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, Z) :- ','(!, eq(Z, 0)).
f(X, Y, Z) :- ','(p(X, P), ','(f(P, Y, U), f(U, Y, Z))).
p(0, 0).
p(s(X), X).
eq(X, X).

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="A: 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: ','(!_1, eq(T8, 0)) | 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\nX3 is free\nX4 is free\nf(T1, T2, T3) !=? f(0, X3, X4)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: eq(T8, 0)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: eq(T8, 0), SCOPE: 2, CLAUSE: 4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(p(T15, X14), ','(f(X14, T18, X15), f(X15, T18, T19)))\n\nT15 is ground\nX3 is free\nX4 is free\nX14 is free\nX15 is free\nf(T15, T2, T3) !=? f(0, X3, X4)", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(p(T15, X14), ','(f(X14, T18, X15), f(X15, T18, T19))), SCOPE: 3, CLAUSE: 2 | ','(p(T15, X14), ','(f(X14, T18, X15), f(X15, T18, T19))), SCOPE: 3, CLAUSE: 3\n\nT15 is ground\nX3 is free\nX4 is free\nX14 is free\nX15 is free\nf(T15, T2, T3) !=? f(0, X3, X4)", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(p(T15, X14), ','(f(X14, T18, X15), f(X15, T18, T19))), SCOPE: 3, CLAUSE: 3\n\nT15 is ground\nX3 is free\nX4 is free\nX14 is free\nX15 is free\nf(T15, T2, T3) !=? f(0, X3, X4)", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(f(T22, T18, X15), f(X15, T18, T19))\n\nT22 is ground\nX15 is free", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="B: f(T22, T18, X15)\n\nT22 is ground\nX15 is free", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: f(T24, T25, T26)\n\nT24 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: f(T22, T18, X15), SCOPE: 4, CLAUSE: 0 | f(T22, T18, X15), SCOPE: 4, CLAUSE: 1\n\nT22 is ground\nX15 is free", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: ','(!_4, eq(X33, 0)) | f(0, T18, X15), SCOPE: 4, CLAUSE: 1\n\nX15 is free\nX33 is free", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: f(T22, T18, X15), SCOPE: 4, CLAUSE: 1\n\nT22 is ground\nX15 is free\nX31 is free\nX32 is free\nf(T22, T18, X15) !=? f(0, X31, X32)", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: eq(X33, 0)\n\nX33 is free", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: eq(X33, 0), SCOPE: 5, CLAUSE: 4\n\nX33 is free", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: ','(p(T37, X56), ','(f(X56, T39, X57), f(X57, T39, X58)))\n\nT37 is ground\nX15 is free\nX31 is free\nX32 is free\nX58 is free\nX56 is free\nX57 is free\nf(T37, T18, X15) !=? f(0, X31, X32)", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: ','(p(T37, X56), ','(f(X56, T39, X57), f(X57, T39, X58))), SCOPE: 6, CLAUSE: 2 | ','(p(T37, X56), ','(f(X56, T39, X57), f(X57, T39, X58))), SCOPE: 6, CLAUSE: 3\n\nT37 is ground\nX15 is free\nX31 is free\nX32 is free\nX58 is free\nX56 is free\nX57 is free\nf(T37, T18, X15) !=? f(0, X31, X32)", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: ','(p(T37, X56), ','(f(X56, T39, X57), f(X57, T39, X58))), SCOPE: 6, CLAUSE: 3\n\nT37 is ground\nX15 is free\nX31 is free\nX32 is free\nX58 is free\nX56 is free\nX57 is free\nf(T37, T18, X15) !=? f(0, X31, X32)", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: ','(f(T43, T39, X57), f(X57, T39, X58))\n\nT43 is ground\nX58 is free\nX57 is free", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="29: f(T43, T39, X57)\n\nT43 is ground\nX57 is free", fontsize=16, style=filled, fillcolor=lightyellow];
30 [label="30: f(T45, T46, X58)\n\nT45 is ground\nX58 is free", fontsize=16, style=filled, fillcolor=lightyellow];
31 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 31 [arrowhead = none ];
31 -> 2;

32 [label="EVAL with clause\nf(0, X3, X4) :- ','(!_1, eq(X4, 0)).\nand substitutionT1 -> 0,\nT2 -> T6,\nX3 -> T6,\nT3 -> T8,\nX4 -> T8,\nT7 -> T8", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 32 [arrowhead = none ];
32 -> 3;

33 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 33 [arrowhead = none ];
33 -> 4;

