Left Termination proof of /tmp/tmp0lm6jj/pl4.4.6a.pl

Left Termination of the query pattern perm(b,f) w.r.t. the given Prolog program could successfully be proven:

↳ PROLOG

  ↳ PrologToPiTRSProof

perm₂({}₀, {}₀).
perm₂(.₂(X, L), Z) :- perm₂(L, Y), insert₃(X, Y, Z).
insert₃(X, {}₀, .₂(X, {}₀)).
insert₃(X, L, .₂(X, L)).
insert₃(X, .₂(H, L1), .₂(H, L2)) :- insert₃(X, L1, L2).

With regard to the inferred argument filtering the predicates were used in the following modes:
perm₂: (b,f)
insert₃: (b,b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(._2₂(X, L), Z) -> if_perm_2_in_1_ga₄(X, L, Z, perm_2_in_ga₂(L, Y))
if_perm_2_in_1_ga₄(X, L, Z, perm_2_out_ga₂(L, Y)) -> if_perm_2_in_2_ga₅(X, L, Z, Y, insert_3_in_gga₃(X, Y, Z))
insert_3_in_gga₃(X, []_0, ._2₂(X, []_0)) -> insert_3_out_gga₃(X, []_0, ._2₂(X, []_0))
insert_3_in_gga₃(X, L, ._2₂(X, L)) -> insert_3_out_gga₃(X, L, ._2₂(X, L))
insert_3_in_gga₃(X, ._2₂(H, L1), ._2₂(H, L2)) -> if_insert_3_in_1_gga₅(X, H, L1, L2, insert_3_in_gga₃(X, L1, L2))
if_insert_3_in_1_gga₅(X, H, L1, L2, insert_3_out_gga₃(X, L1, L2)) -> insert_3_out_gga₃(X, ._2₂(H, L1), ._2₂(H, L2))
if_perm_2_in_2_ga₅(X, L, Z, Y, insert_3_out_gga₃(X, Y, Z)) -> perm_2_out_ga₂(._2₂(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_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(._2₂(X, L), Z) -> if_perm_2_in_1_ga₄(X, L, Z, perm_2_in_ga₂(L, Y))
if_perm_2_in_1_ga₄(X, L, Z, perm_2_out_ga₂(L, Y)) -> if_perm_2_in_2_ga₅(X, L, Z, Y, insert_3_in_gga₃(X, Y, Z))
insert_3_in_gga₃(X, []_0, ._2₂(X, []_0)) -> insert_3_out_gga₃(X, []_0, ._2₂(X, []_0))
insert_3_in_gga₃(X, L, ._2₂(X, L)) -> insert_3_out_gga₃(X, L, ._2₂(X, L))
insert_3_in_gga₃(X, ._2₂(H, L1), ._2₂(H, L2)) -> if_insert_3_in_1_gga₅(X, H, L1, L2, insert_3_in_gga₃(X, L1, L2))
if_insert_3_in_1_gga₅(X, H, L1, L2, insert_3_out_gga₃(X, L1, L2)) -> insert_3_out_gga₃(X, ._2₂(H, L1), ._2₂(H, L2))
if_perm_2_in_2_ga₅(X, L, Z, Y, insert_3_out_gga₃(X, Y, Z)) -> perm_2_out_ga₂(._2₂(X, L), Z)

PERM_2_IN_GA₂(._2₂(X, L), Z) -> IF_PERM_2_IN_1_GA₄(X, L, Z, perm_2_in_ga₂(L, Y))
PERM_2_IN_GA₂(._2₂(X, L), Z) -> PERM_2_IN_GA₂(L, Y)
IF_PERM_2_IN_1_GA₄(X, L, Z, perm_2_out_ga₂(L, Y)) -> IF_PERM_2_IN_2_GA₅(X, L, Z, Y, insert_3_in_gga₃(X, Y, Z))
IF_PERM_2_IN_1_GA₄(X, L, Z, perm_2_out_ga₂(L, Y)) -> INSERT_3_IN_GGA₃(X, Y, Z)
INSERT_3_IN_GGA₃(X, ._2₂(H, L1), ._2₂(H, L2)) -> IF_INSERT_3_IN_1_GGA₅(X, H, L1, L2, insert_3_in_gga₃(X, L1, L2))
INSERT_3_IN_GGA₃(X, ._2₂(H, L1), ._2₂(H, L2)) -> INSERT_3_IN_GGA₃(X, L1, L2)

