Left Termination proof of /tmp/tmpIaZA6

Left Termination of the query pattern perm1_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:

↳ Prolog

  ↳ PrologToPiTRSProof

Clauses:

perm1([], []).
perm1(Xs, .(X, Ys)) :- ','(select(X, Xs, Zs), perm1(Zs, Ys)).
select(X, .(X, Xs), Xs).
select(X, .(Y, Xs), .(Y, Zs)) :- select(X, Xs, Zs).

Queries:

perm1(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:

perm1_in(Xs, .(X, Ys)) → U1(Xs, X, Ys, select_in(X, Xs, Zs))
select_in(X, .(Y, Xs), .(Y, Zs)) → U3(X, Y, Xs, Zs, select_in(X, Xs, Zs))
select_in(X, .(X, Xs), Xs) → select_out(X, .(X, Xs), Xs)
U3(X, Y, Xs, Zs, select_out(X, Xs, Zs)) → select_out(X, .(Y, Xs), .(Y, Zs))
U1(Xs, X, Ys, select_out(X, Xs, Zs)) → U2(Xs, X, Ys, perm1_in(Zs, Ys))
perm1_in([], []) → perm1_out([], [])
U2(Xs, X, Ys, perm1_out(Zs, Ys)) → perm1_out(Xs, .(X, Ys))

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:

perm1_in(Xs, .(X, Ys)) → U1(Xs, X, Ys, select_in(X, Xs, Zs))
select_in(X, .(Y, Xs), .(Y, Zs)) → U3(X, Y, Xs, Zs, select_in(X, Xs, Zs))
select_in(X, .(X, Xs), Xs) → select_out(X, .(X, Xs), Xs)
U3(X, Y, Xs, Zs, select_out(X, Xs, Zs)) → select_out(X, .(Y, Xs), .(Y, Zs))
U1(Xs, X, Ys, select_out(X, Xs, Zs)) → U2(Xs, X, Ys, perm1_in(Zs, Ys))
perm1_in([], []) → perm1_out([], [])
U2(Xs, X, Ys, perm1_out(Zs, Ys)) → perm1_out(Xs, .(X, Ys))

The argument filtering Pi contains the following mapping:
perm1_in(x1, x2) = perm1_in(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x4)
select_in(x1, x2, x3) = select_in(x2)
U3(x1, x2, x3, x4, x5) = U3(x2, x5)
select_out(x1, x2, x3) = select_out(x1, x3)
U2(x1, x2, x3, x4) = U2(x2, x4)
[] = []
perm1_out(x1, x2) = perm1_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:

PERM1_IN(Xs, .(X, Ys)) → U1¹(Xs, X, Ys, select_in(X, Xs, Zs))
PERM1_IN(Xs, .(X, Ys)) → SELECT_IN(X, Xs, Zs)
SELECT_IN(X, .(Y, Xs), .(Y, Zs)) → U3¹(X, Y, Xs, Zs, select_in(X, Xs, Zs))
SELECT_IN(X, .(Y, Xs), .(Y, Zs)) → SELECT_IN(X, Xs, Zs)
U1¹(Xs, X, Ys, select_out(X, Xs, Zs)) → U2¹(Xs, X, Ys, perm1_in(Zs, Ys))
U1¹(Xs, X, Ys, select_out(X, Xs, Zs)) → PERM1_IN(Zs, Ys)

The TRS R consists of the following rules:

perm1_in(Xs, .(X, Ys)) → U1(Xs, X, Ys, select_in(X, Xs, Zs))
select_in(X, .(Y, Xs), .(Y, Zs)) → U3(X, Y, Xs, Zs, select_in(X, Xs, Zs))
select_in(X, .(X, Xs), Xs) → select_out(X, .(X, Xs), Xs)
U3(X, Y, Xs, Zs, select_out(X, Xs, Zs)) → select_out(X, .(Y, Xs), .(Y, Zs))
U1(Xs, X, Ys, select_out(X, Xs, Zs)) → U2(Xs, X, Ys, perm1_in(Zs, Ys))
perm1_in([], []) → perm1_out([], [])
U2(Xs, X, Ys, perm1_out(Zs, Ys)) → perm1_out(Xs, .(X, Ys))