34 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 34 [arrowhead = none ];
34 -> 5;

35 [label="ONLY EVAL with clause\nf(X11, X12, X13) :- ','(p(X11, X14), ','(f(X14, X12, X15), f(X15, X12, X13))).\nand substitutionT1 -> T15,\nX11 -> T15,\nT2 -> T18,\nX12 -> T18,\nT3 -> T19,\nX13 -> T19,\nT16 -> T18,\nT17 -> T19", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 35 [arrowhead = none ];
35 -> 10;

36 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
5 -> 36 [arrowhead = none ];
36 -> 6;

37 [label="EVAL with clause\neq(X7, X7).\nand substitutionT8 -> 0,\nX7 -> 0,\nT11 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 37 [arrowhead = none ];
37 -> 7;

38 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 38 [arrowhead = none ];
38 -> 8;

39 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
7 -> 39 [arrowhead = none ];
39 -> 9;

40 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
10 -> 40 [arrowhead = none ];
40 -> 11;

41 [label="BACKTRACK\nfor clause: p(0, 0)\nwith clash: (f(T15, T2, T3), f(0, X3, X4))", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 41 [arrowhead = none ];
41 -> 12;

42 [label="EVAL with clause\np(s(X18), X18).\nand substitutionX18 -> T22,\nT15 -> s(T22),\nX14 -> T22", fontsize=14, style = filled, fillcolor = lightblue];
12 -> 42 [arrowhead = none ];
42 -> 13;

43 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 43 [arrowhead = none ];
43 -> 14;

44 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
13 -> 44 [arrowhead = none ];
44 -> 15;

45 [label="SPLIT 2\nnew knowledge:\nT22 is ground\nT24 is ground\nreplacements:X15 -> T24,\nT18 -> T25,\nT19 -> T26", fontsize=14, style = filled, fillcolor = violet];
13 -> 45 [arrowhead = none ];
45 -> 16;

46 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
15 -> 46 [arrowhead = none ];
46 -> 17;

47 [label="INSTANCE with matching:\nT1 -> T24\nT2 -> T25\nT3 -> T26", fontsize=14, style = filled, fillcolor = lightgrey];
16 -> 47 [arrowhead = none , style=dashed];
47 -> 1[style=dashed];

48 [label="EVAL with clause\nf(0, X31, X32) :- ','(!_4, eq(X32, 0)).\nand substitutionT22 -> 0,\nT18 -> T30,\nX31 -> T30,\nX15 -> X33,\nX32 -> X33", fontsize=14, style = filled, fillcolor = lightblue];
17 -> 48 [arrowhead = none ];
48 -> 18;

49 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 49 [arrowhead = none ];
49 -> 19;

50 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
18 -> 50 [arrowhead = none ];
50 -> 20;

51 [label="ONLY EVAL with clause\nf(X53, X54, X55) :- ','(p(X53, X56), ','(f(X56, X54, X57), f(X57, X54, X55))).\nand substitutionT22 -> T37,\nX53 -> T37,\nT18 -> T39,\nX54 -> T39,\nX15 -> X58,\nX55 -> X58,\nT38 -> T39", fontsize=14, style = filled, fillcolor = lightblue];
19 -> 51 [arrowhead = none ];
51 -> 24;

52 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
20 -> 52 [arrowhead = none ];
52 -> 21;

53 [label="ONLY EVAL with clause\neq(X38, X38).\nand substitutionX33 -> 0,\nX38 -> 0,\nX39 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
21 -> 53 [arrowhead = none ];
53 -> 22;

54 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
22 -> 54 [arrowhead = none ];
54 -> 23;

55 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
24 -> 55 [arrowhead = none ];
55 -> 25;

56 [label="BACKTRACK\nfor clause: p(0, 0)\nwith clash: (f(T37, T18, X15), f(0, X31, X32))", fontsize=14, style = filled, fillcolor = lightgreen];
25 -> 56 [arrowhead = none ];
56 -> 26;

57 [label="EVAL with clause\np(s(X64), X64).\nand substitutionX64 -> T43,\nT37 -> s(T43),\nX56 -> T43", fontsize=14, style = filled, fillcolor = lightblue];
26 -> 57 [arrowhead = none ];
57 -> 27;

58 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
26 -> 58 [arrowhead = none ];
58 -> 28;

59 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
27 -> 59 [arrowhead = none ];
59 -> 29;

60 [label="SPLIT 2\nnew knowledge:\nT43 is ground\nT45 is ground\nreplacements:X57 -> T45,\nT39 -> T46", fontsize=14, style = filled, fillcolor = violet];
27 -> 60 [arrowhead = none ];
60 -> 30;

