Left Termination of the query pattern
len1(g,a)
w.r.t. the given Prolog program could successfully be proven:

(0) Obligation:

Clauses:

len1([], 0).
len1(.(X1, Ts), N) :- ','(len1(Ts, M), eq(N, s(M))).
eq(X, X).

Query: len1(g,a)

(1) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: len1(T1, T2)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
4 [label="4: len1(T1, T2), SCOPE: 1, CLAUSE: 0 | len1(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: true | len1([], T2), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: len1(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nlen1(T1, T2) !=? len1([], 0)", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: len1([], T2), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: ','(len1(T7, X11), eq(T9, s(X11)))\n\nT7 is ground\nX11 is free", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: ','(len1(T7, X11), eq(T9, s(X11))), SCOPE: 2, CLAUSE: 0 | ','(len1(T7, X11), eq(T9, s(X11))), SCOPE: 2, CLAUSE: 1\n\nT7 is ground\nX11 is free", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: ','(len1(T7, X11), eq(T9, s(X11))), SCOPE: 2, CLAUSE: 0\n\nT7 is ground\nX11 is free", fontsize=16, style=filled, fillcolor=lightyellow];
30 [label="30: ','(len1(T7, X11), eq(T9, s(X11))), SCOPE: 2, CLAUSE: 1\n\nT7 is ground\nX11 is free", fontsize=16, style=filled, fillcolor=lightyellow];
32 [label="32: eq(T9, s(0))\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
34 [label="34: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
36 [label="36: eq(T9, s(0)), SCOPE: 3, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
38 [label="38: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
40 [label="40: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
42 [label="42: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
44 [label="B: ','(','(len1(T18, X27), eq(X28, s(X27))), eq(T9, s(X28)))\n\nT18 is ground\nX28 is free\nX27 is free", fontsize=16, style=filled, fillcolor=lightyellow];
46 [label="46: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
48 [label="C: len1(T18, X27)\n\nT18 is ground\nX27 is free", fontsize=16, style=filled, fillcolor=lightyellow];
50 [label="E: ','(eq(X28, s(T19)), eq(T9, s(X28)))\n\nT19 is ground\nX28 is free", fontsize=16, style=filled, fillcolor=lightyellow];
54 [label="54: len1(T18, X27), SCOPE: 4, CLAUSE: 0 | len1(T18, X27), SCOPE: 4, CLAUSE: 1\n\nT18 is ground\nX27 is free", fontsize=16, style=filled, fillcolor=lightyellow];
57 [label="57: len1(T18, X27), SCOPE: 4, CLAUSE: 0\n\nT18 is ground\nX27 is free", fontsize=16, style=filled, fillcolor=lightyellow];
59 [label="59: len1(T18, X27), SCOPE: 4, CLAUSE: 1\n\nT18 is ground\nX27 is free", fontsize=16, style=filled, fillcolor=lightyellow];
63 [label="63: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
66 [label="66: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
68 [label="68: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
72 [label="D: ','(len1(T25, X41), eq(X42, s(X41)))\n\nT25 is ground\nX42 is free\nX41 is free", fontsize=16, style=filled, fillcolor=lightyellow];
74 [label="74: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
78 [label="78: len1(T25, X41)\n\nT25 is ground\nX41 is free", fontsize=16, style=filled, fillcolor=lightyellow];
79 [label="F: eq(X42, s(T26))\n\nT26 is ground\nX42 is free", fontsize=16, style=filled, fillcolor=lightyellow];
83 [label="83: eq(X42, s(T26)), SCOPE: 5, CLAUSE: 2\n\nT26 is ground\nX42 is free", fontsize=16, style=filled, fillcolor=lightyellow];
88 [label="88: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
90 [label="90: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
93 [label="93: ','(eq(X28, s(T19)), eq(T9, s(X28))), SCOPE: 6, CLAUSE: 2\n\nT19 is ground\nX28 is free", fontsize=16, style=filled, fillcolor=lightyellow];
99 [label="99: eq(T9, s(s(T32)))\n\nT32 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
101 [label="101: eq(T9, s(s(T32))), SCOPE: 7, CLAUSE: 2\n\nT32 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
103 [label="103: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
104 [label="104: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
106 [label="106: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
107 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 107 [arrowhead = none ];
107 -> 4;

