Left Termination proof of /tmp/tmpvOtssD/permutation2.pl

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

↳ PROLOG

  ↳ PrologToPiTRSProof

perm₂({}₀, {}₀).
perm₂(.₂(X, Y), .₂(U, V)) :- delete₃(U, .₂(X, Y), W), perm₂(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:
perm₂: (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:

perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(._2₂(X, Y), ._2₂(U, V)) -> if_perm_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_perm_2_in_1_ga₅(X, Y, U, V, delete_3_out_aga₃(U, ._2₂(X, Y), W)) -> if_perm_2_in_2_ga₆(X, Y, U, V, W, perm_2_in_ga₂(W, V))
if_perm_2_in_2_ga₆(X, Y, U, V, W, perm_2_out_ga₂(W, V)) -> perm_2_out_ga₂(._2₂(X, Y), ._2₂(U, V))

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:

perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(._2₂(X, Y), ._2₂(U, V)) -> if_perm_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_perm_2_in_1_ga₅(X, Y, U, V, delete_3_out_aga₃(U, ._2₂(X, Y), W)) -> if_perm_2_in_2_ga₆(X, Y, U, V, W, perm_2_in_ga₂(W, V))
if_perm_2_in_2_ga₆(X, Y, U, V, W, perm_2_out_ga₂(W, V)) -> perm_2_out_ga₂(._2₂(X, Y), ._2₂(U, V))

PERM_2_IN_GA₂(._2₂(X, Y), ._2₂(U, V)) -> IF_PERM_2_IN_1_GA₅(X, Y, U, V, delete_3_in_aga₃(U, ._2₂(X, Y), W))
PERM_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_PERM_2_IN_1_GA₅(X, Y, U, V, delete_3_out_aga₃(U, ._2₂(X, Y), W)) -> IF_PERM_2_IN_2_GA₆(X, Y, U, V, W, perm_2_in_ga₂(W, V))
IF_PERM_2_IN_1_GA₅(X, Y, U, V, delete_3_out_aga₃(U, ._2₂(X, Y), W)) -> PERM_2_IN_GA₂(W, V)

The TRS R consists of the following rules:

perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(._2₂(X, Y), ._2₂(U, V)) -> if_perm_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_perm_2_in_1_ga₅(X, Y, U, V, delete_3_out_aga₃(U, ._2₂(X, Y), W)) -> if_perm_2_in_2_ga₆(X, Y, U, V, W, perm_2_in_ga₂(W, V))
if_perm_2_in_2_ga₆(X, Y, U, V, W, perm_2_out_ga₂(W, V)) -> perm_2_out_ga₂(._2₂(X, Y), ._2₂(U, V))

The argument filtering Pi contains the following mapping:
perm_2_in_ga₂(x1, x2) = perm_2_in_ga₁(x1)
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
perm_2_out_ga₂(x1, x2) = perm_2_out_ga₁(x2)
if_perm_2_in_1_ga₅(x1, x2, x3, x4, x5) = if_perm_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_perm_2_in_2_ga₆(x1, x2, x3, x4, x5, x6) = if_perm_2_in_2_ga₂(x3, x6)
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)
IF_PERM_2_IN_1_GA₅(x1, x2, x3, x4, x5) = IF_PERM_2_IN_1_GA₁(x5)
PERM_2_IN_GA₂(x1, x2) = PERM_2_IN_GA₁(x1)
IF_PERM_2_IN_2_GA₆(x1, x2, x3, x4, x5, x6) = IF_PERM_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:

PERM_2_IN_GA₂(._2₂(X, Y), ._2₂(U, V)) -> IF_PERM_2_IN_1_GA₅(X, Y, U, V, delete_3_in_aga₃(U, ._2₂(X, Y), W))
PERM_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_PERM_2_IN_1_GA₅(X, Y, U, V, delete_3_out_aga₃(U, ._2₂(X, Y), W)) -> IF_PERM_2_IN_2_GA₆(X, Y, U, V, W, perm_2_in_ga₂(W, V))
IF_PERM_2_IN_1_GA₅(X, Y, U, V, delete_3_out_aga₃(U, ._2₂(X, Y), W)) -> PERM_2_IN_GA₂(W, V)

The TRS R consists of the following rules:

perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(._2₂(X, Y), ._2₂(U, V)) -> if_perm_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_perm_2_in_1_ga₅(X, Y, U, V, delete_3_out_aga₃(U, ._2₂(X, Y), W)) -> if_perm_2_in_2_ga₆(X, Y, U, V, W, perm_2_in_ga₂(W, V))
if_perm_2_in_2_ga₆(X, Y, U, V, W, perm_2_out_ga₂(W, V)) -> perm_2_out_ga₂(._2₂(X, Y), ._2₂(U, V))

↳ 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:

perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(._2₂(X, Y), ._2₂(U, V)) -> if_perm_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_perm_2_in_1_ga₅(X, Y, U, V, delete_3_out_aga₃(U, ._2₂(X, Y), W)) -> if_perm_2_in_2_ga₆(X, Y, U, V, W, perm_2_in_ga₂(W, V))
if_perm_2_in_2_ga₆(X, Y, U, V, W, perm_2_out_ga₂(W, V)) -> perm_2_out_ga₂(._2₂(X, Y), ._2₂(U, V))

