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



PROLOG
  ↳ PrologToPiTRSProof

permute2({}0, {}0).
permute2(.2(X, Y), .2(U, V)) :- delete3(U, .2(X, Y), W), permute2(W, V).
delete3(X, .2(X, Y), Y).
delete3(U, .2(X, Y), .2(X, Z)) :- delete3(U, Y, Z).


With regard to the inferred argument filtering the predicates were used in the following modes:
permute2: (b,f)
delete3: (f,b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


permute_2_in_ga2([]_0, []_0) -> permute_2_out_ga2([]_0, []_0)
permute_2_in_ga2(._22(X, Y), ._22(U, V)) -> if_permute_2_in_1_ga5(X, Y, U, V, delete_3_in_aga3(U, ._22(X, Y), W))
delete_3_in_aga3(X, ._22(X, Y), Y) -> delete_3_out_aga3(X, ._22(X, Y), Y)
delete_3_in_aga3(U, ._22(X, Y), ._22(X, Z)) -> if_delete_3_in_1_aga5(U, X, Y, Z, delete_3_in_aga3(U, Y, Z))
if_delete_3_in_1_aga5(U, X, Y, Z, delete_3_out_aga3(U, Y, Z)) -> delete_3_out_aga3(U, ._22(X, Y), ._22(X, Z))
if_permute_2_in_1_ga5(X, Y, U, V, delete_3_out_aga3(U, ._22(X, Y), W)) -> if_permute_2_in_2_ga6(X, Y, U, V, W, permute_2_in_ga2(W, V))
if_permute_2_in_2_ga6(X, Y, U, V, W, permute_2_out_ga2(W, V)) -> permute_2_out_ga2(._22(X, Y), ._22(U, V))

The argument filtering Pi contains the following mapping:
permute_2_in_ga2(x1, x2)  =  permute_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
permute_2_out_ga2(x1, x2)  =  permute_2_out_ga1(x2)
if_permute_2_in_1_ga5(x1, x2, x3, x4, x5)  =  if_permute_2_in_1_ga1(x5)
delete_3_in_aga3(x1, x2, x3)  =  delete_3_in_aga1(x2)
delete_3_out_aga3(x1, x2, x3)  =  delete_3_out_aga2(x1, x3)
if_delete_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_delete_3_in_1_aga2(x2, x5)
if_permute_2_in_2_ga6(x1, x2, x3, x4, x5, x6)  =  if_permute_2_in_2_ga2(x3, x6)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG



↳ PROLOG
  ↳ PrologToPiTRSProof
PiTRS
      ↳ DependencyPairsProof

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

permute_2_in_ga2([]_0, []_0) -> permute_2_out_ga2([]_0, []_0)
permute_2_in_ga2(._22(X, Y), ._22(U, V)) -> if_permute_2_in_1_ga5(X, Y, U, V, delete_3_in_aga3(U, ._22(X, Y), W))
delete_3_in_aga3(X, ._22(X, Y), Y) -> delete_3_out_aga3(X, ._22(X, Y), Y)
delete_3_in_aga3(U, ._22(X, Y), ._22(X, Z)) -> if_delete_3_in_1_aga5(U, X, Y, Z, delete_3_in_aga3(U, Y, Z))
if_delete_3_in_1_aga5(U, X, Y, Z, delete_3_out_aga3(U, Y, Z)) -> delete_3_out_aga3(U, ._22(X, Y), ._22(X, Z))
if_permute_2_in_1_ga5(X, Y, U, V, delete_3_out_aga3(U, ._22(X, Y), W)) -> if_permute_2_in_2_ga6(X, Y, U, V, W, permute_2_in_ga2(W, V))
if_permute_2_in_2_ga6(X, Y, U, V, W, permute_2_out_ga2(W, V)) -> permute_2_out_ga2(._22(X, Y), ._22(U, V))

The argument filtering Pi contains the following mapping:
permute_2_in_ga2(x1, x2)  =  permute_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
permute_2_out_ga2(x1, x2)  =  permute_2_out_ga1(x2)
if_permute_2_in_1_ga5(x1, x2, x3, x4, x5)  =  if_permute_2_in_1_ga1(x5)
delete_3_in_aga3(x1, x2, x3)  =  delete_3_in_aga1(x2)
delete_3_out_aga3(x1, x2, x3)  =  delete_3_out_aga2(x1, x3)
if_delete_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_delete_3_in_1_aga2(x2, x5)
if_permute_2_in_2_ga6(x1, x2, x3, x4, x5, x6)  =  if_permute_2_in_2_ga2(x3, x6)


