Left Termination proof of /tmp/tmpdH57l7/slowsort-oi.pl

Left Termination of the query pattern slowsort_in_2(a, g) w.r.t. the given Prolog program could not be shown:

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

Clauses:

slowsort(X, Y) :- ','(perm(X, Y), sorted(Y)).
sorted([]).
sorted(.(X, [])).
sorted(.(X, .(Y, Z))) :- ','(le(X, Y), sorted(.(Y, Z))).
perm([], []).
perm(.(X, .(Y, [])), .(U, .(V, []))) :- ','(delete(U, .(X, Y), Z), perm(Z, V)).
delete(X, .(X, Y), Y).
delete(X, .(Y, Z), W) :- delete(X, Z, W).
le(s(X), s(Y)) :- le(X, Y).
le(0, s(X)).
le(0, 0).

Queries:

slowsort(a,g).

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:

slowsort_in(X, Y) → U1(X, Y, perm_in(X, Y))
perm_in(.(X, .(Y, [])), .(U, .(V, []))) → U5(X, Y, U, V, delete_in(U, .(X, Y), Z))
delete_in(X, .(Y, Z), W) → U7(X, Y, Z, W, delete_in(X, Z, W))
delete_in(X, .(X, Y), Y) → delete_out(X, .(X, Y), Y)
U7(X, Y, Z, W, delete_out(X, Z, W)) → delete_out(X, .(Y, Z), W)
U5(X, Y, U, V, delete_out(U, .(X, Y), Z)) → U6(X, Y, U, V, perm_in(Z, V))
perm_in([], []) → perm_out([], [])
U6(X, Y, U, V, perm_out(Z, V)) → perm_out(.(X, .(Y, [])), .(U, .(V, [])))
U1(X, Y, perm_out(X, Y)) → U2(X, Y, sorted_in(Y))
sorted_in(.(X, .(Y, Z))) → U3(X, Y, Z, le_in(X, Y))
le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U8(X, Y, le_in(X, Y))
U8(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))
U3(X, Y, Z, le_out(X, Y)) → U4(X, Y, Z, sorted_in(.(Y, Z)))
sorted_in(.(X, [])) → sorted_out(.(X, []))
sorted_in([]) → sorted_out([])
U4(X, Y, Z, sorted_out(.(Y, Z))) → sorted_out(.(X, .(Y, Z)))
U2(X, Y, sorted_out(Y)) → slowsort_out(X, Y)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

  ↳ PrologToPiTRSProof

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

slowsort_in(X, Y) → U1(X, Y, perm_in(X, Y))
perm_in(.(X, .(Y, [])), .(U, .(V, []))) → U5(X, Y, U, V, delete_in(U, .(X, Y), Z))
delete_in(X, .(Y, Z), W) → U7(X, Y, Z, W, delete_in(X, Z, W))
delete_in(X, .(X, Y), Y) → delete_out(X, .(X, Y), Y)
U7(X, Y, Z, W, delete_out(X, Z, W)) → delete_out(X, .(Y, Z), W)
U5(X, Y, U, V, delete_out(U, .(X, Y), Z)) → U6(X, Y, U, V, perm_in(Z, V))
perm_in([], []) → perm_out([], [])
U6(X, Y, U, V, perm_out(Z, V)) → perm_out(.(X, .(Y, [])), .(U, .(V, [])))
U1(X, Y, perm_out(X, Y)) → U2(X, Y, sorted_in(Y))
sorted_in(.(X, .(Y, Z))) → U3(X, Y, Z, le_in(X, Y))
le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U8(X, Y, le_in(X, Y))
U8(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))
U3(X, Y, Z, le_out(X, Y)) → U4(X, Y, Z, sorted_in(.(Y, Z)))
sorted_in(.(X, [])) → sorted_out(.(X, []))
sorted_in([]) → sorted_out([])
U4(X, Y, Z, sorted_out(.(Y, Z))) → sorted_out(.(X, .(Y, Z)))
U2(X, Y, sorted_out(Y)) → slowsort_out(X, Y)

SLOWSORT_IN(X, Y) → U1¹(X, Y, perm_in(X, Y))
SLOWSORT_IN(X, Y) → PERM_IN(X, Y)
PERM_IN(.(X, .(Y, [])), .(U, .(V, []))) → U5¹(X, Y, U, V, delete_in(U, .(X, Y), Z))
PERM_IN(.(X, .(Y, [])), .(U, .(V, []))) → DELETE_IN(U, .(X, Y), Z)
DELETE_IN(X, .(Y, Z), W) → U7¹(X, Y, Z, W, delete_in(X, Z, W))
DELETE_IN(X, .(Y, Z), W) → DELETE_IN(X, Z, W)
U5¹(X, Y, U, V, delete_out(U, .(X, Y), Z)) → U6¹(X, Y, U, V, perm_in(Z, V))
U5¹(X, Y, U, V, delete_out(U, .(X, Y), Z)) → PERM_IN(Z, V)
U1¹(X, Y, perm_out(X, Y)) → U2¹(X, Y, sorted_in(Y))
U1¹(X, Y, perm_out(X, Y)) → SORTED_IN(Y)
SORTED_IN(.(X, .(Y, Z))) → U3¹(X, Y, Z, le_in(X, Y))
SORTED_IN(.(X, .(Y, Z))) → LE_IN(X, Y)
LE_IN(s(X), s(Y)) → U8¹(X, Y, le_in(X, Y))
LE_IN(s(X), s(Y)) → LE_IN(X, Y)
U3¹(X, Y, Z, le_out(X, Y)) → U4¹(X, Y, Z, sorted_in(.(Y, Z)))
U3¹(X, Y, Z, le_out(X, Y)) → SORTED_IN(.(Y, Z))

