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

(0) Obligation:

Clauses:

f(0, Y, 0).
f(s(X), Y, Z) :- ','(f(X, Y, U), f(U, Y, Z)).

Query: f(g,a,a)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="C: f(T1, T2, T3)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: f(T1, T2, T3), SCOPE: 1, CLAUSE: 0 | f(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: true | f(0, T2, T3), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: f(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nX2 is free\nf(T1, T2, T3) !=? f(0, X2, 0)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: f(0, T2, T3), 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="A: ','(f(T9, T12, X12), f(X12, T12, T13))\n\nT9 is ground\nX12 is free", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(f(T9, T12, X12), f(X12, T12, T13)), SCOPE: 2, CLAUSE: 0 | ','(f(T9, T12, X12), f(X12, T12, T13)), SCOPE: 2, CLAUSE: 1\n\nT9 is ground\nX12 is free", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(f(T9, T12, X12), f(X12, T12, T13)), SCOPE: 2, CLAUSE: 0\n\nT9 is ground\nX12 is free", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(f(T9, T12, X12), f(X12, T12, T13)), SCOPE: 2, CLAUSE: 1\n\nT9 is ground\nX12 is free", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: f(0, T19, T20)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: f(0, T19, T20), SCOPE: 3, CLAUSE: 0 | f(0, T19, T20), SCOPE: 3, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: f(0, T19, T20), SCOPE: 3, CLAUSE: 0\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: f(0, T19, T20), SCOPE: 3, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: ','(','(f(T32, T34, X40), f(X40, T34, X41)), f(X41, T34, T35))\n\nT32 is ground\nX41 is free\nX40 is free", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="B: f(T32, T34, X40)\n\nT32 is ground\nX40 is free", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: ','(f(T38, T39, X41), f(X41, T39, T40))\n\nT38 is ground\nX41 is free", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: f(T32, T34, X40), SCOPE: 4, CLAUSE: 0 | f(T32, T34, X40), SCOPE: 4, CLAUSE: 1\n\nT32 is ground\nX40 is free", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: f(T32, T34, X40), SCOPE: 4, CLAUSE: 0\n\nT32 is ground\nX40 is free", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: f(T32, T34, X40), SCOPE: 4, CLAUSE: 1\n\nT32 is ground\nX40 is free", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="29: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
30 [label="30: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
31 [label="31: ','(f(T52, T54, X63), f(X63, T54, X64))\n\nT52 is ground\nX64 is free\nX63 is free", fontsize=16, style=filled, fillcolor=lightyellow];
32 [label="32: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
33 [label="33: f(T52, T54, X63)\n\nT52 is ground\nX63 is free", fontsize=16, style=filled, fillcolor=lightyellow];
34 [label="34: f(T57, T58, X64)\n\nT57 is ground\nX64 is free", fontsize=16, style=filled, fillcolor=lightyellow];
35 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 35 [arrowhead = none ];
35 -> 2;

36 [label="EVAL with clause\nf(0, X2, 0).\nand substitutionT1 -> 0,\nT2 -> T5,\nX2 -> T5,\nT3 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 36 [arrowhead = none ];
36 -> 3;

37 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 37 [arrowhead = none ];
37 -> 4;

38 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 38 [arrowhead = none ];
38 -> 5;

39 [label="EVAL with clause\nf(s(X9), X10, X11) :- ','(f(X9, X10, X12), f(X12, X10, X11)).\nand substitutionX9 -> T9,\nT1 -> s(T9),\nT2 -> T12,\nX10 -> T12,\nT3 -> T13,\nX11 -> T13,\nT10 -> T12,\nT11 -> T13", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 39 [arrowhead = none ];
39 -> 7;

40 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 40 [arrowhead = none ];
40 -> 8;

41 [label="BACKTRACK\nfor clause: f(s(X), Y, Z) :- ','(f(X, Y, U), f(U, Y, Z))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 41 [arrowhead = none ];
41 -> 6;

42 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 42 [arrowhead = none ];
42 -> 9;

43 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
9 -> 43 [arrowhead = none ];
43 -> 10;

44 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
9 -> 44 [arrowhead = none ];
44 -> 11;

