Left Termination of the query pattern
perm1(g,a)
w.r.t. the given Prolog program could successfully be proven:

(0) Obligation:

Clauses:

perm1([], []).
perm1(Xs, .(X, Ys)) :- ','(select(X, Xs, Zs), perm1(Zs, Ys)).
select(X, .(X, Xs), Xs).
select(X, .(Y, Xs), .(Y, Zs)) :- select(X, Xs, Zs).

Query: perm1(g,a)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: perm1(T1, T2)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: perm1(T1, T2), SCOPE: 1, CLAUSE: 0 | perm1(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: true | perm1([], T2), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: perm1(T1, T2), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nperm1(T1, T2) !=? perm1([], [])", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: perm1([], T2), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: ','(select(T7, [], X7), perm1(X7, T8))\n\nX7 is free", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: ','(select(T7, [], X7), perm1(X7, T8)), SCOPE: 2, CLAUSE: 2 | ','(select(T7, [], X7), perm1(X7, T8)), SCOPE: 2, CLAUSE: 3\n\nX7 is free", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(select(T7, [], X7), perm1(X7, T8)), SCOPE: 2, CLAUSE: 3\n\nX7 is free", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(select(T17, T14, X20), perm1(X20, T18))\n\nT14 is ground\nX20 is free\nperm1(T14, T2) !=? perm1([], [])", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(select(T17, T14, X20), perm1(X20, T18)), SCOPE: 3, CLAUSE: 2 | ','(select(T17, T14, X20), perm1(X20, T18)), SCOPE: 3, CLAUSE: 3\n\nT14 is ground\nX20 is free\nperm1(T14, T2) !=? perm1([], [])", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(select(T17, T14, X20), perm1(X20, T18)), SCOPE: 3, CLAUSE: 2\n\nT14 is ground\nX20 is free\nperm1(T14, T2) !=? perm1([], [])", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: ','(select(T17, T14, X20), perm1(X20, T18)), SCOPE: 3, CLAUSE: 3\n\nT14 is ground\nX20 is free\nperm1(T14, T2) !=? perm1([], [])", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: perm1(T28, T29)\n\nT28 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: ','(select(T39, T38, X45), perm1(.(T37, X45), T40))\n\nT37 is ground\nT38 is ground\nX45 is free", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="B: select(T39, T38, X45)\n\nT38 is ground\nX45 is free", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: perm1(.(T37, T45), T46)\n\nT37 is ground\nT45 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: select(T39, T38, X45), SCOPE: 4, CLAUSE: 2 | select(T39, T38, X45), SCOPE: 4, CLAUSE: 3\n\nT38 is ground\nX45 is free", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: select(T39, T38, X45), SCOPE: 4, CLAUSE: 2\n\nT38 is ground\nX45 is free", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: select(T39, T38, X45), SCOPE: 4, CLAUSE: 3\n\nT38 is ground\nX45 is free", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: select(T70, T69, X78)\n\nT69 is ground\nX78 is free", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="29: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
30 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 30 [arrowhead = none ];
30 -> 2;

31 [label="EVAL with clause\nperm1([], []).\nand substitutionT1 -> [],\nT2 -> []", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 31 [arrowhead = none ];
31 -> 3;

32 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 32 [arrowhead = none ];
32 -> 4;

33 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 33 [arrowhead = none ];
33 -> 5;

34 [label="EVAL with clause\nperm1(X17, .(X18, X19)) :- ','(select(X18, X17, X20), perm1(X20, X19)).\nand substitutionT1 -> T14,\nX17 -> T14,\nX18 -> T17,\nX19 -> T18,\nT2 -> .(T17, T18),\nT15 -> T17,\nT16 -> T18", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 34 [arrowhead = none ];
34 -> 11;

35 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 35 [arrowhead = none ];
35 -> 12;