The TRS R consists of the following rules:

slowsort_in(X, Y) → U1(X, Y, perm_in(X, Y))
perm_in(.(X, .(Y, [])), .(U, .(V, []))) → U5(X, Y, U, V, delete_in(U, .(X, Y), Z))
delete_in(X, .(Y, Z), W) → U7(X, Y, Z, W, delete_in(X, Z, W))
delete_in(X, .(X, Y), Y) → delete_out(X, .(X, Y), Y)
U7(X, Y, Z, W, delete_out(X, Z, W)) → delete_out(X, .(Y, Z), W)
U5(X, Y, U, V, delete_out(U, .(X, Y), Z)) → U6(X, Y, U, V, perm_in(Z, V))
perm_in([], []) → perm_out([], [])
U6(X, Y, U, V, perm_out(Z, V)) → perm_out(.(X, .(Y, [])), .(U, .(V, [])))
U1(X, Y, perm_out(X, Y)) → U2(X, Y, sorted_in(Y))
sorted_in(.(X, .(Y, Z))) → U3(X, Y, Z, le_in(X, Y))
le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U8(X, Y, le_in(X, Y))
U8(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))
U3(X, Y, Z, le_out(X, Y)) → U4(X, Y, Z, sorted_in(.(Y, Z)))
sorted_in(.(X, [])) → sorted_out(.(X, []))
sorted_in([]) → sorted_out([])
U4(X, Y, Z, sorted_out(.(Y, Z))) → sorted_out(.(X, .(Y, Z)))
U2(X, Y, sorted_out(Y)) → slowsort_out(X, Y)

The argument filtering Pi contains the following mapping:
slowsort_in(x1, x2) = slowsort_in(x2)
U1(x1, x2, x3) = U1(x2, x3)
perm_in(x1, x2) = perm_in(x2)
.(x1, x2) = .(x1, x2)
[] = []
U5(x1, x2, x3, x4, x5) = U5(x4, x5)
delete_in(x1, x2, x3) = delete_in(x1)
U7(x1, x2, x3, x4, x5) = U7(x5)
delete_out(x1, x2, x3) = delete_out
U6(x1, x2, x3, x4, x5) = U6(x5)
perm_out(x1, x2) = perm_out
U2(x1, x2, x3) = U2(x3)
sorted_in(x1) = sorted_in(x1)
U3(x1, x2, x3, x4) = U3(x2, x3, x4)
le_in(x1, x2) = le_in(x1, x2)
0 = 0
le_out(x1, x2) = le_out
s(x1) = s(x1)
U8(x1, x2, x3) = U8(x3)
U4(x1, x2, x3, x4) = U4(x4)
sorted_out(x1) = sorted_out
slowsort_out(x1, x2) = slowsort_out
DELETE_IN(x1, x2, x3) = DELETE_IN(x1)
SORTED_IN(x1) = SORTED_IN(x1)
U4¹(x1, x2, x3, x4) = U4¹(x4)
U8¹(x1, x2, x3) = U8¹(x3)
SLOWSORT_IN(x1, x2) = SLOWSORT_IN(x2)
PERM_IN(x1, x2) = PERM_IN(x2)
U6¹(x1, x2, x3, x4, x5) = U6¹(x5)
LE_IN(x1, x2) = LE_IN(x1, x2)
U1¹(x1, x2, x3) = U1¹(x2, x3)
U5¹(x1, x2, x3, x4, x5) = U5¹(x4, x5)
U3¹(x1, x2, x3, x4) = U3¹(x2, x3, x4)
U7¹(x1, x2, x3, x4, x5) = U7¹(x5)
U2¹(x1, x2, x3) = U2¹(x3)

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

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

  ↳ PrologToPiTRSProof

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

SLOWSORT_IN(X, Y) → U1¹(X, Y, perm_in(X, Y))
SLOWSORT_IN(X, Y) → PERM_IN(X, Y)
PERM_IN(.(X, .(Y, [])), .(U, .(V, []))) → U5¹(X, Y, U, V, delete_in(U, .(X, Y), Z))
PERM_IN(.(X, .(Y, [])), .(U, .(V, []))) → DELETE_IN(U, .(X, Y), Z)
DELETE_IN(X, .(Y, Z), W) → U7¹(X, Y, Z, W, delete_in(X, Z, W))
DELETE_IN(X, .(Y, Z), W) → DELETE_IN(X, Z, W)
U5¹(X, Y, U, V, delete_out(U, .(X, Y), Z)) → U6¹(X, Y, U, V, perm_in(Z, V))
U5¹(X, Y, U, V, delete_out(U, .(X, Y), Z)) → PERM_IN(Z, V)
U1¹(X, Y, perm_out(X, Y)) → U2¹(X, Y, sorted_in(Y))
U1¹(X, Y, perm_out(X, Y)) → SORTED_IN(Y)
SORTED_IN(.(X, .(Y, Z))) → U3¹(X, Y, Z, le_in(X, Y))
SORTED_IN(.(X, .(Y, Z))) → LE_IN(X, Y)
LE_IN(s(X), s(Y)) → U8¹(X, Y, le_in(X, Y))
LE_IN(s(X), s(Y)) → LE_IN(X, Y)
U3¹(X, Y, Z, le_out(X, Y)) → U4¹(X, Y, Z, sorted_in(.(Y, Z)))
U3¹(X, Y, Z, le_out(X, Y)) → SORTED_IN(.(Y, Z))

The TRS R consists of the following rules:

slowsort_in(X, Y) → U1(X, Y, perm_in(X, Y))
perm_in(.(X, .(Y, [])), .(U, .(V, []))) → U5(X, Y, U, V, delete_in(U, .(X, Y), Z))
delete_in(X, .(Y, Z), W) → U7(X, Y, Z, W, delete_in(X, Z, W))
delete_in(X, .(X, Y), Y) → delete_out(X, .(X, Y), Y)
U7(X, Y, Z, W, delete_out(X, Z, W)) → delete_out(X, .(Y, Z), W)
U5(X, Y, U, V, delete_out(U, .(X, Y), Z)) → U6(X, Y, U, V, perm_in(Z, V))
perm_in([], []) → perm_out([], [])
U6(X, Y, U, V, perm_out(Z, V)) → perm_out(.(X, .(Y, [])), .(U, .(V, [])))
U1(X, Y, perm_out(X, Y)) → U2(X, Y, sorted_in(Y))
sorted_in(.(X, .(Y, Z))) → U3(X, Y, Z, le_in(X, Y))
le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U8(X, Y, le_in(X, Y))
U8(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))
U3(X, Y, Z, le_out(X, Y)) → U4(X, Y, Z, sorted_in(.(Y, Z)))
sorted_in(.(X, [])) → sorted_out(.(X, []))
sorted_in([]) → sorted_out([])
U4(X, Y, Z, sorted_out(.(Y, Z))) → sorted_out(.(X, .(Y, Z)))
U2(X, Y, sorted_out(Y)) → slowsort_out(X, Y)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

