Left Termination proof of /tmp/tmpE79KJm/permutation2.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:

perm([], []).
perm(.(X, Y), .(U, V)) :- ','(delete(U, .(X, Y), W), perm(W, V)).
delete(X, .(X, Y), Y).
delete(U, .(X, Y), .(X, Z)) :- delete(U, Y, Z).

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(.(X, Y), .(U, V)) → U1(X, Y, U, V, delete_in(U, .(X, Y), W))
delete_in(U, .(X, Y), .(X, Z)) → U3(U, X, Y, Z, delete_in(U, Y, Z))
delete_in(X, .(X, Y), Y) → delete_out(X, .(X, Y), Y)
U3(U, X, Y, Z, delete_out(U, Y, Z)) → delete_out(U, .(X, Y), .(X, Z))
U1(X, Y, U, V, delete_out(U, .(X, Y), W)) → U2(X, Y, U, V, perm_in(W, V))
perm_in([], []) → perm_out([], [])
U2(X, Y, U, V, perm_out(W, V)) → perm_out(.(X, Y), .(U, V))

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

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

perm_in(.(X, Y), .(U, V)) → U1(X, Y, U, V, delete_in(U, .(X, Y), W))
delete_in(U, .(X, Y), .(X, Z)) → U3(U, X, Y, Z, delete_in(U, Y, Z))
delete_in(X, .(X, Y), Y) → delete_out(X, .(X, Y), Y)
U3(U, X, Y, Z, delete_out(U, Y, Z)) → delete_out(U, .(X, Y), .(X, Z))
U1(X, Y, U, V, delete_out(U, .(X, Y), W)) → U2(X, Y, U, V, perm_in(W, V))
perm_in([], []) → perm_out([], [])
U2(X, Y, U, V, perm_out(W, V)) → perm_out(.(X, Y), .(U, V))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2) = perm_in(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x5)
delete_in(x1, x2, x3) = delete_in(x2)
U3(x1, x2, x3, x4, x5) = U3(x2, x5)
delete_out(x1, x2, x3) = delete_out(x1, x3)
U2(x1, x2, x3, x4, x5) = U2(x3, x5)
[] = []
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(.(X, Y), .(U, V)) → U1¹(X, Y, U, V, delete_in(U, .(X, Y), W))
PERM_IN(.(X, Y), .(U, V)) → DELETE_IN(U, .(X, Y), W)
DELETE_IN(U, .(X, Y), .(X, Z)) → U3¹(U, X, Y, Z, delete_in(U, Y, Z))
DELETE_IN(U, .(X, Y), .(X, Z)) → DELETE_IN(U, Y, Z)
U1¹(X, Y, U, V, delete_out(U, .(X, Y), W)) → U2¹(X, Y, U, V, perm_in(W, V))
U1¹(X, Y, U, V, delete_out(U, .(X, Y), W)) → PERM_IN(W, V)

The TRS R consists of the following rules:

perm_in(.(X, Y), .(U, V)) → U1(X, Y, U, V, delete_in(U, .(X, Y), W))
delete_in(U, .(X, Y), .(X, Z)) → U3(U, X, Y, Z, delete_in(U, Y, Z))
delete_in(X, .(X, Y), Y) → delete_out(X, .(X, Y), Y)
U3(U, X, Y, Z, delete_out(U, Y, Z)) → delete_out(U, .(X, Y), .(X, Z))
U1(X, Y, U, V, delete_out(U, .(X, Y), W)) → U2(X, Y, U, V, perm_in(W, V))
perm_in([], []) → perm_out([], [])
U2(X, Y, U, V, perm_out(W, V)) → perm_out(.(X, Y), .(U, V))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2) = perm_in(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x5)
delete_in(x1, x2, x3) = delete_in(x2)
U3(x1, x2, x3, x4, x5) = U3(x2, x5)
delete_out(x1, x2, x3) = delete_out(x1, x3)
U2(x1, x2, x3, x4, x5) = U2(x3, x5)
[] = []
perm_out(x1, x2) = perm_out(x2)
DELETE_IN(x1, x2, x3) = DELETE_IN(x2)
U3¹(x1, x2, x3, x4, x5) = U3¹(x2, x5)
PERM_IN(x1, x2) = PERM_IN(x1)
U2¹(x1, x2, x3, x4, x5) = U2¹(x3, x5)
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(.(X, Y), .(U, V)) → U1¹(X, Y, U, V, delete_in(U, .(X, Y), W))
PERM_IN(.(X, Y), .(U, V)) → DELETE_IN(U, .(X, Y), W)
DELETE_IN(U, .(X, Y), .(X, Z)) → U3¹(U, X, Y, Z, delete_in(U, Y, Z))
DELETE_IN(U, .(X, Y), .(X, Z)) → DELETE_IN(U, Y, Z)
U1¹(X, Y, U, V, delete_out(U, .(X, Y), W)) → U2¹(X, Y, U, V, perm_in(W, V))
U1¹(X, Y, U, V, delete_out(U, .(X, Y), W)) → PERM_IN(W, V)

