Left Termination proof of /tmp/tmpOMf_De/pl1.2

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:

append2(parts([], Y), is(sum(Y))).
append2(parts(.(H, X), Y), is(sum(.(H, Z)))) :- append2(parts(X, Y), is(sum(Z))).
append1(parts([], Y), is(sum(Y))).
append1(parts(.(H, X), Y), is(sum(.(H, Z)))) :- append1(parts(X, Y), is(sum(Z))).
perm([], []).
perm(L, .(H, T)) :- ','(append2(parts(V, .(H, U)), is(sum(L))), ','(append1(parts(V, U), is(sum(W))), perm(W, T))).

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(L, .(H, T)) → U3(L, H, T, append2_in(parts(V, .(H, U)), is(sum(L))))
append2_in(parts(.(H, X), Y), is(sum(.(H, Z)))) → U1(H, X, Y, Z, append2_in(parts(X, Y), is(sum(Z))))
append2_in(parts([], Y), is(sum(Y))) → append2_out(parts([], Y), is(sum(Y)))
U1(H, X, Y, Z, append2_out(parts(X, Y), is(sum(Z)))) → append2_out(parts(.(H, X), Y), is(sum(.(H, Z))))
U3(L, H, T, append2_out(parts(V, .(H, U)), is(sum(L)))) → U4(L, H, T, V, U, append1_in(parts(V, U), is(sum(W))))
append1_in(parts(.(H, X), Y), is(sum(.(H, Z)))) → U2(H, X, Y, Z, append1_in(parts(X, Y), is(sum(Z))))
append1_in(parts([], Y), is(sum(Y))) → append1_out(parts([], Y), is(sum(Y)))
U2(H, X, Y, Z, append1_out(parts(X, Y), is(sum(Z)))) → append1_out(parts(.(H, X), Y), is(sum(.(H, Z))))
U4(L, H, T, V, U, append1_out(parts(V, U), is(sum(W)))) → U5(L, H, T, perm_in(W, T))
perm_in([], []) → perm_out([], [])
U5(L, H, T, perm_out(W, T)) → perm_out(L, .(H, T))

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(L, .(H, T)) → U3(L, H, T, append2_in(parts(V, .(H, U)), is(sum(L))))
append2_in(parts(.(H, X), Y), is(sum(.(H, Z)))) → U1(H, X, Y, Z, append2_in(parts(X, Y), is(sum(Z))))
append2_in(parts([], Y), is(sum(Y))) → append2_out(parts([], Y), is(sum(Y)))
U1(H, X, Y, Z, append2_out(parts(X, Y), is(sum(Z)))) → append2_out(parts(.(H, X), Y), is(sum(.(H, Z))))
U3(L, H, T, append2_out(parts(V, .(H, U)), is(sum(L)))) → U4(L, H, T, V, U, append1_in(parts(V, U), is(sum(W))))
append1_in(parts(.(H, X), Y), is(sum(.(H, Z)))) → U2(H, X, Y, Z, append1_in(parts(X, Y), is(sum(Z))))
append1_in(parts([], Y), is(sum(Y))) → append1_out(parts([], Y), is(sum(Y)))
U2(H, X, Y, Z, append1_out(parts(X, Y), is(sum(Z)))) → append1_out(parts(.(H, X), Y), is(sum(.(H, Z))))
U4(L, H, T, V, U, append1_out(parts(V, U), is(sum(W)))) → U5(L, H, T, perm_in(W, T))
perm_in([], []) → perm_out([], [])
U5(L, H, T, perm_out(W, T)) → perm_out(L, .(H, T))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2) = perm_in(x1)
.(x1, x2) = .(x1, x2)
U3(x1, x2, x3, x4) = U3(x4)
append2_in(x1, x2) = append2_in(x2)
parts(x1, x2) = parts(x1, x2)
is(x1) = is(x1)
sum(x1) = sum(x1)
U1(x1, x2, x3, x4, x5) = U1(x1, x5)
[] = []
append2_out(x1, x2) = append2_out(x1)
U4(x1, x2, x3, x4, x5, x6) = U4(x2, x6)
append1_in(x1, x2) = append1_in(x1)
U2(x1, x2, x3, x4, x5) = U2(x1, x5)
append1_out(x1, x2) = append1_out(x2)
U5(x1, x2, x3, x4) = U5(x2, x4)
perm_out(x1, x2) = perm_out(x2)

Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

PERM_IN(L, .(H, T)) → U3¹(L, H, T, append2_in(parts(V, .(H, U)), is(sum(L))))
PERM_IN(L, .(H, T)) → APPEND2_IN(parts(V, .(H, U)), is(sum(L)))
APPEND2_IN(parts(.(H, X), Y), is(sum(.(H, Z)))) → U1¹(H, X, Y, Z, append2_in(parts(X, Y), is(sum(Z))))
APPEND2_IN(parts(.(H, X), Y), is(sum(.(H, Z)))) → APPEND2_IN(parts(X, Y), is(sum(Z)))
U3¹(L, H, T, append2_out(parts(V, .(H, U)), is(sum(L)))) → U4¹(L, H, T, V, U, append1_in(parts(V, U), is(sum(W))))
U3¹(L, H, T, append2_out(parts(V, .(H, U)), is(sum(L)))) → APPEND1_IN(parts(V, U), is(sum(W)))
APPEND1_IN(parts(.(H, X), Y), is(sum(.(H, Z)))) → U2¹(H, X, Y, Z, append1_in(parts(X, Y), is(sum(Z))))
APPEND1_IN(parts(.(H, X), Y), is(sum(.(H, Z)))) → APPEND1_IN(parts(X, Y), is(sum(Z)))
U4¹(L, H, T, V, U, append1_out(parts(V, U), is(sum(W)))) → U5¹(L, H, T, perm_in(W, T))
U4¹(L, H, T, V, U, append1_out(parts(V, U), is(sum(W)))) → PERM_IN(W, T)

The TRS R consists of the following rules:

perm_in(L, .(H, T)) → U3(L, H, T, append2_in(parts(V, .(H, U)), is(sum(L))))
append2_in(parts(.(H, X), Y), is(sum(.(H, Z)))) → U1(H, X, Y, Z, append2_in(parts(X, Y), is(sum(Z))))
append2_in(parts([], Y), is(sum(Y))) → append2_out(parts([], Y), is(sum(Y)))
U1(H, X, Y, Z, append2_out(parts(X, Y), is(sum(Z)))) → append2_out(parts(.(H, X), Y), is(sum(.(H, Z))))
U3(L, H, T, append2_out(parts(V, .(H, U)), is(sum(L)))) → U4(L, H, T, V, U, append1_in(parts(V, U), is(sum(W))))
append1_in(parts(.(H, X), Y), is(sum(.(H, Z)))) → U2(H, X, Y, Z, append1_in(parts(X, Y), is(sum(Z))))
append1_in(parts([], Y), is(sum(Y))) → append1_out(parts([], Y), is(sum(Y)))
U2(H, X, Y, Z, append1_out(parts(X, Y), is(sum(Z)))) → append1_out(parts(.(H, X), Y), is(sum(.(H, Z))))
U4(L, H, T, V, U, append1_out(parts(V, U), is(sum(W)))) → U5(L, H, T, perm_in(W, T))
perm_in([], []) → perm_out([], [])
U5(L, H, T, perm_out(W, T)) → perm_out(L, .(H, T))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2) = perm_in(x1)
.(x1, x2) = .(x1, x2)
U3(x1, x2, x3, x4) = U3(x4)
append2_in(x1, x2) = append2_in(x2)
parts(x1, x2) = parts(x1, x2)
is(x1) = is(x1)
sum(x1) = sum(x1)
U1(x1, x2, x3, x4, x5) = U1(x1, x5)
[] = []
append2_out(x1, x2) = append2_out(x1)
U4(x1, x2, x3, x4, x5, x6) = U4(x2, x6)
append1_in(x1, x2) = append1_in(x1)
U2(x1, x2, x3, x4, x5) = U2(x1, x5)
append1_out(x1, x2) = append1_out(x2)
U5(x1, x2, x3, x4) = U5(x2, x4)
perm_out(x1, x2) = perm_out(x2)
U5¹(x1, x2, x3, x4) = U5¹(x2, x4)
APPEND1_IN(x1, x2) = APPEND1_IN(x1)
U3¹(x1, x2, x3, x4) = U3¹(x4)
U4¹(x1, x2, x3, x4, x5, x6) = U4¹(x2, x6)
PERM_IN(x1, x2) = PERM_IN(x1)
U2¹(x1, x2, x3, x4, x5) = U2¹(x1, x5)
U1¹(x1, x2, x3, x4, x5) = U1¹(x1, x5)
APPEND2_IN(x1, x2) = APPEND2_IN(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(L, .(H, T)) → U3¹(L, H, T, append2_in(parts(V, .(H, U)), is(sum(L))))
PERM_IN(L, .(H, T)) → APPEND2_IN(parts(V, .(H, U)), is(sum(L)))
APPEND2_IN(parts(.(H, X), Y), is(sum(.(H, Z)))) → U1¹(H, X, Y, Z, append2_in(parts(X, Y), is(sum(Z))))
APPEND2_IN(parts(.(H, X), Y), is(sum(.(H, Z)))) → APPEND2_IN(parts(X, Y), is(sum(Z)))
U3¹(L, H, T, append2_out(parts(V, .(H, U)), is(sum(L)))) → U4¹(L, H, T, V, U, append1_in(parts(V, U), is(sum(W))))
U3¹(L, H, T, append2_out(parts(V, .(H, U)), is(sum(L)))) → APPEND1_IN(parts(V, U), is(sum(W)))
APPEND1_IN(parts(.(H, X), Y), is(sum(.(H, Z)))) → U2¹(H, X, Y, Z, append1_in(parts(X, Y), is(sum(Z))))
APPEND1_IN(parts(.(H, X), Y), is(sum(.(H, Z)))) → APPEND1_IN(parts(X, Y), is(sum(Z)))
U4¹(L, H, T, V, U, append1_out(parts(V, U), is(sum(W)))) → U5¹(L, H, T, perm_in(W, T))
U4¹(L, H, T, V, U, append1_out(parts(V, U), is(sum(W)))) → PERM_IN(W, T)

The TRS R consists of the following rules:

perm_in(L, .(H, T)) → U3(L, H, T, append2_in(parts(V, .(H, U)), is(sum(L))))
append2_in(parts(.(H, X), Y), is(sum(.(H, Z)))) → U1(H, X, Y, Z, append2_in(parts(X, Y), is(sum(Z))))
append2_in(parts([], Y), is(sum(Y))) → append2_out(parts([], Y), is(sum(Y)))
U1(H, X, Y, Z, append2_out(parts(X, Y), is(sum(Z)))) → append2_out(parts(.(H, X), Y), is(sum(.(H, Z))))
U3(L, H, T, append2_out(parts(V, .(H, U)), is(sum(L)))) → U4(L, H, T, V, U, append1_in(parts(V, U), is(sum(W))))
append1_in(parts(.(H, X), Y), is(sum(.(H, Z)))) → U2(H, X, Y, Z, append1_in(parts(X, Y), is(sum(Z))))
append1_in(parts([], Y), is(sum(Y))) → append1_out(parts([], Y), is(sum(Y)))
U2(H, X, Y, Z, append1_out(parts(X, Y), is(sum(Z)))) → append1_out(parts(.(H, X), Y), is(sum(.(H, Z))))
U4(L, H, T, V, U, append1_out(parts(V, U), is(sum(W)))) → U5(L, H, T, perm_in(W, T))
perm_in([], []) → perm_out([], [])
U5(L, H, T, perm_out(W, T)) → perm_out(L, .(H, T))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2) = perm_in(x1)
.(x1, x2) = .(x1, x2)
U3(x1, x2, x3, x4) = U3(x4)
append2_in(x1, x2) = append2_in(x2)
parts(x1, x2) = parts(x1, x2)
is(x1) = is(x1)
sum(x1) = sum(x1)
U1(x1, x2, x3, x4, x5) = U1(x1, x5)
[] = []
append2_out(x1, x2) = append2_out(x1)
U4(x1, x2, x3, x4, x5, x6) = U4(x2, x6)
append1_in(x1, x2) = append1_in(x1)
U2(x1, x2, x3, x4, x5) = U2(x1, x5)
append1_out(x1, x2) = append1_out(x2)
U5(x1, x2, x3, x4) = U5(x2, x4)
perm_out(x1, x2) = perm_out(x2)
U5¹(x1, x2, x3, x4) = U5¹(x2, x4)
APPEND1_IN(x1, x2) = APPEND1_IN(x1)
U3¹(x1, x2, x3, x4) = U3¹(x4)
U4¹(x1, x2, x3, x4, x5, x6) = U4¹(x2, x6)
PERM_IN(x1, x2) = PERM_IN(x1)
U2¹(x1, x2, x3, x4, x5) = U2¹(x1, x5)
U1¹(x1, x2, x3, x4, x5) = U1¹(x1, x5)
APPEND2_IN(x1, x2) = APPEND2_IN(x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 3 SCCs with 5 less nodes.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