108 [label="EVAL with clause\nlen1([], 0).\nand substitutionT1 -> [],\nT2 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 108 [arrowhead = none ];
108 -> 10;

109 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 109 [arrowhead = none ];
109 -> 13;

110 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 110 [arrowhead = none ];
110 -> 16;

111 [label="EVAL with clause\nlen1(.(X8, X9), X10) :- ','(len1(X9, X11), eq(X10, s(X11))).\nand substitutionX8 -> T6,\nX9 -> T7,\nT1 -> .(T6, T7),\nT2 -> T9,\nX10 -> T9,\nT8 -> T9", fontsize=14, style = filled, fillcolor = lightblue];
13 -> 111 [arrowhead = none ];
111 -> 20;

112 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
13 -> 112 [arrowhead = none ];
112 -> 22;

113 [label="BACKTRACK\nfor clause: len1(.(X1, Ts), N) :- ','(len1(Ts, M), eq(N, s(M)))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
16 -> 113 [arrowhead = none ];
113 -> 18;

114 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
20 -> 114 [arrowhead = none ];
114 -> 25;

115 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
25 -> 115 [arrowhead = none ];
115 -> 28;

116 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
25 -> 116 [arrowhead = none ];
116 -> 30;

117 [label="EVAL with clause\nlen1([], 0).\nand substitutionT7 -> [],\nX11 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
28 -> 117 [arrowhead = none ];
117 -> 32;

118 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
28 -> 118 [arrowhead = none ];
118 -> 34;

119 [label="EVAL with clause\nlen1(.(X24, X25), X26) :- ','(len1(X25, X27), eq(X26, s(X27))).\nand substitutionX24 -> T17,\nX25 -> T18,\nT7 -> .(T17, T18),\nX11 -> X28,\nX26 -> X28", fontsize=14, style = filled, fillcolor = lightblue];
30 -> 119 [arrowhead = none ];
119 -> 44;

120 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
30 -> 120 [arrowhead = none ];
120 -> 46;

121 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
32 -> 121 [arrowhead = none ];
121 -> 36;

122 [label="EVAL with clause\neq(X14, X14).\nand substitutionT9 -> s(0),\nX14 -> s(0),\nT12 -> s(0)", fontsize=14, style = filled, fillcolor = lightblue];
36 -> 122 [arrowhead = none ];
122 -> 38;

123 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
36 -> 123 [arrowhead = none ];
123 -> 40;

124 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
38 -> 124 [arrowhead = none ];
124 -> 42;

125 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
44 -> 125 [arrowhead = none ];
125 -> 48;

126 [label="SPLIT 2\nnew knowledge:\nT18 is ground\nT19 is ground\nreplacements:X27 -> T19", fontsize=14, style = filled, fillcolor = violet];
44 -> 126 [arrowhead = none ];
126 -> 50;

127 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
48 -> 127 [arrowhead = none ];
127 -> 54;

128 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
50 -> 128 [arrowhead = none ];
128 -> 93;

129 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
54 -> 129 [arrowhead = none ];
129 -> 57;

130 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
54 -> 130 [arrowhead = none ];
130 -> 59;

131 [label="EVAL with clause\nlen1([], 0).\nand substitutionT18 -> [],\nX27 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
57 -> 131 [arrowhead = none ];
131 -> 63;

132 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
57 -> 132 [arrowhead = none ];
132 -> 66;

133 [label="EVAL with clause\nlen1(.(X38, X39), X40) :- ','(len1(X39, X41), eq(X40, s(X41))).\nand substitutionX38 -> T24,\nX39 -> T25,\nT18 -> .(T24, T25),\nX27 -> X42,\nX40 -> X42", fontsize=14, style = filled, fillcolor = lightblue];
59 -> 133 [arrowhead = none ];
133 -> 72;