45 [label="EVAL with clause\nf(0, X17, 0).\nand substitutionT9 -> 0,\nT12 -> T19,\nX17 -> T19,\nX12 -> 0,\nT18 -> T19,\nT13 -> T20", fontsize=14, style = filled, fillcolor = lightblue];
10 -> 45 [arrowhead = none ];
45 -> 12;

46 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 46 [arrowhead = none ];
46 -> 13;

47 [label="EVAL with clause\nf(s(X37), X38, X39) :- ','(f(X37, X38, X40), f(X40, X38, X39)).\nand substitutionX37 -> T32,\nT9 -> s(T32),\nT12 -> T34,\nX38 -> T34,\nX12 -> X41,\nX39 -> X41,\nT33 -> T34,\nT13 -> T35", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 47 [arrowhead = none ];
47 -> 21;

48 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 48 [arrowhead = none ];
48 -> 22;

49 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
12 -> 49 [arrowhead = none ];
49 -> 14;

50 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
14 -> 50 [arrowhead = none ];
50 -> 15;

51 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
14 -> 51 [arrowhead = none ];
51 -> 16;

52 [label="EVAL with clause\nf(0, X24, 0).\nand substitutionT19 -> T27,\nX24 -> T27,\nT20 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
15 -> 52 [arrowhead = none ];
52 -> 17;

53 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 53 [arrowhead = none ];
53 -> 18;

54 [label="BACKTRACK\nfor clause: f(s(X), Y, Z) :- ','(f(X, Y, U), f(U, Y, Z))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
16 -> 54 [arrowhead = none ];
54 -> 20;

55 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 55 [arrowhead = none ];
55 -> 19;

56 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
21 -> 56 [arrowhead = none ];
56 -> 23;

57 [label="SPLIT 2\nnew knowledge:\nT32 is ground\nT38 is ground\nreplacements:X40 -> T38,\nT34 -> T39,\nT35 -> T40", fontsize=14, style = filled, fillcolor = violet];
21 -> 57 [arrowhead = none ];
57 -> 24;

58 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
23 -> 58 [arrowhead = none ];
58 -> 25;

59 [label="INSTANCE with matching:\nT9 -> T38\nT12 -> T39\nX12 -> X41\nT13 -> T40", fontsize=14, style = filled, fillcolor = lightgrey];
24 -> 59 [arrowhead = none , style=dashed];
59 -> 7[style=dashed];

60 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
25 -> 60 [arrowhead = none ];
60 -> 26;

61 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
25 -> 61 [arrowhead = none ];
61 -> 27;

62 [label="EVAL with clause\nf(0, X50, 0).\nand substitutionT32 -> 0,\nT34 -> T47,\nX50 -> T47,\nX40 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
26 -> 62 [arrowhead = none ];
62 -> 28;

63 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
26 -> 63 [arrowhead = none ];
63 -> 29;

64 [label="EVAL with clause\nf(s(X60), X61, X62) :- ','(f(X60, X61, X63), f(X63, X61, X62)).\nand substitutionX60 -> T52,\nT32 -> s(T52),\nT34 -> T54,\nX61 -> T54,\nX40 -> X64,\nX62 -> X64,\nT53 -> T54", fontsize=14, style = filled, fillcolor = lightblue];
27 -> 64 [arrowhead = none ];
64 -> 31;

65 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
27 -> 65 [arrowhead = none ];
65 -> 32;

66 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
28 -> 66 [arrowhead = none ];
66 -> 30;

67 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
31 -> 67 [arrowhead = none ];
67 -> 33;

68 [label="SPLIT 2\nnew knowledge:\nT52 is ground\nT57 is ground\nreplacements:X63 -> T57,\nT54 -> T58", fontsize=14, style = filled, fillcolor = violet];
31 -> 68 [arrowhead = none ];
68 -> 34;

69 [label="INSTANCE with matching:\nT32 -> T52\nT34 -> T54\nX40 -> X63", fontsize=14, style = filled, fillcolor = lightgrey];
33 -> 69 [arrowhead = none , style=dashed];
69 -> 23[style=dashed];

70 [label="INSTANCE with matching:\nT32 -> T57\nT34 -> T58\nX40 -> X64", fontsize=14, style = filled, fillcolor = lightgrey];
34 -> 70 [arrowhead = none , style=dashed];
70 -> 23[style=dashed];

}

(2) Obligation:

Triples:

pA(s(X1), X2, X3, X4) :- fB(X1, X2, X5).
pA(s(X1), X2, X3, X4) :- ','(fcB(X1, X2, X5), pA(X5, X2, X3, X4)).
fB(s(X1), X2, X3) :- fB(X1, X2, X4).
fB(s(X1), X2, X3) :- ','(fcB(X1, X2, X4), fB(X4, X2, X3)).
fC(s(X1), X2, X3) :- pA(X1, X2, X4, X3).