The TRS R consists of the following rules:

perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(._2₂(X, L), Z) -> if_perm_2_in_1_ga₄(X, L, Z, perm_2_in_ga₂(L, Y))
if_perm_2_in_1_ga₄(X, L, Z, perm_2_out_ga₂(L, Y)) -> if_perm_2_in_2_ga₅(X, L, Z, Y, insert_3_in_gga₃(X, Y, Z))
insert_3_in_gga₃(X, []_0, ._2₂(X, []_0)) -> insert_3_out_gga₃(X, []_0, ._2₂(X, []_0))
insert_3_in_gga₃(X, L, ._2₂(X, L)) -> insert_3_out_gga₃(X, L, ._2₂(X, L))
insert_3_in_gga₃(X, ._2₂(H, L1), ._2₂(H, L2)) -> if_insert_3_in_1_gga₅(X, H, L1, L2, insert_3_in_gga₃(X, L1, L2))
if_insert_3_in_1_gga₅(X, H, L1, L2, insert_3_out_gga₃(X, L1, L2)) -> insert_3_out_gga₃(X, ._2₂(H, L1), ._2₂(H, L2))
if_perm_2_in_2_ga₅(X, L, Z, Y, insert_3_out_gga₃(X, Y, Z)) -> perm_2_out_ga₂(._2₂(X, L), Z)

The argument filtering Pi contains the following mapping:
perm_2_in_ga₂(x1, x2) = perm_2_in_ga₁(x1)
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
perm_2_out_ga₂(x1, x2) = perm_2_out_ga₁(x2)
if_perm_2_in_1_ga₄(x1, x2, x3, x4) = if_perm_2_in_1_ga₂(x1, x4)
if_perm_2_in_2_ga₅(x1, x2, x3, x4, x5) = if_perm_2_in_2_ga₁(x5)
insert_3_in_gga₃(x1, x2, x3) = insert_3_in_gga₂(x1, x2)
insert_3_out_gga₃(x1, x2, x3) = insert_3_out_gga₁(x3)
if_insert_3_in_1_gga₅(x1, x2, x3, x4, x5) = if_insert_3_in_1_gga₂(x2, x5)
INSERT_3_IN_GGA₃(x1, x2, x3) = INSERT_3_IN_GGA₂(x1, x2)
IF_INSERT_3_IN_1_GGA₅(x1, x2, x3, x4, x5) = IF_INSERT_3_IN_1_GGA₂(x2, x5)
PERM_2_IN_GA₂(x1, x2) = PERM_2_IN_GA₁(x1)
IF_PERM_2_IN_2_GA₅(x1, x2, x3, x4, x5) = IF_PERM_2_IN_2_GA₁(x5)
IF_PERM_2_IN_1_GA₄(x1, x2, x3, x4) = IF_PERM_2_IN_1_GA₂(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_2_IN_GA₂(._2₂(X, L), Z) -> IF_PERM_2_IN_1_GA₄(X, L, Z, perm_2_in_ga₂(L, Y))
PERM_2_IN_GA₂(._2₂(X, L), Z) -> PERM_2_IN_GA₂(L, Y)
IF_PERM_2_IN_1_GA₄(X, L, Z, perm_2_out_ga₂(L, Y)) -> IF_PERM_2_IN_2_GA₅(X, L, Z, Y, insert_3_in_gga₃(X, Y, Z))
IF_PERM_2_IN_1_GA₄(X, L, Z, perm_2_out_ga₂(L, Y)) -> INSERT_3_IN_GGA₃(X, Y, Z)
INSERT_3_IN_GGA₃(X, ._2₂(H, L1), ._2₂(H, L2)) -> IF_INSERT_3_IN_1_GGA₅(X, H, L1, L2, insert_3_in_gga₃(X, L1, L2))
INSERT_3_IN_GGA₃(X, ._2₂(H, L1), ._2₂(H, L2)) -> INSERT_3_IN_GGA₃(X, L1, L2)

