Left Termination of the query pattern
sum(g,g,a)
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,g,a)

(1) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: sum(T1, T2, T3)\n\nT1 is ground\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
6 [label="6: sum(T1, T2, T3), SCOPE: 1, CLAUSE: 0 | sum(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: true | sum([], [], T3), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: sum(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nT2 is ground\nsum(T1, T2, T3) !=? sum([], [], [])", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: sum([], [], T3), 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="B: ','(add(T10, T12, T16), sum(T11, T13, T17))\n\nT10 is ground\nT11 is ground\nT12 is ground\nT13 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: ','(add(T10, T12, T16), sum(T11, T13, T17)), SCOPE: 2, CLAUSE: 2 | ','(add(T10, T12, T16), sum(T11, T13, T17)), SCOPE: 2, CLAUSE: 3\n\nT10 is ground\nT11 is ground\nT12 is ground\nT13 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: ','(add(T10, T12, T16), sum(T11, T13, T17)), SCOPE: 2, CLAUSE: 2\n\nT10 is ground\nT11 is ground\nT12 is ground\nT13 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="29: ','(add(T10, T12, T16), sum(T11, T13, T17)), SCOPE: 2, CLAUSE: 3\n\nT10 is ground\nT11 is ground\nT12 is ground\nT13 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
32 [label="32: sum(T11, T13, T23)\n\nT11 is ground\nT13 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
33 [label="33: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
46 [label="46: ','(add(T30, T31, T33), sum(T11, T13, T34))\n\nT11 is ground\nT13 is ground\nT30 is ground\nT31 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
47 [label="47: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
48 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 48 [arrowhead = none ];
48 -> 6;

49 [label="EVAL with clause\nsum([], [], []).\nand substitutionT1 -> [],\nT2 -> [],\nT3 -> []", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 49 [arrowhead = none ];
49 -> 10;

50 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 50 [arrowhead = none ];
50 -> 13;

51 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 51 [arrowhead = none ];
51 -> 16;

52 [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 -> T12,\nX19 -> T13,\nT2 -> .(T12, T13),\nX20 -> T16,\nX21 -> T17,\nT3 -> .(T16, T17),\nT14 -> T16,\nT15 -> T17", fontsize=14, style = filled, fillcolor = lightblue];
13 -> 52 [arrowhead = none ];
52 -> 20;

53 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
13 -> 53 [arrowhead = none ];
53 -> 22;

54 [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];
16 -> 54 [arrowhead = none ];
54 -> 18;

55 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
20 -> 55 [arrowhead = none ];
55 -> 25;

56 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
25 -> 56 [arrowhead = none ];
56 -> 28;

57 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
25 -> 57 [arrowhead = none ];
57 -> 29;

58 [label="EVAL with clause\nadd(0, X26, X26).\nand substitutionT10 -> 0,\nT12 -> T22,\nX26 -> T22,\nT16 -> T22,\nT17 -> T23", fontsize=14, style = filled, fillcolor = lightblue];
28 -> 58 [arrowhead = none ];
58 -> 32;

59 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
28 -> 59 [arrowhead = none ];
59 -> 33;

60 [label="EVAL with clause\nadd(s(X33), X34, s(X35)) :- add(X33, X34, X35).\nand substitutionX33 -> T30,\nT10 -> s(T30),\nT12 -> T31,\nX34 -> T31,\nX35 -> T33,\nT16 -> s(T33),\nT32 -> T33,\nT17 -> T34", fontsize=14, style = filled, fillcolor = lightblue];
29 -> 60 [arrowhead = none ];
60 -> 46;

61 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
29 -> 61 [arrowhead = none ];
61 -> 47;

62 [label="INSTANCE with matching:\nT1 -> T11\nT2 -> T13\nT3 -> T23", fontsize=14, style = filled, fillcolor = lightgrey];
32 -> 62 [arrowhead = none , style=dashed];
62 -> 1[style=dashed];