134 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
59 -> 134 [arrowhead = none ];
134 -> 74;

135 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
63 -> 135 [arrowhead = none ];
135 -> 68;

136 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
72 -> 136 [arrowhead = none ];
136 -> 78;

137 [label="SPLIT 2\nnew knowledge:\nT25 is ground\nT26 is ground\nreplacements:X41 -> T26", fontsize=14, style = filled, fillcolor = violet];
72 -> 137 [arrowhead = none ];
137 -> 79;

138 [label="INSTANCE with matching:\nT18 -> T25\nX27 -> X41", fontsize=14, style = filled, fillcolor = lightgrey];
78 -> 138 [arrowhead = none , style=dashed];
138 -> 48[style=dashed];

139 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
79 -> 139 [arrowhead = none ];
139 -> 83;

140 [label="ONLY EVAL with clause\neq(X47, X47).\nand substitutionX42 -> s(T29),\nX47 -> s(T29),\nT26 -> T29,\nX48 -> s(T29)", fontsize=14, style = filled, fillcolor = lightblue];
83 -> 140 [arrowhead = none ];
140 -> 88;

141 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
88 -> 141 [arrowhead = none ];
141 -> 90;

142 [label="ONLY EVAL with clause\neq(X53, X53).\nand substitutionX28 -> s(T32),\nX53 -> s(T32),\nT19 -> T32,\nX54 -> s(T32)", fontsize=14, style = filled, fillcolor = lightblue];
93 -> 142 [arrowhead = none ];
142 -> 99;

143 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
99 -> 143 [arrowhead = none ];
143 -> 101;

144 [label="EVAL with clause\neq(X57, X57).\nand substitutionT9 -> s(s(T38)),\nX57 -> s(s(T38)),\nT32 -> T38,\nT37 -> s(s(T38))", fontsize=14, style = filled, fillcolor = lightblue];
101 -> 144 [arrowhead = none ];
144 -> 103;

145 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
101 -> 145 [arrowhead = none ];
145 -> 104;

146 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
103 -> 146 [arrowhead = none ];
146 -> 106;

}

(2) Obligation:

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

len1A_in_ga([], 0) → len1A_out_ga([], 0)
len1A_in_ga(.(T6, []), s(0)) → len1A_out_ga(.(T6, []), s(0))
len1A_in_ga(.(T6, .(T17, T18)), T9) → U1_ga(T6, T17, T18, T9, pB_in_gaaa(T18, X27, X28, T9))
pB_in_gaaa(T18, T19, X28, T9) → U3_gaaa(T18, T19, X28, T9, len1C_in_ga(T18, T19))
len1C_in_ga([], 0) → len1C_out_ga([], 0)
len1C_in_ga(.(T24, T25), X42) → U2_ga(T24, T25, X42, pD_in_gaa(T25, X41, X42))
pD_in_gaa(T25, T26, X42) → U5_gaa(T25, T26, X42, len1C_in_ga(T25, T26))
U5_gaa(T25, T26, X42, len1C_out_ga(T25, T26)) → U6_gaa(T25, T26, X42, eqF_in_ag(X42, T26))
eqF_in_ag(s(T29), T29) → eqF_out_ag(s(T29), T29)
U6_gaa(T25, T26, X42, eqF_out_ag(X42, T26)) → pD_out_gaa(T25, T26, X42)
U2_ga(T24, T25, X42, pD_out_gaa(T25, X41, X42)) → len1C_out_ga(.(T24, T25), X42)
U3_gaaa(T18, T19, X28, T9, len1C_out_ga(T18, T19)) → U4_gaaa(T18, T19, X28, T9, pE_in_aga(X28, T19, T9))
pE_in_aga(s(T38), T38, s(s(T38))) → pE_out_aga(s(T38), T38, s(s(T38)))
U4_gaaa(T18, T19, X28, T9, pE_out_aga(X28, T19, T9)) → pB_out_gaaa(T18, T19, X28, T9)
U1_ga(T6, T17, T18, T9, pB_out_gaaa(T18, X27, X28, T9)) → len1A_out_ga(.(T6, .(T17, T18)), T9)

