)), SCOPE: 2, CLAUSE: 1 | ','(half(s(T5), X7), log(X7, T7)), SCOPE: 2, CLAUSE: 2\n\nT5 is ground\nX7 is free", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: ','(half(s(T5), X7), log(X7, T7)), SCOPE: 2, CLAUSE: 1\n\nT5 is ground\nX7 is free", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: ','(half(s(T5), X7), log(X7, T7)), SCOPE: 2, CLAUSE: 2\n\nT5 is ground\nX7 is free", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: log(0, T7)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="29: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
39 [label="39: log(0, T7), SCOPE: 3, CLAUSE: 3 | log(0, T7), SCOPE: 3, CLAUSE: 4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
41 [label="41: log(0, T7), SCOPE: 3, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
43 [label="43: log(0, T7), SCOPE: 3, CLAUSE: 4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
47 [label="47: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
49 [label="49: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
51 [label="51: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
53 [label="53: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
59 [label="B: ','(half(T10, X18), log(s(X18), T7))\n\nT10 is ground\nX18 is free", fontsize=16, style=filled, fillcolor=lightyellow];
60 [label="60: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
66 [label="C: half(T10, X18)\n\nT10 is ground\nX18 is free", fontsize=16, style=filled, fillcolor=lightyellow];
68 [label="68: log(s(T11), T7)\n\nT11 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
70 [label="70: half(T10, X18), SCOPE: 4, CLAUSE: 0 | half(T10, X18), SCOPE: 4, CLAUSE: 1 | half(T10, X18), SCOPE: 4, CLAUSE: 2\n\nT10 is ground\nX18 is free", fontsize=16, style=filled, fillcolor=lightyellow];
72 [label="72: half(T10, X18), SCOPE: 4, CLAUSE: 0\n\nT10 is ground\nX18 is free", fontsize=16, style=filled, fillcolor=lightyellow];
74 [label="74: half(T10, X18), SCOPE: 4, CLAUSE: 1 | half(T10, X18), SCOPE: 4, CLAUSE: 2\n\nT10 is ground\nX18 is free", fontsize=16, style=filled, fillcolor=lightyellow];
78 [label="78: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
81 [label="81: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
82 [label="82: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
85 [label="85: half(T10, X18), SCOPE: 4, CLAUSE: 1\n\nT10 is ground\nX18 is free", fontsize=16, style=filled, fillcolor=lightyellow];
86 [label="86: half(T10, X18), SCOPE: 4, CLAUSE: 2\n\nT10 is ground\nX18 is free", fontsize=16, style=filled, fillcolor=lightyellow];
90 [label="90: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
93 [label="93: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
94 [label="94: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
96 [label="96: half(T14, X27)\n\nT14 is ground\nX27 is free", fontsize=16, style=filled, fillcolor=lightyellow];
98 [label="98: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
99 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
2 -> 99 [arrowhead = none ];
99 -> 5;

100 [label="EVAL with clause\nlog(0, s(0)).\nand substitutionT1 -> 0,\nT2 -> s(0)", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 100 [arrowhead = none ];
100 -> 10;

101 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 101 [arrowhead = none ];
101 -> 12;

102 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 102 [arrowhead = none ];
102 -> 15;

103 [label="EVAL with clause\nlog(s(X5), s(X6)) :- ','(half(s(X5), X7), log(X7, X6)).\nand substitutionX5 -> T5,\nT1 -> s(T5),\nX6 -> T7,\nT2 -> s(T7),\nT6 -> T7", fontsize=14, style = filled, fillcolor = lightblue];
12 -> 103 [arrowhead = none ];
103 -> 17;

104 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 104 [arrowhead = none ];
104 -> 20;

105 [label="BACKTRACK\nfor clause: log(s(X), s(Y)) :- ','(half(s(X), Z), log(Z, Y))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 105 [arrowhead = none ];
105 -> 16;

106 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
17 -> 106 [arrowhead = none ];
106 -> 22;

107 [label="BACKTRACK\nfor clause: half(0, 0)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
22 -> 107 [arrowhead = none ];
107 -> 23;

108 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
23 -> 108 [arrowhead = none ];
108 -> 24;

109 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
23 -> 109 [arrowhead = none ];
109 -> 25;

110 [label="EVAL with clause\nhalf(s(0), 0).\nand substitutionT5 -> 0,\nX7 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
24 -> 110 [arrowhead = none ];
110 -> 28;

111 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
24 -> 111 [arrowhead = none ];
111 -> 29;