The argument filtering Pi contains the following mapping:
perm_2_in_ga₂(x1, x2) = perm_2_in_ga₁(x1)
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
perm_2_out_ga₂(x1, x2) = perm_2_out_ga₁(x2)
if_perm_2_in_1_ga₅(x1, x2, x3, x4, x5) = if_perm_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_perm_2_in_2_ga₆(x1, x2, x3, x4, x5, x6) = if_perm_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:

DELETE_3_IN_AGA₁(._2₂(X, Y)) -> DELETE_3_IN_AGA₁(Y)
The graph contains the following edges 1 > 1

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

IF_PERM_2_IN_1_GA₅(X, Y, U, V, delete_3_out_aga₃(U, ._2₂(X, Y), W)) -> PERM_2_IN_GA₂(W, V)
PERM_2_IN_GA₂(._2₂(X, Y), ._2₂(U, V)) -> IF_PERM_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:

perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(._2₂(X, Y), ._2₂(U, V)) -> if_perm_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_perm_2_in_1_ga₅(X, Y, U, V, delete_3_out_aga₃(U, ._2₂(X, Y), W)) -> if_perm_2_in_2_ga₆(X, Y, U, V, W, perm_2_in_ga₂(W, V))
if_perm_2_in_2_ga₆(X, Y, U, V, W, perm_2_out_ga₂(W, V)) -> perm_2_out_ga₂(._2₂(X, Y), ._2₂(U, V))

The argument filtering Pi contains the following mapping:
perm_2_in_ga₂(x1, x2) = perm_2_in_ga₁(x1)
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
perm_2_out_ga₂(x1, x2) = perm_2_out_ga₁(x2)
if_perm_2_in_1_ga₅(x1, x2, x3, x4, x5) = if_perm_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_perm_2_in_2_ga₆(x1, x2, x3, x4, x5, x6) = if_perm_2_in_2_ga₂(x3, x6)
IF_PERM_2_IN_1_GA₅(x1, x2, x3, x4, x5) = IF_PERM_2_IN_1_GA₁(x5)
PERM_2_IN_GA₂(x1, x2) = PERM_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_PERM_2_IN_1_GA₅(X, Y, U, V, delete_3_out_aga₃(U, ._2₂(X, Y), W)) -> PERM_2_IN_GA₂(W, V)
PERM_2_IN_GA₂(._2₂(X, Y), ._2₂(U, V)) -> IF_PERM_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_PERM_2_IN_1_GA₅(x1, x2, x3, x4, x5) = IF_PERM_2_IN_1_GA₁(x5)
PERM_2_IN_GA₂(x1, x2) = PERM_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

                        ↳ RuleRemovalProof

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

IF_PERM_2_IN_1_GA₁(delete_3_out_aga₂(U, W)) -> PERM_2_IN_GA₁(W)
PERM_2_IN_GA₁(._2₂(X, Y)) -> IF_PERM_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 {PERM_2_IN_GA₁, IF_PERM_2_IN_1_GA₁}.
By using a polynomial ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

IF_PERM_2_IN_1_GA₁(delete_3_out_aga₂(U, W)) -> PERM_2_IN_GA₁(W)

Strictly oriented rules of the TRS R:

delete_3_in_aga₁(._2₂(X, Y)) -> delete_3_out_aga₂(X, Y)

Used ordering: POLO with Polynomial interpretation:

POL(._2₂(x₁, x₂)) = 1 + x₁ + x₂
POL(if_delete_3_in_1_aga₂(x₁, x₂)) = 2 + 2·x₁ + x₂
POL(IF_PERM_2_IN_1_GA₁(x₁)) = x₁
POL(delete_3_in_aga₁(x₁)) = 2·x₁
POL(delete_3_out_aga₂(x₁, x₂)) = 1 + x₁ + 2·x₂
POL(PERM_2_IN_GA₁(x₁)) = 2·x₁

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

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

PERM_2_IN_GA₁(._2₂(X, Y)) -> IF_PERM_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)) -> 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 {IF_PERM_2_IN_1_GA₁, PERM_2_IN_GA₁}.
By using a polynomial ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

delete_3_in_aga₁(._2₂(X, Y)) -> if_delete_3_in_1_aga₂(X, delete_3_in_aga₁(Y))

Used ordering: POLO with Polynomial interpretation:

POL(._2₂(x₁, x₂)) = 1 + x₁ + x₂
POL(if_delete_3_in_1_aga₂(x₁, x₂)) = 1 + x₁ + x₂
POL(IF_PERM_2_IN_1_GA₁(x₁)) = x₁
POL(delete_3_in_aga₁(x₁)) = 1 + 2·x₁
POL(delete_3_out_aga₂(x₁, x₂)) = x₁ + x₂
POL(PERM_2_IN_GA₁(x₁)) = 1 + 2·x₁

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

                              ↳ QDP

                                ↳ DependencyGraphProof

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

PERM_2_IN_GA₁(._2₂(X, Y)) -> IF_PERM_2_IN_1_GA₁(delete_3_in_aga₁(._2₂(X, Y)))

The TRS R consists of the following rules:

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 {IF_PERM_2_IN_1_GA₁, PERM_2_IN_GA₁}.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.