LE_IN(s(X), s(Y)) → LE_IN(X, Y)

The TRS R consists of the following rules:

slowsort_in(X, Y) → U1(X, Y, perm_in(X, Y))
perm_in(.(X, .(Y, [])), .(U, .(V, []))) → U5(X, Y, U, V, delete_in(U, .(X, Y), Z))
delete_in(X, .(Y, Z), W) → U7(X, Y, Z, W, delete_in(X, Z, W))
delete_in(X, .(X, Y), Y) → delete_out(X, .(X, Y), Y)
U7(X, Y, Z, W, delete_out(X, Z, W)) → delete_out(X, .(Y, Z), W)
U5(X, Y, U, V, delete_out(U, .(X, Y), Z)) → U6(X, Y, U, V, perm_in(Z, V))
perm_in([], []) → perm_out([], [])
U6(X, Y, U, V, perm_out(Z, V)) → perm_out(.(X, .(Y, [])), .(U, .(V, [])))
U1(X, Y, perm_out(X, Y)) → U2(X, Y, sorted_in(Y))
sorted_in(.(X, .(Y, Z))) → U3(X, Y, Z, le_in(X, Y))
le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U8(X, Y, le_in(X, Y))
U8(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))
U3(X, Y, Z, le_out(X, Y)) → U4(X, Y, Z, sorted_in(.(Y, Z)))
sorted_in(.(X, [])) → sorted_out(.(X, []))
sorted_in([]) → sorted_out([])
U4(X, Y, Z, sorted_out(.(Y, Z))) → sorted_out(.(X, .(Y, Z)))
U2(X, Y, sorted_out(Y)) → slowsort_out(X, Y)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

LE_IN(s(X), s(Y)) → LE_IN(X, Y)

R is empty.
Pi is empty.
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

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

LE_IN(s(X), s(Y)) → LE_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:

LE_IN(s(X), s(Y)) → LE_IN(X, Y)
The graph contains the following edges 1 > 1, 2 > 2

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

SORTED_IN(.(X, .(Y, Z))) → U3¹(X, Y, Z, le_in(X, Y))
U3¹(X, Y, Z, le_out(X, Y)) → SORTED_IN(.(Y, Z))

The TRS R consists of the following rules:

slowsort_in(X, Y) → U1(X, Y, perm_in(X, Y))
perm_in(.(X, .(Y, [])), .(U, .(V, []))) → U5(X, Y, U, V, delete_in(U, .(X, Y), Z))
delete_in(X, .(Y, Z), W) → U7(X, Y, Z, W, delete_in(X, Z, W))
delete_in(X, .(X, Y), Y) → delete_out(X, .(X, Y), Y)
U7(X, Y, Z, W, delete_out(X, Z, W)) → delete_out(X, .(Y, Z), W)
U5(X, Y, U, V, delete_out(U, .(X, Y), Z)) → U6(X, Y, U, V, perm_in(Z, V))
perm_in([], []) → perm_out([], [])
U6(X, Y, U, V, perm_out(Z, V)) → perm_out(.(X, .(Y, [])), .(U, .(V, [])))
U1(X, Y, perm_out(X, Y)) → U2(X, Y, sorted_in(Y))
sorted_in(.(X, .(Y, Z))) → U3(X, Y, Z, le_in(X, Y))
le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U8(X, Y, le_in(X, Y))
U8(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))
U3(X, Y, Z, le_out(X, Y)) → U4(X, Y, Z, sorted_in(.(Y, Z)))
sorted_in(.(X, [])) → sorted_out(.(X, []))
sorted_in([]) → sorted_out([])
U4(X, Y, Z, sorted_out(.(Y, Z))) → sorted_out(.(X, .(Y, Z)))
U2(X, Y, sorted_out(Y)) → slowsort_out(X, Y)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

SORTED_IN(.(X, .(Y, Z))) → U3¹(X, Y, Z, le_in(X, Y))
U3¹(X, Y, Z, le_out(X, Y)) → SORTED_IN(.(Y, Z))

The TRS R consists of the following rules:

le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U8(X, Y, le_in(X, Y))
U8(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))

The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
le_in(x1, x2) = le_in(x1, x2)
0 = 0
le_out(x1, x2) = le_out
s(x1) = s(x1)
U8(x1, x2, x3) = U8(x3)
SORTED_IN(x1) = SORTED_IN(x1)
U3¹(x1, x2, x3, x4) = U3¹(x2, x3, x4)

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

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

SORTED_IN(.(X, .(Y, Z))) → U3¹(Y, Z, le_in(X, Y))
U3¹(Y, Z, le_out) → SORTED_IN(.(Y, Z))

The TRS R consists of the following rules:

le_in(0, 0) → le_out
le_in(0, s(X)) → le_out
le_in(s(X), s(Y)) → U8(le_in(X, Y))
U8(le_out) → le_out

The set Q consists of the following terms:

le_in(x0, x1)
U8(x0)

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:

SORTED_IN(.(X, .(Y, Z))) → U3¹(Y, Z, le_in(X, Y))
U3¹(Y, Z, le_out) → SORTED_IN(.(Y, Z))

The following rules are removed from R:

le_in(0, 0) → le_out
le_in(0, s(X)) → le_out
le_in(s(X), s(Y)) → U8(le_in(X, Y))
U8(le_out) → le_out

Used ordering: POLO with Polynomial interpretation [25]:

POL(.(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(0) = 2
POL(SORTED_IN(x₁)) = 1 + x₁
POL(U3¹(x₁, x₂, x₃)) = 2 + 2·x₁ + 2·x₂ + x₃
POL(U8(x₁)) = 1 + x₁
POL(le_in(x₁, x₂)) = 2 + x₁ + 2·x₂
POL(le_out) = 2
POL(s(x₁)) = 1 + x₁

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                          ↳ QDP

                            ↳ PisEmptyProof

              ↳ PiDP

              ↳ PiDP

  ↳ PrologToPiTRSProof

Q DP problem:
P is empty.
R is empty.
The set Q consists of the following terms:

le_in(x0, x1)
U8(x0)

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

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

DELETE_IN(X, .(Y, Z), W) → DELETE_IN(X, Z, W)