36 [label="EVAL with clause\nperm1(X4, .(X5, X6)) :- ','(select(X5, X4, X7), perm1(X7, X6)).\nand substitutionX4 -> [],\nX5 -> T7,\nX6 -> T8,\nT2 -> .(T7, T8),\nT5 -> T7,\nT6 -> T8", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 36 [arrowhead = none ];
36 -> 6;

37 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 37 [arrowhead = none ];
37 -> 7;

38 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
6 -> 38 [arrowhead = none ];
38 -> 8;

39 [label="BACKTRACK\nfor clause: select(X, .(X, Xs), Xs)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
8 -> 39 [arrowhead = none ];
39 -> 9;

40 [label="BACKTRACK\nfor clause: select(X, .(Y, Xs), .(Y, Zs)) :- select(X, Xs, Zs)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 40 [arrowhead = none ];
40 -> 10;

41 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
11 -> 41 [arrowhead = none ];
41 -> 13;

42 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
13 -> 42 [arrowhead = none ];
42 -> 14;

43 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
13 -> 43 [arrowhead = none ];
43 -> 15;

44 [label="EVAL with clause\nselect(X29, .(X29, X30), X30).\nand substitutionT17 -> T27,\nX29 -> T27,\nX30 -> T28,\nT14 -> .(T27, T28),\nX20 -> T28,\nT18 -> T29", fontsize=14, style = filled, fillcolor = lightblue];
14 -> 44 [arrowhead = none ];
44 -> 16;

45 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
14 -> 45 [arrowhead = none ];
45 -> 17;

46 [label="EVAL with clause\nselect(X41, .(X42, X43), .(X42, X44)) :- select(X41, X43, X44).\nand substitutionT17 -> T39,\nX41 -> T39,\nX42 -> T37,\nX43 -> T38,\nT14 -> .(T37, T38),\nX44 -> X45,\nX20 -> .(T37, X45),\nT36 -> T39,\nT18 -> T40", fontsize=14, style = filled, fillcolor = lightblue];
15 -> 46 [arrowhead = none ];
46 -> 18;

47 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 47 [arrowhead = none ];
47 -> 19;

48 [label="INSTANCE with matching:\nT1 -> T28\nT2 -> T29", fontsize=14, style = filled, fillcolor = lightgrey];
16 -> 48 [arrowhead = none , style=dashed];
48 -> 1[style=dashed];

49 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
18 -> 49 [arrowhead = none ];
49 -> 20;

50 [label="SPLIT 2\nnew knowledge:\nT39 is ground\nT38 is ground\nT45 is ground\nreplacements:X45 -> T45,\nT40 -> T46", fontsize=14, style = filled, fillcolor = violet];
18 -> 50 [arrowhead = none ];
50 -> 21;

51 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
20 -> 51 [arrowhead = none ];
51 -> 22;

52 [label="INSTANCE with matching:\nT1 -> .(T37, T45)\nT2 -> T46", fontsize=14, style = filled, fillcolor = lightgrey];
21 -> 52 [arrowhead = none , style=dashed];
52 -> 1[style=dashed];

53 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
22 -> 53 [arrowhead = none ];
53 -> 23;

54 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
22 -> 54 [arrowhead = none ];
54 -> 24;

55 [label="EVAL with clause\nselect(X62, .(X62, X63), X63).\nand substitutionT39 -> T59,\nX62 -> T59,\nX63 -> T60,\nT38 -> .(T59, T60),\nX45 -> T60", fontsize=14, style = filled, fillcolor = lightblue];
23 -> 55 [arrowhead = none ];
55 -> 25;

56 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
23 -> 56 [arrowhead = none ];
56 -> 26;

57 [label="EVAL with clause\nselect(X74, .(X75, X76), .(X75, X77)) :- select(X74, X76, X77).\nand substitutionT39 -> T70,\nX74 -> T70,\nX75 -> T68,\nX76 -> T69,\nT38 -> .(T68, T69),\nX77 -> X78,\nX45 -> .(T68, X78),\nT67 -> T70", fontsize=14, style = filled, fillcolor = lightblue];
24 -> 57 [arrowhead = none ];
57 -> 28;

