Left Termination proof of /tmp/tmpPBh4ln/permute1.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:

app1(.(X, Xs), Ys, .(X, Zs)) :- app1(Xs, Ys, Zs).
app1([], Ys, Ys).
app2(.(X, Xs), Ys, .(X, Zs)) :- app2(Xs, Ys, Zs).
app2([], Ys, Ys).
perm(Xs, .(X, Ys)) :- ','(app2(X1s, .(X, X2s), Xs), ','(app1(X1s, X2s, Zs), perm(Zs, Ys))).
perm([], []).

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([], []) → perm_out([], [])
perm_in(Xs, .(X, Ys)) → U3(Xs, X, Ys, app2_in(X1s, .(X, X2s), Xs))
app2_in([], Ys, Ys) → app2_out([], Ys, Ys)
app2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, app2_in(Xs, Ys, Zs))
U2(X, Xs, Ys, Zs, app2_out(Xs, Ys, Zs)) → app2_out(.(X, Xs), Ys, .(X, Zs))
U3(Xs, X, Ys, app2_out(X1s, .(X, X2s), Xs)) → U4(Xs, X, Ys, X1s, X2s, app1_in(X1s, X2s, Zs))
app1_in([], Ys, Ys) → app1_out([], Ys, Ys)
app1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app1_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app1_out(Xs, Ys, Zs)) → app1_out(.(X, Xs), Ys, .(X, Zs))
U4(Xs, X, Ys, X1s, X2s, app1_out(X1s, X2s, Zs)) → U5(Xs, X, Ys, perm_in(Zs, Ys))
U5(Xs, X, Ys, perm_out(Zs, Ys)) → perm_out(Xs, .(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([], []) → perm_out([], [])
perm_in(Xs, .(X, Ys)) → U3(Xs, X, Ys, app2_in(X1s, .(X, X2s), Xs))
app2_in([], Ys, Ys) → app2_out([], Ys, Ys)
app2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, app2_in(Xs, Ys, Zs))
U2(X, Xs, Ys, Zs, app2_out(Xs, Ys, Zs)) → app2_out(.(X, Xs), Ys, .(X, Zs))
U3(Xs, X, Ys, app2_out(X1s, .(X, X2s), Xs)) → U4(Xs, X, Ys, X1s, X2s, app1_in(X1s, X2s, Zs))
app1_in([], Ys, Ys) → app1_out([], Ys, Ys)
app1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app1_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app1_out(Xs, Ys, Zs)) → app1_out(.(X, Xs), Ys, .(X, Zs))
U4(Xs, X, Ys, X1s, X2s, app1_out(X1s, X2s, Zs)) → U5(Xs, X, Ys, perm_in(Zs, Ys))
U5(Xs, X, Ys, perm_out(Zs, Ys)) → perm_out(Xs, .(X, Ys))

PERM_IN(Xs, .(X, Ys)) → U3¹(Xs, X, Ys, app2_in(X1s, .(X, X2s), Xs))
PERM_IN(Xs, .(X, Ys)) → APP2_IN(X1s, .(X, X2s), Xs)
APP2_IN(.(X, Xs), Ys, .(X, Zs)) → U2¹(X, Xs, Ys, Zs, app2_in(Xs, Ys, Zs))
APP2_IN(.(X, Xs), Ys, .(X, Zs)) → APP2_IN(Xs, Ys, Zs)
U3¹(Xs, X, Ys, app2_out(X1s, .(X, X2s), Xs)) → U4¹(Xs, X, Ys, X1s, X2s, app1_in(X1s, X2s, Zs))
U3¹(Xs, X, Ys, app2_out(X1s, .(X, X2s), Xs)) → APP1_IN(X1s, X2s, Zs)
APP1_IN(.(X, Xs), Ys, .(X, Zs)) → U1¹(X, Xs, Ys, Zs, app1_in(Xs, Ys, Zs))
APP1_IN(.(X, Xs), Ys, .(X, Zs)) → APP1_IN(Xs, Ys, Zs)
U4¹(Xs, X, Ys, X1s, X2s, app1_out(X1s, X2s, Zs)) → U5¹(Xs, X, Ys, perm_in(Zs, Ys))
U4¹(Xs, X, Ys, X1s, X2s, app1_out(X1s, X2s, Zs)) → PERM_IN(Zs, Ys)

