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

(0) Obligation:

Clauses:

sum([], [], []).
sum(.(X1, Y1), .(X2, Y2), .(X3, Y3)) :- ','(add(X1, X2, X3), sum(Y1, Y2, Y3)).
add(0, X, X).
add(s(X), Y, s(Z)) :- add(X, Y, Z).

Query: sum(g,a,g)

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: sum(T1, T2, T3)\n\nT1 is ground\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: sum(T1, T2, T3), SCOPE: 1, CLAUSE: 0 | sum(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: true | sum([], T2, []), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: sum(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nT3 is ground\nsum(T1, T2, T3) !=? sum([], [], [])", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: sum([], T2, []), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: ','(add(T10, T16, T14), sum(T11, T17, T15))\n\nT10 is ground\nT11 is ground\nT14 is ground\nT15 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(add(T10, T16, T14), sum(T11, T17, T15)), SCOPE: 2, CLAUSE: 2 | ','(add(T10, T16, T14), sum(T11, T17, T15)), SCOPE: 2, CLAUSE: 3\n\nT10 is ground\nT11 is ground\nT14 is ground\nT15 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(add(T10, T16, T14), sum(T11, T17, T15)), SCOPE: 2, CLAUSE: 2\n\nT10 is ground\nT11 is ground\nT14 is ground\nT15 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(add(T10, T16, T14), sum(T11, T17, T15)), SCOPE: 2, CLAUSE: 3\n\nT10 is ground\nT11 is ground\nT14 is ground\nT15 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: sum(T11, T23, T15)\n\nT11 is ground\nT15 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(add(T30, T33, T32), sum(T11, T34, T15))\n\nT11 is ground\nT15 is ground\nT30 is ground\nT32 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 16 [arrowhead = none ];
16 -> 2;

17 [label="EVAL with clause\nsum([], [], []).\nand substitutionT1 -> [],\nT2 -> [],\nT3 -> []", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 17 [arrowhead = none ];
17 -> 3;

18 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 18 [arrowhead = none ];
18 -> 4;

19 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 19 [arrowhead = none ];
19 -> 5;

20 [label="EVAL with clause\nsum(.(X16, X17), .(X18, X19), .(X20, X21)) :- ','(add(X16, X18, X20), sum(X17, X19, X21)).\nand substitutionX16 -> T10,\nX17 -> T11,\nT1 -> .(T10, T11),\nX18 -> T16,\nX19 -> T17,\nT2 -> .(T16, T17),\nX20 -> T14,\nX21 -> T15,\nT3 -> .(T14, T15),\nT12 -> T16,\nT13 -> T17", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 20 [arrowhead = none ];
20 -> 7;

21 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 21 [arrowhead = none ];
21 -> 8;

22 [label="BACKTRACK\nfor clause: sum(.(X1, Y1), .(X2, Y2), .(X3, Y3)) :- ','(add(X1, X2, X3), sum(Y1, Y2, Y3))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 22 [arrowhead = none ];
22 -> 6;

23 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 23 [arrowhead = none ];
23 -> 9;

24 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
9 -> 24 [arrowhead = none ];
24 -> 10;

25 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
9 -> 25 [arrowhead = none ];
25 -> 11;

26 [label="EVAL with clause\nadd(0, X26, X26).\nand substitutionT10 -> 0,\nT16 -> T22,\nX26 -> T22,\nT14 -> T22,\nT17 -> T23", fontsize=14, style = filled, fillcolor = lightblue];
10 -> 26 [arrowhead = none ];
26 -> 12;

27 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 27 [arrowhead = none ];
27 -> 13;

28 [label="EVAL with clause\nadd(s(X33), X34, s(X35)) :- add(X33, X34, X35).\nand substitutionX33 -> T30,\nT10 -> s(T30),\nT16 -> T33,\nX34 -> T33,\nX35 -> T32,\nT14 -> s(T32),\nT31 -> T33,\nT17 -> T34", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 28 [arrowhead = none ];
28 -> 14;

29 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 29 [arrowhead = none ];
29 -> 15;

30 [label="INSTANCE with matching:\nT1 -> T11\nT2 -> T23\nT3 -> T15", fontsize=14, style = filled, fillcolor = lightgrey];
12 -> 30 [arrowhead = none , style=dashed];
30 -> 1[style=dashed];

31 [label="INSTANCE with matching:\nT10 -> T30\nT16 -> T33\nT14 -> T32\nT17 -> T34", fontsize=14, style = filled, fillcolor = lightgrey];
14 -> 31 [arrowhead = none , style=dashed];
31 -> 7[style=dashed];

}

(2) Obligation:

Clauses:

pA(0, T22, T22, T11, T23, T15) :- sumB(T11, T23, T15).
pA(s(T30), T33, s(T32), T11, T34, T15) :- pA(T30, T33, T32, T11, T34, T15).
sumB([], [], []).
sumB(.(T10, T11), .(T16, T17), .(T14, T15)) :- pA(T10, T16, T14, T11, T17, T15).