The argument filtering Pi contains the following mapping:
perm1_in(x1, x2) = perm1_in(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x4)
select_in(x1, x2, x3) = select_in(x2)
U3(x1, x2, x3, x4, x5) = U3(x2, x5)
select_out(x1, x2, x3) = select_out(x1, x3)
U2(x1, x2, x3, x4) = U2(x2, x4)
[] = []
perm1_out(x1, x2) = perm1_out(x2)
U3¹(x1, x2, x3, x4, x5) = U3¹(x2, x5)
U2¹(x1, x2, x3, x4) = U2¹(x2, x4)
SELECT_IN(x1, x2, x3) = SELECT_IN(x2)
PERM1_IN(x1, x2) = PERM1_IN(x1)
U1¹(x1, x2, x3, x4) = U1¹(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:

PERM1_IN(Xs, .(X, Ys)) → U1¹(Xs, X, Ys, select_in(X, Xs, Zs))
PERM1_IN(Xs, .(X, Ys)) → SELECT_IN(X, Xs, Zs)
SELECT_IN(X, .(Y, Xs), .(Y, Zs)) → U3¹(X, Y, Xs, Zs, select_in(X, Xs, Zs))
SELECT_IN(X, .(Y, Xs), .(Y, Zs)) → SELECT_IN(X, Xs, Zs)
U1¹(Xs, X, Ys, select_out(X, Xs, Zs)) → U2¹(Xs, X, Ys, perm1_in(Zs, Ys))
U1¹(Xs, X, Ys, select_out(X, Xs, Zs)) → PERM1_IN(Zs, Ys)

The TRS R consists of the following rules:

perm1_in(Xs, .(X, Ys)) → U1(Xs, X, Ys, select_in(X, Xs, Zs))
select_in(X, .(Y, Xs), .(Y, Zs)) → U3(X, Y, Xs, Zs, select_in(X, Xs, Zs))
select_in(X, .(X, Xs), Xs) → select_out(X, .(X, Xs), Xs)
U3(X, Y, Xs, Zs, select_out(X, Xs, Zs)) → select_out(X, .(Y, Xs), .(Y, Zs))
U1(Xs, X, Ys, select_out(X, Xs, Zs)) → U2(Xs, X, Ys, perm1_in(Zs, Ys))
perm1_in([], []) → perm1_out([], [])
U2(Xs, X, Ys, perm1_out(Zs, Ys)) → perm1_out(Xs, .(X, Ys))

The argument filtering Pi contains the following mapping:
perm1_in(x1, x2) = perm1_in(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x4)
select_in(x1, x2, x3) = select_in(x2)
U3(x1, x2, x3, x4, x5) = U3(x2, x5)
select_out(x1, x2, x3) = select_out(x1, x3)
U2(x1, x2, x3, x4) = U2(x2, x4)
[] = []
perm1_out(x1, x2) = perm1_out(x2)
U3¹(x1, x2, x3, x4, x5) = U3¹(x2, x5)
U2¹(x1, x2, x3, x4) = U2¹(x2, x4)
SELECT_IN(x1, x2, x3) = SELECT_IN(x2)
PERM1_IN(x1, x2) = PERM1_IN(x1)
U1¹(x1, x2, x3, x4) = U1¹(x4)

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:

SELECT_IN(X, .(Y, Xs), .(Y, Zs)) → SELECT_IN(X, Xs, Zs)

The TRS R consists of the following rules:

perm1_in(Xs, .(X, Ys)) → U1(Xs, X, Ys, select_in(X, Xs, Zs))
select_in(X, .(Y, Xs), .(Y, Zs)) → U3(X, Y, Xs, Zs, select_in(X, Xs, Zs))
select_in(X, .(X, Xs), Xs) → select_out(X, .(X, Xs), Xs)
U3(X, Y, Xs, Zs, select_out(X, Xs, Zs)) → select_out(X, .(Y, Xs), .(Y, Zs))
U1(Xs, X, Ys, select_out(X, Xs, Zs)) → U2(Xs, X, Ys, perm1_in(Zs, Ys))
perm1_in([], []) → perm1_out([], [])
U2(Xs, X, Ys, perm1_out(Zs, Ys)) → perm1_out(Xs, .(X, Ys))

The argument filtering Pi contains the following mapping:
perm1_in(x1, x2) = perm1_in(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x4)
select_in(x1, x2, x3) = select_in(x2)
U3(x1, x2, x3, x4, x5) = U3(x2, x5)
select_out(x1, x2, x3) = select_out(x1, x3)
U2(x1, x2, x3, x4) = U2(x2, x4)
[] = []
perm1_out(x1, x2) = perm1_out(x2)
SELECT_IN(x1, x2, x3) = SELECT_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:

SELECT_IN(X, .(Y, Xs), .(Y, Zs)) → SELECT_IN(X, Xs, Zs)

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

SELECT_IN(.(Y, Xs)) → SELECT_IN(Xs)

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:

SELECT_IN(.(Y, Xs)) → SELECT_IN(Xs)
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:

PERM1_IN(Xs, .(X, Ys)) → U1¹(Xs, X, Ys, select_in(X, Xs, Zs))
U1¹(Xs, X, Ys, select_out(X, Xs, Zs)) → PERM1_IN(Zs, Ys)

The TRS R consists of the following rules:

perm1_in(Xs, .(X, Ys)) → U1(Xs, X, Ys, select_in(X, Xs, Zs))
select_in(X, .(Y, Xs), .(Y, Zs)) → U3(X, Y, Xs, Zs, select_in(X, Xs, Zs))
select_in(X, .(X, Xs), Xs) → select_out(X, .(X, Xs), Xs)
U3(X, Y, Xs, Zs, select_out(X, Xs, Zs)) → select_out(X, .(Y, Xs), .(Y, Zs))
U1(Xs, X, Ys, select_out(X, Xs, Zs)) → U2(Xs, X, Ys, perm1_in(Zs, Ys))
perm1_in([], []) → perm1_out([], [])
U2(Xs, X, Ys, perm1_out(Zs, Ys)) → perm1_out(Xs, .(X, Ys))

The argument filtering Pi contains the following mapping:
perm1_in(x1, x2) = perm1_in(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x4)
select_in(x1, x2, x3) = select_in(x2)
U3(x1, x2, x3, x4, x5) = U3(x2, x5)
select_out(x1, x2, x3) = select_out(x1, x3)
U2(x1, x2, x3, x4) = U2(x2, x4)
[] = []
perm1_out(x1, x2) = perm1_out(x2)
PERM1_IN(x1, x2) = PERM1_IN(x1)
U1¹(x1, x2, x3, x4) = U1¹(x4)

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:

PERM1_IN(Xs, .(X, Ys)) → U1¹(Xs, X, Ys, select_in(X, Xs, Zs))
U1¹(Xs, X, Ys, select_out(X, Xs, Zs)) → PERM1_IN(Zs, Ys)

The TRS R consists of the following rules:

select_in(X, .(Y, Xs), .(Y, Zs)) → U3(X, Y, Xs, Zs, select_in(X, Xs, Zs))
select_in(X, .(X, Xs), Xs) → select_out(X, .(X, Xs), Xs)
U3(X, Y, Xs, Zs, select_out(X, Xs, Zs)) → select_out(X, .(Y, Xs), .(Y, Zs))

The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
select_in(x1, x2, x3) = select_in(x2)
U3(x1, x2, x3, x4, x5) = U3(x2, x5)
select_out(x1, x2, x3) = select_out(x1, x3)
PERM1_IN(x1, x2) = PERM1_IN(x1)
U1¹(x1, x2, x3, x4) = U1¹(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

                        ↳ RuleRemovalProof

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

PERM1_IN(Xs) → U1¹(select_in(Xs))
U1¹(select_out(X, Zs)) → PERM1_IN(Zs)

The TRS R consists of the following rules:

select_in(.(Y, Xs)) → U3(Y, select_in(Xs))
select_in(.(X, Xs)) → select_out(X, Xs)
U3(Y, select_out(X, Zs)) → select_out(X, .(Y, Zs))

The set Q consists of the following terms:

select_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 dependency pairs:

PERM1_IN(Xs) → U1¹(select_in(Xs))

Strictly oriented rules of the TRS R:

select_in(.(X, Xs)) → select_out(X, Xs)
U3(Y, select_out(X, Zs)) → select_out(X, .(Y, Zs))

Used ordering: POLO with Polynomial interpretation [25]:

POL(.(x₁, x₂)) = 2 + x₁ + 2·x₂
POL(PERM1_IN(x₁)) = 2 + 2·x₁
POL(U1¹(x₁)) = 2·x₁
POL(U3(x₁, x₂)) = 2 + x₁ + 2·x₂
POL(select_in(x₁)) = x₁
POL(select_out(x₁, x₂)) = 1 + x₁ + x₂

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ DependencyGraphProof

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

U1¹(select_out(X, Zs)) → PERM1_IN(Zs)

The TRS R consists of the following rules:

select_in(.(Y, Xs)) → U3(Y, select_in(Xs))

The set Q consists of the following terms:

select_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.