Clauses:

qcA(0, X1, 0, 0).
qcA(s(X1), X2, X3, X4) :- ','(fcB(X1, X2, X5), qcA(X5, X2, X3, X4)).
fcB(0, X1, 0).
fcB(s(X1), X2, X3) :- ','(fcB(X1, X2, X4), fcB(X4, X2, X3)).

Afs:

fC(x1, x2, x3)  =  fC(x1)

(3) TriplesToPiDPProof (SOUND transformation)

We use the technique of [DT09]. With regard to the inferred argument filtering the predicates were used in the following modes:
fC_in: (b,f,f)
pA_in: (b,f,f,f)
fB_in: (b,f,f)
fcB_in: (b,f,f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

FC_IN_GAA(s(X1), X2, X3) → U7_GAA(X1, X2, X3, pA_in_gaaa(X1, X2, X4, X3))
FC_IN_GAA(s(X1), X2, X3) → PA_IN_GAAA(X1, X2, X4, X3)
PA_IN_GAAA(s(X1), X2, X3, X4) → U1_GAAA(X1, X2, X3, X4, fB_in_gaa(X1, X2, X5))
PA_IN_GAAA(s(X1), X2, X3, X4) → FB_IN_GAA(X1, X2, X5)
FB_IN_GAA(s(X1), X2, X3) → U4_GAA(X1, X2, X3, fB_in_gaa(X1, X2, X4))
FB_IN_GAA(s(X1), X2, X3) → FB_IN_GAA(X1, X2, X4)
FB_IN_GAA(s(X1), X2, X3) → U5_GAA(X1, X2, X3, fcB_in_gaa(X1, X2, X4))
U5_GAA(X1, X2, X3, fcB_out_gaa(X1, X2, X4)) → U6_GAA(X1, X2, X3, fB_in_gaa(X4, X2, X3))
U5_GAA(X1, X2, X3, fcB_out_gaa(X1, X2, X4)) → FB_IN_GAA(X4, X2, X3)
PA_IN_GAAA(s(X1), X2, X3, X4) → U2_GAAA(X1, X2, X3, X4, fcB_in_gaa(X1, X2, X5))
U2_GAAA(X1, X2, X3, X4, fcB_out_gaa(X1, X2, X5)) → U3_GAAA(X1, X2, X3, X4, pA_in_gaaa(X5, X2, X3, X4))
U2_GAAA(X1, X2, X3, X4, fcB_out_gaa(X1, X2, X5)) → PA_IN_GAAA(X5, X2, X3, X4)

The TRS R consists of the following rules:

fcB_in_gaa(0, X1, 0) → fcB_out_gaa(0, X1, 0)
fcB_in_gaa(s(X1), X2, X3) → U11_gaa(X1, X2, X3, fcB_in_gaa(X1, X2, X4))
U11_gaa(X1, X2, X3, fcB_out_gaa(X1, X2, X4)) → U12_gaa(X1, X2, X3, X4, fcB_in_gaa(X4, X2, X3))
U12_gaa(X1, X2, X3, X4, fcB_out_gaa(X4, X2, X3)) → fcB_out_gaa(s(X1), X2, X3)