Query: sumB(g,a,g)

(3) PrologToPiTRSProof (SOUND transformation)

We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
sumB_in: (b,f,b)
pA_in: (b,f,b,b,f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

sumB_in_gag([], [], []) → sumB_out_gag([], [], [])
sumB_in_gag(.(T10, T11), .(T16, T17), .(T14, T15)) → U3_gag(T10, T11, T16, T17, T14, T15, pA_in_gaggag(T10, T16, T14, T11, T17, T15))
pA_in_gaggag(0, T22, T22, T11, T23, T15) → U1_gaggag(T22, T11, T23, T15, sumB_in_gag(T11, T23, T15))
U1_gaggag(T22, T11, T23, T15, sumB_out_gag(T11, T23, T15)) → pA_out_gaggag(0, T22, T22, T11, T23, T15)
pA_in_gaggag(s(T30), T33, s(T32), T11, T34, T15) → U2_gaggag(T30, T33, T32, T11, T34, T15, pA_in_gaggag(T30, T33, T32, T11, T34, T15))
U2_gaggag(T30, T33, T32, T11, T34, T15, pA_out_gaggag(T30, T33, T32, T11, T34, T15)) → pA_out_gaggag(s(T30), T33, s(T32), T11, T34, T15)
U3_gag(T10, T11, T16, T17, T14, T15, pA_out_gaggag(T10, T16, T14, T11, T17, T15)) → sumB_out_gag(.(T10, T11), .(T16, T17), .(T14, T15))

The argument filtering Pi contains the following mapping:
sumB_in_gag(x1, x2, x3)  =  sumB_in_gag(x1, x3)
[]  =  []
sumB_out_gag(x1, x2, x3)  =  sumB_out_gag(x2)
.(x1, x2)  =  .(x1, x2)
U3_gag(x1, x2, x3, x4, x5, x6, x7)  =  U3_gag(x7)
pA_in_gaggag(x1, x2, x3, x4, x5, x6)  =  pA_in_gaggag(x1, x3, x4, x6)
0  =  0
U1_gaggag(x1, x2, x3, x4, x5)  =  U1_gaggag(x1, x5)
pA_out_gaggag(x1, x2, x3, x4, x5, x6)  =  pA_out_gaggag(x2, x5)
s(x1)  =  s(x1)
U2_gaggag(x1, x2, x3, x4, x5, x6, x7)  =  U2_gaggag(x7)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

sumB_in_gag([], [], []) → sumB_out_gag([], [], [])
sumB_in_gag(.(T10, T11), .(T16, T17), .(T14, T15)) → U3_gag(T10, T11, T16, T17, T14, T15, pA_in_gaggag(T10, T16, T14, T11, T17, T15))
pA_in_gaggag(0, T22, T22, T11, T23, T15) → U1_gaggag(T22, T11, T23, T15, sumB_in_gag(T11, T23, T15))
U1_gaggag(T22, T11, T23, T15, sumB_out_gag(T11, T23, T15)) → pA_out_gaggag(0, T22, T22, T11, T23, T15)
pA_in_gaggag(s(T30), T33, s(T32), T11, T34, T15) → U2_gaggag(T30, T33, T32, T11, T34, T15, pA_in_gaggag(T30, T33, T32, T11, T34, T15))
U2_gaggag(T30, T33, T32, T11, T34, T15, pA_out_gaggag(T30, T33, T32, T11, T34, T15)) → pA_out_gaggag(s(T30), T33, s(T32), T11, T34, T15)
U3_gag(T10, T11, T16, T17, T14, T15, pA_out_gaggag(T10, T16, T14, T11, T17, T15)) → sumB_out_gag(.(T10, T11), .(T16, T17), .(T14, T15))

