Left Termination proof of /tmp/tmp81f6Ke/perm.pl

Left Termination of the query pattern perm_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:

↳ Prolog

  ↳ PrologToPiTRSProof

Clauses:

ap1(nil, X, X).
ap1(cons(H, X), Y, cons(H, Z)) :- ap1(X, Y, Z).
ap2(nil, X, X).
ap2(cons(H, X), Y, cons(H, Z)) :- ap2(X, Y, Z).
perm(nil, nil).
perm(Xs, cons(X, Ys)) :- ','(ap1(X1s, cons(X, X2s), Xs), ','(ap2(X1s, X2s, Zs), perm(Zs, Ys))).

Queries:

perm(g,a).

We use the technique of [30].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

perm_in(Xs, cons(X, Ys)) → U3(Xs, X, Ys, ap1_in(X1s, cons(X, X2s), Xs))
ap1_in(cons(H, X), Y, cons(H, Z)) → U1(H, X, Y, Z, ap1_in(X, Y, Z))
ap1_in(nil, X, X) → ap1_out(nil, X, X)
U1(H, X, Y, Z, ap1_out(X, Y, Z)) → ap1_out(cons(H, X), Y, cons(H, Z))
U3(Xs, X, Ys, ap1_out(X1s, cons(X, X2s), Xs)) → U4(Xs, X, Ys, X1s, X2s, ap2_in(X1s, X2s, Zs))
ap2_in(cons(H, X), Y, cons(H, Z)) → U2(H, X, Y, Z, ap2_in(X, Y, Z))
ap2_in(nil, X, X) → ap2_out(nil, X, X)
U2(H, X, Y, Z, ap2_out(X, Y, Z)) → ap2_out(cons(H, X), Y, cons(H, Z))
U4(Xs, X, Ys, X1s, X2s, ap2_out(X1s, X2s, Zs)) → U5(Xs, X, Ys, perm_in(Zs, Ys))
perm_in(nil, nil) → perm_out(nil, nil)
U5(Xs, X, Ys, perm_out(Zs, Ys)) → perm_out(Xs, cons(X, Ys))

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_in(Xs, cons(X, Ys)) → U3(Xs, X, Ys, ap1_in(X1s, cons(X, X2s), Xs))
ap1_in(cons(H, X), Y, cons(H, Z)) → U1(H, X, Y, Z, ap1_in(X, Y, Z))
ap1_in(nil, X, X) → ap1_out(nil, X, X)
U1(H, X, Y, Z, ap1_out(X, Y, Z)) → ap1_out(cons(H, X), Y, cons(H, Z))
U3(Xs, X, Ys, ap1_out(X1s, cons(X, X2s), Xs)) → U4(Xs, X, Ys, X1s, X2s, ap2_in(X1s, X2s, Zs))
ap2_in(cons(H, X), Y, cons(H, Z)) → U2(H, X, Y, Z, ap2_in(X, Y, Z))
ap2_in(nil, X, X) → ap2_out(nil, X, X)
U2(H, X, Y, Z, ap2_out(X, Y, Z)) → ap2_out(cons(H, X), Y, cons(H, Z))
U4(Xs, X, Ys, X1s, X2s, ap2_out(X1s, X2s, Zs)) → U5(Xs, X, Ys, perm_in(Zs, Ys))
perm_in(nil, nil) → perm_out(nil, nil)
U5(Xs, X, Ys, perm_out(Zs, Ys)) → perm_out(Xs, cons(X, Ys))