61 [label="INSTANCE with matching:\nT22 -> T43\nT18 -> T39\nX15 -> X57", fontsize=14, style = filled, fillcolor = lightgrey];
29 -> 61 [arrowhead = none , style=dashed];
61 -> 15[style=dashed];

62 [label="INSTANCE with matching:\nT22 -> T45\nT18 -> T46\nX15 -> X58", fontsize=14, style = filled, fillcolor = lightgrey];
30 -> 62 [arrowhead = none , style=dashed];
62 -> 15[style=dashed];

}

(2) Obligation:

Triples:

fB(s(X1), X2, X3) :- fB(X1, X2, X4).
fB(s(X1), X2, X3) :- ','(fcB(X1, X2, X4), fB(X4, X2, X3)).
fA(s(X1), X2, X3) :- fB(X1, X2, X4).
fA(s(X1), X2, X3) :- ','(fcB(X1, X2, X4), fA(X4, X2, X3)).

Clauses:

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

Afs:

fA(x1, x2, x3)  =  fA(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:
fA_in: (b,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:

FA_IN_GAA(s(X1), X2, X3) → U4_GAA(X1, X2, X3, fB_in_gaa(X1, X2, X4))
FA_IN_GAA(s(X1), X2, X3) → FB_IN_GAA(X1, X2, X4)
FB_IN_GAA(s(X1), X2, X3) → U1_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) → U2_GAA(X1, X2, X3, fcB_in_gaa(X1, X2, X4))
U2_GAA(X1, X2, X3, fcB_out_gaa(X1, X2, X4)) → U3_GAA(X1, X2, X3, fB_in_gaa(X4, X2, X3))
U2_GAA(X1, X2, X3, fcB_out_gaa(X1, X2, X4)) → FB_IN_GAA(X4, X2, X3)
FA_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, fA_in_gaa(X4, X2, X3))
U5_GAA(X1, X2, X3, fcB_out_gaa(X1, X2, X4)) → FA_IN_GAA(X4, X2, X3)

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) → U10_gaa(X1, X2, X3, fcB_in_gaa(X1, X2, X4))
U10_gaa(X1, X2, X3, fcB_out_gaa(X1, X2, X4)) → U11_gaa(X1, X2, X3, X4, fcB_in_gaa(X4, X2, X3))
U11_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:
fA_in_gaa(x1, x2, x3)  =  fA_in_gaa(x1)
s(x1)  =  s(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)
U10_gaa(x1, x2, x3, x4)  =  U10_gaa(x1, x4)
U11_gaa(x1, x2, x3, x4, x5)  =  U11_gaa(x1, x5)
FA_IN_GAA(x1, x2, x3)  =  FA_IN_GAA(x1)
U4_GAA(x1, x2, x3, x4)  =  U4_GAA(x1, x4)
FB_IN_GAA(x1, x2, x3)  =  FB_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4)  =  U1_GAA(x1, x4)
U2_GAA(x1, x2, x3, x4)  =  U2_GAA(x1, x4)
U3_GAA(x1, x2, x3, x4)  =  U3_GAA(x1, x4)
U5_GAA(x1, x2, x3, x4)  =  U5_GAA(x1, x4)
U6_GAA(x1, x2, x3, x4)  =  U6_GAA(x1, x4)

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:

FA_IN_GAA(s(X1), X2, X3) → U4_GAA(X1, X2, X3, fB_in_gaa(X1, X2, X4))
FA_IN_GAA(s(X1), X2, X3) → FB_IN_GAA(X1, X2, X4)
FB_IN_GAA(s(X1), X2, X3) → U1_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) → U2_GAA(X1, X2, X3, fcB_in_gaa(X1, X2, X4))
U2_GAA(X1, X2, X3, fcB_out_gaa(X1, X2, X4)) → U3_GAA(X1, X2, X3, fB_in_gaa(X4, X2, X3))
U2_GAA(X1, X2, X3, fcB_out_gaa(X1, X2, X4)) → FB_IN_GAA(X4, X2, X3)
FA_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, fA_in_gaa(X4, X2, X3))
U5_GAA(X1, X2, X3, fcB_out_gaa(X1, X2, X4)) → FA_IN_GAA(X4, X2, X3)