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

PERMUTE_2_IN_GA2(._22(X, Y), ._22(U, V)) -> IF_PERMUTE_2_IN_1_GA5(X, Y, U, V, delete_3_in_aga3(U, ._22(X, Y), W))
PERMUTE_2_IN_GA2(._22(X, Y), ._22(U, V)) -> DELETE_3_IN_AGA3(U, ._22(X, Y), W)
DELETE_3_IN_AGA3(U, ._22(X, Y), ._22(X, Z)) -> IF_DELETE_3_IN_1_AGA5(U, X, Y, Z, delete_3_in_aga3(U, Y, Z))
DELETE_3_IN_AGA3(U, ._22(X, Y), ._22(X, Z)) -> DELETE_3_IN_AGA3(U, Y, Z)
IF_PERMUTE_2_IN_1_GA5(X, Y, U, V, delete_3_out_aga3(U, ._22(X, Y), W)) -> IF_PERMUTE_2_IN_2_GA6(X, Y, U, V, W, permute_2_in_ga2(W, V))
IF_PERMUTE_2_IN_1_GA5(X, Y, U, V, delete_3_out_aga3(U, ._22(X, Y), W)) -> PERMUTE_2_IN_GA2(W, V)

The TRS R consists of the following rules:

permute_2_in_ga2([]_0, []_0) -> permute_2_out_ga2([]_0, []_0)
permute_2_in_ga2(._22(X, Y), ._22(U, V)) -> if_permute_2_in_1_ga5(X, Y, U, V, delete_3_in_aga3(U, ._22(X, Y), W))
delete_3_in_aga3(X, ._22(X, Y), Y) -> delete_3_out_aga3(X, ._22(X, Y), Y)
delete_3_in_aga3(U, ._22(X, Y), ._22(X, Z)) -> if_delete_3_in_1_aga5(U, X, Y, Z, delete_3_in_aga3(U, Y, Z))
if_delete_3_in_1_aga5(U, X, Y, Z, delete_3_out_aga3(U, Y, Z)) -> delete_3_out_aga3(U, ._22(X, Y), ._22(X, Z))
if_permute_2_in_1_ga5(X, Y, U, V, delete_3_out_aga3(U, ._22(X, Y), W)) -> if_permute_2_in_2_ga6(X, Y, U, V, W, permute_2_in_ga2(W, V))
if_permute_2_in_2_ga6(X, Y, U, V, W, permute_2_out_ga2(W, V)) -> permute_2_out_ga2(._22(X, Y), ._22(U, V))

The argument filtering Pi contains the following mapping:
permute_2_in_ga2(x1, x2)  =  permute_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
permute_2_out_ga2(x1, x2)  =  permute_2_out_ga1(x2)
if_permute_2_in_1_ga5(x1, x2, x3, x4, x5)  =  if_permute_2_in_1_ga1(x5)
delete_3_in_aga3(x1, x2, x3)  =  delete_3_in_aga1(x2)
delete_3_out_aga3(x1, x2, x3)  =  delete_3_out_aga2(x1, x3)
if_delete_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_delete_3_in_1_aga2(x2, x5)
if_permute_2_in_2_ga6(x1, x2, x3, x4, x5, x6)  =  if_permute_2_in_2_ga2(x3, x6)
IF_PERMUTE_2_IN_1_GA5(x1, x2, x3, x4, x5)  =  IF_PERMUTE_2_IN_1_GA1(x5)
DELETE_3_IN_AGA3(x1, x2, x3)  =  DELETE_3_IN_AGA1(x2)
IF_DELETE_3_IN_1_AGA5(x1, x2, x3, x4, x5)  =  IF_DELETE_3_IN_1_AGA2(x2, x5)
PERMUTE_2_IN_GA2(x1, x2)  =  PERMUTE_2_IN_GA1(x1)
IF_PERMUTE_2_IN_2_GA6(x1, x2, x3, x4, x5, x6)  =  IF_PERMUTE_2_IN_2_GA2(x3, x6)

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

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
PiDP
          ↳ DependencyGraphProof