PERM_IN(Xs, cons(X, Ys)) → U3¹(Xs, X, Ys, ap1_in(X1s, cons(X, X2s), Xs))
PERM_IN(Xs, cons(X, Ys)) → AP1_IN(X1s, cons(X, X2s), Xs)
AP1_IN(cons(H, X), Y, cons(H, Z)) → U1¹(H, X, Y, Z, ap1_in(X, Y, Z))
AP1_IN(cons(H, X), Y, cons(H, Z)) → AP1_IN(X, Y, Z)
U3¹(Xs, X, Ys, ap1_out(X1s, cons(X, X2s), Xs)) → U4¹(Xs, X, Ys, X1s, X2s, ap2_in(X1s, X2s, Zs))
U3¹(Xs, X, Ys, ap1_out(X1s, cons(X, X2s), Xs)) → AP2_IN(X1s, X2s, Zs)
AP2_IN(cons(H, X), Y, cons(H, Z)) → U2¹(H, X, Y, Z, ap2_in(X, Y, Z))
AP2_IN(cons(H, X), Y, cons(H, Z)) → AP2_IN(X, Y, Z)
U4¹(Xs, X, Ys, X1s, X2s, ap2_out(X1s, X2s, Zs)) → U5¹(Xs, X, Ys, perm_in(Zs, Ys))
U4¹(Xs, X, Ys, X1s, X2s, ap2_out(X1s, X2s, Zs)) → PERM_IN(Zs, Ys)

The TRS R consists of the following rules:

perm_in(Xs, cons(X, Ys)) → U3(Xs, X, Ys, ap1_in(X1s, cons(X, X2s), Xs))
ap1_in(cons(H, X), Y, cons(H, Z)) → U1(H, X, Y, Z, ap1_in(X, Y, Z))
ap1_in(nil, X, X) → ap1_out(nil, X, X)
U1(H, X, Y, Z, ap1_out(X, Y, Z)) → ap1_out(cons(H, X), Y, cons(H, Z))
U3(Xs, X, Ys, ap1_out(X1s, cons(X, X2s), Xs)) → U4(Xs, X, Ys, X1s, X2s, ap2_in(X1s, X2s, Zs))
ap2_in(cons(H, X), Y, cons(H, Z)) → U2(H, X, Y, Z, ap2_in(X, Y, Z))
ap2_in(nil, X, X) → ap2_out(nil, X, X)
U2(H, X, Y, Z, ap2_out(X, Y, Z)) → ap2_out(cons(H, X), Y, cons(H, Z))
U4(Xs, X, Ys, X1s, X2s, ap2_out(X1s, X2s, Zs)) → U5(Xs, X, Ys, perm_in(Zs, Ys))
perm_in(nil, nil) → perm_out(nil, nil)
U5(Xs, X, Ys, perm_out(Zs, Ys)) → perm_out(Xs, cons(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2) = perm_in(x1)
cons(x1, x2) = cons(x2)
U3(x1, x2, x3, x4) = U3(x4)
ap1_in(x1, x2, x3) = ap1_in(x3)
U1(x1, x2, x3, x4, x5) = U1(x5)
nil = nil
ap1_out(x1, x2, x3) = ap1_out(x1, x2)
U4(x1, x2, x3, x4, x5, x6) = U4(x6)
ap2_in(x1, x2, x3) = ap2_in(x1, x2)
U2(x1, x2, x3, x4, x5) = U2(x5)
ap2_out(x1, x2, x3) = ap2_out(x3)
U5(x1, x2, x3, x4) = U5(x4)
perm_out(x1, x2) = perm_out(x2)
U5¹(x1, x2, x3, x4) = U5¹(x4)
U3¹(x1, x2, x3, x4) = U3¹(x4)
AP2_IN(x1, x2, x3) = AP2_IN(x1, x2)
U4¹(x1, x2, x3, x4, x5, x6) = U4¹(x6)
PERM_IN(x1, x2) = PERM_IN(x1)
U2¹(x1, x2, x3, x4, x5) = U2¹(x5)
AP1_IN(x1, x2, x3) = AP1_IN(x3)
U1¹(x1, x2, x3, x4, x5) = U1¹(x5)

