Left Termination proof of /tmp/tmpgYtm-q/subset-fb.pl

Left Termination of the query pattern subset_in_2(a, g) w.r.t. the given Prolog program could not be shown:

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

Clauses:

subset([], X).
subset(.(X, Xs), Ys) :- ','(member(X, Ys), subset(Xs, Ys)).
member(X, .(X, X1)).
member(X, .(X1, Xs)) :- member(X, Xs).

Queries:

subset(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:

subset_in(.(X, Xs), Ys) → U1(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X1, Xs)) → U3(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U3(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))
U1(X, Xs, Ys, member_out(X, Ys)) → U2(X, Xs, Ys, subset_in(Xs, Ys))
subset_in([], X) → subset_out([], X)
U2(X, Xs, Ys, subset_out(Xs, Ys)) → subset_out(.(X, Xs), Ys)

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:

subset_in(.(X, Xs), Ys) → U1(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X1, Xs)) → U3(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U3(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))
U1(X, Xs, Ys, member_out(X, Ys)) → U2(X, Xs, Ys, subset_in(Xs, Ys))
subset_in([], X) → subset_out([], X)
U2(X, Xs, Ys, subset_out(Xs, Ys)) → subset_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in(x1, x2) = subset_in(x2)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x3, x4)
member_in(x1, x2) = member_in(x2)
U3(x1, x2, x3, x4) = U3(x4)
member_out(x1, x2) = member_out(x1)
U2(x1, x2, x3, x4) = U2(x1, x4)
[] = []
subset_out(x1, x2) = subset_out(x1)

Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

SUBSET_IN(.(X, Xs), Ys) → U1¹(X, Xs, Ys, member_in(X, Ys))
SUBSET_IN(.(X, Xs), Ys) → MEMBER_IN(X, Ys)
MEMBER_IN(X, .(X1, Xs)) → U3¹(X, X1, Xs, member_in(X, Xs))
MEMBER_IN(X, .(X1, Xs)) → MEMBER_IN(X, Xs)
U1¹(X, Xs, Ys, member_out(X, Ys)) → U2¹(X, Xs, Ys, subset_in(Xs, Ys))
U1¹(X, Xs, Ys, member_out(X, Ys)) → SUBSET_IN(Xs, Ys)

The TRS R consists of the following rules:

subset_in(.(X, Xs), Ys) → U1(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X1, Xs)) → U3(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U3(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))
U1(X, Xs, Ys, member_out(X, Ys)) → U2(X, Xs, Ys, subset_in(Xs, Ys))
subset_in([], X) → subset_out([], X)
U2(X, Xs, Ys, subset_out(Xs, Ys)) → subset_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in(x1, x2) = subset_in(x2)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x3, x4)
member_in(x1, x2) = member_in(x2)
U3(x1, x2, x3, x4) = U3(x4)
member_out(x1, x2) = member_out(x1)
U2(x1, x2, x3, x4) = U2(x1, x4)
[] = []
subset_out(x1, x2) = subset_out(x1)
U3¹(x1, x2, x3, x4) = U3¹(x4)
MEMBER_IN(x1, x2) = MEMBER_IN(x2)
SUBSET_IN(x1, x2) = SUBSET_IN(x2)
U2¹(x1, x2, x3, x4) = U2¹(x1, x4)
U1¹(x1, x2, x3, x4) = U1¹(x3, x4)

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:

SUBSET_IN(.(X, Xs), Ys) → U1¹(X, Xs, Ys, member_in(X, Ys))
SUBSET_IN(.(X, Xs), Ys) → MEMBER_IN(X, Ys)
MEMBER_IN(X, .(X1, Xs)) → U3¹(X, X1, Xs, member_in(X, Xs))
MEMBER_IN(X, .(X1, Xs)) → MEMBER_IN(X, Xs)
U1¹(X, Xs, Ys, member_out(X, Ys)) → U2¹(X, Xs, Ys, subset_in(Xs, Ys))
U1¹(X, Xs, Ys, member_out(X, Ys)) → SUBSET_IN(Xs, Ys)