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) → U10_gaa(X1, X2, X3, fcB_in_gaa(X1, X2, X4))
U10_gaa(X1, X2, X3, fcB_out_gaa(X1, X2, X4)) → U11_gaa(X1, X2, X3, X4, fcB_in_gaa(X4, X2, X3))
U11_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:
fA_in_gaa(x1, x2, x3)  =  fA_in_gaa(x1)
s(x1)  =  s(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)
U10_gaa(x1, x2, x3, x4)  =  U10_gaa(x1, x4)
U11_gaa(x1, x2, x3, x4, x5)  =  U11_gaa(x1, x5)
FA_IN_GAA(x1, x2, x3)  =  FA_IN_GAA(x1)
U4_GAA(x1, x2, x3, x4)  =  U4_GAA(x1, x4)
FB_IN_GAA(x1, x2, x3)  =  FB_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4)  =  U1_GAA(x1, x4)
U2_GAA(x1, x2, x3, x4)  =  U2_GAA(x1, x4)
U3_GAA(x1, x2, x3, x4)  =  U3_GAA(x1, x4)
U5_GAA(x1, x2, x3, x4)  =  U5_GAA(x1, x4)
U6_GAA(x1, x2, x3, x4)  =  U6_GAA(x1, 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 5 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) → U2_GAA(X1, X2, X3, fcB_in_gaa(X1, X2, X4))
U2_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) → U10_gaa(X1, X2, X3, fcB_in_gaa(X1, X2, X4))
U10_gaa(X1, X2, X3, fcB_out_gaa(X1, X2, X4)) → U11_gaa(X1, X2, X3, X4, fcB_in_gaa(X4, X2, X3))
U11_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)
U10_gaa(x1, x2, x3, x4)  =  U10_gaa(x1, x4)
U11_gaa(x1, x2, x3, x4, x5)  =  U11_gaa(x1, x5)
FB_IN_GAA(x1, x2, x3)  =  FB_IN_GAA(x1)
U2_GAA(x1, x2, x3, x4)  =  U2_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)) → U2_GAA(X1, fcB_in_gaa(X1))
U2_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)) → U10_gaa(X1, fcB_in_gaa(X1))
U10_gaa(X1, fcB_out_gaa(X1, X4)) → U11_gaa(X1, fcB_in_gaa(X4))
U11_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)
U10_gaa(x0, x1)
U11_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)) → U2_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(U10_gaa(x₁, x₂)) = 1 + x₁ + x₂   
POL(U11_gaa(x₁, x₂)) = 1 + 2·x₁ + x₂   
POL(U2_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:

U2_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)) → U10_gaa(X1, fcB_in_gaa(X1))
U10_gaa(X1, fcB_out_gaa(X1, X4)) → U11_gaa(X1, fcB_in_gaa(X4))
U11_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)
U10_gaa(x0, x1)
U11_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:

FA_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)) → FA_IN_GAA(X4, X2, X3)

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) → U10_gaa(X1, X2, X3, fcB_in_gaa(X1, X2, X4))
U10_gaa(X1, X2, X3, fcB_out_gaa(X1, X2, X4)) → U11_gaa(X1, X2, X3, X4, fcB_in_gaa(X4, X2, X3))
U11_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)
U10_gaa(x1, x2, x3, x4)  =  U10_gaa(x1, x4)
U11_gaa(x1, x2, x3, x4, x5)  =  U11_gaa(x1, x5)
FA_IN_GAA(x1, x2, x3)  =  FA_IN_GAA(x1)
U5_GAA(x1, x2, x3, x4)  =  U5_GAA(x1, x4)

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:

FA_IN_GAA(s(X1)) → U5_GAA(X1, fcB_in_gaa(X1))
U5_GAA(X1, fcB_out_gaa(X1, X4)) → FA_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)) → U10_gaa(X1, fcB_in_gaa(X1))
U10_gaa(X1, fcB_out_gaa(X1, X4)) → U11_gaa(X1, fcB_in_gaa(X4))
U11_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)
U10_gaa(x0, x1)
U11_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:

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

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(FA_IN_GAA(x₁)) = 2 + 2·x₁   
POL(U10_gaa(x₁, x₂)) = 1 + x₁ + x₂   
POL(U11_gaa(x₁, x₂)) = 1 + 2·x₁ + 2·x₂   
POL(U5_GAA(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:

U5_GAA(X1, fcB_out_gaa(X1, X4)) → FA_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)) → U10_gaa(X1, fcB_in_gaa(X1))
U10_gaa(X1, fcB_out_gaa(X1, X4)) → U11_gaa(X1, fcB_in_gaa(X4))
U11_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)
U10_gaa(x0, x1)
U11_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