The TRS R consists of the following rules:

slowsort_in(X, Y) → U1(X, Y, perm_in(X, Y))
perm_in(.(X, .(Y, [])), .(U, .(V, []))) → U5(X, Y, U, V, delete_in(U, .(X, Y), Z))
delete_in(X, .(Y, Z), W) → U7(X, Y, Z, W, delete_in(X, Z, W))
delete_in(X, .(X, Y), Y) → delete_out(X, .(X, Y), Y)
U7(X, Y, Z, W, delete_out(X, Z, W)) → delete_out(X, .(Y, Z), W)
U5(X, Y, U, V, delete_out(U, .(X, Y), Z)) → U6(X, Y, U, V, perm_in(Z, V))
perm_in([], []) → perm_out([], [])
U6(X, Y, U, V, perm_out(Z, V)) → perm_out(.(X, .(Y, [])), .(U, .(V, [])))
U1(X, Y, perm_out(X, Y)) → U2(X, Y, sorted_in(Y))
sorted_in(.(X, .(Y, Z))) → U3(X, Y, Z, le_in(X, Y))
le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U8(X, Y, le_in(X, Y))
U8(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))
U3(X, Y, Z, le_out(X, Y)) → U4(X, Y, Z, sorted_in(.(Y, Z)))
sorted_in(.(X, [])) → sorted_out(.(X, []))
sorted_in([]) → sorted_out([])
U4(X, Y, Z, sorted_out(.(Y, Z))) → sorted_out(.(X, .(Y, Z)))
U2(X, Y, sorted_out(Y)) → slowsort_out(X, Y)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

DELETE_IN(X, .(Y, Z), W) → DELETE_IN(X, Z, W)

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

                        ↳ NonTerminationProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

DELETE_IN(X) → DELETE_IN(X)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

DELETE_IN(X) → DELETE_IN(X)

The TRS R consists of the following rules:none

s = DELETE_IN(X) evaluates to t =DELETE_IN(X)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from DELETE_IN(X) to DELETE_IN(X).

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

  ↳ PrologToPiTRSProof

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

PERM_IN(.(X, .(Y, [])), .(U, .(V, []))) → U5¹(X, Y, U, V, delete_in(U, .(X, Y), Z))
U5¹(X, Y, U, V, delete_out(U, .(X, Y), Z)) → PERM_IN(Z, V)

The TRS R consists of the following rules:

slowsort_in(X, Y) → U1(X, Y, perm_in(X, Y))
perm_in(.(X, .(Y, [])), .(U, .(V, []))) → U5(X, Y, U, V, delete_in(U, .(X, Y), Z))
delete_in(X, .(Y, Z), W) → U7(X, Y, Z, W, delete_in(X, Z, W))
delete_in(X, .(X, Y), Y) → delete_out(X, .(X, Y), Y)
U7(X, Y, Z, W, delete_out(X, Z, W)) → delete_out(X, .(Y, Z), W)
U5(X, Y, U, V, delete_out(U, .(X, Y), Z)) → U6(X, Y, U, V, perm_in(Z, V))
perm_in([], []) → perm_out([], [])
U6(X, Y, U, V, perm_out(Z, V)) → perm_out(.(X, .(Y, [])), .(U, .(V, [])))
U1(X, Y, perm_out(X, Y)) → U2(X, Y, sorted_in(Y))
sorted_in(.(X, .(Y, Z))) → U3(X, Y, Z, le_in(X, Y))
le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U8(X, Y, le_in(X, Y))
U8(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))
U3(X, Y, Z, le_out(X, Y)) → U4(X, Y, Z, sorted_in(.(Y, Z)))
sorted_in(.(X, [])) → sorted_out(.(X, []))
sorted_in([]) → sorted_out([])
U4(X, Y, Z, sorted_out(.(Y, Z))) → sorted_out(.(X, .(Y, Z)))
U2(X, Y, sorted_out(Y)) → slowsort_out(X, Y)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

  ↳ PrologToPiTRSProof

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

PERM_IN(.(X, .(Y, [])), .(U, .(V, []))) → U5¹(X, Y, U, V, delete_in(U, .(X, Y), Z))
U5¹(X, Y, U, V, delete_out(U, .(X, Y), Z)) → PERM_IN(Z, V)

The TRS R consists of the following rules:

delete_in(X, .(Y, Z), W) → U7(X, Y, Z, W, delete_in(X, Z, W))
delete_in(X, .(X, Y), Y) → delete_out(X, .(X, Y), Y)
U7(X, Y, Z, W, delete_out(X, Z, W)) → delete_out(X, .(Y, Z), W)

The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
[] = []
delete_in(x1, x2, x3) = delete_in(x1)
U7(x1, x2, x3, x4, x5) = U7(x5)
delete_out(x1, x2, x3) = delete_out
PERM_IN(x1, x2) = PERM_IN(x2)
U5¹(x1, x2, x3, x4, x5) = U5¹(x4, x5)

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

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

  ↳ PrologToPiTRSProof

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

U5¹(V, delete_out) → PERM_IN(V)
PERM_IN(.(U, .(V, []))) → U5¹(V, delete_in(U))

The TRS R consists of the following rules:

delete_in(X) → U7(delete_in(X))
delete_in(X) → delete_out
U7(delete_out) → delete_out

The set Q consists of the following terms:

delete_in(x0)
U7(x0)

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:

U5¹(V, delete_out) → PERM_IN(V)
The graph contains the following edges 1 >= 1
PERM_IN(.(U, .(V, []))) → U5¹(V, delete_in(U))
The graph contains the following edges 1 > 1