58 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
24 -> 58 [arrowhead = none ];
58 -> 29;

59 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
25 -> 59 [arrowhead = none ];
59 -> 27;

60 [label="INSTANCE with matching:\nT39 -> T70\nT38 -> T69\nX45 -> X78", fontsize=14, style = filled, fillcolor = lightgrey];
28 -> 60 [arrowhead = none , style=dashed];
60 -> 20[style=dashed];

}

(2) Obligation:

Triples:

selectB(X1, .(X2, X3), .(X2, X4)) :- selectB(X1, X3, X4).
perm1A(.(X1, X2), .(X1, X3)) :- perm1A(X2, X3).
perm1A(.(X1, X2), .(X3, X4)) :- selectB(X3, X2, X5).
perm1A(.(X1, X2), .(X3, X4)) :- ','(selectcB(X3, X2, X5), perm1A(.(X1, X5), X4)).

Clauses:

perm1cA([], []).
perm1cA(.(X1, X2), .(X1, X3)) :- perm1cA(X2, X3).
perm1cA(.(X1, X2), .(X3, X4)) :- ','(selectcB(X3, X2, X5), perm1cA(.(X1, X5), X4)).
selectcB(X1, .(X1, X2), X2).
selectcB(X1, .(X2, X3), .(X2, X4)) :- selectcB(X1, X3, X4).

Afs:

perm1A(x1, x2)  =  perm1A(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:
perm1A_in: (b,f)
selectB_in: (f,b,f)
selectcB_in: (f,b,f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

PERM1A_IN_GA(.(X1, X2), .(X1, X3)) → U2_GA(X1, X2, X3, perm1A_in_ga(X2, X3))
PERM1A_IN_GA(.(X1, X2), .(X1, X3)) → PERM1A_IN_GA(X2, X3)
PERM1A_IN_GA(.(X1, X2), .(X3, X4)) → U3_GA(X1, X2, X3, X4, selectB_in_aga(X3, X2, X5))
PERM1A_IN_GA(.(X1, X2), .(X3, X4)) → SELECTB_IN_AGA(X3, X2, X5)
SELECTB_IN_AGA(X1, .(X2, X3), .(X2, X4)) → U1_AGA(X1, X2, X3, X4, selectB_in_aga(X1, X3, X4))
SELECTB_IN_AGA(X1, .(X2, X3), .(X2, X4)) → SELECTB_IN_AGA(X1, X3, X4)
PERM1A_IN_GA(.(X1, X2), .(X3, X4)) → U4_GA(X1, X2, X3, X4, selectcB_in_aga(X3, X2, X5))
U4_GA(X1, X2, X3, X4, selectcB_out_aga(X3, X2, X5)) → U5_GA(X1, X2, X3, X4, perm1A_in_ga(.(X1, X5), X4))
U4_GA(X1, X2, X3, X4, selectcB_out_aga(X3, X2, X5)) → PERM1A_IN_GA(.(X1, X5), X4)

The TRS R consists of the following rules:

selectcB_in_aga(X1, .(X1, X2), X2) → selectcB_out_aga(X1, .(X1, X2), X2)
selectcB_in_aga(X1, .(X2, X3), .(X2, X4)) → U10_aga(X1, X2, X3, X4, selectcB_in_aga(X1, X3, X4))
U10_aga(X1, X2, X3, X4, selectcB_out_aga(X1, X3, X4)) → selectcB_out_aga(X1, .(X2, X3), .(X2, X4))