The TRS R consists of the following rules:

subset_in(.(X, Xs), Ys) → U1(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X1, Xs)) → U3(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U3(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))
U1(X, Xs, Ys, member_out(X, Ys)) → U2(X, Xs, Ys, subset_in(Xs, Ys))
subset_in([], X) → subset_out([], X)
U2(X, Xs, Ys, subset_out(Xs, Ys)) → subset_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in(x1, x2) = subset_in(x2)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x3, x4)
member_in(x1, x2) = member_in(x2)
U3(x1, x2, x3, x4) = U3(x4)
member_out(x1, x2) = member_out(x1)
U2(x1, x2, x3, x4) = U2(x1, x4)
[] = []
subset_out(x1, x2) = subset_out(x1)
U3¹(x1, x2, x3, x4) = U3¹(x4)
MEMBER_IN(x1, x2) = MEMBER_IN(x2)
SUBSET_IN(x1, x2) = SUBSET_IN(x2)
U2¹(x1, x2, x3, x4) = U2¹(x1, x4)
U1¹(x1, x2, x3, x4) = U1¹(x3, 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

  ↳ PrologToPiTRSProof

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

MEMBER_IN(X, .(X1, Xs)) → MEMBER_IN(X, Xs)

The TRS R consists of the following rules:

subset_in(.(X, Xs), Ys) → U1(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X1, Xs)) → U3(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U3(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))
U1(X, Xs, Ys, member_out(X, Ys)) → U2(X, Xs, Ys, subset_in(Xs, Ys))
subset_in([], X) → subset_out([], X)
U2(X, Xs, Ys, subset_out(Xs, Ys)) → subset_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in(x1, x2) = subset_in(x2)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x3, x4)
member_in(x1, x2) = member_in(x2)
U3(x1, x2, x3, x4) = U3(x4)
member_out(x1, x2) = member_out(x1)
U2(x1, x2, x3, x4) = U2(x1, x4)
[] = []
subset_out(x1, x2) = subset_out(x1)
MEMBER_IN(x1, x2) = MEMBER_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

  ↳ PrologToPiTRSProof

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

MEMBER_IN(X, .(X1, Xs)) → MEMBER_IN(X, Xs)

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

  ↳ PrologToPiTRSProof

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

MEMBER_IN(.(X1, Xs)) → MEMBER_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:

MEMBER_IN(.(X1, Xs)) → MEMBER_IN(Xs)
The graph contains the following edges 1 > 1

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

  ↳ PrologToPiTRSProof

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

SUBSET_IN(.(X, Xs), Ys) → U1¹(X, Xs, Ys, member_in(X, Ys))
U1¹(X, Xs, Ys, member_out(X, Ys)) → SUBSET_IN(Xs, Ys)

The TRS R consists of the following rules:

subset_in(.(X, Xs), Ys) → U1(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X1, Xs)) → U3(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U3(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))
U1(X, Xs, Ys, member_out(X, Ys)) → U2(X, Xs, Ys, subset_in(Xs, Ys))
subset_in([], X) → subset_out([], X)
U2(X, Xs, Ys, subset_out(Xs, Ys)) → subset_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in(x1, x2) = subset_in(x2)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x3, x4)
member_in(x1, x2) = member_in(x2)
U3(x1, x2, x3, x4) = U3(x4)
member_out(x1, x2) = member_out(x1)
U2(x1, x2, x3, x4) = U2(x1, x4)
[] = []
subset_out(x1, x2) = subset_out(x1)
SUBSET_IN(x1, x2) = SUBSET_IN(x2)
U1¹(x1, x2, x3, x4) = U1¹(x3, 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

  ↳ PrologToPiTRSProof

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

SUBSET_IN(.(X, Xs), Ys) → U1¹(X, Xs, Ys, member_in(X, Ys))
U1¹(X, Xs, Ys, member_out(X, Ys)) → SUBSET_IN(Xs, Ys)

The TRS R consists of the following rules:

member_in(X, .(X1, Xs)) → U3(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U3(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))