63 [label="INSTANCE with matching:\nT10 -> T30\nT12 -> T31\nT16 -> T33\nT17 -> T34", fontsize=14, style = filled, fillcolor = lightgrey];
46 -> 63 [arrowhead = none , style=dashed];
63 -> 20[style=dashed];

}

(2) Obligation:

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

sumA_in_gga([], [], []) → sumA_out_gga([], [], [])
sumA_in_gga(.(T10, T11), .(T12, T13), .(T16, T17)) → U1_gga(T10, T11, T12, T13, T16, T17, pB_in_ggagga(T10, T12, T16, T11, T13, T17))
pB_in_ggagga(0, T22, T22, T11, T13, T23) → U2_ggagga(T22, T11, T13, T23, sumA_in_gga(T11, T13, T23))
U2_ggagga(T22, T11, T13, T23, sumA_out_gga(T11, T13, T23)) → pB_out_ggagga(0, T22, T22, T11, T13, T23)
pB_in_ggagga(s(T30), T31, s(T33), T11, T13, T34) → U3_ggagga(T30, T31, T33, T11, T13, T34, pB_in_ggagga(T30, T31, T33, T11, T13, T34))
U3_ggagga(T30, T31, T33, T11, T13, T34, pB_out_ggagga(T30, T31, T33, T11, T13, T34)) → pB_out_ggagga(s(T30), T31, s(T33), T11, T13, T34)
U1_gga(T10, T11, T12, T13, T16, T17, pB_out_ggagga(T10, T12, T16, T11, T13, T17)) → sumA_out_gga(.(T10, T11), .(T12, T13), .(T16, T17))

The argument filtering Pi contains the following mapping:
sumA_in_gga(x1, x2, x3)  =  sumA_in_gga(x1, x2)
[]  =  []
sumA_out_gga(x1, x2, x3)  =  sumA_out_gga(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_gga(x1, x2, x3, x4, x5, x6, x7)  =  U1_gga(x1, x2, x3, x4, x7)
pB_in_ggagga(x1, x2, x3, x4, x5, x6)  =  pB_in_ggagga(x1, x2, x4, x5)
0  =  0
U2_ggagga(x1, x2, x3, x4, x5)  =  U2_ggagga(x1, x2, x3, x5)
pB_out_ggagga(x1, x2, x3, x4, x5, x6)  =  pB_out_ggagga(x1, x2, x3, x4, x5, x6)
s(x1)  =  s(x1)
U3_ggagga(x1, x2, x3, x4, x5, x6, x7)  =  U3_ggagga(x1, x2, x4, x5, x7)

(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:

SUMA_IN_GGA(.(T10, T11), .(T12, T13), .(T16, T17)) → U1_GGA(T10, T11, T12, T13, T16, T17, pB_in_ggagga(T10, T12, T16, T11, T13, T17))
SUMA_IN_GGA(.(T10, T11), .(T12, T13), .(T16, T17)) → PB_IN_GGAGGA(T10, T12, T16, T11, T13, T17)
PB_IN_GGAGGA(0, T22, T22, T11, T13, T23) → U2_GGAGGA(T22, T11, T13, T23, sumA_in_gga(T11, T13, T23))
PB_IN_GGAGGA(0, T22, T22, T11, T13, T23) → SUMA_IN_GGA(T11, T13, T23)
PB_IN_GGAGGA(s(T30), T31, s(T33), T11, T13, T34) → U3_GGAGGA(T30, T31, T33, T11, T13, T34, pB_in_ggagga(T30, T31, T33, T11, T13, T34))
PB_IN_GGAGGA(s(T30), T31, s(T33), T11, T13, T34) → PB_IN_GGAGGA(T30, T31, T33, T11, T13, T34)

