Left Termination proof of /tmp/tmpGf93oW/pl4.4.6a.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, L), Z) :- ','(perm(L, Y), insert(X, Y, Z)).
insert(X, [], .(X, [])).
insert(X, L, .(X, L)).
insert(X, .(H, L1), .(H, L2)) :- insert(X, L1, L2).

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, L), Z) → U1(X, L, Z, perm_in(L, Y))
perm_in([], []) → perm_out([], [])
U1(X, L, Z, perm_out(L, Y)) → U2(X, L, Z, insert_in(X, Y, Z))
insert_in(X, .(H, L1), .(H, L2)) → U3(X, H, L1, L2, insert_in(X, L1, L2))
insert_in(X, L, .(X, L)) → insert_out(X, L, .(X, L))
insert_in(X, [], .(X, [])) → insert_out(X, [], .(X, []))
U3(X, H, L1, L2, insert_out(X, L1, L2)) → insert_out(X, .(H, L1), .(H, L2))
U2(X, L, Z, insert_out(X, Y, Z)) → perm_out(.(X, L), Z)

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, L), Z) → U1(X, L, Z, perm_in(L, Y))
perm_in([], []) → perm_out([], [])
U1(X, L, Z, perm_out(L, Y)) → U2(X, L, Z, insert_in(X, Y, Z))
insert_in(X, .(H, L1), .(H, L2)) → U3(X, H, L1, L2, insert_in(X, L1, L2))
insert_in(X, L, .(X, L)) → insert_out(X, L, .(X, L))
insert_in(X, [], .(X, [])) → insert_out(X, [], .(X, []))
U3(X, H, L1, L2, insert_out(X, L1, L2)) → insert_out(X, .(H, L1), .(H, L2))
U2(X, L, Z, insert_out(X, Y, Z)) → perm_out(.(X, L), Z)

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

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, L), Z) → U1¹(X, L, Z, perm_in(L, Y))
PERM_IN(.(X, L), Z) → PERM_IN(L, Y)
U1¹(X, L, Z, perm_out(L, Y)) → U2¹(X, L, Z, insert_in(X, Y, Z))
U1¹(X, L, Z, perm_out(L, Y)) → INSERT_IN(X, Y, Z)
INSERT_IN(X, .(H, L1), .(H, L2)) → U3¹(X, H, L1, L2, insert_in(X, L1, L2))
INSERT_IN(X, .(H, L1), .(H, L2)) → INSERT_IN(X, L1, L2)

The TRS R consists of the following rules:

perm_in(.(X, L), Z) → U1(X, L, Z, perm_in(L, Y))
perm_in([], []) → perm_out([], [])
U1(X, L, Z, perm_out(L, Y)) → U2(X, L, Z, insert_in(X, Y, Z))
insert_in(X, .(H, L1), .(H, L2)) → U3(X, H, L1, L2, insert_in(X, L1, L2))
insert_in(X, L, .(X, L)) → insert_out(X, L, .(X, L))
insert_in(X, [], .(X, [])) → insert_out(X, [], .(X, []))
U3(X, H, L1, L2, insert_out(X, L1, L2)) → insert_out(X, .(H, L1), .(H, L2))
U2(X, L, Z, insert_out(X, Y, Z)) → perm_out(.(X, L), Z)

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

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, L), Z) → U1¹(X, L, Z, perm_in(L, Y))
PERM_IN(.(X, L), Z) → PERM_IN(L, Y)
U1¹(X, L, Z, perm_out(L, Y)) → U2¹(X, L, Z, insert_in(X, Y, Z))
U1¹(X, L, Z, perm_out(L, Y)) → INSERT_IN(X, Y, Z)
INSERT_IN(X, .(H, L1), .(H, L2)) → U3¹(X, H, L1, L2, insert_in(X, L1, L2))
INSERT_IN(X, .(H, L1), .(H, L2)) → INSERT_IN(X, L1, L2)

The TRS R consists of the following rules:

perm_in(.(X, L), Z) → U1(X, L, Z, perm_in(L, Y))
perm_in([], []) → perm_out([], [])
U1(X, L, Z, perm_out(L, Y)) → U2(X, L, Z, insert_in(X, Y, Z))
insert_in(X, .(H, L1), .(H, L2)) → U3(X, H, L1, L2, insert_in(X, L1, L2))
insert_in(X, L, .(X, L)) → insert_out(X, L, .(X, L))
insert_in(X, [], .(X, [])) → insert_out(X, [], .(X, []))
U3(X, H, L1, L2, insert_out(X, L1, L2)) → insert_out(X, .(H, L1), .(H, L2))
U2(X, L, Z, insert_out(X, Y, Z)) → perm_out(.(X, L), Z)

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

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 4 less nodes.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

INSERT_IN(X, .(H, L1), .(H, L2)) → INSERT_IN(X, L1, L2)

The TRS R consists of the following rules:

perm_in(.(X, L), Z) → U1(X, L, Z, perm_in(L, Y))
perm_in([], []) → perm_out([], [])
U1(X, L, Z, perm_out(L, Y)) → U2(X, L, Z, insert_in(X, Y, Z))
insert_in(X, .(H, L1), .(H, L2)) → U3(X, H, L1, L2, insert_in(X, L1, L2))
insert_in(X, L, .(X, L)) → insert_out(X, L, .(X, L))
insert_in(X, [], .(X, [])) → insert_out(X, [], .(X, []))
U3(X, H, L1, L2, insert_out(X, L1, L2)) → insert_out(X, .(H, L1), .(H, L2))
U2(X, L, Z, insert_out(X, Y, Z)) → perm_out(.(X, L), Z)

The argument filtering Pi contains the following mapping:
perm_in(x1, x2) = perm_in(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x1, x4)
[] = []
perm_out(x1, x2) = perm_out(x2)
U2(x1, x2, x3, x4) = U2(x4)
insert_in(x1, x2, x3) = insert_in(x1, x2)
U3(x1, x2, x3, x4, x5) = U3(x2, x5)
insert_out(x1, x2, x3) = insert_out(x3)
INSERT_IN(x1, x2, x3) = INSERT_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

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

INSERT_IN(X, .(H, L1), .(H, L2)) → INSERT_IN(X, L1, L2)

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

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

INSERT_IN(X, .(H, L1)) → INSERT_IN(X, L1)

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:

INSERT_IN(X, .(H, L1)) → INSERT_IN(X, L1)
The graph contains the following edges 1 >= 1, 2 > 2

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

PERM_IN(.(X, L), Z) → PERM_IN(L, Y)

The TRS R consists of the following rules:

perm_in(.(X, L), Z) → U1(X, L, Z, perm_in(L, Y))
perm_in([], []) → perm_out([], [])
U1(X, L, Z, perm_out(L, Y)) → U2(X, L, Z, insert_in(X, Y, Z))
insert_in(X, .(H, L1), .(H, L2)) → U3(X, H, L1, L2, insert_in(X, L1, L2))
insert_in(X, L, .(X, L)) → insert_out(X, L, .(X, L))
insert_in(X, [], .(X, [])) → insert_out(X, [], .(X, []))
U3(X, H, L1, L2, insert_out(X, L1, L2)) → insert_out(X, .(H, L1), .(H, L2))
U2(X, L, Z, insert_out(X, Y, Z)) → perm_out(.(X, L), Z)

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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

PERM_IN(.(X, L), Z) → PERM_IN(L, Y)

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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

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

PERM_IN(.(X, L)) → PERM_IN(L)

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

PERM_IN(.(X, L)) → PERM_IN(L)
The graph contains the following edges 1 > 1