The argument filtering Pi contains the following mapping:
len1A_in_ga(x1, x2)  =  len1A_in_ga(x1)
[]  =  []
len1A_out_ga(x1, x2)  =  len1A_out_ga(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x1, x2, x3, x5)
pB_in_gaaa(x1, x2, x3, x4)  =  pB_in_gaaa(x1)
U3_gaaa(x1, x2, x3, x4, x5)  =  U3_gaaa(x1, x5)
len1C_in_ga(x1, x2)  =  len1C_in_ga(x1)
len1C_out_ga(x1, x2)  =  len1C_out_ga(x1, x2)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x1, x2, x4)
pD_in_gaa(x1, x2, x3)  =  pD_in_gaa(x1)
U5_gaa(x1, x2, x3, x4)  =  U5_gaa(x1, x4)
U6_gaa(x1, x2, x3, x4)  =  U6_gaa(x1, x2, x4)
eqF_in_ag(x1, x2)  =  eqF_in_ag(x2)
eqF_out_ag(x1, x2)  =  eqF_out_ag(x1, x2)
pD_out_gaa(x1, x2, x3)  =  pD_out_gaa(x1, x2, x3)
U4_gaaa(x1, x2, x3, x4, x5)  =  U4_gaaa(x1, x2, x5)
pE_in_aga(x1, x2, x3)  =  pE_in_aga(x2)
pE_out_aga(x1, x2, x3)  =  pE_out_aga(x1, x2, x3)
pB_out_gaaa(x1, x2, x3, x4)  =  pB_out_gaaa(x1, x2, x3, 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:

LEN1A_IN_GA(.(T6, .(T17, T18)), T9) → U1_GA(T6, T17, T18, T9, pB_in_gaaa(T18, X27, X28, T9))
LEN1A_IN_GA(.(T6, .(T17, T18)), T9) → PB_IN_GAAA(T18, X27, X28, T9)
PB_IN_GAAA(T18, T19, X28, T9) → U3_GAAA(T18, T19, X28, T9, len1C_in_ga(T18, T19))
PB_IN_GAAA(T18, T19, X28, T9) → LEN1C_IN_GA(T18, T19)
LEN1C_IN_GA(.(T24, T25), X42) → U2_GA(T24, T25, X42, pD_in_gaa(T25, X41, X42))
LEN1C_IN_GA(.(T24, T25), X42) → PD_IN_GAA(T25, X41, X42)
PD_IN_GAA(T25, T26, X42) → U5_GAA(T25, T26, X42, len1C_in_ga(T25, T26))
PD_IN_GAA(T25, T26, X42) → LEN1C_IN_GA(T25, T26)
U5_GAA(T25, T26, X42, len1C_out_ga(T25, T26)) → U6_GAA(T25, T26, X42, eqF_in_ag(X42, T26))
U5_GAA(T25, T26, X42, len1C_out_ga(T25, T26)) → EQF_IN_AG(X42, T26)
U3_GAAA(T18, T19, X28, T9, len1C_out_ga(T18, T19)) → U4_GAAA(T18, T19, X28, T9, pE_in_aga(X28, T19, T9))
U3_GAAA(T18, T19, X28, T9, len1C_out_ga(T18, T19)) → PE_IN_AGA(X28, T19, T9)