The TRS R consists of the following rules:

perm_in([], []) → perm_out([], [])
perm_in(Xs, .(X, Ys)) → U3(Xs, X, Ys, app2_in(X1s, .(X, X2s), Xs))
app2_in([], Ys, Ys) → app2_out([], Ys, Ys)
app2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, app2_in(Xs, Ys, Zs))
U2(X, Xs, Ys, Zs, app2_out(Xs, Ys, Zs)) → app2_out(.(X, Xs), Ys, .(X, Zs))
U3(Xs, X, Ys, app2_out(X1s, .(X, X2s), Xs)) → U4(Xs, X, Ys, X1s, X2s, app1_in(X1s, X2s, Zs))
app1_in([], Ys, Ys) → app1_out([], Ys, Ys)
app1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app1_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app1_out(Xs, Ys, Zs)) → app1_out(.(X, Xs), Ys, .(X, Zs))
U4(Xs, X, Ys, X1s, X2s, app1_out(X1s, X2s, Zs)) → U5(Xs, X, Ys, perm_in(Zs, Ys))
U5(Xs, X, Ys, perm_out(Zs, Ys)) → perm_out(Xs, .(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2) = perm_in(x1)
[] = []
perm_out(x1, x2) = perm_out(x2)
.(x1, x2) = .(x2)
U3(x1, x2, x3, x4) = U3(x4)
app2_in(x1, x2, x3) = app2_in(x3)
app2_out(x1, x2, x3) = app2_out(x1, x2)
U2(x1, x2, x3, x4, x5) = U2(x5)
U4(x1, x2, x3, x4, x5, x6) = U4(x6)
app1_in(x1, x2, x3) = app1_in(x1, x2)
app1_out(x1, x2, x3) = app1_out(x3)
U1(x1, x2, x3, x4, x5) = U1(x5)
U5(x1, x2, x3, x4) = U5(x4)
U5¹(x1, x2, x3, x4) = U5¹(x4)
U3¹(x1, x2, x3, x4) = U3¹(x4)
APP2_IN(x1, x2, x3) = APP2_IN(x3)
U4¹(x1, x2, x3, x4, x5, x6) = U4¹(x6)
PERM_IN(x1, x2) = PERM_IN(x1)
U2¹(x1, x2, x3, x4, x5) = U2¹(x5)
U1¹(x1, x2, x3, x4, x5) = U1¹(x5)
APP1_IN(x1, x2, x3) = APP1_IN(x1, x2)

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, .(X, Ys)) → U3¹(Xs, X, Ys, app2_in(X1s, .(X, X2s), Xs))
PERM_IN(Xs, .(X, Ys)) → APP2_IN(X1s, .(X, X2s), Xs)
APP2_IN(.(X, Xs), Ys, .(X, Zs)) → U2¹(X, Xs, Ys, Zs, app2_in(Xs, Ys, Zs))
APP2_IN(.(X, Xs), Ys, .(X, Zs)) → APP2_IN(Xs, Ys, Zs)
U3¹(Xs, X, Ys, app2_out(X1s, .(X, X2s), Xs)) → U4¹(Xs, X, Ys, X1s, X2s, app1_in(X1s, X2s, Zs))
U3¹(Xs, X, Ys, app2_out(X1s, .(X, X2s), Xs)) → APP1_IN(X1s, X2s, Zs)
APP1_IN(.(X, Xs), Ys, .(X, Zs)) → U1¹(X, Xs, Ys, Zs, app1_in(Xs, Ys, Zs))
APP1_IN(.(X, Xs), Ys, .(X, Zs)) → APP1_IN(Xs, Ys, Zs)
U4¹(Xs, X, Ys, X1s, X2s, app1_out(X1s, X2s, Zs)) → U5¹(Xs, X, Ys, perm_in(Zs, Ys))
U4¹(Xs, X, Ys, X1s, X2s, app1_out(X1s, X2s, Zs)) → PERM_IN(Zs, Ys)