The TRS R consists of the following rules:

perm_in(.(X, Y), .(U, V)) → U1(X, Y, U, V, delete_in(U, .(X, Y), W))
delete_in(U, .(X, Y), .(X, Z)) → U3(U, X, Y, Z, delete_in(U, Y, Z))
delete_in(X, .(X, Y), Y) → delete_out(X, .(X, Y), Y)
U3(U, X, Y, Z, delete_out(U, Y, Z)) → delete_out(U, .(X, Y), .(X, Z))
U1(X, Y, U, V, delete_out(U, .(X, Y), W)) → U2(X, Y, U, V, perm_in(W, V))
perm_in([], []) → perm_out([], [])
U2(X, Y, U, V, perm_out(W, V)) → perm_out(.(X, Y), .(U, V))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2) = perm_in(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x5)
delete_in(x1, x2, x3) = delete_in(x2)
U3(x1, x2, x3, x4, x5) = U3(x2, x5)
delete_out(x1, x2, x3) = delete_out(x1, x3)
U2(x1, x2, x3, x4, x5) = U2(x3, x5)
[] = []
perm_out(x1, x2) = perm_out(x2)
DELETE_IN(x1, x2, x3) = DELETE_IN(x2)
U3¹(x1, x2, x3, x4, x5) = U3¹(x2, x5)
PERM_IN(x1, x2) = PERM_IN(x1)
U2¹(x1, x2, x3, x4, x5) = U2¹(x3, x5)
U1¹(x1, x2, x3, x4, x5) = U1¹(x5)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 3 less nodes.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

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

The TRS R consists of the following rules:

perm_in(.(X, Y), .(U, V)) → U1(X, Y, U, V, delete_in(U, .(X, Y), W))
delete_in(U, .(X, Y), .(X, Z)) → U3(U, X, Y, Z, delete_in(U, Y, Z))
delete_in(X, .(X, Y), Y) → delete_out(X, .(X, Y), Y)
U3(U, X, Y, Z, delete_out(U, Y, Z)) → delete_out(U, .(X, Y), .(X, Z))
U1(X, Y, U, V, delete_out(U, .(X, Y), W)) → U2(X, Y, U, V, perm_in(W, V))
perm_in([], []) → perm_out([], [])
U2(X, Y, U, V, perm_out(W, V)) → perm_out(.(X, Y), .(U, V))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2) = perm_in(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x5)
delete_in(x1, x2, x3) = delete_in(x2)
U3(x1, x2, x3, x4, x5) = U3(x2, x5)
delete_out(x1, x2, x3) = delete_out(x1, x3)
U2(x1, x2, x3, x4, x5) = U2(x3, x5)
[] = []
perm_out(x1, x2) = perm_out(x2)
DELETE_IN(x1, x2, x3) = DELETE_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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

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

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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

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

DELETE_IN(.(X, Y)) → DELETE_IN(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:

DELETE_IN(.(X, Y)) → DELETE_IN(Y)
The graph contains the following edges 1 > 1

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

U1¹(X, Y, U, V, delete_out(U, .(X, Y), W)) → PERM_IN(W, V)
PERM_IN(.(X, Y), .(U, V)) → U1¹(X, Y, U, V, delete_in(U, .(X, Y), W))