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_IN(Xs, cons(X, Ys)) → U3¹(Xs, X, Ys, ap1_in(X1s, cons(X, X2s), Xs))
PERM_IN(Xs, cons(X, Ys)) → AP1_IN(X1s, cons(X, X2s), Xs)
AP1_IN(cons(H, X), Y, cons(H, Z)) → U1¹(H, X, Y, Z, ap1_in(X, Y, Z))
AP1_IN(cons(H, X), Y, cons(H, Z)) → AP1_IN(X, Y, Z)
U3¹(Xs, X, Ys, ap1_out(X1s, cons(X, X2s), Xs)) → U4¹(Xs, X, Ys, X1s, X2s, ap2_in(X1s, X2s, Zs))
U3¹(Xs, X, Ys, ap1_out(X1s, cons(X, X2s), Xs)) → AP2_IN(X1s, X2s, Zs)
AP2_IN(cons(H, X), Y, cons(H, Z)) → U2¹(H, X, Y, Z, ap2_in(X, Y, Z))
AP2_IN(cons(H, X), Y, cons(H, Z)) → AP2_IN(X, Y, Z)
U4¹(Xs, X, Ys, X1s, X2s, ap2_out(X1s, X2s, Zs)) → U5¹(Xs, X, Ys, perm_in(Zs, Ys))
U4¹(Xs, X, Ys, X1s, X2s, ap2_out(X1s, X2s, Zs)) → PERM_IN(Zs, Ys)

The TRS R consists of the following rules:

perm_in(Xs, cons(X, Ys)) → U3(Xs, X, Ys, ap1_in(X1s, cons(X, X2s), Xs))
ap1_in(cons(H, X), Y, cons(H, Z)) → U1(H, X, Y, Z, ap1_in(X, Y, Z))
ap1_in(nil, X, X) → ap1_out(nil, X, X)
U1(H, X, Y, Z, ap1_out(X, Y, Z)) → ap1_out(cons(H, X), Y, cons(H, Z))
U3(Xs, X, Ys, ap1_out(X1s, cons(X, X2s), Xs)) → U4(Xs, X, Ys, X1s, X2s, ap2_in(X1s, X2s, Zs))
ap2_in(cons(H, X), Y, cons(H, Z)) → U2(H, X, Y, Z, ap2_in(X, Y, Z))
ap2_in(nil, X, X) → ap2_out(nil, X, X)
U2(H, X, Y, Z, ap2_out(X, Y, Z)) → ap2_out(cons(H, X), Y, cons(H, Z))
U4(Xs, X, Ys, X1s, X2s, ap2_out(X1s, X2s, Zs)) → U5(Xs, X, Ys, perm_in(Zs, Ys))
perm_in(nil, nil) → perm_out(nil, nil)
U5(Xs, X, Ys, perm_out(Zs, Ys)) → perm_out(Xs, cons(X, Ys))

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

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

AP2_IN(cons(H, X), Y, cons(H, Z)) → AP2_IN(X, Y, Z)

The TRS R consists of the following rules:

perm_in(Xs, cons(X, Ys)) → U3(Xs, X, Ys, ap1_in(X1s, cons(X, X2s), Xs))
ap1_in(cons(H, X), Y, cons(H, Z)) → U1(H, X, Y, Z, ap1_in(X, Y, Z))
ap1_in(nil, X, X) → ap1_out(nil, X, X)
U1(H, X, Y, Z, ap1_out(X, Y, Z)) → ap1_out(cons(H, X), Y, cons(H, Z))
U3(Xs, X, Ys, ap1_out(X1s, cons(X, X2s), Xs)) → U4(Xs, X, Ys, X1s, X2s, ap2_in(X1s, X2s, Zs))
ap2_in(cons(H, X), Y, cons(H, Z)) → U2(H, X, Y, Z, ap2_in(X, Y, Z))
ap2_in(nil, X, X) → ap2_out(nil, X, X)
U2(H, X, Y, Z, ap2_out(X, Y, Z)) → ap2_out(cons(H, X), Y, cons(H, Z))
U4(Xs, X, Ys, X1s, X2s, ap2_out(X1s, X2s, Zs)) → U5(Xs, X, Ys, perm_in(Zs, Ys))
perm_in(nil, nil) → perm_out(nil, nil)
U5(Xs, X, Ys, perm_out(Zs, Ys)) → perm_out(Xs, cons(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2) = perm_in(x1)
cons(x1, x2) = cons(x2)
U3(x1, x2, x3, x4) = U3(x4)
ap1_in(x1, x2, x3) = ap1_in(x3)
U1(x1, x2, x3, x4, x5) = U1(x5)
nil = nil
ap1_out(x1, x2, x3) = ap1_out(x1, x2)
U4(x1, x2, x3, x4, x5, x6) = U4(x6)
ap2_in(x1, x2, x3) = ap2_in(x1, x2)
U2(x1, x2, x3, x4, x5) = U2(x5)
ap2_out(x1, x2, x3) = ap2_out(x3)
U5(x1, x2, x3, x4) = U5(x4)
perm_out(x1, x2) = perm_out(x2)
AP2_IN(x1, x2, x3) = AP2_IN(x1, x2)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

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

AP2_IN(cons(H, X), Y, cons(H, Z)) → AP2_IN(X, Y, Z)

R is empty.
The argument filtering Pi contains the following mapping:
cons(x1, x2) = cons(x2)
AP2_IN(x1, x2, x3) = AP2_IN(x1, x2)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

              ↳ PiDP

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

AP2_IN(cons(X), Y) → AP2_IN(X, Y)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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:

AP2_IN(cons(X), Y) → AP2_IN(X, Y)
The graph contains the following edges 1 > 1, 2 >= 2

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

AP1_IN(cons(H, X), Y, cons(H, Z)) → AP1_IN(X, Y, Z)

The TRS R consists of the following rules:

perm_in(Xs, cons(X, Ys)) → U3(Xs, X, Ys, ap1_in(X1s, cons(X, X2s), Xs))
ap1_in(cons(H, X), Y, cons(H, Z)) → U1(H, X, Y, Z, ap1_in(X, Y, Z))
ap1_in(nil, X, X) → ap1_out(nil, X, X)
U1(H, X, Y, Z, ap1_out(X, Y, Z)) → ap1_out(cons(H, X), Y, cons(H, Z))
U3(Xs, X, Ys, ap1_out(X1s, cons(X, X2s), Xs)) → U4(Xs, X, Ys, X1s, X2s, ap2_in(X1s, X2s, Zs))
ap2_in(cons(H, X), Y, cons(H, Z)) → U2(H, X, Y, Z, ap2_in(X, Y, Z))
ap2_in(nil, X, X) → ap2_out(nil, X, X)
U2(H, X, Y, Z, ap2_out(X, Y, Z)) → ap2_out(cons(H, X), Y, cons(H, Z))
U4(Xs, X, Ys, X1s, X2s, ap2_out(X1s, X2s, Zs)) → U5(Xs, X, Ys, perm_in(Zs, Ys))
perm_in(nil, nil) → perm_out(nil, nil)
U5(Xs, X, Ys, perm_out(Zs, Ys)) → perm_out(Xs, cons(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2) = perm_in(x1)
cons(x1, x2) = cons(x2)
U3(x1, x2, x3, x4) = U3(x4)
ap1_in(x1, x2, x3) = ap1_in(x3)
U1(x1, x2, x3, x4, x5) = U1(x5)
nil = nil
ap1_out(x1, x2, x3) = ap1_out(x1, x2)
U4(x1, x2, x3, x4, x5, x6) = U4(x6)
ap2_in(x1, x2, x3) = ap2_in(x1, x2)
U2(x1, x2, x3, x4, x5) = U2(x5)
ap2_out(x1, x2, x3) = ap2_out(x3)
U5(x1, x2, x3, x4) = U5(x4)
perm_out(x1, x2) = perm_out(x2)
AP1_IN(x1, x2, x3) = AP1_IN(x3)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