The TRS R consists of the following rules:

perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(._2₂(X, L), Z) -> if_perm_2_in_1_ga₄(X, L, Z, perm_2_in_ga₂(L, Y))
if_perm_2_in_1_ga₄(X, L, Z, perm_2_out_ga₂(L, Y)) -> if_perm_2_in_2_ga₅(X, L, Z, Y, insert_3_in_gga₃(X, Y, Z))
insert_3_in_gga₃(X, []_0, ._2₂(X, []_0)) -> insert_3_out_gga₃(X, []_0, ._2₂(X, []_0))
insert_3_in_gga₃(X, L, ._2₂(X, L)) -> insert_3_out_gga₃(X, L, ._2₂(X, L))
insert_3_in_gga₃(X, ._2₂(H, L1), ._2₂(H, L2)) -> if_insert_3_in_1_gga₅(X, H, L1, L2, insert_3_in_gga₃(X, L1, L2))
if_insert_3_in_1_gga₅(X, H, L1, L2, insert_3_out_gga₃(X, L1, L2)) -> insert_3_out_gga₃(X, ._2₂(H, L1), ._2₂(H, L2))
if_perm_2_in_2_ga₅(X, L, Z, Y, insert_3_out_gga₃(X, Y, Z)) -> perm_2_out_ga₂(._2₂(X, L), Z)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

INSERT_3_IN_GGA₃(X, ._2₂(H, L1), ._2₂(H, L2)) -> INSERT_3_IN_GGA₃(X, L1, L2)

The TRS R consists of the following rules:

perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(._2₂(X, L), Z) -> if_perm_2_in_1_ga₄(X, L, Z, perm_2_in_ga₂(L, Y))
if_perm_2_in_1_ga₄(X, L, Z, perm_2_out_ga₂(L, Y)) -> if_perm_2_in_2_ga₅(X, L, Z, Y, insert_3_in_gga₃(X, Y, Z))
insert_3_in_gga₃(X, []_0, ._2₂(X, []_0)) -> insert_3_out_gga₃(X, []_0, ._2₂(X, []_0))
insert_3_in_gga₃(X, L, ._2₂(X, L)) -> insert_3_out_gga₃(X, L, ._2₂(X, L))
insert_3_in_gga₃(X, ._2₂(H, L1), ._2₂(H, L2)) -> if_insert_3_in_1_gga₅(X, H, L1, L2, insert_3_in_gga₃(X, L1, L2))
if_insert_3_in_1_gga₅(X, H, L1, L2, insert_3_out_gga₃(X, L1, L2)) -> insert_3_out_gga₃(X, ._2₂(H, L1), ._2₂(H, L2))
if_perm_2_in_2_ga₅(X, L, Z, Y, insert_3_out_gga₃(X, Y, Z)) -> perm_2_out_ga₂(._2₂(X, L), Z)

The argument filtering Pi contains the following mapping:
perm_2_in_ga₂(x1, x2) = perm_2_in_ga₁(x1)
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
perm_2_out_ga₂(x1, x2) = perm_2_out_ga₁(x2)
if_perm_2_in_1_ga₄(x1, x2, x3, x4) = if_perm_2_in_1_ga₂(x1, x4)
if_perm_2_in_2_ga₅(x1, x2, x3, x4, x5) = if_perm_2_in_2_ga₁(x5)
insert_3_in_gga₃(x1, x2, x3) = insert_3_in_gga₂(x1, x2)
insert_3_out_gga₃(x1, x2, x3) = insert_3_out_gga₁(x3)
if_insert_3_in_1_gga₅(x1, x2, x3, x4, x5) = if_insert_3_in_1_gga₂(x2, x5)
INSERT_3_IN_GGA₃(x1, x2, x3) = INSERT_3_IN_GGA₂(x1, x2)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting 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_3_IN_GGA₃(X, ._2₂(H, L1), ._2₂(H, L2)) -> INSERT_3_IN_GGA₃(X, L1, L2)