The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
member_in(x1, x2) = member_in(x2)
U3(x1, x2, x3, x4) = U3(x4)
member_out(x1, x2) = member_out(x1)
SUBSET_IN(x1, x2) = SUBSET_IN(x2)
U1¹(x1, x2, x3, x4) = U1¹(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

                        ↳ Narrowing

  ↳ PrologToPiTRSProof

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

SUBSET_IN(Ys) → U1¹(Ys, member_in(Ys))
U1¹(Ys, member_out(X)) → SUBSET_IN(Ys)

The TRS R consists of the following rules:

member_in(.(X1, Xs)) → U3(member_in(Xs))
member_in(.(X, X1)) → member_out(X)
U3(member_out(X)) → member_out(X)

The set Q consists of the following terms:

member_in(x0)
U3(x0)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule SUBSET_IN(Ys) → U1¹(Ys, member_in(Ys)) at position [1] we obtained the following new rules:

SUBSET_IN(.(x0, x1)) → U1¹(.(x0, x1), member_out(x0))
SUBSET_IN(.(x0, x1)) → U1¹(.(x0, x1), U3(member_in(x1)))

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Instantiation

  ↳ PrologToPiTRSProof

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

SUBSET_IN(.(x0, x1)) → U1¹(.(x0, x1), U3(member_in(x1)))
SUBSET_IN(.(x0, x1)) → U1¹(.(x0, x1), member_out(x0))
U1¹(Ys, member_out(X)) → SUBSET_IN(Ys)

The TRS R consists of the following rules:

member_in(.(X1, Xs)) → U3(member_in(Xs))
member_in(.(X, X1)) → member_out(X)
U3(member_out(X)) → member_out(X)

The set Q consists of the following terms:

member_in(x0)
U3(x0)

We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule U1¹(Ys, member_out(X)) → SUBSET_IN(Ys) we obtained the following new rules:

U1¹(.(z0, z1), member_out(x1)) → SUBSET_IN(.(z0, z1))
U1¹(.(z0, z1), member_out(z0)) → SUBSET_IN(.(z0, z1))

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ NonTerminationProof

  ↳ PrologToPiTRSProof

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

SUBSET_IN(.(x0, x1)) → U1¹(.(x0, x1), U3(member_in(x1)))
SUBSET_IN(.(x0, x1)) → U1¹(.(x0, x1), member_out(x0))
U1¹(.(z0, z1), member_out(z0)) → SUBSET_IN(.(z0, z1))
U1¹(.(z0, z1), member_out(x1)) → SUBSET_IN(.(z0, z1))

The TRS R consists of the following rules:

member_in(.(X1, Xs)) → U3(member_in(Xs))
member_in(.(X, X1)) → member_out(X)
U3(member_out(X)) → member_out(X)

The set Q consists of the following terms:

member_in(x0)
U3(x0)

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 narrowing to the left:

The TRS P consists of the following rules:

SUBSET_IN(.(x0, x1)) → U1¹(.(x0, x1), U3(member_in(x1)))
SUBSET_IN(.(x0, x1)) → U1¹(.(x0, x1), member_out(x0))
U1¹(.(z0, z1), member_out(z0)) → SUBSET_IN(.(z0, z1))
U1¹(.(z0, z1), member_out(x1)) → SUBSET_IN(.(z0, z1))

The TRS R consists of the following rules:

member_in(.(X1, Xs)) → U3(member_in(Xs))
member_in(.(X, X1)) → member_out(X)
U3(member_out(X)) → member_out(X)

s = U1¹(.(z0, z1), member_out(z0)) evaluates to t =U1¹(.(z0, z1), member_out(z0))

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

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

U1¹(.(z0, z1), member_out(z0)) → SUBSET_IN(.(z0, z1))
with rule U1¹(.(z0', z1'), member_out(z0')) → SUBSET_IN(.(z0', z1')) at position [] and matcher [z1' / z1, z0' / z0]

SUBSET_IN(.(z0, z1)) → U1¹(.(z0, z1), member_out(z0))
with rule SUBSET_IN(.(x0, x1)) → U1¹(.(x0, x1), member_out(x0))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence

All these steps are and every following step will be a correct step w.r.t to Q.