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:

slowsort_in(X, Y) → U1(X, Y, perm_in(X, Y))
perm_in(.(X, .(Y, [])), .(U, .(V, []))) → U5(X, Y, U, V, delete_in(U, .(X, Y), Z))
delete_in(X, .(Y, Z), W) → U7(X, Y, Z, W, delete_in(X, Z, W))
delete_in(X, .(X, Y), Y) → delete_out(X, .(X, Y), Y)
U7(X, Y, Z, W, delete_out(X, Z, W)) → delete_out(X, .(Y, Z), W)
U5(X, Y, U, V, delete_out(U, .(X, Y), Z)) → U6(X, Y, U, V, perm_in(Z, V))
perm_in([], []) → perm_out([], [])
U6(X, Y, U, V, perm_out(Z, V)) → perm_out(.(X, .(Y, [])), .(U, .(V, [])))
U1(X, Y, perm_out(X, Y)) → U2(X, Y, sorted_in(Y))
sorted_in(.(X, .(Y, Z))) → U3(X, Y, Z, le_in(X, Y))
le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U8(X, Y, le_in(X, Y))
U8(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))
U3(X, Y, Z, le_out(X, Y)) → U4(X, Y, Z, sorted_in(.(Y, Z)))
sorted_in(.(X, [])) → sorted_out(.(X, []))
sorted_in([]) → sorted_out([])
U4(X, Y, Z, sorted_out(.(Y, Z))) → sorted_out(.(X, .(Y, Z)))
U2(X, Y, sorted_out(Y)) → slowsort_out(X, Y)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

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

slowsort_in(X, Y) → U1(X, Y, perm_in(X, Y))
perm_in(.(X, .(Y, [])), .(U, .(V, []))) → U5(X, Y, U, V, delete_in(U, .(X, Y), Z))
delete_in(X, .(Y, Z), W) → U7(X, Y, Z, W, delete_in(X, Z, W))
delete_in(X, .(X, Y), Y) → delete_out(X, .(X, Y), Y)
U7(X, Y, Z, W, delete_out(X, Z, W)) → delete_out(X, .(Y, Z), W)
U5(X, Y, U, V, delete_out(U, .(X, Y), Z)) → U6(X, Y, U, V, perm_in(Z, V))
perm_in([], []) → perm_out([], [])
U6(X, Y, U, V, perm_out(Z, V)) → perm_out(.(X, .(Y, [])), .(U, .(V, [])))
U1(X, Y, perm_out(X, Y)) → U2(X, Y, sorted_in(Y))
sorted_in(.(X, .(Y, Z))) → U3(X, Y, Z, le_in(X, Y))
le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U8(X, Y, le_in(X, Y))
U8(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))
U3(X, Y, Z, le_out(X, Y)) → U4(X, Y, Z, sorted_in(.(Y, Z)))
sorted_in(.(X, [])) → sorted_out(.(X, []))
sorted_in([]) → sorted_out([])
U4(X, Y, Z, sorted_out(.(Y, Z))) → sorted_out(.(X, .(Y, Z)))
U2(X, Y, sorted_out(Y)) → slowsort_out(X, Y)

SLOWSORT_IN(X, Y) → U1¹(X, Y, perm_in(X, Y))
SLOWSORT_IN(X, Y) → PERM_IN(X, Y)
PERM_IN(.(X, .(Y, [])), .(U, .(V, []))) → U5¹(X, Y, U, V, delete_in(U, .(X, Y), Z))
PERM_IN(.(X, .(Y, [])), .(U, .(V, []))) → DELETE_IN(U, .(X, Y), Z)
DELETE_IN(X, .(Y, Z), W) → U7¹(X, Y, Z, W, delete_in(X, Z, W))
DELETE_IN(X, .(Y, Z), W) → DELETE_IN(X, Z, W)
U5¹(X, Y, U, V, delete_out(U, .(X, Y), Z)) → U6¹(X, Y, U, V, perm_in(Z, V))
U5¹(X, Y, U, V, delete_out(U, .(X, Y), Z)) → PERM_IN(Z, V)
U1¹(X, Y, perm_out(X, Y)) → U2¹(X, Y, sorted_in(Y))
U1¹(X, Y, perm_out(X, Y)) → SORTED_IN(Y)
SORTED_IN(.(X, .(Y, Z))) → U3¹(X, Y, Z, le_in(X, Y))
SORTED_IN(.(X, .(Y, Z))) → LE_IN(X, Y)
LE_IN(s(X), s(Y)) → U8¹(X, Y, le_in(X, Y))
LE_IN(s(X), s(Y)) → LE_IN(X, Y)
U3¹(X, Y, Z, le_out(X, Y)) → U4¹(X, Y, Z, sorted_in(.(Y, Z)))
U3¹(X, Y, Z, le_out(X, Y)) → SORTED_IN(.(Y, Z))

The TRS R consists of the following rules:

slowsort_in(X, Y) → U1(X, Y, perm_in(X, Y))
perm_in(.(X, .(Y, [])), .(U, .(V, []))) → U5(X, Y, U, V, delete_in(U, .(X, Y), Z))
delete_in(X, .(Y, Z), W) → U7(X, Y, Z, W, delete_in(X, Z, W))
delete_in(X, .(X, Y), Y) → delete_out(X, .(X, Y), Y)
U7(X, Y, Z, W, delete_out(X, Z, W)) → delete_out(X, .(Y, Z), W)
U5(X, Y, U, V, delete_out(U, .(X, Y), Z)) → U6(X, Y, U, V, perm_in(Z, V))
perm_in([], []) → perm_out([], [])
U6(X, Y, U, V, perm_out(Z, V)) → perm_out(.(X, .(Y, [])), .(U, .(V, [])))
U1(X, Y, perm_out(X, Y)) → U2(X, Y, sorted_in(Y))
sorted_in(.(X, .(Y, Z))) → U3(X, Y, Z, le_in(X, Y))
le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U8(X, Y, le_in(X, Y))
U8(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))
U3(X, Y, Z, le_out(X, Y)) → U4(X, Y, Z, sorted_in(.(Y, Z)))
sorted_in(.(X, [])) → sorted_out(.(X, []))
sorted_in([]) → sorted_out([])
U4(X, Y, Z, sorted_out(.(Y, Z))) → sorted_out(.(X, .(Y, Z)))
U2(X, Y, sorted_out(Y)) → slowsort_out(X, Y)