The TRS R consists of the following rules:

len1A_in_ga([], 0) → len1A_out_ga([], 0)
len1A_in_ga(.(T6, []), s(0)) → len1A_out_ga(.(T6, []), s(0))
len1A_in_ga(.(T6, .(T17, T18)), T9) → U1_ga(T6, T17, T18, T9, pB_in_gaaa(T18, X27, X28, T9))
pB_in_gaaa(T18, T19, X28, T9) → U3_gaaa(T18, T19, X28, T9, len1C_in_ga(T18, T19))
len1C_in_ga([], 0) → len1C_out_ga([], 0)
len1C_in_ga(.(T24, T25), X42) → U2_ga(T24, T25, X42, pD_in_gaa(T25, X41, X42))
pD_in_gaa(T25, T26, X42) → U5_gaa(T25, T26, X42, len1C_in_ga(T25, T26))
U5_gaa(T25, T26, X42, len1C_out_ga(T25, T26)) → U6_gaa(T25, T26, X42, eqF_in_ag(X42, T26))
eqF_in_ag(s(T29), T29) → eqF_out_ag(s(T29), T29)
U6_gaa(T25, T26, X42, eqF_out_ag(X42, T26)) → pD_out_gaa(T25, T26, X42)
U2_ga(T24, T25, X42, pD_out_gaa(T25, X41, X42)) → len1C_out_ga(.(T24, T25), X42)
U3_gaaa(T18, T19, X28, T9, len1C_out_ga(T18, T19)) → U4_gaaa(T18, T19, X28, T9, pE_in_aga(X28, T19, T9))
pE_in_aga(s(T38), T38, s(s(T38))) → pE_out_aga(s(T38), T38, s(s(T38)))
U4_gaaa(T18, T19, X28, T9, pE_out_aga(X28, T19, T9)) → pB_out_gaaa(T18, T19, X28, T9)
U1_ga(T6, T17, T18, T9, pB_out_gaaa(T18, X27, X28, T9)) → len1A_out_ga(.(T6, .(T17, T18)), T9)

The argument filtering Pi contains the following mapping:
len1A_in_ga(x1, x2)  =  len1A_in_ga(x1)
[]  =  []
len1A_out_ga(x1, x2)  =  len1A_out_ga(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x1, x2, x3, x5)
pB_in_gaaa(x1, x2, x3, x4)  =  pB_in_gaaa(x1)
U3_gaaa(x1, x2, x3, x4, x5)  =  U3_gaaa(x1, x5)
len1C_in_ga(x1, x2)  =  len1C_in_ga(x1)
len1C_out_ga(x1, x2)  =  len1C_out_ga(x1, x2)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x1, x2, x4)
pD_in_gaa(x1, x2, x3)  =  pD_in_gaa(x1)
U5_gaa(x1, x2, x3, x4)  =  U5_gaa(x1, x4)
U6_gaa(x1, x2, x3, x4)  =  U6_gaa(x1, x2, x4)
eqF_in_ag(x1, x2)  =  eqF_in_ag(x2)
eqF_out_ag(x1, x2)  =  eqF_out_ag(x1, x2)
pD_out_gaa(x1, x2, x3)  =  pD_out_gaa(x1, x2, x3)
U4_gaaa(x1, x2, x3, x4, x5)  =  U4_gaaa(x1, x2, x5)
pE_in_aga(x1, x2, x3)  =  pE_in_aga(x2)
pE_out_aga(x1, x2, x3)  =  pE_out_aga(x1, x2, x3)
pB_out_gaaa(x1, x2, x3, x4)  =  pB_out_gaaa(x1, x2, x3, x4)
LEN1A_IN_GA(x1, x2)  =  LEN1A_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5)  =  U1_GA(x1, x2, x3, x5)
PB_IN_GAAA(x1, x2, x3, x4)  =  PB_IN_GAAA(x1)
U3_GAAA(x1, x2, x3, x4, x5)  =  U3_GAAA(x1, x5)
LEN1C_IN_GA(x1, x2)  =  LEN1C_IN_GA(x1)
U2_GA(x1, x2, x3, x4)  =  U2_GA(x1, x2, x4)
PD_IN_GAA(x1, x2, x3)  =  PD_IN_GAA(x1)
U5_GAA(x1, x2, x3, x4)  =  U5_GAA(x1, x4)
U6_GAA(x1, x2, x3, x4)  =  U6_GAA(x1, x2, x4)
EQF_IN_AG(x1, x2)  =  EQF_IN_AG(x2)
U4_GAAA(x1, x2, x3, x4, x5)  =  U4_GAAA(x1, x2, x5)
PE_IN_AGA(x1, x2, x3)  =  PE_IN_AGA(x2)