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:

subset_in(.(X, Xs), Ys) → U1(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X1, Xs)) → U3(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U3(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))
U1(X, Xs, Ys, member_out(X, Ys)) → U2(X, Xs, Ys, subset_in(Xs, Ys))
subset_in([], X) → subset_out([], X)
U2(X, Xs, Ys, subset_out(Xs, Ys)) → subset_out(.(X, Xs), Ys)

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:

subset_in(.(X, Xs), Ys) → U1(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X1, Xs)) → U3(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U3(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))
U1(X, Xs, Ys, member_out(X, Ys)) → U2(X, Xs, Ys, subset_in(Xs, Ys))
subset_in([], X) → subset_out([], X)
U2(X, Xs, Ys, subset_out(Xs, Ys)) → subset_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in(x1, x2) = subset_in(x2)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x3, x4)
member_in(x1, x2) = member_in(x2)
U3(x1, x2, x3, x4) = U3(x2, x3, x4)
member_out(x1, x2) = member_out(x1, x2)
U2(x1, x2, x3, x4) = U2(x1, x3, x4)
[] = []
subset_out(x1, x2) = subset_out(x1, 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:

SUBSET_IN(.(X, Xs), Ys) → U1¹(X, Xs, Ys, member_in(X, Ys))
SUBSET_IN(.(X, Xs), Ys) → MEMBER_IN(X, Ys)
MEMBER_IN(X, .(X1, Xs)) → U3¹(X, X1, Xs, member_in(X, Xs))
MEMBER_IN(X, .(X1, Xs)) → MEMBER_IN(X, Xs)
U1¹(X, Xs, Ys, member_out(X, Ys)) → U2¹(X, Xs, Ys, subset_in(Xs, Ys))
U1¹(X, Xs, Ys, member_out(X, Ys)) → SUBSET_IN(Xs, Ys)

The TRS R consists of the following rules:

subset_in(.(X, Xs), Ys) → U1(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X1, Xs)) → U3(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U3(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))
U1(X, Xs, Ys, member_out(X, Ys)) → U2(X, Xs, Ys, subset_in(Xs, Ys))
subset_in([], X) → subset_out([], X)
U2(X, Xs, Ys, subset_out(Xs, Ys)) → subset_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in(x1, x2) = subset_in(x2)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x3, x4)
member_in(x1, x2) = member_in(x2)
U3(x1, x2, x3, x4) = U3(x2, x3, x4)
member_out(x1, x2) = member_out(x1, x2)
U2(x1, x2, x3, x4) = U2(x1, x3, x4)
[] = []
subset_out(x1, x2) = subset_out(x1, x2)
U3¹(x1, x2, x3, x4) = U3¹(x2, x3, x4)
MEMBER_IN(x1, x2) = MEMBER_IN(x2)
SUBSET_IN(x1, x2) = SUBSET_IN(x2)
U2¹(x1, x2, x3, x4) = U2¹(x1, x3, x4)
U1¹(x1, x2, x3, x4) = U1¹(x3, x4)

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:

SUBSET_IN(.(X, Xs), Ys) → U1¹(X, Xs, Ys, member_in(X, Ys))
SUBSET_IN(.(X, Xs), Ys) → MEMBER_IN(X, Ys)
MEMBER_IN(X, .(X1, Xs)) → U3¹(X, X1, Xs, member_in(X, Xs))
MEMBER_IN(X, .(X1, Xs)) → MEMBER_IN(X, Xs)
U1¹(X, Xs, Ys, member_out(X, Ys)) → U2¹(X, Xs, Ys, subset_in(Xs, Ys))
U1¹(X, Xs, Ys, member_out(X, Ys)) → SUBSET_IN(Xs, Ys)