The argument filtering Pi contains the following mapping:
slowsort_in(x1, x2) = slowsort_in(x2)
U1(x1, x2, x3) = U1(x2, x3)
perm_in(x1, x2) = perm_in(x2)
.(x1, x2) = .(x1, x2)
[] = []
U5(x1, x2, x3, x4, x5) = U5(x3, x4, x5)
delete_in(x1, x2, x3) = delete_in(x1)
U7(x1, x2, x3, x4, x5) = U7(x1, x5)
delete_out(x1, x2, x3) = delete_out(x1)
U6(x1, x2, x3, x4, x5) = U6(x3, x4, x5)
perm_out(x1, x2) = perm_out(x2)
U2(x1, x2, x3) = U2(x2, x3)
sorted_in(x1) = sorted_in(x1)
U3(x1, x2, x3, x4) = U3(x1, x2, x3, x4)
le_in(x1, x2) = le_in(x1, x2)
0 = 0
le_out(x1, x2) = le_out(x1, x2)
s(x1) = s(x1)
U8(x1, x2, x3) = U8(x1, x2, x3)
U4(x1, x2, x3, x4) = U4(x1, x2, x3, x4)
sorted_out(x1) = sorted_out(x1)
slowsort_out(x1, x2) = slowsort_out(x2)
DELETE_IN(x1, x2, x3) = DELETE_IN(x1)
SORTED_IN(x1) = SORTED_IN(x1)
U4¹(x1, x2, x3, x4) = U4¹(x1, x2, x3, x4)
U8¹(x1, x2, x3) = U8¹(x1, x2, x3)
SLOWSORT_IN(x1, x2) = SLOWSORT_IN(x2)
PERM_IN(x1, x2) = PERM_IN(x2)
U6¹(x1, x2, x3, x4, x5) = U6¹(x3, x4, x5)
LE_IN(x1, x2) = LE_IN(x1, x2)
U1¹(x1, x2, x3) = U1¹(x2, x3)
U5¹(x1, x2, x3, x4, x5) = U5¹(x3, x4, x5)
U3¹(x1, x2, x3, x4) = U3¹(x1, x2, x3, x4)
U7¹(x1, x2, x3, x4, x5) = U7¹(x1, x5)
U2¹(x1, x2, x3) = U2¹(x2, x3)

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

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

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

SLOWSORT_IN(X, Y) → U1¹(X, Y, perm_in(X, Y))
SLOWSORT_IN(X, Y) → PERM_IN(X, Y)
PERM_IN(.(X, .(Y, [])), .(U, .(V, []))) → U5¹(X, Y, U, V, delete_in(U, .(X, Y), Z))
PERM_IN(.(X, .(Y, [])), .(U, .(V, []))) → DELETE_IN(U, .(X, Y), Z)
DELETE_IN(X, .(Y, Z), W) → U7¹(X, Y, Z, W, delete_in(X, Z, W))
DELETE_IN(X, .(Y, Z), W) → DELETE_IN(X, Z, W)
U5¹(X, Y, U, V, delete_out(U, .(X, Y), Z)) → U6¹(X, Y, U, V, perm_in(Z, V))
U5¹(X, Y, U, V, delete_out(U, .(X, Y), Z)) → PERM_IN(Z, V)
U1¹(X, Y, perm_out(X, Y)) → U2¹(X, Y, sorted_in(Y))
U1¹(X, Y, perm_out(X, Y)) → SORTED_IN(Y)
SORTED_IN(.(X, .(Y, Z))) → U3¹(X, Y, Z, le_in(X, Y))
SORTED_IN(.(X, .(Y, Z))) → LE_IN(X, Y)
LE_IN(s(X), s(Y)) → U8¹(X, Y, le_in(X, Y))
LE_IN(s(X), s(Y)) → LE_IN(X, Y)
U3¹(X, Y, Z, le_out(X, Y)) → U4¹(X, Y, Z, sorted_in(.(Y, Z)))
U3¹(X, Y, Z, le_out(X, Y)) → SORTED_IN(.(Y, Z))

The TRS R consists of the following rules:

slowsort_in(X, Y) → U1(X, Y, perm_in(X, Y))
perm_in(.(X, .(Y, [])), .(U, .(V, []))) → U5(X, Y, U, V, delete_in(U, .(X, Y), Z))
delete_in(X, .(Y, Z), W) → U7(X, Y, Z, W, delete_in(X, Z, W))
delete_in(X, .(X, Y), Y) → delete_out(X, .(X, Y), Y)
U7(X, Y, Z, W, delete_out(X, Z, W)) → delete_out(X, .(Y, Z), W)
U5(X, Y, U, V, delete_out(U, .(X, Y), Z)) → U6(X, Y, U, V, perm_in(Z, V))
perm_in([], []) → perm_out([], [])
U6(X, Y, U, V, perm_out(Z, V)) → perm_out(.(X, .(Y, [])), .(U, .(V, [])))
U1(X, Y, perm_out(X, Y)) → U2(X, Y, sorted_in(Y))
sorted_in(.(X, .(Y, Z))) → U3(X, Y, Z, le_in(X, Y))
le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U8(X, Y, le_in(X, Y))
U8(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))
U3(X, Y, Z, le_out(X, Y)) → U4(X, Y, Z, sorted_in(.(Y, Z)))
sorted_in(.(X, [])) → sorted_out(.(X, []))
sorted_in([]) → sorted_out([])
U4(X, Y, Z, sorted_out(.(Y, Z))) → sorted_out(.(X, .(Y, Z)))
U2(X, Y, sorted_out(Y)) → slowsort_out(X, Y)

The argument filtering Pi contains the following mapping:
slowsort_in(x1, x2) = slowsort_in(x2)
U1(x1, x2, x3) = U1(x2, x3)
perm_in(x1, x2) = perm_in(x2)
.(x1, x2) = .(x1, x2)
[] = []
U5(x1, x2, x3, x4, x5) = U5(x3, x4, x5)
delete_in(x1, x2, x3) = delete_in(x1)
U7(x1, x2, x3, x4, x5) = U7(x1, x5)
delete_out(x1, x2, x3) = delete_out(x1)
U6(x1, x2, x3, x4, x5) = U6(x3, x4, x5)
perm_out(x1, x2) = perm_out(x2)
U2(x1, x2, x3) = U2(x2, x3)
sorted_in(x1) = sorted_in(x1)
U3(x1, x2, x3, x4) = U3(x1, x2, x3, x4)
le_in(x1, x2) = le_in(x1, x2)
0 = 0
le_out(x1, x2) = le_out(x1, x2)
s(x1) = s(x1)
U8(x1, x2, x3) = U8(x1, x2, x3)
U4(x1, x2, x3, x4) = U4(x1, x2, x3, x4)
sorted_out(x1) = sorted_out(x1)
slowsort_out(x1, x2) = slowsort_out(x2)
DELETE_IN(x1, x2, x3) = DELETE_IN(x1)
SORTED_IN(x1) = SORTED_IN(x1)
U4¹(x1, x2, x3, x4) = U4¹(x1, x2, x3, x4)
U8¹(x1, x2, x3) = U8¹(x1, x2, x3)
SLOWSORT_IN(x1, x2) = SLOWSORT_IN(x2)
PERM_IN(x1, x2) = PERM_IN(x2)
U6¹(x1, x2, x3, x4, x5) = U6¹(x3, x4, x5)
LE_IN(x1, x2) = LE_IN(x1, x2)
U1¹(x1, x2, x3) = U1¹(x2, x3)
U5¹(x1, x2, x3, x4, x5) = U5¹(x3, x4, x5)
U3¹(x1, x2, x3, x4) = U3¹(x1, x2, x3, x4)
U7¹(x1, x2, x3, x4, x5) = U7¹(x1, x5)
U2¹(x1, x2, x3) = U2¹(x2, x3)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 4 SCCs with 10 less nodes.