The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
pA_in_gaaa(x1, x2, x3, x4)  =  pA_in_gaaa(x1)
fB_in_gaa(x1, x2, x3)  =  fB_in_gaa(x1)
fcB_in_gaa(x1, x2, x3)  =  fcB_in_gaa(x1)
0  =  0
fcB_out_gaa(x1, x2, x3)  =  fcB_out_gaa(x1, x3)
U11_gaa(x1, x2, x3, x4)  =  U11_gaa(x1, x4)
U12_gaa(x1, x2, x3, x4, x5)  =  U12_gaa(x1, x5)
FC_IN_GAA(x1, x2, x3)  =  FC_IN_GAA(x1)
U7_GAA(x1, x2, x3, x4)  =  U7_GAA(x1, x4)
PA_IN_GAAA(x1, x2, x3, x4)  =  PA_IN_GAAA(x1)
U1_GAAA(x1, x2, x3, x4, x5)  =  U1_GAAA(x1, x5)
FB_IN_GAA(x1, x2, x3)  =  FB_IN_GAA(x1)
U4_GAA(x1, x2, x3, x4)  =  U4_GAA(x1, x4)
U5_GAA(x1, x2, x3, x4)  =  U5_GAA(x1, x4)
U6_GAA(x1, x2, x3, x4)  =  U6_GAA(x1, x4)
U2_GAAA(x1, x2, x3, x4, x5)  =  U2_GAAA(x1, x5)
U3_GAAA(x1, x2, x3, x4, x5)  =  U3_GAAA(x1, x5)

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

Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES

(4) Obligation:

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

FC_IN_GAA(s(X1), X2, X3) → U7_GAA(X1, X2, X3, pA_in_gaaa(X1, X2, X4, X3))
FC_IN_GAA(s(X1), X2, X3) → PA_IN_GAAA(X1, X2, X4, X3)
PA_IN_GAAA(s(X1), X2, X3, X4) → U1_GAAA(X1, X2, X3, X4, fB_in_gaa(X1, X2, X5))
PA_IN_GAAA(s(X1), X2, X3, X4) → FB_IN_GAA(X1, X2, X5)
FB_IN_GAA(s(X1), X2, X3) → U4_GAA(X1, X2, X3, fB_in_gaa(X1, X2, X4))
FB_IN_GAA(s(X1), X2, X3) → FB_IN_GAA(X1, X2, X4)
FB_IN_GAA(s(X1), X2, X3) → U5_GAA(X1, X2, X3, fcB_in_gaa(X1, X2, X4))
U5_GAA(X1, X2, X3, fcB_out_gaa(X1, X2, X4)) → U6_GAA(X1, X2, X3, fB_in_gaa(X4, X2, X3))
U5_GAA(X1, X2, X3, fcB_out_gaa(X1, X2, X4)) → FB_IN_GAA(X4, X2, X3)
PA_IN_GAAA(s(X1), X2, X3, X4) → U2_GAAA(X1, X2, X3, X4, fcB_in_gaa(X1, X2, X5))
U2_GAAA(X1, X2, X3, X4, fcB_out_gaa(X1, X2, X5)) → U3_GAAA(X1, X2, X3, X4, pA_in_gaaa(X5, X2, X3, X4))
U2_GAAA(X1, X2, X3, X4, fcB_out_gaa(X1, X2, X5)) → PA_IN_GAAA(X5, X2, X3, X4)

The TRS R consists of the following rules:

fcB_in_gaa(0, X1, 0) → fcB_out_gaa(0, X1, 0)
fcB_in_gaa(s(X1), X2, X3) → U11_gaa(X1, X2, X3, fcB_in_gaa(X1, X2, X4))
U11_gaa(X1, X2, X3, fcB_out_gaa(X1, X2, X4)) → U12_gaa(X1, X2, X3, X4, fcB_in_gaa(X4, X2, X3))
U12_gaa(X1, X2, X3, X4, fcB_out_gaa(X4, X2, X3)) → fcB_out_gaa(s(X1), X2, X3)

The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
pA_in_gaaa(x1, x2, x3, x4)  =  pA_in_gaaa(x1)
fB_in_gaa(x1, x2, x3)  =  fB_in_gaa(x1)
fcB_in_gaa(x1, x2, x3)  =  fcB_in_gaa(x1)
0  =  0
fcB_out_gaa(x1, x2, x3)  =  fcB_out_gaa(x1, x3)
U11_gaa(x1, x2, x3, x4)  =  U11_gaa(x1, x4)
U12_gaa(x1, x2, x3, x4, x5)  =  U12_gaa(x1, x5)
FC_IN_GAA(x1, x2, x3)  =  FC_IN_GAA(x1)
U7_GAA(x1, x2, x3, x4)  =  U7_GAA(x1, x4)
PA_IN_GAAA(x1, x2, x3, x4)  =  PA_IN_GAAA(x1)
U1_GAAA(x1, x2, x3, x4, x5)  =  U1_GAAA(x1, x5)
FB_IN_GAA(x1, x2, x3)  =  FB_IN_GAA(x1)
U4_GAA(x1, x2, x3, x4)  =  U4_GAA(x1, x4)
U5_GAA(x1, x2, x3, x4)  =  U5_GAA(x1, x4)
U6_GAA(x1, x2, x3, x4)  =  U6_GAA(x1, x4)
U2_GAAA(x1, x2, x3, x4, x5)  =  U2_GAAA(x1, x5)
U3_GAAA(x1, x2, x3, x4, x5)  =  U3_GAAA(x1, 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 7 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