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

APPEND1_IN(parts(.(H, X), Y), is(sum(.(H, Z)))) → APPEND1_IN(parts(X, Y), is(sum(Z)))

The TRS R consists of the following rules:

perm_in(L, .(H, T)) → U3(L, H, T, append2_in(parts(V, .(H, U)), is(sum(L))))
append2_in(parts(.(H, X), Y), is(sum(.(H, Z)))) → U1(H, X, Y, Z, append2_in(parts(X, Y), is(sum(Z))))
append2_in(parts([], Y), is(sum(Y))) → append2_out(parts([], Y), is(sum(Y)))
U1(H, X, Y, Z, append2_out(parts(X, Y), is(sum(Z)))) → append2_out(parts(.(H, X), Y), is(sum(.(H, Z))))
U3(L, H, T, append2_out(parts(V, .(H, U)), is(sum(L)))) → U4(L, H, T, V, U, append1_in(parts(V, U), is(sum(W))))
append1_in(parts(.(H, X), Y), is(sum(.(H, Z)))) → U2(H, X, Y, Z, append1_in(parts(X, Y), is(sum(Z))))
append1_in(parts([], Y), is(sum(Y))) → append1_out(parts([], Y), is(sum(Y)))
U2(H, X, Y, Z, append1_out(parts(X, Y), is(sum(Z)))) → append1_out(parts(.(H, X), Y), is(sum(.(H, Z))))
U4(L, H, T, V, U, append1_out(parts(V, U), is(sum(W)))) → U5(L, H, T, perm_in(W, T))
perm_in([], []) → perm_out([], [])
U5(L, H, T, perm_out(W, T)) → perm_out(L, .(H, T))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2) = perm_in(x1)
.(x1, x2) = .(x1, x2)
U3(x1, x2, x3, x4) = U3(x4)
append2_in(x1, x2) = append2_in(x2)
parts(x1, x2) = parts(x1, x2)
is(x1) = is(x1)
sum(x1) = sum(x1)
U1(x1, x2, x3, x4, x5) = U1(x1, x5)
[] = []
append2_out(x1, x2) = append2_out(x1)
U4(x1, x2, x3, x4, x5, x6) = U4(x2, x6)
append1_in(x1, x2) = append1_in(x1)
U2(x1, x2, x3, x4, x5) = U2(x1, x5)
append1_out(x1, x2) = append1_out(x2)
U5(x1, x2, x3, x4) = U5(x2, x4)
perm_out(x1, x2) = perm_out(x2)
APPEND1_IN(x1, x2) = APPEND1_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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