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

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

LE_IN(s(X), s(Y)) → LE_IN(X, Y)

The TRS R consists of the following rules:

slowsort_in(X, Y) → U1(X, Y, perm_in(X, Y))
perm_in(.(X, .(Y, [])), .(U, .(V, []))) → U5(X, Y, U, V, delete_in(U, .(X, Y), Z))
delete_in(X, .(Y, Z), W) → U7(X, Y, Z, W, delete_in(X, Z, W))
delete_in(X, .(X, Y), Y) → delete_out(X, .(X, Y), Y)
U7(X, Y, Z, W, delete_out(X, Z, W)) → delete_out(X, .(Y, Z), W)
U5(X, Y, U, V, delete_out(U, .(X, Y), Z)) → U6(X, Y, U, V, perm_in(Z, V))
perm_in([], []) → perm_out([], [])
U6(X, Y, U, V, perm_out(Z, V)) → perm_out(.(X, .(Y, [])), .(U, .(V, [])))
U1(X, Y, perm_out(X, Y)) → U2(X, Y, sorted_in(Y))
sorted_in(.(X, .(Y, Z))) → U3(X, Y, Z, le_in(X, Y))
le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U8(X, Y, le_in(X, Y))
U8(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))
U3(X, Y, Z, le_out(X, Y)) → U4(X, Y, Z, sorted_in(.(Y, Z)))
sorted_in(.(X, [])) → sorted_out(.(X, []))
sorted_in([]) → sorted_out([])
U4(X, Y, Z, sorted_out(.(Y, Z))) → sorted_out(.(X, .(Y, Z)))
U2(X, Y, sorted_out(Y)) → slowsort_out(X, Y)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

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

LE_IN(s(X), s(Y)) → LE_IN(X, Y)

R is empty.
Pi is empty.
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

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

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

LE_IN(s(X), s(Y)) → LE_IN(X, Y)

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

LE_IN(s(X), s(Y)) → LE_IN(X, Y)
The graph contains the following edges 1 > 1, 2 > 2

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

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

SORTED_IN(.(X, .(Y, Z))) → U3¹(X, Y, Z, le_in(X, Y))
U3¹(X, Y, Z, le_out(X, Y)) → SORTED_IN(.(Y, Z))

The TRS R consists of the following rules:

slowsort_in(X, Y) → U1(X, Y, perm_in(X, Y))
perm_in(.(X, .(Y, [])), .(U, .(V, []))) → U5(X, Y, U, V, delete_in(U, .(X, Y), Z))
delete_in(X, .(Y, Z), W) → U7(X, Y, Z, W, delete_in(X, Z, W))
delete_in(X, .(X, Y), Y) → delete_out(X, .(X, Y), Y)
U7(X, Y, Z, W, delete_out(X, Z, W)) → delete_out(X, .(Y, Z), W)
U5(X, Y, U, V, delete_out(U, .(X, Y), Z)) → U6(X, Y, U, V, perm_in(Z, V))
perm_in([], []) → perm_out([], [])
U6(X, Y, U, V, perm_out(Z, V)) → perm_out(.(X, .(Y, [])), .(U, .(V, [])))
U1(X, Y, perm_out(X, Y)) → U2(X, Y, sorted_in(Y))
sorted_in(.(X, .(Y, Z))) → U3(X, Y, Z, le_in(X, Y))
le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U8(X, Y, le_in(X, Y))
U8(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))
U3(X, Y, Z, le_out(X, Y)) → U4(X, Y, Z, sorted_in(.(Y, Z)))
sorted_in(.(X, [])) → sorted_out(.(X, []))
sorted_in([]) → sorted_out([])
U4(X, Y, Z, sorted_out(.(Y, Z))) → sorted_out(.(X, .(Y, Z)))
U2(X, Y, sorted_out(Y)) → slowsort_out(X, Y)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

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

SORTED_IN(.(X, .(Y, Z))) → U3¹(X, Y, Z, le_in(X, Y))
U3¹(X, Y, Z, le_out(X, Y)) → SORTED_IN(.(Y, Z))

The TRS R consists of the following rules:

le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U8(X, Y, le_in(X, Y))
U8(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))

Pi is empty.
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

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

              ↳ PiDP

              ↳ PiDP

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

SORTED_IN(.(X, .(Y, Z))) → U3¹(X, Y, Z, le_in(X, Y))
U3¹(X, Y, Z, le_out(X, Y)) → SORTED_IN(.(Y, Z))

The TRS R consists of the following rules:

le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U8(X, Y, le_in(X, Y))
U8(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))

The set Q consists of the following terms:

le_in(x0, x1)
U8(x0, x1, x2)

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:

SORTED_IN(.(X, .(Y, Z))) → U3¹(X, Y, Z, le_in(X, Y))

Strictly oriented rules of the TRS R:

le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))

Used ordering: POLO with Polynomial interpretation [25]:

POL(.(x₁, x₂)) = 1 + 2·x₁ + x₂
POL(0) = 1
POL(SORTED_IN(x₁)) = 2·x₁
POL(U3¹(x₁, x₂, x₃, x₄)) = 2 + x₁ + 2·x₂ + 2·x₃ + x₄
POL(U8(x₁, x₂, x₃)) = 2·x₁ + 2·x₂ + x₃
POL(le_in(x₁, x₂)) = 2·x₁ + 2·x₂
POL(le_out(x₁, x₂)) = x₁ + 2·x₂
POL(s(x₁)) = 2·x₁

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ DependencyGraphProof

              ↳ PiDP

              ↳ PiDP

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

U3¹(X, Y, Z, le_out(X, Y)) → SORTED_IN(.(Y, Z))

The TRS R consists of the following rules:

le_in(s(X), s(Y)) → U8(X, Y, le_in(X, Y))
U8(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))

The set Q consists of the following terms:

le_in(x0, x1)
U8(x0, x1, x2)

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.