The argument filtering Pi contains the following mapping:
sumB_in_gag(x1, x2, x3)  =  sumB_in_gag(x1, x3)
[]  =  []
sumB_out_gag(x1, x2, x3)  =  sumB_out_gag(x2)
.(x1, x2)  =  .(x1, x2)
U3_gag(x1, x2, x3, x4, x5, x6, x7)  =  U3_gag(x7)
pA_in_gaggag(x1, x2, x3, x4, x5, x6)  =  pA_in_gaggag(x1, x3, x4, x6)
0  =  0
U1_gaggag(x1, x2, x3, x4, x5)  =  U1_gaggag(x1, x5)
pA_out_gaggag(x1, x2, x3, x4, x5, x6)  =  pA_out_gaggag(x2, x5)
s(x1)  =  s(x1)
U2_gaggag(x1, x2, x3, x4, x5, x6, x7)  =  U2_gaggag(x7)

(5) 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:

SUMB_IN_GAG(.(T10, T11), .(T16, T17), .(T14, T15)) → U3_GAG(T10, T11, T16, T17, T14, T15, pA_in_gaggag(T10, T16, T14, T11, T17, T15))
SUMB_IN_GAG(.(T10, T11), .(T16, T17), .(T14, T15)) → PA_IN_GAGGAG(T10, T16, T14, T11, T17, T15)
PA_IN_GAGGAG(0, T22, T22, T11, T23, T15) → U1_GAGGAG(T22, T11, T23, T15, sumB_in_gag(T11, T23, T15))
PA_IN_GAGGAG(0, T22, T22, T11, T23, T15) → SUMB_IN_GAG(T11, T23, T15)
PA_IN_GAGGAG(s(T30), T33, s(T32), T11, T34, T15) → U2_GAGGAG(T30, T33, T32, T11, T34, T15, pA_in_gaggag(T30, T33, T32, T11, T34, T15))
PA_IN_GAGGAG(s(T30), T33, s(T32), T11, T34, T15) → PA_IN_GAGGAG(T30, T33, T32, T11, T34, T15)

The TRS R consists of the following rules:

sumB_in_gag([], [], []) → sumB_out_gag([], [], [])
sumB_in_gag(.(T10, T11), .(T16, T17), .(T14, T15)) → U3_gag(T10, T11, T16, T17, T14, T15, pA_in_gaggag(T10, T16, T14, T11, T17, T15))
pA_in_gaggag(0, T22, T22, T11, T23, T15) → U1_gaggag(T22, T11, T23, T15, sumB_in_gag(T11, T23, T15))
U1_gaggag(T22, T11, T23, T15, sumB_out_gag(T11, T23, T15)) → pA_out_gaggag(0, T22, T22, T11, T23, T15)
pA_in_gaggag(s(T30), T33, s(T32), T11, T34, T15) → U2_gaggag(T30, T33, T32, T11, T34, T15, pA_in_gaggag(T30, T33, T32, T11, T34, T15))
U2_gaggag(T30, T33, T32, T11, T34, T15, pA_out_gaggag(T30, T33, T32, T11, T34, T15)) → pA_out_gaggag(s(T30), T33, s(T32), T11, T34, T15)
U3_gag(T10, T11, T16, T17, T14, T15, pA_out_gaggag(T10, T16, T14, T11, T17, T15)) → sumB_out_gag(.(T10, T11), .(T16, T17), .(T14, T15))

The argument filtering Pi contains the following mapping:
sumB_in_gag(x1, x2, x3)  =  sumB_in_gag(x1, x3)
[]  =  []
sumB_out_gag(x1, x2, x3)  =  sumB_out_gag(x2)
.(x1, x2)  =  .(x1, x2)
U3_gag(x1, x2, x3, x4, x5, x6, x7)  =  U3_gag(x7)
pA_in_gaggag(x1, x2, x3, x4, x5, x6)  =  pA_in_gaggag(x1, x3, x4, x6)
0  =  0
U1_gaggag(x1, x2, x3, x4, x5)  =  U1_gaggag(x1, x5)
pA_out_gaggag(x1, x2, x3, x4, x5, x6)  =  pA_out_gaggag(x2, x5)
s(x1)  =  s(x1)
U2_gaggag(x1, x2, x3, x4, x5, x6, x7)  =  U2_gaggag(x7)
SUMB_IN_GAG(x1, x2, x3)  =  SUMB_IN_GAG(x1, x3)
U3_GAG(x1, x2, x3, x4, x5, x6, x7)  =  U3_GAG(x7)
PA_IN_GAGGAG(x1, x2, x3, x4, x5, x6)  =  PA_IN_GAGGAG(x1, x3, x4, x6)
U1_GAGGAG(x1, x2, x3, x4, x5)  =  U1_GAGGAG(x1, x5)
U2_GAGGAG(x1, x2, x3, x4, x5, x6, x7)  =  U2_GAGGAG(x7)

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

(6) Obligation:

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

SUMB_IN_GAG(.(T10, T11), .(T16, T17), .(T14, T15)) → U3_GAG(T10, T11, T16, T17, T14, T15, pA_in_gaggag(T10, T16, T14, T11, T17, T15))
SUMB_IN_GAG(.(T10, T11), .(T16, T17), .(T14, T15)) → PA_IN_GAGGAG(T10, T16, T14, T11, T17, T15)
PA_IN_GAGGAG(0, T22, T22, T11, T23, T15) → U1_GAGGAG(T22, T11, T23, T15, sumB_in_gag(T11, T23, T15))
PA_IN_GAGGAG(0, T22, T22, T11, T23, T15) → SUMB_IN_GAG(T11, T23, T15)
PA_IN_GAGGAG(s(T30), T33, s(T32), T11, T34, T15) → U2_GAGGAG(T30, T33, T32, T11, T34, T15, pA_in_gaggag(T30, T33, T32, T11, T34, T15))
PA_IN_GAGGAG(s(T30), T33, s(T32), T11, T34, T15) → PA_IN_GAGGAG(T30, T33, T32, T11, T34, T15)

The TRS R consists of the following rules:

sumB_in_gag([], [], []) → sumB_out_gag([], [], [])
sumB_in_gag(.(T10, T11), .(T16, T17), .(T14, T15)) → U3_gag(T10, T11, T16, T17, T14, T15, pA_in_gaggag(T10, T16, T14, T11, T17, T15))
pA_in_gaggag(0, T22, T22, T11, T23, T15) → U1_gaggag(T22, T11, T23, T15, sumB_in_gag(T11, T23, T15))
U1_gaggag(T22, T11, T23, T15, sumB_out_gag(T11, T23, T15)) → pA_out_gaggag(0, T22, T22, T11, T23, T15)
pA_in_gaggag(s(T30), T33, s(T32), T11, T34, T15) → U2_gaggag(T30, T33, T32, T11, T34, T15, pA_in_gaggag(T30, T33, T32, T11, T34, T15))
U2_gaggag(T30, T33, T32, T11, T34, T15, pA_out_gaggag(T30, T33, T32, T11, T34, T15)) → pA_out_gaggag(s(T30), T33, s(T32), T11, T34, T15)
U3_gag(T10, T11, T16, T17, T14, T15, pA_out_gaggag(T10, T16, T14, T11, T17, T15)) → sumB_out_gag(.(T10, T11), .(T16, T17), .(T14, T15))

The argument filtering Pi contains the following mapping:
sumB_in_gag(x1, x2, x3)  =  sumB_in_gag(x1, x3)
[]  =  []
sumB_out_gag(x1, x2, x3)  =  sumB_out_gag(x2)
.(x1, x2)  =  .(x1, x2)
U3_gag(x1, x2, x3, x4, x5, x6, x7)  =  U3_gag(x7)
pA_in_gaggag(x1, x2, x3, x4, x5, x6)  =  pA_in_gaggag(x1, x3, x4, x6)
0  =  0
U1_gaggag(x1, x2, x3, x4, x5)  =  U1_gaggag(x1, x5)
pA_out_gaggag(x1, x2, x3, x4, x5, x6)  =  pA_out_gaggag(x2, x5)
s(x1)  =  s(x1)
U2_gaggag(x1, x2, x3, x4, x5, x6, x7)  =  U2_gaggag(x7)
SUMB_IN_GAG(x1, x2, x3)  =  SUMB_IN_GAG(x1, x3)
U3_GAG(x1, x2, x3, x4, x5, x6, x7)  =  U3_GAG(x7)
PA_IN_GAGGAG(x1, x2, x3, x4, x5, x6)  =  PA_IN_GAGGAG(x1, x3, x4, x6)
U1_GAGGAG(x1, x2, x3, x4, x5)  =  U1_GAGGAG(x1, x5)
U2_GAGGAG(x1, x2, x3, x4, x5, x6, x7)  =  U2_GAGGAG(x7)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes.