The argument filtering Pi contains the following mapping:
perm1A_in_ga(x1, x2)  =  perm1A_in_ga(x1)
.(x1, x2)  =  .(x1, x2)
selectB_in_aga(x1, x2, x3)  =  selectB_in_aga(x2)
selectcB_in_aga(x1, x2, x3)  =  selectcB_in_aga(x2)
selectcB_out_aga(x1, x2, x3)  =  selectcB_out_aga(x1, x2, x3)
U10_aga(x1, x2, x3, x4, x5)  =  U10_aga(x2, x3, x5)
PERM1A_IN_GA(x1, x2)  =  PERM1A_IN_GA(x1)
U2_GA(x1, x2, x3, x4)  =  U2_GA(x1, x2, x4)
U3_GA(x1, x2, x3, x4, x5)  =  U3_GA(x1, x2, x5)
SELECTB_IN_AGA(x1, x2, x3)  =  SELECTB_IN_AGA(x2)
U1_AGA(x1, x2, x3, x4, x5)  =  U1_AGA(x2, x3, x5)
U4_GA(x1, x2, x3, x4, x5)  =  U4_GA(x1, x2, x5)
U5_GA(x1, x2, x3, x4, x5)  =  U5_GA(x1, x2, 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:

PERM1A_IN_GA(.(X1, X2), .(X1, X3)) → U2_GA(X1, X2, X3, perm1A_in_ga(X2, X3))
PERM1A_IN_GA(.(X1, X2), .(X1, X3)) → PERM1A_IN_GA(X2, X3)
PERM1A_IN_GA(.(X1, X2), .(X3, X4)) → U3_GA(X1, X2, X3, X4, selectB_in_aga(X3, X2, X5))
PERM1A_IN_GA(.(X1, X2), .(X3, X4)) → SELECTB_IN_AGA(X3, X2, X5)
SELECTB_IN_AGA(X1, .(X2, X3), .(X2, X4)) → U1_AGA(X1, X2, X3, X4, selectB_in_aga(X1, X3, X4))
SELECTB_IN_AGA(X1, .(X2, X3), .(X2, X4)) → SELECTB_IN_AGA(X1, X3, X4)
PERM1A_IN_GA(.(X1, X2), .(X3, X4)) → U4_GA(X1, X2, X3, X4, selectcB_in_aga(X3, X2, X5))
U4_GA(X1, X2, X3, X4, selectcB_out_aga(X3, X2, X5)) → U5_GA(X1, X2, X3, X4, perm1A_in_ga(.(X1, X5), X4))
U4_GA(X1, X2, X3, X4, selectcB_out_aga(X3, X2, X5)) → PERM1A_IN_GA(.(X1, X5), X4)

The TRS R consists of the following rules:

selectcB_in_aga(X1, .(X1, X2), X2) → selectcB_out_aga(X1, .(X1, X2), X2)
selectcB_in_aga(X1, .(X2, X3), .(X2, X4)) → U10_aga(X1, X2, X3, X4, selectcB_in_aga(X1, X3, X4))
U10_aga(X1, X2, X3, X4, selectcB_out_aga(X1, X3, X4)) → selectcB_out_aga(X1, .(X2, X3), .(X2, X4))

The argument filtering Pi contains the following mapping:
perm1A_in_ga(x1, x2)  =  perm1A_in_ga(x1)
.(x1, x2)  =  .(x1, x2)
selectB_in_aga(x1, x2, x3)  =  selectB_in_aga(x2)
selectcB_in_aga(x1, x2, x3)  =  selectcB_in_aga(x2)
selectcB_out_aga(x1, x2, x3)  =  selectcB_out_aga(x1, x2, x3)
U10_aga(x1, x2, x3, x4, x5)  =  U10_aga(x2, x3, x5)
PERM1A_IN_GA(x1, x2)  =  PERM1A_IN_GA(x1)
U2_GA(x1, x2, x3, x4)  =  U2_GA(x1, x2, x4)
U3_GA(x1, x2, x3, x4, x5)  =  U3_GA(x1, x2, x5)
SELECTB_IN_AGA(x1, x2, x3)  =  SELECTB_IN_AGA(x2)
U1_AGA(x1, x2, x3, x4, x5)  =  U1_AGA(x2, x3, x5)
U4_GA(x1, x2, x3, x4, x5)  =  U4_GA(x1, x2, x5)
U5_GA(x1, x2, x3, x4, x5)  =  U5_GA(x1, x2, 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 5 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