112 [label="EVAL with clause\nhalf(s(s(X16)), s(X17)) :- half(X16, X17).\nand substitutionX16 -> T10,\nT5 -> s(T10),\nX17 -> X18,\nX7 -> s(X18)", fontsize=14, style = filled, fillcolor = lightblue];
25 -> 112 [arrowhead = none ];
112 -> 59;

113 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
25 -> 113 [arrowhead = none ];
113 -> 60;

114 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
28 -> 114 [arrowhead = none ];
114 -> 39;

115 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
39 -> 115 [arrowhead = none ];
115 -> 41;

116 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
39 -> 116 [arrowhead = none ];
116 -> 43;

117 [label="EVAL with clause\nlog(0, s(0)).\nand substitutionT7 -> s(0)", fontsize=14, style = filled, fillcolor = lightblue];
41 -> 117 [arrowhead = none ];
117 -> 47;

118 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
41 -> 118 [arrowhead = none ];
118 -> 49;

119 [label="BACKTRACK\nfor clause: log(s(X), s(Y)) :- ','(half(s(X), Z), log(Z, Y))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
43 -> 119 [arrowhead = none ];
119 -> 53;

120 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
47 -> 120 [arrowhead = none ];
120 -> 51;

121 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
59 -> 121 [arrowhead = none ];
121 -> 66;

122 [label="SPLIT 2\nnew knowledge:\nT10 is ground\nT11 is ground\nreplacements:X18 -> T11", fontsize=14, style = filled, fillcolor = violet];
59 -> 122 [arrowhead = none ];
122 -> 68;

123 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
66 -> 123 [arrowhead = none ];
123 -> 70;

124 [label="INSTANCE with matching:\nT1 -> s(T11)\nT2 -> T7", fontsize=14, style = filled, fillcolor = lightgrey];
68 -> 124 [arrowhead = none , style=dashed];
124 -> 2[style=dashed];

125 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
70 -> 125 [arrowhead = none ];
125 -> 72;

126 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
70 -> 126 [arrowhead = none ];
126 -> 74;

127 [label="EVAL with clause\nhalf(0, 0).\nand substitutionT10 -> 0,\nX18 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
72 -> 127 [arrowhead = none ];
127 -> 78;

128 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
72 -> 128 [arrowhead = none ];
128 -> 81;

129 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
74 -> 129 [arrowhead = none ];
129 -> 85;

130 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
74 -> 130 [arrowhead = none ];
130 -> 86;

131 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
78 -> 131 [arrowhead = none ];
131 -> 82;

132 [label="EVAL with clause\nhalf(s(0), 0).\nand substitutionT10 -> s(0),\nX18 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
85 -> 132 [arrowhead = none ];
132 -> 90;

133 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
85 -> 133 [arrowhead = none ];
133 -> 93;

134 [label="EVAL with clause\nhalf(s(s(X25)), s(X26)) :- half(X25, X26).\nand substitutionX25 -> T14,\nT10 -> s(s(T14)),\nX26 -> X27,\nX18 -> s(X27)", fontsize=14, style = filled, fillcolor = lightblue];
86 -> 134 [arrowhead = none ];
134 -> 96;

135 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
86 -> 135 [arrowhead = none ];
135 -> 98;

136 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
90 -> 136 [arrowhead = none ];
136 -> 94;

137 [label="INSTANCE with matching:\nT10 -> T14\nX18 -> X27", fontsize=14, style = filled, fillcolor = lightgrey];
96 -> 137 [arrowhead = none , style=dashed];
137 -> 66[style=dashed];

}

(2) Obligation:

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

logA_in_ga(0, s(0)) → logA_out_ga(0, s(0))
logA_in_ga(s(0), s(s(0))) → logA_out_ga(s(0), s(s(0)))
logA_in_ga(s(s(T10)), s(T7)) → U1_ga(T10, T7, pB_in_gaa(T10, X18, T7))
pB_in_gaa(T10, T11, T7) → U3_gaa(T10, T11, T7, halfC_in_ga(T10, T11))
halfC_in_ga(0, 0) → halfC_out_ga(0, 0)
halfC_in_ga(s(0), 0) → halfC_out_ga(s(0), 0)
halfC_in_ga(s(s(T14)), s(X27)) → U2_ga(T14, X27, halfC_in_ga(T14, X27))
U2_ga(T14, X27, halfC_out_ga(T14, X27)) → halfC_out_ga(s(s(T14)), s(X27))
U3_gaa(T10, T11, T7, halfC_out_ga(T10, T11)) → U4_gaa(T10, T11, T7, logA_in_ga(s(T11), T7))
U4_gaa(T10, T11, T7, logA_out_ga(s(T11), T7)) → pB_out_gaa(T10, T11, T7)
U1_ga(T10, T7, pB_out_gaa(T10, X18, T7)) → logA_out_ga(s(s(T10)), s(T7))