The TRS R consists of the following rules:

subset_in(.(X, Xs), Ys) → U1(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X1, Xs)) → U3(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U3(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))
U1(X, Xs, Ys, member_out(X, Ys)) → U2(X, Xs, Ys, subset_in(Xs, Ys))
subset_in([], X) → subset_out([], X)
U2(X, Xs, Ys, subset_out(Xs, Ys)) → subset_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in(x1, x2) = subset_in(x2)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x3, x4)
member_in(x1, x2) = member_in(x2)
U3(x1, x2, x3, x4) = U3(x2, x3, x4)
member_out(x1, x2) = member_out(x1, x2)
U2(x1, x2, x3, x4) = U2(x1, x3, x4)
[] = []
subset_out(x1, x2) = subset_out(x1, x2)
U3¹(x1, x2, x3, x4) = U3¹(x2, x3, x4)
MEMBER_IN(x1, x2) = MEMBER_IN(x2)
SUBSET_IN(x1, x2) = SUBSET_IN(x2)
U2¹(x1, x2, x3, x4) = U2¹(x1, x3, x4)
U1¹(x1, x2, x3, x4) = U1¹(x3, 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

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

MEMBER_IN(X, .(X1, Xs)) → MEMBER_IN(X, Xs)

The TRS R consists of the following rules:

subset_in(.(X, Xs), Ys) → U1(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X1, Xs)) → U3(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U3(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))
U1(X, Xs, Ys, member_out(X, Ys)) → U2(X, Xs, Ys, subset_in(Xs, Ys))
subset_in([], X) → subset_out([], X)
U2(X, Xs, Ys, subset_out(Xs, Ys)) → subset_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in(x1, x2) = subset_in(x2)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x3, x4)
member_in(x1, x2) = member_in(x2)
U3(x1, x2, x3, x4) = U3(x2, x3, x4)
member_out(x1, x2) = member_out(x1, x2)
U2(x1, x2, x3, x4) = U2(x1, x3, x4)
[] = []
subset_out(x1, x2) = subset_out(x1, x2)
MEMBER_IN(x1, x2) = MEMBER_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

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

MEMBER_IN(X, .(X1, Xs)) → MEMBER_IN(X, Xs)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

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

MEMBER_IN(.(X1, Xs)) → MEMBER_IN(Xs)

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

MEMBER_IN(.(X1, Xs)) → MEMBER_IN(Xs)
The graph contains the following edges 1 > 1

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

SUBSET_IN(.(X, Xs), Ys) → U1¹(X, Xs, Ys, member_in(X, Ys))
U1¹(X, Xs, Ys, member_out(X, Ys)) → SUBSET_IN(Xs, Ys)

The TRS R consists of the following rules:

subset_in(.(X, Xs), Ys) → U1(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X1, Xs)) → U3(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U3(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))
U1(X, Xs, Ys, member_out(X, Ys)) → U2(X, Xs, Ys, subset_in(Xs, Ys))
subset_in([], X) → subset_out([], X)
U2(X, Xs, Ys, subset_out(Xs, Ys)) → subset_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in(x1, x2) = subset_in(x2)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x3, x4)
member_in(x1, x2) = member_in(x2)
U3(x1, x2, x3, x4) = U3(x2, x3, x4)
member_out(x1, x2) = member_out(x1, x2)
U2(x1, x2, x3, x4) = U2(x1, x3, x4)
[] = []
subset_out(x1, x2) = subset_out(x1, x2)
SUBSET_IN(x1, x2) = SUBSET_IN(x2)
U1¹(x1, x2, x3, x4) = U1¹(x3, 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

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

SUBSET_IN(.(X, Xs), Ys) → U1¹(X, Xs, Ys, member_in(X, Ys))
U1¹(X, Xs, Ys, member_out(X, Ys)) → SUBSET_IN(Xs, Ys)

The TRS R consists of the following rules:

member_in(X, .(X1, Xs)) → U3(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U3(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))

The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
member_in(x1, x2) = member_in(x2)
U3(x1, x2, x3, x4) = U3(x2, x3, x4)
member_out(x1, x2) = member_out(x1, x2)
SUBSET_IN(x1, x2) = SUBSET_IN(x2)
U1¹(x1, x2, x3, x4) = U1¹(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