R is empty.
The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₂(x1, x2)
INSERT_3_IN_GGA₃(x1, x2, x3) = INSERT_3_IN_GGA₂(x1, x2)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem 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_3_IN_GGA₂(X, ._2₂(H, L1)) -> INSERT_3_IN_GGA₂(X, L1)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {INSERT_3_IN_GGA₂}.
By using the subterm criterion together with the size-change analysis 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_3_IN_GGA₂(X, ._2₂(H, L1)) -> INSERT_3_IN_GGA₂(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_2_IN_GA₂(._2₂(X, L), Z) -> PERM_2_IN_GA₂(L, Y)

The TRS R consists of the following rules:

perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(._2₂(X, L), Z) -> if_perm_2_in_1_ga₄(X, L, Z, perm_2_in_ga₂(L, Y))
if_perm_2_in_1_ga₄(X, L, Z, perm_2_out_ga₂(L, Y)) -> if_perm_2_in_2_ga₅(X, L, Z, Y, insert_3_in_gga₃(X, Y, Z))
insert_3_in_gga₃(X, []_0, ._2₂(X, []_0)) -> insert_3_out_gga₃(X, []_0, ._2₂(X, []_0))
insert_3_in_gga₃(X, L, ._2₂(X, L)) -> insert_3_out_gga₃(X, L, ._2₂(X, L))
insert_3_in_gga₃(X, ._2₂(H, L1), ._2₂(H, L2)) -> if_insert_3_in_1_gga₅(X, H, L1, L2, insert_3_in_gga₃(X, L1, L2))
if_insert_3_in_1_gga₅(X, H, L1, L2, insert_3_out_gga₃(X, L1, L2)) -> insert_3_out_gga₃(X, ._2₂(H, L1), ._2₂(H, L2))
if_perm_2_in_2_ga₅(X, L, Z, Y, insert_3_out_gga₃(X, Y, Z)) -> perm_2_out_ga₂(._2₂(X, L), Z)

The argument filtering Pi contains the following mapping:
perm_2_in_ga₂(x1, x2) = perm_2_in_ga₁(x1)
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
perm_2_out_ga₂(x1, x2) = perm_2_out_ga₁(x2)
if_perm_2_in_1_ga₄(x1, x2, x3, x4) = if_perm_2_in_1_ga₂(x1, x4)
if_perm_2_in_2_ga₅(x1, x2, x3, x4, x5) = if_perm_2_in_2_ga₁(x5)
insert_3_in_gga₃(x1, x2, x3) = insert_3_in_gga₂(x1, x2)
insert_3_out_gga₃(x1, x2, x3) = insert_3_out_gga₁(x3)
if_insert_3_in_1_gga₅(x1, x2, x3, x4, x5) = if_insert_3_in_1_gga₂(x2, x5)
PERM_2_IN_GA₂(x1, x2) = PERM_2_IN_GA₁(x1)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting 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_2_IN_GA₂(._2₂(X, L), Z) -> PERM_2_IN_GA₂(L, Y)

R is empty.
The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₂(x1, x2)
PERM_2_IN_GA₂(x1, x2) = PERM_2_IN_GA₁(x1)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem 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_2_IN_GA₁(._2₂(X, L)) -> PERM_2_IN_GA₁(L)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {PERM_2_IN_GA₁}.
By using the subterm criterion together with the size-change analysis 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:

PERM_2_IN_GA₁(._2₂(X, L)) -> PERM_2_IN_GA₁(L)
The graph contains the following edges 1 > 1