The argument filtering Pi contains the following mapping:
logA_in_ga(x1, x2)  =  logA_in_ga(x1)
0  =  0
logA_out_ga(x1, x2)  =  logA_out_ga(x1, x2)
s(x1)  =  s(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
pB_in_gaa(x1, x2, x3)  =  pB_in_gaa(x1)
U3_gaa(x1, x2, x3, x4)  =  U3_gaa(x1, x4)
halfC_in_ga(x1, x2)  =  halfC_in_ga(x1)
halfC_out_ga(x1, x2)  =  halfC_out_ga(x1, x2)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
U4_gaa(x1, x2, x3, x4)  =  U4_gaa(x1, x2, x4)
pB_out_gaa(x1, x2, x3)  =  pB_out_gaa(x1, x2, x3)

(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:

LOGA_IN_GA(s(s(T10)), s(T7)) → U1_GA(T10, T7, pB_in_gaa(T10, X18, T7))
LOGA_IN_GA(s(s(T10)), s(T7)) → PB_IN_GAA(T10, X18, T7)
PB_IN_GAA(T10, T11, T7) → U3_GAA(T10, T11, T7, halfC_in_ga(T10, T11))
PB_IN_GAA(T10, T11, T7) → HALFC_IN_GA(T10, T11)
HALFC_IN_GA(s(s(T14)), s(X27)) → U2_GA(T14, X27, halfC_in_ga(T14, X27))
HALFC_IN_GA(s(s(T14)), s(X27)) → HALFC_IN_GA(T14, X27)
U3_GAA(T10, T11, T7, halfC_out_ga(T10, T11)) → U4_GAA(T10, T11, T7, logA_in_ga(s(T11), T7))
U3_GAA(T10, T11, T7, halfC_out_ga(T10, T11)) → LOGA_IN_GA(s(T11), T7)

The TRS R consists of the following rules:

logA_in_ga(0, s(0)) → logA_out_ga(0, s(0))
logA_in_ga(s(0), s(s(0))) → logA_out_ga(s(0), s(s(0)))
logA_in_ga(s(s(T10)), s(T7)) → U1_ga(T10, T7, pB_in_gaa(T10, X18, T7))
pB_in_gaa(T10, T11, T7) → U3_gaa(T10, T11, T7, halfC_in_ga(T10, T11))
halfC_in_ga(0, 0) → halfC_out_ga(0, 0)
halfC_in_ga(s(0), 0) → halfC_out_ga(s(0), 0)
halfC_in_ga(s(s(T14)), s(X27)) → U2_ga(T14, X27, halfC_in_ga(T14, X27))
U2_ga(T14, X27, halfC_out_ga(T14, X27)) → halfC_out_ga(s(s(T14)), s(X27))
U3_gaa(T10, T11, T7, halfC_out_ga(T10, T11)) → U4_gaa(T10, T11, T7, logA_in_ga(s(T11), T7))
U4_gaa(T10, T11, T7, logA_out_ga(s(T11), T7)) → pB_out_gaa(T10, T11, T7)
U1_ga(T10, T7, pB_out_gaa(T10, X18, T7)) → logA_out_ga(s(s(T10)), s(T7))

The argument filtering Pi contains the following mapping:
logA_in_ga(x1, x2)  =  logA_in_ga(x1)
0  =  0
logA_out_ga(x1, x2)  =  logA_out_ga(x1, x2)
s(x1)  =  s(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
pB_in_gaa(x1, x2, x3)  =  pB_in_gaa(x1)
U3_gaa(x1, x2, x3, x4)  =  U3_gaa(x1, x4)
halfC_in_ga(x1, x2)  =  halfC_in_ga(x1)
halfC_out_ga(x1, x2)  =  halfC_out_ga(x1, x2)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
U4_gaa(x1, x2, x3, x4)  =  U4_gaa(x1, x2, x4)
pB_out_gaa(x1, x2, x3)  =  pB_out_gaa(x1, x2, x3)
LOGA_IN_GA(x1, x2)  =  LOGA_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x1, x3)
PB_IN_GAA(x1, x2, x3)  =  PB_IN_GAA(x1)
U3_GAA(x1, x2, x3, x4)  =  U3_GAA(x1, x4)
HALFC_IN_GA(x1, x2)  =  HALFC_IN_GA(x1)
U2_GA(x1, x2, x3)  =  U2_GA(x1, x3)
U4_GAA(x1, x2, x3, x4)  =  U4_GAA(x1, x2, x4)