The TRS R consists of the following rules:

sumA_in_gga([], [], []) → sumA_out_gga([], [], [])
sumA_in_gga(.(T10, T11), .(T12, T13), .(T16, T17)) → U1_gga(T10, T11, T12, T13, T16, T17, pB_in_ggagga(T10, T12, T16, T11, T13, T17))
pB_in_ggagga(0, T22, T22, T11, T13, T23) → U2_ggagga(T22, T11, T13, T23, sumA_in_gga(T11, T13, T23))
U2_ggagga(T22, T11, T13, T23, sumA_out_gga(T11, T13, T23)) → pB_out_ggagga(0, T22, T22, T11, T13, T23)
pB_in_ggagga(s(T30), T31, s(T33), T11, T13, T34) → U3_ggagga(T30, T31, T33, T11, T13, T34, pB_in_ggagga(T30, T31, T33, T11, T13, T34))
U3_ggagga(T30, T31, T33, T11, T13, T34, pB_out_ggagga(T30, T31, T33, T11, T13, T34)) → pB_out_ggagga(s(T30), T31, s(T33), T11, T13, T34)
U1_gga(T10, T11, T12, T13, T16, T17, pB_out_ggagga(T10, T12, T16, T11, T13, T17)) → sumA_out_gga(.(T10, T11), .(T12, T13), .(T16, T17))

The argument filtering Pi contains the following mapping:
sumA_in_gga(x1, x2, x3)  =  sumA_in_gga(x1, x2)
[]  =  []
sumA_out_gga(x1, x2, x3)  =  sumA_out_gga(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_gga(x1, x2, x3, x4, x5, x6, x7)  =  U1_gga(x1, x2, x3, x4, x7)
pB_in_ggagga(x1, x2, x3, x4, x5, x6)  =  pB_in_ggagga(x1, x2, x4, x5)
0  =  0
U2_ggagga(x1, x2, x3, x4, x5)  =  U2_ggagga(x1, x2, x3, x5)
pB_out_ggagga(x1, x2, x3, x4, x5, x6)  =  pB_out_ggagga(x1, x2, x3, x4, x5, x6)
s(x1)  =  s(x1)
U3_ggagga(x1, x2, x3, x4, x5, x6, x7)  =  U3_ggagga(x1, x2, x4, x5, x7)
SUMA_IN_GGA(x1, x2, x3)  =  SUMA_IN_GGA(x1, x2)
U1_GGA(x1, x2, x3, x4, x5, x6, x7)  =  U1_GGA(x1, x2, x3, x4, x7)
PB_IN_GGAGGA(x1, x2, x3, x4, x5, x6)  =  PB_IN_GGAGGA(x1, x2, x4, x5)
U2_GGAGGA(x1, x2, x3, x4, x5)  =  U2_GGAGGA(x1, x2, x3, x5)
U3_GGAGGA(x1, x2, x3, x4, x5, x6, x7)  =  U3_GGAGGA(x1, x2, x4, x5, x7)

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

(4) Obligation:

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

SUMA_IN_GGA(.(T10, T11), .(T12, T13), .(T16, T17)) → U1_GGA(T10, T11, T12, T13, T16, T17, pB_in_ggagga(T10, T12, T16, T11, T13, T17))
SUMA_IN_GGA(.(T10, T11), .(T12, T13), .(T16, T17)) → PB_IN_GGAGGA(T10, T12, T16, T11, T13, T17)
PB_IN_GGAGGA(0, T22, T22, T11, T13, T23) → U2_GGAGGA(T22, T11, T13, T23, sumA_in_gga(T11, T13, T23))
PB_IN_GGAGGA(0, T22, T22, T11, T13, T23) → SUMA_IN_GGA(T11, T13, T23)
PB_IN_GGAGGA(s(T30), T31, s(T33), T11, T13, T34) → U3_GGAGGA(T30, T31, T33, T11, T13, T34, pB_in_ggagga(T30, T31, T33, T11, T13, T34))
PB_IN_GGAGGA(s(T30), T31, s(T33), T11, T13, T34) → PB_IN_GGAGGA(T30, T31, T33, T11, T13, T34)