SELECTB_IN_AGA(X1, .(X2, X3), .(X2, X4)) → SELECTB_IN_AGA(X1, X3, X4)

The TRS R consists of the following rules:

selectcB_in_aga(X1, .(X1, X2), X2) → selectcB_out_aga(X1, .(X1, X2), X2)
selectcB_in_aga(X1, .(X2, X3), .(X2, X4)) → U10_aga(X1, X2, X3, X4, selectcB_in_aga(X1, X3, X4))
U10_aga(X1, X2, X3, X4, selectcB_out_aga(X1, X3, X4)) → selectcB_out_aga(X1, .(X2, X3), .(X2, X4))

The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
selectcB_in_aga(x1, x2, x3)  =  selectcB_in_aga(x2)
selectcB_out_aga(x1, x2, x3)  =  selectcB_out_aga(x1, x2, x3)
U10_aga(x1, x2, x3, x4, x5)  =  U10_aga(x2, x3, x5)
SELECTB_IN_AGA(x1, x2, x3)  =  SELECTB_IN_AGA(x2)

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

(8) UsableRulesProof (EQUIVALENT transformation)

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

(9) Obligation:

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

SELECTB_IN_AGA(X1, .(X2, X3), .(X2, X4)) → SELECTB_IN_AGA(X1, X3, X4)

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

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

(10) PiDPToQDPProof (SOUND transformation)

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

(11) Obligation:

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

SELECTB_IN_AGA(.(X2, X3)) → SELECTB_IN_AGA(X3)

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

(12) 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:

• SELECTB_IN_AGA(.(X2, X3)) → SELECTB_IN_AGA(X3)
  The graph contains the following edges 1 > 1

(13) YES

(14) Obligation:

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

PERM1A_IN_GA(.(X1, X2), .(X3, X4)) → U4_GA(X1, X2, X3, X4, selectcB_in_aga(X3, X2, X5))
U4_GA(X1, X2, X3, X4, selectcB_out_aga(X3, X2, X5)) → PERM1A_IN_GA(.(X1, X5), X4)
PERM1A_IN_GA(.(X1, X2), .(X1, X3)) → PERM1A_IN_GA(X2, X3)

The TRS R consists of the following rules:

selectcB_in_aga(X1, .(X1, X2), X2) → selectcB_out_aga(X1, .(X1, X2), X2)
selectcB_in_aga(X1, .(X2, X3), .(X2, X4)) → U10_aga(X1, X2, X3, X4, selectcB_in_aga(X1, X3, X4))
U10_aga(X1, X2, X3, X4, selectcB_out_aga(X1, X3, X4)) → selectcB_out_aga(X1, .(X2, X3), .(X2, X4))

The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
selectcB_in_aga(x1, x2, x3)  =  selectcB_in_aga(x2)
selectcB_out_aga(x1, x2, x3)  =  selectcB_out_aga(x1, x2, x3)
U10_aga(x1, x2, x3, x4, x5)  =  U10_aga(x2, x3, x5)
PERM1A_IN_GA(x1, x2)  =  PERM1A_IN_GA(x1)
U4_GA(x1, x2, x3, x4, x5)  =  U4_GA(x1, x2, 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:

PERM1A_IN_GA(.(X1, X2)) → U4_GA(X1, X2, selectcB_in_aga(X2))
U4_GA(X1, X2, selectcB_out_aga(X3, X2, X5)) → PERM1A_IN_GA(.(X1, X5))
PERM1A_IN_GA(.(X1, X2)) → PERM1A_IN_GA(X2)