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

(4) Obligation:

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

LOGA_IN_GA(s(s(T10)), s(T7)) → U1_GA(T10, T7, pB_in_gaa(T10, X18, T7))
LOGA_IN_GA(s(s(T10)), s(T7)) → PB_IN_GAA(T10, X18, T7)
PB_IN_GAA(T10, T11, T7) → U3_GAA(T10, T11, T7, halfC_in_ga(T10, T11))
PB_IN_GAA(T10, T11, T7) → HALFC_IN_GA(T10, T11)
HALFC_IN_GA(s(s(T14)), s(X27)) → U2_GA(T14, X27, halfC_in_ga(T14, X27))
HALFC_IN_GA(s(s(T14)), s(X27)) → HALFC_IN_GA(T14, X27)
U3_GAA(T10, T11, T7, halfC_out_ga(T10, T11)) → U4_GAA(T10, T11, T7, logA_in_ga(s(T11), T7))
U3_GAA(T10, T11, T7, halfC_out_ga(T10, T11)) → LOGA_IN_GA(s(T11), T7)

The TRS R consists of the following rules:

logA_in_ga(0, s(0)) → logA_out_ga(0, s(0))
logA_in_ga(s(0), s(s(0))) → logA_out_ga(s(0), s(s(0)))
logA_in_ga(s(s(T10)), s(T7)) → U1_ga(T10, T7, pB_in_gaa(T10, X18, T7))
pB_in_gaa(T10, T11, T7) → U3_gaa(T10, T11, T7, halfC_in_ga(T10, T11))
halfC_in_ga(0, 0) → halfC_out_ga(0, 0)
halfC_in_ga(s(0), 0) → halfC_out_ga(s(0), 0)
halfC_in_ga(s(s(T14)), s(X27)) → U2_ga(T14, X27, halfC_in_ga(T14, X27))
U2_ga(T14, X27, halfC_out_ga(T14, X27)) → halfC_out_ga(s(s(T14)), s(X27))
U3_gaa(T10, T11, T7, halfC_out_ga(T10, T11)) → U4_gaa(T10, T11, T7, logA_in_ga(s(T11), T7))
U4_gaa(T10, T11, T7, logA_out_ga(s(T11), T7)) → pB_out_gaa(T10, T11, T7)
U1_ga(T10, T7, pB_out_gaa(T10, X18, T7)) → logA_out_ga(s(s(T10)), s(T7))

The argument filtering Pi contains the following mapping:
logA_in_ga(x1, x2)  =  logA_in_ga(x1)
0  =  0
logA_out_ga(x1, x2)  =  logA_out_ga(x1, x2)
s(x1)  =  s(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
pB_in_gaa(x1, x2, x3)  =  pB_in_gaa(x1)
U3_gaa(x1, x2, x3, x4)  =  U3_gaa(x1, x4)
halfC_in_ga(x1, x2)  =  halfC_in_ga(x1)
halfC_out_ga(x1, x2)  =  halfC_out_ga(x1, x2)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
U4_gaa(x1, x2, x3, x4)  =  U4_gaa(x1, x2, x4)
pB_out_gaa(x1, x2, x3)  =  pB_out_gaa(x1, x2, x3)
LOGA_IN_GA(x1, x2)  =  LOGA_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x1, x3)
PB_IN_GAA(x1, x2, x3)  =  PB_IN_GAA(x1)
U3_GAA(x1, x2, x3, x4)  =  U3_GAA(x1, x4)
HALFC_IN_GA(x1, x2)  =  HALFC_IN_GA(x1)
U2_GA(x1, x2, x3)  =  U2_GA(x1, x3)
U4_GAA(x1, x2, x3, x4)  =  U4_GAA(x1, x2, 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:

HALFC_IN_GA(s(s(T14)), s(X27)) → HALFC_IN_GA(T14, X27)