The TRS R consists of the following rules:

sumA_in_gga([], [], []) → sumA_out_gga([], [], [])
sumA_in_gga(.(T10, T11), .(T12, T13), .(T16, T17)) → U1_gga(T10, T11, T12, T13, T16, T17, pB_in_ggagga(T10, T12, T16, T11, T13, T17))
pB_in_ggagga(0, T22, T22, T11, T13, T23) → U2_ggagga(T22, T11, T13, T23, sumA_in_gga(T11, T13, T23))
U2_ggagga(T22, T11, T13, T23, sumA_out_gga(T11, T13, T23)) → pB_out_ggagga(0, T22, T22, T11, T13, T23)
pB_in_ggagga(s(T30), T31, s(T33), T11, T13, T34) → U3_ggagga(T30, T31, T33, T11, T13, T34, pB_in_ggagga(T30, T31, T33, T11, T13, T34))
U3_ggagga(T30, T31, T33, T11, T13, T34, pB_out_ggagga(T30, T31, T33, T11, T13, T34)) → pB_out_ggagga(s(T30), T31, s(T33), T11, T13, T34)
U1_gga(T10, T11, T12, T13, T16, T17, pB_out_ggagga(T10, T12, T16, T11, T13, T17)) → sumA_out_gga(.(T10, T11), .(T12, T13), .(T16, T17))

The argument filtering Pi contains the following mapping:
sumA_in_gga(x1, x2, x3)  =  sumA_in_gga(x1, x2)
[]  =  []
sumA_out_gga(x1, x2, x3)  =  sumA_out_gga(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_gga(x1, x2, x3, x4, x5, x6, x7)  =  U1_gga(x1, x2, x3, x4, x7)
pB_in_ggagga(x1, x2, x3, x4, x5, x6)  =  pB_in_ggagga(x1, x2, x4, x5)
0  =  0
U2_ggagga(x1, x2, x3, x4, x5)  =  U2_ggagga(x1, x2, x3, x5)
pB_out_ggagga(x1, x2, x3, x4, x5, x6)  =  pB_out_ggagga(x1, x2, x3, x4, x5, x6)
s(x1)  =  s(x1)
U3_ggagga(x1, x2, x3, x4, x5, x6, x7)  =  U3_ggagga(x1, x2, x4, x5, x7)
SUMA_IN_GGA(x1, x2, x3)  =  SUMA_IN_GGA(x1, x2)
U1_GGA(x1, x2, x3, x4, x5, x6, x7)  =  U1_GGA(x1, x2, x3, x4, x7)
PB_IN_GGAGGA(x1, x2, x3, x4, x5, x6)  =  PB_IN_GGAGGA(x1, x2, x4, x5)
U2_GGAGGA(x1, x2, x3, x4, x5)  =  U2_GGAGGA(x1, x2, x3, x5)
U3_GGAGGA(x1, x2, x3, x4, x5, x6, x7)  =  U3_GGAGGA(x1, x2, x4, x5, x7)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Obligation:

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

SUMA_IN_GGA(.(T10, T11), .(T12, T13), .(T16, T17)) → PB_IN_GGAGGA(T10, T12, T16, T11, T13, T17)
PB_IN_GGAGGA(0, T22, T22, T11, T13, T23) → SUMA_IN_GGA(T11, T13, T23)
PB_IN_GGAGGA(s(T30), T31, s(T33), T11, T13, T34) → PB_IN_GGAGGA(T30, T31, T33, T11, T13, T34)