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

APPEND1_IN(parts(.(H, X), Y), is(sum(.(H, Z)))) → APPEND1_IN(parts(X, Y), is(sum(Z)))

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
parts(x1, x2) = parts(x1, x2)
is(x1) = is(x1)
sum(x1) = sum(x1)
APPEND1_IN(x1, x2) = APPEND1_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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

              ↳ PiDP

              ↳ PiDP

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

APPEND1_IN(parts(.(H, X), Y)) → APPEND1_IN(parts(X, Y))

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

APPEND1_IN(parts(.(H, X), Y)) → APPEND1_IN(parts(X, Y))

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(.(x₁, x₂)) = x₁ + 2·x₂
POL(APPEND1_IN(x₁)) = 2·x₁
POL(parts(x₁, x₂)) = 2·x₁ + x₂

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                          ↳ QDP

                            ↳ PisEmptyProof

              ↳ PiDP

              ↳ PiDP

Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

APPEND2_IN(parts(.(H, X), Y), is(sum(.(H, Z)))) → APPEND2_IN(parts(X, Y), is(sum(Z)))

The TRS R consists of the following rules:

perm_in(L, .(H, T)) → U3(L, H, T, append2_in(parts(V, .(H, U)), is(sum(L))))
append2_in(parts(.(H, X), Y), is(sum(.(H, Z)))) → U1(H, X, Y, Z, append2_in(parts(X, Y), is(sum(Z))))
append2_in(parts([], Y), is(sum(Y))) → append2_out(parts([], Y), is(sum(Y)))
U1(H, X, Y, Z, append2_out(parts(X, Y), is(sum(Z)))) → append2_out(parts(.(H, X), Y), is(sum(.(H, Z))))
U3(L, H, T, append2_out(parts(V, .(H, U)), is(sum(L)))) → U4(L, H, T, V, U, append1_in(parts(V, U), is(sum(W))))
append1_in(parts(.(H, X), Y), is(sum(.(H, Z)))) → U2(H, X, Y, Z, append1_in(parts(X, Y), is(sum(Z))))
append1_in(parts([], Y), is(sum(Y))) → append1_out(parts([], Y), is(sum(Y)))
U2(H, X, Y, Z, append1_out(parts(X, Y), is(sum(Z)))) → append1_out(parts(.(H, X), Y), is(sum(.(H, Z))))
U4(L, H, T, V, U, append1_out(parts(V, U), is(sum(W)))) → U5(L, H, T, perm_in(W, T))
perm_in([], []) → perm_out([], [])
U5(L, H, T, perm_out(W, T)) → perm_out(L, .(H, T))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2) = perm_in(x1)
.(x1, x2) = .(x1, x2)
U3(x1, x2, x3, x4) = U3(x4)
append2_in(x1, x2) = append2_in(x2)
parts(x1, x2) = parts(x1, x2)
is(x1) = is(x1)
sum(x1) = sum(x1)
U1(x1, x2, x3, x4, x5) = U1(x1, x5)
[] = []
append2_out(x1, x2) = append2_out(x1)
U4(x1, x2, x3, x4, x5, x6) = U4(x2, x6)
append1_in(x1, x2) = append1_in(x1)
U2(x1, x2, x3, x4, x5) = U2(x1, x5)
append1_out(x1, x2) = append1_out(x2)
U5(x1, x2, x3, x4) = U5(x2, x4)
perm_out(x1, x2) = perm_out(x2)
APPEND2_IN(x1, x2) = APPEND2_IN(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

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

APPEND2_IN(parts(.(H, X), Y), is(sum(.(H, Z)))) → APPEND2_IN(parts(X, Y), is(sum(Z)))

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
parts(x1, x2) = parts(x1, x2)
is(x1) = is(x1)
sum(x1) = sum(x1)
APPEND2_IN(x1, x2) = APPEND2_IN(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

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

              ↳ PiDP

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

APPEND2_IN(is(sum(.(H, Z)))) → APPEND2_IN(is(sum(Z)))