FB_IN_GAA(s(X1), X2, X3) → U5_GAA(X1, X2, X3, fcB_in_gaa(X1, X2, X4))
U5_GAA(X1, X2, X3, fcB_out_gaa(X1, X2, X4)) → FB_IN_GAA(X4, X2, X3)
FB_IN_GAA(s(X1), X2, X3) → FB_IN_GAA(X1, X2, X4)

The TRS R consists of the following rules:

fcB_in_gaa(0, X1, 0) → fcB_out_gaa(0, X1, 0)
fcB_in_gaa(s(X1), X2, X3) → U11_gaa(X1, X2, X3, fcB_in_gaa(X1, X2, X4))
U11_gaa(X1, X2, X3, fcB_out_gaa(X1, X2, X4)) → U12_gaa(X1, X2, X3, X4, fcB_in_gaa(X4, X2, X3))
U12_gaa(X1, X2, X3, X4, fcB_out_gaa(X4, X2, X3)) → fcB_out_gaa(s(X1), X2, X3)

The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
fcB_in_gaa(x1, x2, x3)  =  fcB_in_gaa(x1)
0  =  0
fcB_out_gaa(x1, x2, x3)  =  fcB_out_gaa(x1, x3)
U11_gaa(x1, x2, x3, x4)  =  U11_gaa(x1, x4)
U12_gaa(x1, x2, x3, x4, x5)  =  U12_gaa(x1, x5)
FB_IN_GAA(x1, x2, x3)  =  FB_IN_GAA(x1)
U5_GAA(x1, x2, x3, x4)  =  U5_GAA(x1, x4)

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

(8) PiDPToQDPProof (SOUND transformation)

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

(9) Obligation:

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

FB_IN_GAA(s(X1)) → U5_GAA(X1, fcB_in_gaa(X1))
U5_GAA(X1, fcB_out_gaa(X1, X4)) → FB_IN_GAA(X4)
FB_IN_GAA(s(X1)) → FB_IN_GAA(X1)

The TRS R consists of the following rules:

fcB_in_gaa(0) → fcB_out_gaa(0, 0)
fcB_in_gaa(s(X1)) → U11_gaa(X1, fcB_in_gaa(X1))
U11_gaa(X1, fcB_out_gaa(X1, X4)) → U12_gaa(X1, fcB_in_gaa(X4))
U12_gaa(X1, fcB_out_gaa(X4, X3)) → fcB_out_gaa(s(X1), X3)

The set Q consists of the following terms:

fcB_in_gaa(x0)
U11_gaa(x0, x1)
U12_gaa(x0, x1)

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

(10) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

FB_IN_GAA(s(X1)) → U5_GAA(X1, fcB_in_gaa(X1))
FB_IN_GAA(s(X1)) → FB_IN_GAA(X1)

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(FB_IN_GAA(x₁)) = 2 + 2·x₁   
POL(U11_gaa(x₁, x₂)) = 1 + x₁ + x₂   
POL(U12_gaa(x₁, x₂)) = 1 + 2·x₁ + x₂   
POL(U5_GAA(x₁, x₂)) = 2 + 2·x₁ + 2·x₂   
POL(fcB_in_gaa(x₁)) = x₁   
POL(fcB_out_gaa(x₁, x₂)) = x₁ + x₂   
POL(s(x₁)) = 1 + 2·x₁   

(11) Obligation:

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

U5_GAA(X1, fcB_out_gaa(X1, X4)) → FB_IN_GAA(X4)

The TRS R consists of the following rules:

fcB_in_gaa(0) → fcB_out_gaa(0, 0)
fcB_in_gaa(s(X1)) → U11_gaa(X1, fcB_in_gaa(X1))
U11_gaa(X1, fcB_out_gaa(X1, X4)) → U12_gaa(X1, fcB_in_gaa(X4))
U12_gaa(X1, fcB_out_gaa(X4, X3)) → fcB_out_gaa(s(X1), X3)