The TRS R consists of the following rules:

sumA_in_gga([], [], []) → sumA_out_gga([], [], [])
sumA_in_gga(.(T10, T11), .(T12, T13), .(T16, T17)) → U1_gga(T10, T11, T12, T13, T16, T17, pB_in_ggagga(T10, T12, T16, T11, T13, T17))
pB_in_ggagga(0, T22, T22, T11, T13, T23) → U2_ggagga(T22, T11, T13, T23, sumA_in_gga(T11, T13, T23))
U2_ggagga(T22, T11, T13, T23, sumA_out_gga(T11, T13, T23)) → pB_out_ggagga(0, T22, T22, T11, T13, T23)
pB_in_ggagga(s(T30), T31, s(T33), T11, T13, T34) → U3_ggagga(T30, T31, T33, T11, T13, T34, pB_in_ggagga(T30, T31, T33, T11, T13, T34))
U3_ggagga(T30, T31, T33, T11, T13, T34, pB_out_ggagga(T30, T31, T33, T11, T13, T34)) → pB_out_ggagga(s(T30), T31, s(T33), T11, T13, T34)
U1_gga(T10, T11, T12, T13, T16, T17, pB_out_ggagga(T10, T12, T16, T11, T13, T17)) → sumA_out_gga(.(T10, T11), .(T12, T13), .(T16, T17))

The argument filtering Pi contains the following mapping:
sumA_in_gga(x1, x2, x3)  =  sumA_in_gga(x1, x2)
[]  =  []
sumA_out_gga(x1, x2, x3)  =  sumA_out_gga(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_gga(x1, x2, x3, x4, x5, x6, x7)  =  U1_gga(x1, x2, x3, x4, x7)
pB_in_ggagga(x1, x2, x3, x4, x5, x6)  =  pB_in_ggagga(x1, x2, x4, x5)
0  =  0
U2_ggagga(x1, x2, x3, x4, x5)  =  U2_ggagga(x1, x2, x3, x5)
pB_out_ggagga(x1, x2, x3, x4, x5, x6)  =  pB_out_ggagga(x1, x2, x3, x4, x5, x6)
s(x1)  =  s(x1)
U3_ggagga(x1, x2, x3, x4, x5, x6, x7)  =  U3_ggagga(x1, x2, x4, x5, x7)
SUMA_IN_GGA(x1, x2, x3)  =  SUMA_IN_GGA(x1, x2)
PB_IN_GGAGGA(x1, x2, x3, x4, x5, x6)  =  PB_IN_GGAGGA(x1, x2, x4, x5)

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:

SUMA_IN_GGA(.(T10, T11), .(T12, T13), .(T16, T17)) → PB_IN_GGAGGA(T10, T12, T16, T11, T13, T17)
PB_IN_GGAGGA(0, T22, T22, T11, T13, T23) → SUMA_IN_GGA(T11, T13, T23)
PB_IN_GGAGGA(s(T30), T31, s(T33), T11, T13, T34) → PB_IN_GGAGGA(T30, T31, T33, T11, T13, T34)

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

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:

SUMA_IN_GGA(.(T10, T11), .(T12, T13)) → PB_IN_GGAGGA(T10, T12, T11, T13)
PB_IN_GGAGGA(0, T22, T11, T13) → SUMA_IN_GGA(T11, T13)
PB_IN_GGAGGA(s(T30), T31, T11, T13) → PB_IN_GGAGGA(T30, T31, T11, T13)

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:

• PB_IN_GGAGGA(0, T22, T11, T13) → SUMA_IN_GGA(T11, T13)
  The graph contains the following edges 3 >= 1, 4 >= 2

• PB_IN_GGAGGA(s(T30), T31, T11, T13) → PB_IN_GGAGGA(T30, T31, T11, T13)
  The graph contains the following edges 1 > 1, 2 >= 2, 3 >= 3, 4 >= 4

• SUMA_IN_GGA(.(T10, T11), .(T12, T13)) → PB_IN_GGAGGA(T10, T12, T11, T13)
  The graph contains the following edges 1 > 1, 2 > 2, 1 > 3, 2 > 4

(12) YES