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(.(x₁, x₂)) = x₁ + 2·x₂
POL(APPEND2_IN(x₁)) = 2·x₁
POL(is(x₁)) = 2·x₁
POL(sum(x₁)) = 2·x₁

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                          ↳ QDP

                            ↳ PisEmptyProof

              ↳ PiDP

Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

PERM_IN(L, .(H, T)) → U3¹(L, H, T, append2_in(parts(V, .(H, U)), is(sum(L))))
U4¹(L, H, T, V, U, append1_out(parts(V, U), is(sum(W)))) → PERM_IN(W, T)
U3¹(L, H, T, append2_out(parts(V, .(H, U)), is(sum(L)))) → U4¹(L, H, T, V, U, append1_in(parts(V, U), is(sum(W))))

The TRS R consists of the following rules:

perm_in(L, .(H, T)) → U3(L, H, T, append2_in(parts(V, .(H, U)), is(sum(L))))
append2_in(parts(.(H, X), Y), is(sum(.(H, Z)))) → U1(H, X, Y, Z, append2_in(parts(X, Y), is(sum(Z))))
append2_in(parts([], Y), is(sum(Y))) → append2_out(parts([], Y), is(sum(Y)))
U1(H, X, Y, Z, append2_out(parts(X, Y), is(sum(Z)))) → append2_out(parts(.(H, X), Y), is(sum(.(H, Z))))
U3(L, H, T, append2_out(parts(V, .(H, U)), is(sum(L)))) → U4(L, H, T, V, U, append1_in(parts(V, U), is(sum(W))))
append1_in(parts(.(H, X), Y), is(sum(.(H, Z)))) → U2(H, X, Y, Z, append1_in(parts(X, Y), is(sum(Z))))
append1_in(parts([], Y), is(sum(Y))) → append1_out(parts([], Y), is(sum(Y)))
U2(H, X, Y, Z, append1_out(parts(X, Y), is(sum(Z)))) → append1_out(parts(.(H, X), Y), is(sum(.(H, Z))))
U4(L, H, T, V, U, append1_out(parts(V, U), is(sum(W)))) → U5(L, H, T, perm_in(W, T))
perm_in([], []) → perm_out([], [])
U5(L, H, T, perm_out(W, T)) → perm_out(L, .(H, T))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2) = perm_in(x1)
.(x1, x2) = .(x1, x2)
U3(x1, x2, x3, x4) = U3(x4)
append2_in(x1, x2) = append2_in(x2)
parts(x1, x2) = parts(x1, x2)
is(x1) = is(x1)
sum(x1) = sum(x1)
U1(x1, x2, x3, x4, x5) = U1(x1, x5)
[] = []
append2_out(x1, x2) = append2_out(x1)
U4(x1, x2, x3, x4, x5, x6) = U4(x2, x6)
append1_in(x1, x2) = append1_in(x1)
U2(x1, x2, x3, x4, x5) = U2(x1, x5)
append1_out(x1, x2) = append1_out(x2)
U5(x1, x2, x3, x4) = U5(x2, x4)
perm_out(x1, x2) = perm_out(x2)
U3¹(x1, x2, x3, x4) = U3¹(x4)
U4¹(x1, x2, x3, x4, x5, x6) = U4¹(x2, 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(L, .(H, T)) → U3¹(L, H, T, append2_in(parts(V, .(H, U)), is(sum(L))))
U4¹(L, H, T, V, U, append1_out(parts(V, U), is(sum(W)))) → PERM_IN(W, T)
U3¹(L, H, T, append2_out(parts(V, .(H, U)), is(sum(L)))) → U4¹(L, H, T, V, U, append1_in(parts(V, U), is(sum(W))))