The TRS R consists of the following rules:

logA_in_ga(0, s(0)) → logA_out_ga(0, s(0))
logA_in_ga(s(0), s(s(0))) → logA_out_ga(s(0), s(s(0)))
logA_in_ga(s(s(T10)), s(T7)) → U1_ga(T10, T7, pB_in_gaa(T10, X18, T7))
pB_in_gaa(T10, T11, T7) → U3_gaa(T10, T11, T7, halfC_in_ga(T10, T11))
halfC_in_ga(0, 0) → halfC_out_ga(0, 0)
halfC_in_ga(s(0), 0) → halfC_out_ga(s(0), 0)
halfC_in_ga(s(s(T14)), s(X27)) → U2_ga(T14, X27, halfC_in_ga(T14, X27))
U2_ga(T14, X27, halfC_out_ga(T14, X27)) → halfC_out_ga(s(s(T14)), s(X27))
U3_gaa(T10, T11, T7, halfC_out_ga(T10, T11)) → U4_gaa(T10, T11, T7, logA_in_ga(s(T11), T7))
U4_gaa(T10, T11, T7, logA_out_ga(s(T11), T7)) → pB_out_gaa(T10, T11, T7)
U1_ga(T10, T7, pB_out_gaa(T10, X18, T7)) → logA_out_ga(s(s(T10)), s(T7))

The argument filtering Pi contains the following mapping:
logA_in_ga(x1, x2)  =  logA_in_ga(x1)
0  =  0
logA_out_ga(x1, x2)  =  logA_out_ga(x1, x2)
s(x1)  =  s(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
pB_in_gaa(x1, x2, x3)  =  pB_in_gaa(x1)
U3_gaa(x1, x2, x3, x4)  =  U3_gaa(x1, x4)
halfC_in_ga(x1, x2)  =  halfC_in_ga(x1)
halfC_out_ga(x1, x2)  =  halfC_out_ga(x1, x2)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
U4_gaa(x1, x2, x3, x4)  =  U4_gaa(x1, x2, x4)
pB_out_gaa(x1, x2, x3)  =  pB_out_gaa(x1, x2, x3)
HALFC_IN_GA(x1, x2)  =  HALFC_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:

HALFC_IN_GA(s(s(T14)), s(X27)) → HALFC_IN_GA(T14, X27)

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

HALFC_IN_GA(s(s(T14))) → HALFC_IN_GA(T14)

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:

• HALFC_IN_GA(s(s(T14))) → HALFC_IN_GA(T14)
  The graph contains the following edges 1 > 1

(13) YES

(14) Obligation:

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

LOGA_IN_GA(s(s(T10)), s(T7)) → PB_IN_GAA(T10, X18, T7)
PB_IN_GAA(T10, T11, T7) → U3_GAA(T10, T11, T7, halfC_in_ga(T10, T11))
U3_GAA(T10, T11, T7, halfC_out_ga(T10, T11)) → LOGA_IN_GA(s(T11), T7)

The TRS R consists of the following rules:

logA_in_ga(0, s(0)) → logA_out_ga(0, s(0))
logA_in_ga(s(0), s(s(0))) → logA_out_ga(s(0), s(s(0)))
logA_in_ga(s(s(T10)), s(T7)) → U1_ga(T10, T7, pB_in_gaa(T10, X18, T7))
pB_in_gaa(T10, T11, T7) → U3_gaa(T10, T11, T7, halfC_in_ga(T10, T11))
halfC_in_ga(0, 0) → halfC_out_ga(0, 0)
halfC_in_ga(s(0), 0) → halfC_out_ga(s(0), 0)
halfC_in_ga(s(s(T14)), s(X27)) → U2_ga(T14, X27, halfC_in_ga(T14, X27))
U2_ga(T14, X27, halfC_out_ga(T14, X27)) → halfC_out_ga(s(s(T14)), s(X27))
U3_gaa(T10, T11, T7, halfC_out_ga(T10, T11)) → U4_gaa(T10, T11, T7, logA_in_ga(s(T11), T7))
U4_gaa(T10, T11, T7, logA_out_ga(s(T11), T7)) → pB_out_gaa(T10, T11, T7)
U1_ga(T10, T7, pB_out_gaa(T10, X18, T7)) → logA_out_ga(s(s(T10)), s(T7))