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

(4) Obligation:

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

LEN1A_IN_GA(.(T6, .(T17, T18)), T9) → U1_GA(T6, T17, T18, T9, pB_in_gaaa(T18, X27, X28, T9))
LEN1A_IN_GA(.(T6, .(T17, T18)), T9) → PB_IN_GAAA(T18, X27, X28, T9)
PB_IN_GAAA(T18, T19, X28, T9) → U3_GAAA(T18, T19, X28, T9, len1C_in_ga(T18, T19))
PB_IN_GAAA(T18, T19, X28, T9) → LEN1C_IN_GA(T18, T19)
LEN1C_IN_GA(.(T24, T25), X42) → U2_GA(T24, T25, X42, pD_in_gaa(T25, X41, X42))
LEN1C_IN_GA(.(T24, T25), X42) → PD_IN_GAA(T25, X41, X42)
PD_IN_GAA(T25, T26, X42) → U5_GAA(T25, T26, X42, len1C_in_ga(T25, T26))
PD_IN_GAA(T25, T26, X42) → LEN1C_IN_GA(T25, T26)
U5_GAA(T25, T26, X42, len1C_out_ga(T25, T26)) → U6_GAA(T25, T26, X42, eqF_in_ag(X42, T26))
U5_GAA(T25, T26, X42, len1C_out_ga(T25, T26)) → EQF_IN_AG(X42, T26)
U3_GAAA(T18, T19, X28, T9, len1C_out_ga(T18, T19)) → U4_GAAA(T18, T19, X28, T9, pE_in_aga(X28, T19, T9))
U3_GAAA(T18, T19, X28, T9, len1C_out_ga(T18, T19)) → PE_IN_AGA(X28, T19, T9)

The TRS R consists of the following rules:

len1A_in_ga([], 0) → len1A_out_ga([], 0)
len1A_in_ga(.(T6, []), s(0)) → len1A_out_ga(.(T6, []), s(0))
len1A_in_ga(.(T6, .(T17, T18)), T9) → U1_ga(T6, T17, T18, T9, pB_in_gaaa(T18, X27, X28, T9))
pB_in_gaaa(T18, T19, X28, T9) → U3_gaaa(T18, T19, X28, T9, len1C_in_ga(T18, T19))
len1C_in_ga([], 0) → len1C_out_ga([], 0)
len1C_in_ga(.(T24, T25), X42) → U2_ga(T24, T25, X42, pD_in_gaa(T25, X41, X42))
pD_in_gaa(T25, T26, X42) → U5_gaa(T25, T26, X42, len1C_in_ga(T25, T26))
U5_gaa(T25, T26, X42, len1C_out_ga(T25, T26)) → U6_gaa(T25, T26, X42, eqF_in_ag(X42, T26))
eqF_in_ag(s(T29), T29) → eqF_out_ag(s(T29), T29)
U6_gaa(T25, T26, X42, eqF_out_ag(X42, T26)) → pD_out_gaa(T25, T26, X42)
U2_ga(T24, T25, X42, pD_out_gaa(T25, X41, X42)) → len1C_out_ga(.(T24, T25), X42)
U3_gaaa(T18, T19, X28, T9, len1C_out_ga(T18, T19)) → U4_gaaa(T18, T19, X28, T9, pE_in_aga(X28, T19, T9))
pE_in_aga(s(T38), T38, s(s(T38))) → pE_out_aga(s(T38), T38, s(s(T38)))
U4_gaaa(T18, T19, X28, T9, pE_out_aga(X28, T19, T9)) → pB_out_gaaa(T18, T19, X28, T9)
U1_ga(T6, T17, T18, T9, pB_out_gaaa(T18, X27, X28, T9)) → len1A_out_ga(.(T6, .(T17, T18)), T9)