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

DELETE_IN(X, .(Y, Z), W) → DELETE_IN(X, Z, W)

The TRS R consists of the following rules:

slowsort_in(X, Y) → U1(X, Y, perm_in(X, Y))
perm_in(.(X, .(Y, [])), .(U, .(V, []))) → U5(X, Y, U, V, delete_in(U, .(X, Y), Z))
delete_in(X, .(Y, Z), W) → U7(X, Y, Z, W, delete_in(X, Z, W))
delete_in(X, .(X, Y), Y) → delete_out(X, .(X, Y), Y)
U7(X, Y, Z, W, delete_out(X, Z, W)) → delete_out(X, .(Y, Z), W)
U5(X, Y, U, V, delete_out(U, .(X, Y), Z)) → U6(X, Y, U, V, perm_in(Z, V))
perm_in([], []) → perm_out([], [])
U6(X, Y, U, V, perm_out(Z, V)) → perm_out(.(X, .(Y, [])), .(U, .(V, [])))
U1(X, Y, perm_out(X, Y)) → U2(X, Y, sorted_in(Y))
sorted_in(.(X, .(Y, Z))) → U3(X, Y, Z, le_in(X, Y))
le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U8(X, Y, le_in(X, Y))
U8(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))
U3(X, Y, Z, le_out(X, Y)) → U4(X, Y, Z, sorted_in(.(Y, Z)))
sorted_in(.(X, [])) → sorted_out(.(X, []))
sorted_in([]) → sorted_out([])
U4(X, Y, Z, sorted_out(.(Y, Z))) → sorted_out(.(X, .(Y, Z)))
U2(X, Y, sorted_out(Y)) → slowsort_out(X, Y)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

DELETE_IN(X, .(Y, Z), W) → DELETE_IN(X, Z, W)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ NonTerminationProof

              ↳ PiDP

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

DELETE_IN(X) → DELETE_IN(X)

The TRS R consists of the following rules:none

s = DELETE_IN(X) evaluates to t =DELETE_IN(X)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from DELETE_IN(X) to DELETE_IN(X).

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

PERM_IN(.(X, .(Y, [])), .(U, .(V, []))) → U5¹(X, Y, U, V, delete_in(U, .(X, Y), Z))
U5¹(X, Y, U, V, delete_out(U, .(X, Y), Z)) → PERM_IN(Z, V)

The TRS R consists of the following rules:

slowsort_in(X, Y) → U1(X, Y, perm_in(X, Y))
perm_in(.(X, .(Y, [])), .(U, .(V, []))) → U5(X, Y, U, V, delete_in(U, .(X, Y), Z))
delete_in(X, .(Y, Z), W) → U7(X, Y, Z, W, delete_in(X, Z, W))
delete_in(X, .(X, Y), Y) → delete_out(X, .(X, Y), Y)
U7(X, Y, Z, W, delete_out(X, Z, W)) → delete_out(X, .(Y, Z), W)
U5(X, Y, U, V, delete_out(U, .(X, Y), Z)) → U6(X, Y, U, V, perm_in(Z, V))
perm_in([], []) → perm_out([], [])
U6(X, Y, U, V, perm_out(Z, V)) → perm_out(.(X, .(Y, [])), .(U, .(V, [])))
U1(X, Y, perm_out(X, Y)) → U2(X, Y, sorted_in(Y))
sorted_in(.(X, .(Y, Z))) → U3(X, Y, Z, le_in(X, Y))
le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U8(X, Y, le_in(X, Y))
U8(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))
U3(X, Y, Z, le_out(X, Y)) → U4(X, Y, Z, sorted_in(.(Y, Z)))
sorted_in(.(X, [])) → sorted_out(.(X, []))
sorted_in([]) → sorted_out([])
U4(X, Y, Z, sorted_out(.(Y, Z))) → sorted_out(.(X, .(Y, Z)))
U2(X, Y, sorted_out(Y)) → slowsort_out(X, Y)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

PERM_IN(.(X, .(Y, [])), .(U, .(V, []))) → U5¹(X, Y, U, V, delete_in(U, .(X, Y), Z))
U5¹(X, Y, U, V, delete_out(U, .(X, Y), Z)) → PERM_IN(Z, V)

The TRS R consists of the following rules:

delete_in(X, .(Y, Z), W) → U7(X, Y, Z, W, delete_in(X, Z, W))
delete_in(X, .(X, Y), Y) → delete_out(X, .(X, Y), Y)
U7(X, Y, Z, W, delete_out(X, Z, W)) → delete_out(X, .(Y, Z), W)

The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
[] = []
delete_in(x1, x2, x3) = delete_in(x1)
U7(x1, x2, x3, x4, x5) = U7(x1, x5)
delete_out(x1, x2, x3) = delete_out(x1)
PERM_IN(x1, x2) = PERM_IN(x2)
U5¹(x1, x2, x3, x4, x5) = U5¹(x3, x4, x5)

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

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

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

U5¹(U, V, delete_out(U)) → PERM_IN(V)
PERM_IN(.(U, .(V, []))) → U5¹(U, V, delete_in(U))

The TRS R consists of the following rules:

delete_in(X) → U7(X, delete_in(X))
delete_in(X) → delete_out(X)
U7(X, delete_out(X)) → delete_out(X)

The set Q consists of the following terms:

delete_in(x0)
U7(x0, x1)

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:

U5¹(U, V, delete_out(U)) → PERM_IN(V)
The graph contains the following edges 2 >= 1
PERM_IN(.(U, .(V, []))) → U5¹(U, V, delete_in(U))
The graph contains the following edges 1 > 1, 1 > 2