The TRS R consists of the following rules:

perm_in([], []) → perm_out([], [])
perm_in(Xs, .(X, Ys)) → U3(Xs, X, Ys, app2_in(X1s, .(X, X2s), Xs))
app2_in([], Ys, Ys) → app2_out([], Ys, Ys)
app2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, app2_in(Xs, Ys, Zs))
U2(X, Xs, Ys, Zs, app2_out(Xs, Ys, Zs)) → app2_out(.(X, Xs), Ys, .(X, Zs))
U3(Xs, X, Ys, app2_out(X1s, .(X, X2s), Xs)) → U4(Xs, X, Ys, X1s, X2s, app1_in(X1s, X2s, Zs))
app1_in([], Ys, Ys) → app1_out([], Ys, Ys)
app1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app1_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app1_out(Xs, Ys, Zs)) → app1_out(.(X, Xs), Ys, .(X, Zs))
U4(Xs, X, Ys, X1s, X2s, app1_out(X1s, X2s, Zs)) → U5(Xs, X, Ys, perm_in(Zs, Ys))
U5(Xs, X, Ys, perm_out(Zs, Ys)) → perm_out(Xs, .(X, Ys))

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

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

APP1_IN(.(X, Xs), Ys, .(X, Zs)) → APP1_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

perm_in([], []) → perm_out([], [])
perm_in(Xs, .(X, Ys)) → U3(Xs, X, Ys, app2_in(X1s, .(X, X2s), Xs))
app2_in([], Ys, Ys) → app2_out([], Ys, Ys)
app2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, app2_in(Xs, Ys, Zs))
U2(X, Xs, Ys, Zs, app2_out(Xs, Ys, Zs)) → app2_out(.(X, Xs), Ys, .(X, Zs))
U3(Xs, X, Ys, app2_out(X1s, .(X, X2s), Xs)) → U4(Xs, X, Ys, X1s, X2s, app1_in(X1s, X2s, Zs))
app1_in([], Ys, Ys) → app1_out([], Ys, Ys)
app1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app1_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app1_out(Xs, Ys, Zs)) → app1_out(.(X, Xs), Ys, .(X, Zs))
U4(Xs, X, Ys, X1s, X2s, app1_out(X1s, X2s, Zs)) → U5(Xs, X, Ys, perm_in(Zs, Ys))
U5(Xs, X, Ys, perm_out(Zs, Ys)) → perm_out(Xs, .(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2) = perm_in(x1)
[] = []
perm_out(x1, x2) = perm_out(x2)
.(x1, x2) = .(x2)
U3(x1, x2, x3, x4) = U3(x4)
app2_in(x1, x2, x3) = app2_in(x3)
app2_out(x1, x2, x3) = app2_out(x1, x2)
U2(x1, x2, x3, x4, x5) = U2(x5)
U4(x1, x2, x3, x4, x5, x6) = U4(x6)
app1_in(x1, x2, x3) = app1_in(x1, x2)
app1_out(x1, x2, x3) = app1_out(x3)
U1(x1, x2, x3, x4, x5) = U1(x5)
U5(x1, x2, x3, x4) = U5(x4)
APP1_IN(x1, x2, x3) = APP1_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:

APP1_IN(.(X, Xs), Ys, .(X, Zs)) → APP1_IN(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x2)
APP1_IN(x1, x2, x3) = APP1_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:

APP1_IN(.(Xs), Ys) → APP1_IN(Xs, Ys)

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:

APP1_IN(.(Xs), Ys) → APP1_IN(Xs, Ys)
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:

APP2_IN(.(X, Xs), Ys, .(X, Zs)) → APP2_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

perm_in([], []) → perm_out([], [])
perm_in(Xs, .(X, Ys)) → U3(Xs, X, Ys, app2_in(X1s, .(X, X2s), Xs))
app2_in([], Ys, Ys) → app2_out([], Ys, Ys)
app2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, app2_in(Xs, Ys, Zs))
U2(X, Xs, Ys, Zs, app2_out(Xs, Ys, Zs)) → app2_out(.(X, Xs), Ys, .(X, Zs))
U3(Xs, X, Ys, app2_out(X1s, .(X, X2s), Xs)) → U4(Xs, X, Ys, X1s, X2s, app1_in(X1s, X2s, Zs))
app1_in([], Ys, Ys) → app1_out([], Ys, Ys)
app1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app1_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app1_out(Xs, Ys, Zs)) → app1_out(.(X, Xs), Ys, .(X, Zs))
U4(Xs, X, Ys, X1s, X2s, app1_out(X1s, X2s, Zs)) → U5(Xs, X, Ys, perm_in(Zs, Ys))
U5(Xs, X, Ys, perm_out(Zs, Ys)) → perm_out(Xs, .(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2) = perm_in(x1)
[] = []
perm_out(x1, x2) = perm_out(x2)
.(x1, x2) = .(x2)
U3(x1, x2, x3, x4) = U3(x4)
app2_in(x1, x2, x3) = app2_in(x3)
app2_out(x1, x2, x3) = app2_out(x1, x2)
U2(x1, x2, x3, x4, x5) = U2(x5)
U4(x1, x2, x3, x4, x5, x6) = U4(x6)
app1_in(x1, x2, x3) = app1_in(x1, x2)
app1_out(x1, x2, x3) = app1_out(x3)
U1(x1, x2, x3, x4, x5) = U1(x5)
U5(x1, x2, x3, x4) = U5(x4)
APP2_IN(x1, x2, x3) = APP2_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:

APP2_IN(.(X, Xs), Ys, .(X, Zs)) → APP2_IN(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x2)
APP2_IN(x1, x2, x3) = APP2_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:

APP2_IN(.(Zs)) → APP2_IN(Zs)

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

APP2_IN(.(Zs)) → APP2_IN(Zs)
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:

U4¹(Xs, X, Ys, X1s, X2s, app1_out(X1s, X2s, Zs)) → PERM_IN(Zs, Ys)
U3¹(Xs, X, Ys, app2_out(X1s, .(X, X2s), Xs)) → U4¹(Xs, X, Ys, X1s, X2s, app1_in(X1s, X2s, Zs))
PERM_IN(Xs, .(X, Ys)) → U3¹(Xs, X, Ys, app2_in(X1s, .(X, X2s), Xs))

The TRS R consists of the following rules:

perm_in([], []) → perm_out([], [])
perm_in(Xs, .(X, Ys)) → U3(Xs, X, Ys, app2_in(X1s, .(X, X2s), Xs))
app2_in([], Ys, Ys) → app2_out([], Ys, Ys)
app2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, app2_in(Xs, Ys, Zs))
U2(X, Xs, Ys, Zs, app2_out(Xs, Ys, Zs)) → app2_out(.(X, Xs), Ys, .(X, Zs))
U3(Xs, X, Ys, app2_out(X1s, .(X, X2s), Xs)) → U4(Xs, X, Ys, X1s, X2s, app1_in(X1s, X2s, Zs))
app1_in([], Ys, Ys) → app1_out([], Ys, Ys)
app1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app1_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app1_out(Xs, Ys, Zs)) → app1_out(.(X, Xs), Ys, .(X, Zs))
U4(Xs, X, Ys, X1s, X2s, app1_out(X1s, X2s, Zs)) → U5(Xs, X, Ys, perm_in(Zs, Ys))
U5(Xs, X, Ys, perm_out(Zs, Ys)) → perm_out(Xs, .(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2) = perm_in(x1)
[] = []
perm_out(x1, x2) = perm_out(x2)
.(x1, x2) = .(x2)
U3(x1, x2, x3, x4) = U3(x4)
app2_in(x1, x2, x3) = app2_in(x3)
app2_out(x1, x2, x3) = app2_out(x1, x2)
U2(x1, x2, x3, x4, x5) = U2(x5)
U4(x1, x2, x3, x4, x5, x6) = U4(x6)
app1_in(x1, x2, x3) = app1_in(x1, x2)
app1_out(x1, x2, x3) = app1_out(x3)
U1(x1, x2, x3, x4, x5) = U1(x5)
U5(x1, x2, x3, x4) = U5(x4)
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:

U4¹(Xs, X, Ys, X1s, X2s, app1_out(X1s, X2s, Zs)) → PERM_IN(Zs, Ys)
U3¹(Xs, X, Ys, app2_out(X1s, .(X, X2s), Xs)) → U4¹(Xs, X, Ys, X1s, X2s, app1_in(X1s, X2s, Zs))
PERM_IN(Xs, .(X, Ys)) → U3¹(Xs, X, Ys, app2_in(X1s, .(X, X2s), Xs))

The TRS R consists of the following rules:

app1_in([], Ys, Ys) → app1_out([], Ys, Ys)
app1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app1_in(Xs, Ys, Zs))
app2_in([], Ys, Ys) → app2_out([], Ys, Ys)
app2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, app2_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app1_out(Xs, Ys, Zs)) → app1_out(.(X, Xs), Ys, .(X, Zs))
U2(X, Xs, Ys, Zs, app2_out(Xs, Ys, Zs)) → app2_out(.(X, Xs), Ys, .(X, Zs))

The argument filtering Pi contains the following mapping:
[] = []
.(x1, x2) = .(x2)
app2_in(x1, x2, x3) = app2_in(x3)
app2_out(x1, x2, x3) = app2_out(x1, x2)
U2(x1, x2, x3, x4, x5) = U2(x5)
app1_in(x1, x2, x3) = app1_in(x1, x2)
app1_out(x1, x2, x3) = app1_out(x3)
U1(x1, x2, x3, x4, x5) = U1(x5)
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:

U3¹(app2_out(X1s, .(X2s))) → U4¹(app1_in(X1s, X2s))
PERM_IN(Xs) → U3¹(app2_in(Xs))
U4¹(app1_out(Zs)) → PERM_IN(Zs)

The TRS R consists of the following rules:

app1_in([], Ys) → app1_out(Ys)
app1_in(.(Xs), Ys) → U1(app1_in(Xs, Ys))
app2_in(Ys) → app2_out([], Ys)
app2_in(.(Zs)) → U2(app2_in(Zs))
U1(app1_out(Zs)) → app1_out(.(Zs))
U2(app2_out(Xs, Ys)) → app2_out(.(Xs), Ys)

The set Q consists of the following terms:

app1_in(x0, x1)
app2_in(x0)
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:

U3¹(app2_out(X1s, .(X2s))) → U4¹(app1_in(X1s, X2s))

Strictly oriented rules of the TRS R:

app2_in(Ys) → app2_out([], Ys)

Used ordering: POLO with Polynomial interpretation [25]:

POL(.(x₁)) = 1 + x₁
POL(PERM_IN(x₁)) = 1 + 2·x₁
POL(U1(x₁)) = 2 + x₁
POL(U2(x₁)) = 2 + x₁
POL(U3¹(x₁)) = x₁
POL(U4¹(x₁)) = 1 + x₁
POL([]) = 0
POL(app1_in(x₁, x₂)) = 2·x₁ + 2·x₂
POL(app1_out(x₁)) = 2·x₁
POL(app2_in(x₁)) = 1 + 2·x₁
POL(app2_out(x₁, x₂)) = 2·x₁ + 2·x₂

↳ 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:

PERM_IN(Xs) → U3¹(app2_in(Xs))
U4¹(app1_out(Zs)) → PERM_IN(Zs)

The TRS R consists of the following rules:

app1_in([], Ys) → app1_out(Ys)
app1_in(.(Xs), Ys) → U1(app1_in(Xs, Ys))
app2_in(.(Zs)) → U2(app2_in(Zs))
U1(app1_out(Zs)) → app1_out(.(Zs))
U2(app2_out(Xs, Ys)) → app2_out(.(Xs), Ys)

The set Q consists of the following terms:

app1_in(x0, x1)
app2_in(x0)
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.