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Narrowing

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

U1¹(Ys, member_out(X, Ys)) → SUBSET_IN(Ys)
SUBSET_IN(Ys) → U1¹(Ys, member_in(Ys))

The TRS R consists of the following rules:

member_in(.(X1, Xs)) → U3(X1, Xs, member_in(Xs))
member_in(.(X, X1)) → member_out(X, .(X, X1))
U3(X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))

The set Q consists of the following terms:

member_in(x0)
U3(x0, x1, x2)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule SUBSET_IN(Ys) → U1¹(Ys, member_in(Ys)) at position [1] we obtained the following new rules:

SUBSET_IN(.(x0, x1)) → U1¹(.(x0, x1), member_out(x0, .(x0, x1)))
SUBSET_IN(.(x0, x1)) → U1¹(.(x0, x1), U3(x0, x1, member_in(x1)))

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Instantiation

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

SUBSET_IN(.(x0, x1)) → U1¹(.(x0, x1), member_out(x0, .(x0, x1)))
SUBSET_IN(.(x0, x1)) → U1¹(.(x0, x1), U3(x0, x1, member_in(x1)))
U1¹(Ys, member_out(X, Ys)) → SUBSET_IN(Ys)

The TRS R consists of the following rules:

member_in(.(X1, Xs)) → U3(X1, Xs, member_in(Xs))
member_in(.(X, X1)) → member_out(X, .(X, X1))
U3(X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))

The set Q consists of the following terms:

member_in(x0)
U3(x0, x1, x2)

We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule U1¹(Ys, member_out(X, Ys)) → SUBSET_IN(Ys) we obtained the following new rules:

U1¹(.(z0, z1), member_out(z0, .(z0, z1))) → SUBSET_IN(.(z0, z1))
U1¹(.(z0, z1), member_out(x1, .(z0, z1))) → SUBSET_IN(.(z0, z1))

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ NonTerminationProof

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

U1¹(.(z0, z1), member_out(z0, .(z0, z1))) → SUBSET_IN(.(z0, z1))
SUBSET_IN(.(x0, x1)) → U1¹(.(x0, x1), member_out(x0, .(x0, x1)))
U1¹(.(z0, z1), member_out(x1, .(z0, z1))) → SUBSET_IN(.(z0, z1))
SUBSET_IN(.(x0, x1)) → U1¹(.(x0, x1), U3(x0, x1, member_in(x1)))

The TRS R consists of the following rules:

member_in(.(X1, Xs)) → U3(X1, Xs, member_in(Xs))
member_in(.(X, X1)) → member_out(X, .(X, X1))
U3(X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))

The set Q consists of the following terms:

member_in(x0)
U3(x0, x1, x2)

U1¹(.(z0, z1), member_out(z0, .(z0, z1))) → SUBSET_IN(.(z0, z1))
SUBSET_IN(.(x0, x1)) → U1¹(.(x0, x1), member_out(x0, .(x0, x1)))
U1¹(.(z0, z1), member_out(x1, .(z0, z1))) → SUBSET_IN(.(z0, z1))
SUBSET_IN(.(x0, x1)) → U1¹(.(x0, x1), U3(x0, x1, member_in(x1)))

The TRS R consists of the following rules:

member_in(.(X1, Xs)) → U3(X1, Xs, member_in(Xs))
member_in(.(X, X1)) → member_out(X, .(X, X1))
U3(X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))

s = SUBSET_IN(.(x0, x1)) evaluates to t =SUBSET_IN(.(x0, x1))

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

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

SUBSET_IN(.(x0, x1)) → U1¹(.(x0, x1), member_out(x0, .(x0, x1)))
with rule SUBSET_IN(.(x0', x1')) → U1¹(.(x0', x1'), member_out(x0', .(x0', x1'))) at position [] and matcher [x1' / x1, x0' / x0]

U1¹(.(x0, x1), member_out(x0, .(x0, x1))) → SUBSET_IN(.(x0, x1))
with rule U1¹(.(z0, z1), member_out(z0, .(z0, z1))) → SUBSET_IN(.(z0, z1))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence

All these steps are and every following step will be a correct step w.r.t to Q.