The argument filtering Pi contains the following mapping:
len1A_in_ga(x1, x2)  =  len1A_in_ga(x1)
[]  =  []
len1A_out_ga(x1, x2)  =  len1A_out_ga(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x1, x2, x3, x5)
pB_in_gaaa(x1, x2, x3, x4)  =  pB_in_gaaa(x1)
U3_gaaa(x1, x2, x3, x4, x5)  =  U3_gaaa(x1, x5)
len1C_in_ga(x1, x2)  =  len1C_in_ga(x1)
len1C_out_ga(x1, x2)  =  len1C_out_ga(x1, x2)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x1, x2, x4)
pD_in_gaa(x1, x2, x3)  =  pD_in_gaa(x1)
U5_gaa(x1, x2, x3, x4)  =  U5_gaa(x1, x4)
U6_gaa(x1, x2, x3, x4)  =  U6_gaa(x1, x2, x4)
eqF_in_ag(x1, x2)  =  eqF_in_ag(x2)
eqF_out_ag(x1, x2)  =  eqF_out_ag(x1, x2)
pD_out_gaa(x1, x2, x3)  =  pD_out_gaa(x1, x2, x3)
U4_gaaa(x1, x2, x3, x4, x5)  =  U4_gaaa(x1, x2, x5)
pE_in_aga(x1, x2, x3)  =  pE_in_aga(x2)
pE_out_aga(x1, x2, x3)  =  pE_out_aga(x1, x2, x3)
pB_out_gaaa(x1, x2, x3, x4)  =  pB_out_gaaa(x1, x2, x3, x4)
LEN1A_IN_GA(x1, x2)  =  LEN1A_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5)  =  U1_GA(x1, x2, x3, x5)
PB_IN_GAAA(x1, x2, x3, x4)  =  PB_IN_GAAA(x1)
U3_GAAA(x1, x2, x3, x4, x5)  =  U3_GAAA(x1, x5)
LEN1C_IN_GA(x1, x2)  =  LEN1C_IN_GA(x1)
U2_GA(x1, x2, x3, x4)  =  U2_GA(x1, x2, x4)
PD_IN_GAA(x1, x2, x3)  =  PD_IN_GAA(x1)
U5_GAA(x1, x2, x3, x4)  =  U5_GAA(x1, x4)
U6_GAA(x1, x2, x3, x4)  =  U6_GAA(x1, x2, x4)
EQF_IN_AG(x1, x2)  =  EQF_IN_AG(x2)
U4_GAAA(x1, x2, x3, x4, x5)  =  U4_GAAA(x1, x2, x5)
PE_IN_AGA(x1, x2, x3)  =  PE_IN_AGA(x2)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 10 less nodes.

(6) Obligation:

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

LEN1C_IN_GA(.(T24, T25), X42) → PD_IN_GAA(T25, X41, X42)
PD_IN_GAA(T25, T26, X42) → LEN1C_IN_GA(T25, T26)

The TRS R consists of the following rules:

len1A_in_ga([], 0) → len1A_out_ga([], 0)
len1A_in_ga(.(T6, []), s(0)) → len1A_out_ga(.(T6, []), s(0))
len1A_in_ga(.(T6, .(T17, T18)), T9) → U1_ga(T6, T17, T18, T9, pB_in_gaaa(T18, X27, X28, T9))
pB_in_gaaa(T18, T19, X28, T9) → U3_gaaa(T18, T19, X28, T9, len1C_in_ga(T18, T19))
len1C_in_ga([], 0) → len1C_out_ga([], 0)
len1C_in_ga(.(T24, T25), X42) → U2_ga(T24, T25, X42, pD_in_gaa(T25, X41, X42))
pD_in_gaa(T25, T26, X42) → U5_gaa(T25, T26, X42, len1C_in_ga(T25, T26))
U5_gaa(T25, T26, X42, len1C_out_ga(T25, T26)) → U6_gaa(T25, T26, X42, eqF_in_ag(X42, T26))
eqF_in_ag(s(T29), T29) → eqF_out_ag(s(T29), T29)
U6_gaa(T25, T26, X42, eqF_out_ag(X42, T26)) → pD_out_gaa(T25, T26, X42)
U2_ga(T24, T25, X42, pD_out_gaa(T25, X41, X42)) → len1C_out_ga(.(T24, T25), X42)
U3_gaaa(T18, T19, X28, T9, len1C_out_ga(T18, T19)) → U4_gaaa(T18, T19, X28, T9, pE_in_aga(X28, T19, T9))
pE_in_aga(s(T38), T38, s(s(T38))) → pE_out_aga(s(T38), T38, s(s(T38)))
U4_gaaa(T18, T19, X28, T9, pE_out_aga(X28, T19, T9)) → pB_out_gaaa(T18, T19, X28, T9)
U1_ga(T6, T17, T18, T9, pB_out_gaaa(T18, X27, X28, T9)) → len1A_out_ga(.(T6, .(T17, T18)), T9)

The argument filtering Pi contains the following mapping:
len1A_in_ga(x1, x2)  =  len1A_in_ga(x1)
[]  =  []
len1A_out_ga(x1, x2)  =  len1A_out_ga(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x1, x2, x3, x5)
pB_in_gaaa(x1, x2, x3, x4)  =  pB_in_gaaa(x1)
U3_gaaa(x1, x2, x3, x4, x5)  =  U3_gaaa(x1, x5)
len1C_in_ga(x1, x2)  =  len1C_in_ga(x1)
len1C_out_ga(x1, x2)  =  len1C_out_ga(x1, x2)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x1, x2, x4)
pD_in_gaa(x1, x2, x3)  =  pD_in_gaa(x1)
U5_gaa(x1, x2, x3, x4)  =  U5_gaa(x1, x4)
U6_gaa(x1, x2, x3, x4)  =  U6_gaa(x1, x2, x4)
eqF_in_ag(x1, x2)  =  eqF_in_ag(x2)
eqF_out_ag(x1, x2)  =  eqF_out_ag(x1, x2)
pD_out_gaa(x1, x2, x3)  =  pD_out_gaa(x1, x2, x3)
U4_gaaa(x1, x2, x3, x4, x5)  =  U4_gaaa(x1, x2, x5)
pE_in_aga(x1, x2, x3)  =  pE_in_aga(x2)
pE_out_aga(x1, x2, x3)  =  pE_out_aga(x1, x2, x3)
pB_out_gaaa(x1, x2, x3, x4)  =  pB_out_gaaa(x1, x2, x3, x4)
LEN1C_IN_GA(x1, x2)  =  LEN1C_IN_GA(x1)
PD_IN_GAA(x1, x2, x3)  =  PD_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:

LEN1C_IN_GA(.(T24, T25), X42) → PD_IN_GAA(T25, X41, X42)
PD_IN_GAA(T25, T26, X42) → LEN1C_IN_GA(T25, T26)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
LEN1C_IN_GA(x1, x2)  =  LEN1C_IN_GA(x1)
PD_IN_GAA(x1, x2, x3)  =  PD_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:

LEN1C_IN_GA(.(T24, T25)) → PD_IN_GAA(T25)
PD_IN_GAA(T25) → LEN1C_IN_GA(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:

• PD_IN_GAA(T25) → LEN1C_IN_GA(T25)
  The graph contains the following edges 1 >= 1

• LEN1C_IN_GA(.(T24, T25)) → PD_IN_GAA(T25)
  The graph contains the following edges 1 > 1

(12) YES