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

PERMUTE_2_IN_GA2(._22(X, Y), ._22(U, V)) -> IF_PERMUTE_2_IN_1_GA5(X, Y, U, V, delete_3_in_aga3(U, ._22(X, Y), W))
PERMUTE_2_IN_GA2(._22(X, Y), ._22(U, V)) -> DELETE_3_IN_AGA3(U, ._22(X, Y), W)
DELETE_3_IN_AGA3(U, ._22(X, Y), ._22(X, Z)) -> IF_DELETE_3_IN_1_AGA5(U, X, Y, Z, delete_3_in_aga3(U, Y, Z))
DELETE_3_IN_AGA3(U, ._22(X, Y), ._22(X, Z)) -> DELETE_3_IN_AGA3(U, Y, Z)
IF_PERMUTE_2_IN_1_GA5(X, Y, U, V, delete_3_out_aga3(U, ._22(X, Y), W)) -> IF_PERMUTE_2_IN_2_GA6(X, Y, U, V, W, permute_2_in_ga2(W, V))
IF_PERMUTE_2_IN_1_GA5(X, Y, U, V, delete_3_out_aga3(U, ._22(X, Y), W)) -> PERMUTE_2_IN_GA2(W, V)

The TRS R consists of the following rules:

permute_2_in_ga2([]_0, []_0) -> permute_2_out_ga2([]_0, []_0)
permute_2_in_ga2(._22(X, Y), ._22(U, V)) -> if_permute_2_in_1_ga5(X, Y, U, V, delete_3_in_aga3(U, ._22(X, Y), W))
delete_3_in_aga3(X, ._22(X, Y), Y) -> delete_3_out_aga3(X, ._22(X, Y), Y)
delete_3_in_aga3(U, ._22(X, Y), ._22(X, Z)) -> if_delete_3_in_1_aga5(U, X, Y, Z, delete_3_in_aga3(U, Y, Z))
if_delete_3_in_1_aga5(U, X, Y, Z, delete_3_out_aga3(U, Y, Z)) -> delete_3_out_aga3(U, ._22(X, Y), ._22(X, Z))
if_permute_2_in_1_ga5(X, Y, U, V, delete_3_out_aga3(U, ._22(X, Y), W)) -> if_permute_2_in_2_ga6(X, Y, U, V, W, permute_2_in_ga2(W, V))
if_permute_2_in_2_ga6(X, Y, U, V, W, permute_2_out_ga2(W, V)) -> permute_2_out_ga2(._22(X, Y), ._22(U, V))

The argument filtering Pi contains the following mapping:
permute_2_in_ga2(x1, x2)  =  permute_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
permute_2_out_ga2(x1, x2)  =  permute_2_out_ga1(x2)
if_permute_2_in_1_ga5(x1, x2, x3, x4, x5)  =  if_permute_2_in_1_ga1(x5)
delete_3_in_aga3(x1, x2, x3)  =  delete_3_in_aga1(x2)
delete_3_out_aga3(x1, x2, x3)  =  delete_3_out_aga2(x1, x3)
if_delete_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_delete_3_in_1_aga2(x2, x5)
if_permute_2_in_2_ga6(x1, x2, x3, x4, x5, x6)  =  if_permute_2_in_2_ga2(x3, x6)
IF_PERMUTE_2_IN_1_GA5(x1, x2, x3, x4, x5)  =  IF_PERMUTE_2_IN_1_GA1(x5)
DELETE_3_IN_AGA3(x1, x2, x3)  =  DELETE_3_IN_AGA1(x2)
IF_DELETE_3_IN_1_AGA5(x1, x2, x3, x4, x5)  =  IF_DELETE_3_IN_1_AGA2(x2, x5)
PERMUTE_2_IN_GA2(x1, x2)  =  PERMUTE_2_IN_GA1(x1)
IF_PERMUTE_2_IN_2_GA6(x1, x2, x3, x4, x5, x6)  =  IF_PERMUTE_2_IN_2_GA2(x3, x6)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 2 SCCs with 3 less nodes.