The TRS R consists of the following rules:

selectcB_in_aga(.(X1, X2)) → selectcB_out_aga(X1, .(X1, X2), X2)
selectcB_in_aga(.(X2, X3)) → U10_aga(X2, X3, selectcB_in_aga(X3))
U10_aga(X2, X3, selectcB_out_aga(X1, X3, X4)) → selectcB_out_aga(X1, .(X2, X3), .(X2, X4))

The set Q consists of the following terms:

selectcB_in_aga(x0)
U10_aga(x0, x1, x2)

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

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

PERM1A_IN_GA(.(X1, X2)) → PERM1A_IN_GA(X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(.(x₁, x₂)) = 1 + x₂   
POL(PERM1A_IN_GA(x₁)) = 1 + x₁   
POL(U10_aga(x₁, x₂, x₃)) = 1 + x₃   
POL(U4_GA(x₁, x₂, x₃)) = 1 + x₃   
POL(selectcB_in_aga(x₁)) = 1 + x₁   
POL(selectcB_out_aga(x₁, x₂, x₃)) = 1 + x₃   

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

selectcB_in_aga(.(X1, X2)) → selectcB_out_aga(X1, .(X1, X2), X2)
selectcB_in_aga(.(X2, X3)) → U10_aga(X2, X3, selectcB_in_aga(X3))
U10_aga(X2, X3, selectcB_out_aga(X1, X3, X4)) → selectcB_out_aga(X1, .(X2, X3), .(X2, X4))

(18) Obligation:

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

PERM1A_IN_GA(.(X1, X2)) → U4_GA(X1, X2, selectcB_in_aga(X2))
U4_GA(X1, X2, selectcB_out_aga(X3, X2, X5)) → PERM1A_IN_GA(.(X1, X5))

The TRS R consists of the following rules:

selectcB_in_aga(.(X1, X2)) → selectcB_out_aga(X1, .(X1, X2), X2)
selectcB_in_aga(.(X2, X3)) → U10_aga(X2, X3, selectcB_in_aga(X3))
U10_aga(X2, X3, selectcB_out_aga(X1, X3, X4)) → selectcB_out_aga(X1, .(X2, X3), .(X2, X4))

The set Q consists of the following terms:

selectcB_in_aga(x0)
U10_aga(x0, x1, x2)

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

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

PERM1A_IN_GA(.(X1, X2)) → U4_GA(X1, X2, selectcB_in_aga(X2))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(.(x₁, x₂)) = 1 + x₂   
POL(PERM1A_IN_GA(x₁)) = x₁   
POL(U10_aga(x₁, x₂, x₃)) = 1 + x₃   
POL(U4_GA(x₁, x₂, x₃)) = x₃   
POL(selectcB_in_aga(x₁)) = x₁   
POL(selectcB_out_aga(x₁, x₂, x₃)) = 1 + x₃   

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

selectcB_in_aga(.(X1, X2)) → selectcB_out_aga(X1, .(X1, X2), X2)
selectcB_in_aga(.(X2, X3)) → U10_aga(X2, X3, selectcB_in_aga(X3))
U10_aga(X2, X3, selectcB_out_aga(X1, X3, X4)) → selectcB_out_aga(X1, .(X2, X3), .(X2, X4))

(20) Obligation:

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

U4_GA(X1, X2, selectcB_out_aga(X3, X2, X5)) → PERM1A_IN_GA(.(X1, X5))

The TRS R consists of the following rules:

selectcB_in_aga(.(X1, X2)) → selectcB_out_aga(X1, .(X1, X2), X2)
selectcB_in_aga(.(X2, X3)) → U10_aga(X2, X3, selectcB_in_aga(X3))
U10_aga(X2, X3, selectcB_out_aga(X1, X3, X4)) → selectcB_out_aga(X1, .(X2, X3), .(X2, X4))

The set Q consists of the following terms:

selectcB_in_aga(x0)
U10_aga(x0, x1, x2)

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

(21) DependencyGraphProof (EQUIVALENT transformation)

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

(22) TRUE