The argument filtering Pi contains the following mapping:
logA_in_ga(x1, x2)  =  logA_in_ga(x1)
0  =  0
logA_out_ga(x1, x2)  =  logA_out_ga(x1, x2)
s(x1)  =  s(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
pB_in_gaa(x1, x2, x3)  =  pB_in_gaa(x1)
U3_gaa(x1, x2, x3, x4)  =  U3_gaa(x1, x4)
halfC_in_ga(x1, x2)  =  halfC_in_ga(x1)
halfC_out_ga(x1, x2)  =  halfC_out_ga(x1, x2)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
U4_gaa(x1, x2, x3, x4)  =  U4_gaa(x1, x2, x4)
pB_out_gaa(x1, x2, x3)  =  pB_out_gaa(x1, x2, x3)
LOGA_IN_GA(x1, x2)  =  LOGA_IN_GA(x1)
PB_IN_GAA(x1, x2, x3)  =  PB_IN_GAA(x1)
U3_GAA(x1, x2, x3, x4)  =  U3_GAA(x1, x4)

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:

LOGA_IN_GA(s(s(T10)), s(T7)) → PB_IN_GAA(T10, X18, T7)
PB_IN_GAA(T10, T11, T7) → U3_GAA(T10, T11, T7, halfC_in_ga(T10, T11))
U3_GAA(T10, T11, T7, halfC_out_ga(T10, T11)) → LOGA_IN_GA(s(T11), T7)

The TRS R consists of the following rules:

halfC_in_ga(0, 0) → halfC_out_ga(0, 0)
halfC_in_ga(s(0), 0) → halfC_out_ga(s(0), 0)
halfC_in_ga(s(s(T14)), s(X27)) → U2_ga(T14, X27, halfC_in_ga(T14, X27))
U2_ga(T14, X27, halfC_out_ga(T14, X27)) → halfC_out_ga(s(s(T14)), s(X27))

The argument filtering Pi contains the following mapping:
0  =  0
s(x1)  =  s(x1)
halfC_in_ga(x1, x2)  =  halfC_in_ga(x1)
halfC_out_ga(x1, x2)  =  halfC_out_ga(x1, x2)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
LOGA_IN_GA(x1, x2)  =  LOGA_IN_GA(x1)
PB_IN_GAA(x1, x2, x3)  =  PB_IN_GAA(x1)
U3_GAA(x1, x2, x3, x4)  =  U3_GAA(x1, x4)

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:

LOGA_IN_GA(s(s(T10))) → PB_IN_GAA(T10)
PB_IN_GAA(T10) → U3_GAA(T10, halfC_in_ga(T10))
U3_GAA(T10, halfC_out_ga(T10, T11)) → LOGA_IN_GA(s(T11))

The TRS R consists of the following rules:

halfC_in_ga(0) → halfC_out_ga(0, 0)
halfC_in_ga(s(0)) → halfC_out_ga(s(0), 0)
halfC_in_ga(s(s(T14))) → U2_ga(T14, halfC_in_ga(T14))
U2_ga(T14, halfC_out_ga(T14, X27)) → halfC_out_ga(s(s(T14)), s(X27))

The set Q consists of the following terms:

halfC_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.

LOGA_IN_GA(s(s(T10))) → PB_IN_GAA(T10)

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

POL(0) = 0   
POL(LOGA_IN_GA(x₁)) = x₁   
POL(PB_IN_GAA(x₁)) = 1 + x₁   
POL(U2_ga(x₁, x₂)) = 1 + x₂   
POL(U3_GAA(x₁, x₂)) = 1 + x₂   
POL(halfC_in_ga(x₁)) = x₁   
POL(halfC_out_ga(x₁, x₂)) = x₂   
POL(s(x₁)) = 1 + x₁   

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

halfC_in_ga(0) → halfC_out_ga(0, 0)
halfC_in_ga(s(0)) → halfC_out_ga(s(0), 0)
halfC_in_ga(s(s(T14))) → U2_ga(T14, halfC_in_ga(T14))
U2_ga(T14, halfC_out_ga(T14, X27)) → halfC_out_ga(s(s(T14)), s(X27))

(20) Obligation:

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

PB_IN_GAA(T10) → U3_GAA(T10, halfC_in_ga(T10))
U3_GAA(T10, halfC_out_ga(T10, T11)) → LOGA_IN_GA(s(T11))

The TRS R consists of the following rules:

halfC_in_ga(0) → halfC_out_ga(0, 0)
halfC_in_ga(s(0)) → halfC_out_ga(s(0), 0)
halfC_in_ga(s(s(T14))) → U2_ga(T14, halfC_in_ga(T14))
U2_ga(T14, halfC_out_ga(T14, X27)) → halfC_out_ga(s(s(T14)), s(X27))

The set Q consists of the following terms:

halfC_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