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
PiDP
                ↳ UsableRulesProof
              ↳ PiDP

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

DELETE_3_IN_AGA3(U, ._22(X, Y), ._22(X, Z)) -> DELETE_3_IN_AGA3(U, Y, Z)

The TRS R consists of the following rules:

permute_2_in_ga2([]_0, []_0) -> permute_2_out_ga2([]_0, []_0)
permute_2_in_ga2(._22(X, Y), ._22(U, V)) -> if_permute_2_in_1_ga5(X, Y, U, V, delete_3_in_aga3(U, ._22(X, Y), W))
delete_3_in_aga3(X, ._22(X, Y), Y) -> delete_3_out_aga3(X, ._22(X, Y), Y)
delete_3_in_aga3(U, ._22(X, Y), ._22(X, Z)) -> if_delete_3_in_1_aga5(U, X, Y, Z, delete_3_in_aga3(U, Y, Z))
if_delete_3_in_1_aga5(U, X, Y, Z, delete_3_out_aga3(U, Y, Z)) -> delete_3_out_aga3(U, ._22(X, Y), ._22(X, Z))
if_permute_2_in_1_ga5(X, Y, U, V, delete_3_out_aga3(U, ._22(X, Y), W)) -> if_permute_2_in_2_ga6(X, Y, U, V, W, permute_2_in_ga2(W, V))
if_permute_2_in_2_ga6(X, Y, U, V, W, permute_2_out_ga2(W, V)) -> permute_2_out_ga2(._22(X, Y), ._22(U, V))

The argument filtering Pi contains the following mapping:
permute_2_in_ga2(x1, x2)  =  permute_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
permute_2_out_ga2(x1, x2)  =  permute_2_out_ga1(x2)
if_permute_2_in_1_ga5(x1, x2, x3, x4, x5)  =  if_permute_2_in_1_ga1(x5)
delete_3_in_aga3(x1, x2, x3)  =  delete_3_in_aga1(x2)
delete_3_out_aga3(x1, x2, x3)  =  delete_3_out_aga2(x1, x3)
if_delete_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_delete_3_in_1_aga2(x2, x5)
if_permute_2_in_2_ga6(x1, x2, x3, x4, x5, x6)  =  if_permute_2_in_2_ga2(x3, x6)
DELETE_3_IN_AGA3(x1, x2, x3)  =  DELETE_3_IN_AGA1(x2)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting we can delete all non-usable rules from R.

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP

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

DELETE_3_IN_AGA3(U, ._22(X, Y), ._22(X, Z)) -> DELETE_3_IN_AGA3(U, Y, Z)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._22(x1, x2)
DELETE_3_IN_AGA3(x1, x2, x3)  =  DELETE_3_IN_AGA1(x2)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem by application of Pi.

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
QDP
                        ↳ QDPSizeChangeProof
              ↳ PiDP

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

DELETE_3_IN_AGA1(._22(X, Y)) -> DELETE_3_IN_AGA1(Y)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {DELETE_3_IN_AGA1}.
By using the subterm criterion together with the size-change analysis 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:



↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
PiDP
                ↳ UsableRulesProof

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

IF_PERMUTE_2_IN_1_GA5(X, Y, U, V, delete_3_out_aga3(U, ._22(X, Y), W)) -> PERMUTE_2_IN_GA2(W, V)
PERMUTE_2_IN_GA2(._22(X, Y), ._22(U, V)) -> IF_PERMUTE_2_IN_1_GA5(X, Y, U, V, delete_3_in_aga3(U, ._22(X, Y), W))

The TRS R consists of the following rules:

permute_2_in_ga2([]_0, []_0) -> permute_2_out_ga2([]_0, []_0)
permute_2_in_ga2(._22(X, Y), ._22(U, V)) -> if_permute_2_in_1_ga5(X, Y, U, V, delete_3_in_aga3(U, ._22(X, Y), W))
delete_3_in_aga3(X, ._22(X, Y), Y) -> delete_3_out_aga3(X, ._22(X, Y), Y)
delete_3_in_aga3(U, ._22(X, Y), ._22(X, Z)) -> if_delete_3_in_1_aga5(U, X, Y, Z, delete_3_in_aga3(U, Y, Z))
if_delete_3_in_1_aga5(U, X, Y, Z, delete_3_out_aga3(U, Y, Z)) -> delete_3_out_aga3(U, ._22(X, Y), ._22(X, Z))
if_permute_2_in_1_ga5(X, Y, U, V, delete_3_out_aga3(U, ._22(X, Y), W)) -> if_permute_2_in_2_ga6(X, Y, U, V, W, permute_2_in_ga2(W, V))
if_permute_2_in_2_ga6(X, Y, U, V, W, permute_2_out_ga2(W, V)) -> permute_2_out_ga2(._22(X, Y), ._22(U, V))