The TRS R consists of the following rules:

append2_in(parts(.(H, X), Y), is(sum(.(H, Z)))) → U1(H, X, Y, Z, append2_in(parts(X, Y), is(sum(Z))))
append2_in(parts([], Y), is(sum(Y))) → append2_out(parts([], Y), is(sum(Y)))
append1_in(parts(.(H, X), Y), is(sum(.(H, Z)))) → U2(H, X, Y, Z, append1_in(parts(X, Y), is(sum(Z))))
append1_in(parts([], Y), is(sum(Y))) → append1_out(parts([], Y), is(sum(Y)))
U1(H, X, Y, Z, append2_out(parts(X, Y), is(sum(Z)))) → append2_out(parts(.(H, X), Y), is(sum(.(H, Z))))
U2(H, X, Y, Z, append1_out(parts(X, Y), is(sum(Z)))) → append1_out(parts(.(H, X), Y), is(sum(.(H, Z))))

The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
append2_in(x1, x2) = append2_in(x2)
parts(x1, x2) = parts(x1, x2)
is(x1) = is(x1)
sum(x1) = sum(x1)
U1(x1, x2, x3, x4, x5) = U1(x1, x5)
[] = []
append2_out(x1, x2) = append2_out(x1)
append1_in(x1, x2) = append1_in(x1)
U2(x1, x2, x3, x4, x5) = U2(x1, x5)
append1_out(x1, x2) = append1_out(x2)
U3¹(x1, x2, x3, x4) = U3¹(x4)
U4¹(x1, x2, x3, x4, x5, x6) = U4¹(x2, 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(append2_out(parts(V, .(H, U)))) → U4¹(H, append1_in(parts(V, U)))
PERM_IN(L) → U3¹(append2_in(is(sum(L))))
U4¹(H, append1_out(is(sum(W)))) → PERM_IN(W)

The TRS R consists of the following rules:

append2_in(is(sum(.(H, Z)))) → U1(H, append2_in(is(sum(Z))))
append2_in(is(sum(Y))) → append2_out(parts([], Y))
append1_in(parts(.(H, X), Y)) → U2(H, append1_in(parts(X, Y)))
append1_in(parts([], Y)) → append1_out(is(sum(Y)))
U1(H, append2_out(parts(X, Y))) → append2_out(parts(.(H, X), Y))
U2(H, append1_out(is(sum(Z)))) → append1_out(is(sum(.(H, Z))))

The set Q consists of the following terms:

append2_in(x0)
append1_in(x0)
U1(x0, x1)
U2(x0, x1)

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(append2_out(parts(V, .(H, U)))) → U4¹(H, append1_in(parts(V, U)))
PERM_IN(L) → U3¹(append2_in(is(sum(L))))

Used ordering: POLO with Polynomial interpretation [25]:

POL(.(x₁, x₂)) = 1 + x₁ + x₂
POL(PERM_IN(x₁)) = 1 + 2·x₁
POL(U1(x₁, x₂)) = 1 + x₁ + x₂
POL(U2(x₁, x₂)) = 1 + x₁ + x₂
POL(U3¹(x₁)) = 2·x₁
POL(U4¹(x₁, x₂)) = 1 + x₁ + 2·x₂
POL([]) = 0
POL(append1_in(x₁)) = x₁
POL(append1_out(x₁)) = x₁
POL(append2_in(x₁)) = x₁
POL(append2_out(x₁)) = x₁
POL(is(x₁)) = x₁
POL(parts(x₁, x₂)) = x₁ + x₂
POL(sum(x₁)) = x₁

↳ 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(H, append1_out(is(sum(W)))) → PERM_IN(W)

The TRS R consists of the following rules:

append2_in(is(sum(.(H, Z)))) → U1(H, append2_in(is(sum(Z))))
append2_in(is(sum(Y))) → append2_out(parts([], Y))
append1_in(parts(.(H, X), Y)) → U2(H, append1_in(parts(X, Y)))
append1_in(parts([], Y)) → append1_out(is(sum(Y)))
U1(H, append2_out(parts(X, Y))) → append2_out(parts(.(H, X), Y))
U2(H, append1_out(is(sum(Z)))) → append1_out(is(sum(.(H, Z))))

The set Q consists of the following terms:

append2_in(x0)
append1_in(x0)
U1(x0, x1)
U2(x0, x1)

We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.