(8) Obligation:

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

SUMB_IN_GAG(.(T10, T11), .(T16, T17), .(T14, T15)) → PA_IN_GAGGAG(T10, T16, T14, T11, T17, T15)
PA_IN_GAGGAG(0, T22, T22, T11, T23, T15) → SUMB_IN_GAG(T11, T23, T15)
PA_IN_GAGGAG(s(T30), T33, s(T32), T11, T34, T15) → PA_IN_GAGGAG(T30, T33, T32, T11, T34, T15)

The TRS R consists of the following rules:

sumB_in_gag([], [], []) → sumB_out_gag([], [], [])
sumB_in_gag(.(T10, T11), .(T16, T17), .(T14, T15)) → U3_gag(T10, T11, T16, T17, T14, T15, pA_in_gaggag(T10, T16, T14, T11, T17, T15))
pA_in_gaggag(0, T22, T22, T11, T23, T15) → U1_gaggag(T22, T11, T23, T15, sumB_in_gag(T11, T23, T15))
U1_gaggag(T22, T11, T23, T15, sumB_out_gag(T11, T23, T15)) → pA_out_gaggag(0, T22, T22, T11, T23, T15)
pA_in_gaggag(s(T30), T33, s(T32), T11, T34, T15) → U2_gaggag(T30, T33, T32, T11, T34, T15, pA_in_gaggag(T30, T33, T32, T11, T34, T15))
U2_gaggag(T30, T33, T32, T11, T34, T15, pA_out_gaggag(T30, T33, T32, T11, T34, T15)) → pA_out_gaggag(s(T30), T33, s(T32), T11, T34, T15)
U3_gag(T10, T11, T16, T17, T14, T15, pA_out_gaggag(T10, T16, T14, T11, T17, T15)) → sumB_out_gag(.(T10, T11), .(T16, T17), .(T14, T15))

The argument filtering Pi contains the following mapping:
sumB_in_gag(x1, x2, x3)  =  sumB_in_gag(x1, x3)
[]  =  []
sumB_out_gag(x1, x2, x3)  =  sumB_out_gag(x2)
.(x1, x2)  =  .(x1, x2)
U3_gag(x1, x2, x3, x4, x5, x6, x7)  =  U3_gag(x7)
pA_in_gaggag(x1, x2, x3, x4, x5, x6)  =  pA_in_gaggag(x1, x3, x4, x6)
0  =  0
U1_gaggag(x1, x2, x3, x4, x5)  =  U1_gaggag(x1, x5)
pA_out_gaggag(x1, x2, x3, x4, x5, x6)  =  pA_out_gaggag(x2, x5)
s(x1)  =  s(x1)
U2_gaggag(x1, x2, x3, x4, x5, x6, x7)  =  U2_gaggag(x7)
SUMB_IN_GAG(x1, x2, x3)  =  SUMB_IN_GAG(x1, x3)
PA_IN_GAGGAG(x1, x2, x3, x4, x5, x6)  =  PA_IN_GAGGAG(x1, x3, x4, x6)

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

(9) UsableRulesProof (EQUIVALENT transformation)

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

(10) Obligation:

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

SUMB_IN_GAG(.(T10, T11), .(T16, T17), .(T14, T15)) → PA_IN_GAGGAG(T10, T16, T14, T11, T17, T15)
PA_IN_GAGGAG(0, T22, T22, T11, T23, T15) → SUMB_IN_GAG(T11, T23, T15)
PA_IN_GAGGAG(s(T30), T33, s(T32), T11, T34, T15) → PA_IN_GAGGAG(T30, T33, T32, T11, T34, T15)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
0  =  0
s(x1)  =  s(x1)
SUMB_IN_GAG(x1, x2, x3)  =  SUMB_IN_GAG(x1, x3)
PA_IN_GAGGAG(x1, x2, x3, x4, x5, x6)  =  PA_IN_GAGGAG(x1, x3, x4, x6)

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

(11) PiDPToQDPProof (SOUND transformation)

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

(12) Obligation:

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

SUMB_IN_GAG(.(T10, T11), .(T14, T15)) → PA_IN_GAGGAG(T10, T14, T11, T15)
PA_IN_GAGGAG(0, T22, T11, T15) → SUMB_IN_GAG(T11, T15)
PA_IN_GAGGAG(s(T30), s(T32), T11, T15) → PA_IN_GAGGAG(T30, T32, T11, T15)

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

(13) 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:

• PA_IN_GAGGAG(0, T22, T11, T15) → SUMB_IN_GAG(T11, T15)
  The graph contains the following edges 3 >= 1, 4 >= 2

• PA_IN_GAGGAG(s(T30), s(T32), T11, T15) → PA_IN_GAGGAG(T30, T32, T11, T15)
  The graph contains the following edges 1 > 1, 2 > 2, 3 >= 3, 4 >= 4

• SUMB_IN_GAG(.(T10, T11), .(T14, T15)) → PA_IN_GAGGAG(T10, T14, T11, T15)
  The graph contains the following edges 1 > 1, 2 > 2, 1 > 3, 2 > 4

(14) YES