The argument filtering Pi contains the following mapping:
permute_2_in_ga2(x1, x2)  =  permute_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
permute_2_out_ga2(x1, x2)  =  permute_2_out_ga1(x2)
if_permute_2_in_1_ga5(x1, x2, x3, x4, x5)  =  if_permute_2_in_1_ga1(x5)
delete_3_in_aga3(x1, x2, x3)  =  delete_3_in_aga1(x2)
delete_3_out_aga3(x1, x2, x3)  =  delete_3_out_aga2(x1, x3)
if_delete_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_delete_3_in_1_aga2(x2, x5)
if_permute_2_in_2_ga6(x1, x2, x3, x4, x5, x6)  =  if_permute_2_in_2_ga2(x3, x6)
IF_PERMUTE_2_IN_1_GA5(x1, x2, x3, x4, x5)  =  IF_PERMUTE_2_IN_1_GA1(x5)
PERMUTE_2_IN_GA2(x1, x2)  =  PERMUTE_2_IN_GA1(x1)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting we can delete all non-usable rules from R.

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
PiDP
                    ↳ PiDPToQDPProof

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

IF_PERMUTE_2_IN_1_GA5(X, Y, U, V, delete_3_out_aga3(U, ._22(X, Y), W)) -> PERMUTE_2_IN_GA2(W, V)
PERMUTE_2_IN_GA2(._22(X, Y), ._22(U, V)) -> IF_PERMUTE_2_IN_1_GA5(X, Y, U, V, delete_3_in_aga3(U, ._22(X, Y), W))

The TRS R consists of the following rules:

delete_3_in_aga3(X, ._22(X, Y), Y) -> delete_3_out_aga3(X, ._22(X, Y), Y)
delete_3_in_aga3(U, ._22(X, Y), ._22(X, Z)) -> if_delete_3_in_1_aga5(U, X, Y, Z, delete_3_in_aga3(U, Y, Z))
if_delete_3_in_1_aga5(U, X, Y, Z, delete_3_out_aga3(U, Y, Z)) -> delete_3_out_aga3(U, ._22(X, Y), ._22(X, Z))

The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._22(x1, x2)
delete_3_in_aga3(x1, x2, x3)  =  delete_3_in_aga1(x2)
delete_3_out_aga3(x1, x2, x3)  =  delete_3_out_aga2(x1, x3)
if_delete_3_in_1_aga5(x1, x2, x3, x4, x5)  =  if_delete_3_in_1_aga2(x2, x5)
IF_PERMUTE_2_IN_1_GA5(x1, x2, x3, x4, x5)  =  IF_PERMUTE_2_IN_1_GA1(x5)
PERMUTE_2_IN_GA2(x1, x2)  =  PERMUTE_2_IN_GA1(x1)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem by application of Pi.

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
QDP
                        ↳ QDPSizeChangeProof

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

IF_PERMUTE_2_IN_1_GA1(delete_3_out_aga2(U, W)) -> PERMUTE_2_IN_GA1(W)
PERMUTE_2_IN_GA1(._22(X, Y)) -> IF_PERMUTE_2_IN_1_GA1(delete_3_in_aga1(._22(X, Y)))

The TRS R consists of the following rules:

delete_3_in_aga1(._22(X, Y)) -> delete_3_out_aga2(X, Y)
delete_3_in_aga1(._22(X, Y)) -> if_delete_3_in_1_aga2(X, delete_3_in_aga1(Y))
if_delete_3_in_1_aga2(X, delete_3_out_aga2(U, Z)) -> delete_3_out_aga2(U, ._22(X, Z))

The set Q consists of the following terms:

delete_3_in_aga1(x0)
if_delete_3_in_1_aga2(x0, x1)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {PERMUTE_2_IN_GA1, IF_PERMUTE_2_IN_1_GA1}.
We used the following order together with the size-change analysis to show that there are no infinite chains for this DP problem.

Order:Polynomial interpretation:


POL(._22(x1, x2)) = 1 + x2   
POL(if_delete_3_in_1_aga2(x1, x2)) = 1 + x2   
POL(delete_3_in_aga1(x1)) = x1   
POL(delete_3_out_aga2(x1, x2)) = 1 + x2   

From the DPs we obtained the following set of size-change graphs:

We oriented the following set of usable rules.


if_delete_3_in_1_aga2(X, delete_3_out_aga2(U, Z)) -> delete_3_out_aga2(U, ._22(X, Z))
delete_3_in_aga1(._22(X, Y)) -> if_delete_3_in_1_aga2(X, delete_3_in_aga1(Y))
delete_3_in_aga1(._22(X, Y)) -> delete_3_out_aga2(X, Y)