The set Q consists of the following terms:

fcB_in_gaa(x0)
U11_gaa(x0, x1)
U12_gaa(x0, x1)

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

(12) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(13) TRUE

(14) Obligation:

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

PA_IN_GAAA(s(X1), X2, X3, X4) → U2_GAAA(X1, X2, X3, X4, fcB_in_gaa(X1, X2, X5))
U2_GAAA(X1, X2, X3, X4, fcB_out_gaa(X1, X2, X5)) → PA_IN_GAAA(X5, X2, X3, X4)

The TRS R consists of the following rules:

fcB_in_gaa(0, X1, 0) → fcB_out_gaa(0, X1, 0)
fcB_in_gaa(s(X1), X2, X3) → U11_gaa(X1, X2, X3, fcB_in_gaa(X1, X2, X4))
U11_gaa(X1, X2, X3, fcB_out_gaa(X1, X2, X4)) → U12_gaa(X1, X2, X3, X4, fcB_in_gaa(X4, X2, X3))
U12_gaa(X1, X2, X3, X4, fcB_out_gaa(X4, X2, X3)) → fcB_out_gaa(s(X1), X2, X3)

The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
fcB_in_gaa(x1, x2, x3)  =  fcB_in_gaa(x1)
0  =  0
fcB_out_gaa(x1, x2, x3)  =  fcB_out_gaa(x1, x3)
U11_gaa(x1, x2, x3, x4)  =  U11_gaa(x1, x4)
U12_gaa(x1, x2, x3, x4, x5)  =  U12_gaa(x1, x5)
PA_IN_GAAA(x1, x2, x3, x4)  =  PA_IN_GAAA(x1)
U2_GAAA(x1, x2, x3, x4, x5)  =  U2_GAAA(x1, x5)

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

(15) PiDPToQDPProof (SOUND transformation)

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

(16) Obligation:

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

PA_IN_GAAA(s(X1)) → U2_GAAA(X1, fcB_in_gaa(X1))
U2_GAAA(X1, fcB_out_gaa(X1, X5)) → PA_IN_GAAA(X5)

The TRS R consists of the following rules:

fcB_in_gaa(0) → fcB_out_gaa(0, 0)
fcB_in_gaa(s(X1)) → U11_gaa(X1, fcB_in_gaa(X1))
U11_gaa(X1, fcB_out_gaa(X1, X4)) → U12_gaa(X1, fcB_in_gaa(X4))
U12_gaa(X1, fcB_out_gaa(X4, X3)) → fcB_out_gaa(s(X1), X3)

The set Q consists of the following terms:

fcB_in_gaa(x0)
U11_gaa(x0, x1)
U12_gaa(x0, x1)

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

(17) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

PA_IN_GAAA(s(X1)) → U2_GAAA(X1, fcB_in_gaa(X1))

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(PA_IN_GAAA(x₁)) = 2 + 2·x₁   
POL(U11_gaa(x₁, x₂)) = 1 + x₁ + x₂   
POL(U12_gaa(x₁, x₂)) = 1 + 2·x₁ + 2·x₂   
POL(U2_GAAA(x₁, x₂)) = 2 + x₁ + 2·x₂   
POL(fcB_in_gaa(x₁)) = x₁   
POL(fcB_out_gaa(x₁, x₂)) = x₁ + 2·x₂   
POL(s(x₁)) = 1 + 2·x₁   

(18) Obligation:

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

U2_GAAA(X1, fcB_out_gaa(X1, X5)) → PA_IN_GAAA(X5)

The TRS R consists of the following rules:

fcB_in_gaa(0) → fcB_out_gaa(0, 0)
fcB_in_gaa(s(X1)) → U11_gaa(X1, fcB_in_gaa(X1))
U11_gaa(X1, fcB_out_gaa(X1, X4)) → U12_gaa(X1, fcB_in_gaa(X4))
U12_gaa(X1, fcB_out_gaa(X4, X3)) → fcB_out_gaa(s(X1), X3)

The set Q consists of the following terms:

fcB_in_gaa(x0)
U11_gaa(x0, x1)
U12_gaa(x0, x1)

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

(19) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(20) TRUE