The TRS R consists of the following rules:

perm_in(.(X, Y), .(U, V)) → U1(X, Y, U, V, delete_in(U, .(X, Y), W))
delete_in(U, .(X, Y), .(X, Z)) → U3(U, X, Y, Z, delete_in(U, Y, Z))
delete_in(X, .(X, Y), Y) → delete_out(X, .(X, Y), Y)
U3(U, X, Y, Z, delete_out(U, Y, Z)) → delete_out(U, .(X, Y), .(X, Z))
U1(X, Y, U, V, delete_out(U, .(X, Y), W)) → U2(X, Y, U, V, perm_in(W, V))
perm_in([], []) → perm_out([], [])
U2(X, Y, U, V, perm_out(W, V)) → perm_out(.(X, Y), .(U, V))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2) = perm_in(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x5)
delete_in(x1, x2, x3) = delete_in(x2)
U3(x1, x2, x3, x4, x5) = U3(x2, x5)
delete_out(x1, x2, x3) = delete_out(x1, x3)
U2(x1, x2, x3, x4, x5) = U2(x3, x5)
[] = []
perm_out(x1, x2) = perm_out(x2)
PERM_IN(x1, x2) = PERM_IN(x1)
U1¹(x1, x2, x3, x4, x5) = U1¹(x5)

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

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

U1¹(X, Y, U, V, delete_out(U, .(X, Y), W)) → PERM_IN(W, V)
PERM_IN(.(X, Y), .(U, V)) → U1¹(X, Y, U, V, delete_in(U, .(X, Y), W))

The TRS R consists of the following rules:

delete_in(U, .(X, Y), .(X, Z)) → U3(U, X, Y, Z, delete_in(U, Y, Z))
delete_in(X, .(X, Y), Y) → delete_out(X, .(X, Y), Y)
U3(U, X, Y, Z, delete_out(U, Y, Z)) → delete_out(U, .(X, Y), .(X, Z))

The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
delete_in(x1, x2, x3) = delete_in(x2)
U3(x1, x2, x3, x4, x5) = U3(x2, x5)
delete_out(x1, x2, x3) = delete_out(x1, x3)
PERM_IN(x1, x2) = PERM_IN(x1)
U1¹(x1, x2, x3, x4, x5) = U1¹(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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

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

PERM_IN(.(X, Y)) → U1¹(delete_in(.(X, Y)))
U1¹(delete_out(U, W)) → PERM_IN(W)

The TRS R consists of the following rules:

delete_in(.(X, Y)) → U3(X, delete_in(Y))
delete_in(.(X, Y)) → delete_out(X, Y)
U3(X, delete_out(U, Z)) → delete_out(U, .(X, Z))

The set Q consists of the following terms:

delete_in(x0)
U3(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 rules of the TRS R:

delete_in(.(X, Y)) → delete_out(X, Y)

Used ordering: POLO with Polynomial interpretation [25]:

POL(.(x₁, x₂)) = 1 + x₁ + x₂
POL(PERM_IN(x₁)) = 2·x₁
POL(U1¹(x₁)) = 2·x₁
POL(U3(x₁, x₂)) = 1 + x₁ + x₂
POL(delete_in(x₁)) = x₁
POL(delete_out(x₁, x₂)) = x₁ + x₂

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

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

PERM_IN(.(X, Y)) → U1¹(delete_in(.(X, Y)))
U1¹(delete_out(U, W)) → PERM_IN(W)

The TRS R consists of the following rules:

delete_in(.(X, Y)) → U3(X, delete_in(Y))
U3(X, delete_out(U, Z)) → delete_out(U, .(X, Z))

The set Q consists of the following terms:

delete_in(x0)
U3(x0, x1)

U1¹(delete_out(U, W)) → PERM_IN(W)

Used ordering: POLO with Polynomial interpretation [25]:

POL(.(x₁, x₂)) = x₁ + x₂
POL(PERM_IN(x₁)) = 2·x₁
POL(U1¹(x₁)) = x₁
POL(U3(x₁, x₂)) = 2·x₁ + x₂
POL(delete_in(x₁)) = 2·x₁
POL(delete_out(x₁, x₂)) = 2 + x₁ + 2·x₂

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

                              ↳ QDP

                                ↳ DependencyGraphProof

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

PERM_IN(.(X, Y)) → U1¹(delete_in(.(X, Y)))

The TRS R consists of the following rules:

delete_in(.(X, Y)) → U3(X, delete_in(Y))
U3(X, delete_out(U, Z)) → delete_out(U, .(X, Z))

The set Q consists of the following terms:

delete_in(x0)
U3(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.