AP1_IN(cons(H, X), Y, cons(H, Z)) → AP1_IN(X, Y, Z)

R is empty.
The argument filtering Pi contains the following mapping:
cons(x1, x2) = cons(x2)
AP1_IN(x1, x2, x3) = AP1_IN(x3)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

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

AP1_IN(cons(Z)) → AP1_IN(Z)

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

AP1_IN(cons(Z)) → AP1_IN(Z)
The graph contains the following edges 1 > 1

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

PERM_IN(Xs, cons(X, Ys)) → U3¹(Xs, X, Ys, ap1_in(X1s, cons(X, X2s), Xs))
U3¹(Xs, X, Ys, ap1_out(X1s, cons(X, X2s), Xs)) → U4¹(Xs, X, Ys, X1s, X2s, ap2_in(X1s, X2s, Zs))
U4¹(Xs, X, Ys, X1s, X2s, ap2_out(X1s, X2s, Zs)) → PERM_IN(Zs, Ys)

The TRS R consists of the following rules:

perm_in(Xs, cons(X, Ys)) → U3(Xs, X, Ys, ap1_in(X1s, cons(X, X2s), Xs))
ap1_in(cons(H, X), Y, cons(H, Z)) → U1(H, X, Y, Z, ap1_in(X, Y, Z))
ap1_in(nil, X, X) → ap1_out(nil, X, X)
U1(H, X, Y, Z, ap1_out(X, Y, Z)) → ap1_out(cons(H, X), Y, cons(H, Z))
U3(Xs, X, Ys, ap1_out(X1s, cons(X, X2s), Xs)) → U4(Xs, X, Ys, X1s, X2s, ap2_in(X1s, X2s, Zs))
ap2_in(cons(H, X), Y, cons(H, Z)) → U2(H, X, Y, Z, ap2_in(X, Y, Z))
ap2_in(nil, X, X) → ap2_out(nil, X, X)
U2(H, X, Y, Z, ap2_out(X, Y, Z)) → ap2_out(cons(H, X), Y, cons(H, Z))
U4(Xs, X, Ys, X1s, X2s, ap2_out(X1s, X2s, Zs)) → U5(Xs, X, Ys, perm_in(Zs, Ys))
perm_in(nil, nil) → perm_out(nil, nil)
U5(Xs, X, Ys, perm_out(Zs, Ys)) → perm_out(Xs, cons(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2) = perm_in(x1)
cons(x1, x2) = cons(x2)
U3(x1, x2, x3, x4) = U3(x4)
ap1_in(x1, x2, x3) = ap1_in(x3)
U1(x1, x2, x3, x4, x5) = U1(x5)
nil = nil
ap1_out(x1, x2, x3) = ap1_out(x1, x2)
U4(x1, x2, x3, x4, x5, x6) = U4(x6)
ap2_in(x1, x2, x3) = ap2_in(x1, x2)
U2(x1, x2, x3, x4, x5) = U2(x5)
ap2_out(x1, x2, x3) = ap2_out(x3)
U5(x1, x2, x3, x4) = U5(x4)
perm_out(x1, x2) = perm_out(x2)
U3¹(x1, x2, x3, x4) = U3¹(x4)
U4¹(x1, x2, x3, x4, x5, x6) = U4¹(x6)
PERM_IN(x1, x2) = PERM_IN(x1)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

PERM_IN(Xs, cons(X, Ys)) → U3¹(Xs, X, Ys, ap1_in(X1s, cons(X, X2s), Xs))
U3¹(Xs, X, Ys, ap1_out(X1s, cons(X, X2s), Xs)) → U4¹(Xs, X, Ys, X1s, X2s, ap2_in(X1s, X2s, Zs))
U4¹(Xs, X, Ys, X1s, X2s, ap2_out(X1s, X2s, Zs)) → PERM_IN(Zs, Ys)

The TRS R consists of the following rules:

ap1_in(cons(H, X), Y, cons(H, Z)) → U1(H, X, Y, Z, ap1_in(X, Y, Z))
ap1_in(nil, X, X) → ap1_out(nil, X, X)
ap2_in(cons(H, X), Y, cons(H, Z)) → U2(H, X, Y, Z, ap2_in(X, Y, Z))
ap2_in(nil, X, X) → ap2_out(nil, X, X)
U1(H, X, Y, Z, ap1_out(X, Y, Z)) → ap1_out(cons(H, X), Y, cons(H, Z))
U2(H, X, Y, Z, ap2_out(X, Y, Z)) → ap2_out(cons(H, X), Y, cons(H, Z))

The argument filtering Pi contains the following mapping:
cons(x1, x2) = cons(x2)
ap1_in(x1, x2, x3) = ap1_in(x3)
U1(x1, x2, x3, x4, x5) = U1(x5)
nil = nil
ap1_out(x1, x2, x3) = ap1_out(x1, x2)
ap2_in(x1, x2, x3) = ap2_in(x1, x2)
U2(x1, x2, x3, x4, x5) = U2(x5)
ap2_out(x1, x2, x3) = ap2_out(x3)
U3¹(x1, x2, x3, x4) = U3¹(x4)
U4¹(x1, x2, x3, x4, x5, x6) = U4¹(x6)
PERM_IN(x1, x2) = PERM_IN(x1)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

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

U4¹(ap2_out(Zs)) → PERM_IN(Zs)
PERM_IN(Xs) → U3¹(ap1_in(Xs))
U3¹(ap1_out(X1s, cons(X2s))) → U4¹(ap2_in(X1s, X2s))

The TRS R consists of the following rules:

ap1_in(cons(Z)) → U1(ap1_in(Z))
ap1_in(X) → ap1_out(nil, X)
ap2_in(cons(X), Y) → U2(ap2_in(X, Y))
ap2_in(nil, X) → ap2_out(X)
U1(ap1_out(X, Y)) → ap1_out(cons(X), Y)
U2(ap2_out(Z)) → ap2_out(cons(Z))

The set Q consists of the following terms:

ap1_in(x0)
ap2_in(x0, x1)
U1(x0)
U2(x0)

We have to consider all (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

PERM_IN(Xs) → U3¹(ap1_in(Xs))

Used ordering: POLO with Polynomial interpretation [25]:

POL(PERM_IN(x₁)) = 2 + 2·x₁
POL(U1(x₁)) = 1 + x₁
POL(U2(x₁)) = 2 + x₁
POL(U3¹(x₁)) = 2·x₁
POL(U4¹(x₁)) = 2 + x₁
POL(ap1_in(x₁)) = x₁
POL(ap1_out(x₁, x₂)) = x₁ + x₂
POL(ap2_in(x₁, x₂)) = 2·x₁ + 2·x₂
POL(ap2_out(x₁)) = 2·x₁
POL(cons(x₁)) = 1 + x₁
POL(nil) = 0

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ DependencyGraphProof

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

U4¹(ap2_out(Zs)) → PERM_IN(Zs)
U3¹(ap1_out(X1s, cons(X2s))) → U4¹(ap2_in(X1s, X2s))

The TRS R consists of the following rules:

ap1_in(cons(Z)) → U1(ap1_in(Z))
ap1_in(X) → ap1_out(nil, X)
ap2_in(cons(X), Y) → U2(ap2_in(X, Y))
ap2_in(nil, X) → ap2_out(X)
U1(ap1_out(X, Y)) → ap1_out(cons(X), Y)
U2(ap2_out(Z)) → ap2_out(cons(Z))

The set Q consists of the following terms:

ap1_in(x0)
ap2_in(x0, x1)
U1(x0)
U2(x0)

We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.