Left Termination proof of /tmp/tmpEGQYnX/subset-bf.pl

Left Termination of the query pattern subset_in_2(g, a) 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(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:

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(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x2, x4)
member_in(x1, x2) = member_in(x1)
U3(x1, x2, x3, x4) = U3(x4)
member_out(x1, x2) = member_out
U2(x1, x2, x3, x4) = U2(x4)
[] = []
subset_out(x1, x2) = subset_out

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(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x2, x4)
member_in(x1, x2) = member_in(x1)
U3(x1, x2, x3, x4) = U3(x4)
member_out(x1, x2) = member_out
U2(x1, x2, x3, x4) = U2(x4)
[] = []
subset_out(x1, x2) = subset_out
U3¹(x1, x2, x3, x4) = U3¹(x4)
MEMBER_IN(x1, x2) = MEMBER_IN(x1)
SUBSET_IN(x1, x2) = SUBSET_IN(x1)
U2¹(x1, x2, x3, x4) = U2¹(x4)
U1¹(x1, x2, x3, x4) = U1¹(x2, 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(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x2, x4)
member_in(x1, x2) = member_in(x1)
U3(x1, x2, x3, x4) = U3(x4)
member_out(x1, x2) = member_out
U2(x1, x2, x3, x4) = U2(x4)
[] = []
subset_out(x1, x2) = subset_out
U3¹(x1, x2, x3, x4) = U3¹(x4)
MEMBER_IN(x1, x2) = MEMBER_IN(x1)
SUBSET_IN(x1, x2) = SUBSET_IN(x1)
U2¹(x1, x2, x3, x4) = U2¹(x4)
U1¹(x1, x2, x3, x4) = U1¹(x2, 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(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x2, x4)
member_in(x1, x2) = member_in(x1)
U3(x1, x2, x3, x4) = U3(x4)
member_out(x1, x2) = member_out
U2(x1, x2, x3, x4) = U2(x4)
[] = []
subset_out(x1, x2) = subset_out
MEMBER_IN(x1, x2) = MEMBER_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

                ↳ 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(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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ NonTerminationProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

MEMBER_IN(X) → MEMBER_IN(X)

R is empty.
Q is empty.
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 semiunifying a rule from P directly.

The TRS P consists of the following rules:

MEMBER_IN(X) → MEMBER_IN(X)

The TRS R consists of the following rules:none

s = MEMBER_IN(X) evaluates to t =MEMBER_IN(X)

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

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from MEMBER_IN(X) to MEMBER_IN(X).

↳ 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(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x2, x4)
member_in(x1, x2) = member_in(x1)
U3(x1, x2, x3, x4) = U3(x4)
member_out(x1, x2) = member_out
U2(x1, x2, x3, x4) = U2(x4)
[] = []
subset_out(x1, x2) = subset_out
SUBSET_IN(x1, x2) = SUBSET_IN(x1)
U1¹(x1, x2, x3, x4) = U1¹(x2, 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(x1)
U3(x1, x2, x3, x4) = U3(x4)
member_out(x1, x2) = member_out
SUBSET_IN(x1, x2) = SUBSET_IN(x1)
U1¹(x1, x2, x3, x4) = U1¹(x2, 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

                        ↳ QDPSizeChangeProof

  ↳ PrologToPiTRSProof

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

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

The TRS R consists of the following rules:

member_in(X) → U3(member_in(X))
member_in(X) → member_out
U3(member_out) → member_out

The set Q consists of the following terms:

member_in(x0)
U3(x0)

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:

U1¹(Xs, member_out) → SUBSET_IN(Xs)
The graph contains the following edges 1 >= 1
SUBSET_IN(.(X, Xs)) → U1¹(Xs, member_in(X))
The graph contains the following edges 1 > 1

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(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x1, x2, x4)
member_in(x1, x2) = member_in(x1)
U3(x1, x2, x3, x4) = U3(x1, x4)
member_out(x1, x2) = member_out(x1)
U2(x1, x2, x3, x4) = U2(x1, x2, 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(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x1, x2, x4)
member_in(x1, x2) = member_in(x1)
U3(x1, x2, x3, x4) = U3(x1, x4)
member_out(x1, x2) = member_out(x1)
U2(x1, x2, x3, x4) = U2(x1, x2, x4)
[] = []
subset_out(x1, x2) = subset_out(x1)
U3¹(x1, x2, x3, x4) = U3¹(x1, x4)
MEMBER_IN(x1, x2) = MEMBER_IN(x1)
SUBSET_IN(x1, x2) = SUBSET_IN(x1)
U2¹(x1, x2, x3, x4) = U2¹(x1, x2, x4)
U1¹(x1, x2, x3, x4) = U1¹(x1, x2, 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(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x1, x2, x4)
member_in(x1, x2) = member_in(x1)
U3(x1, x2, x3, x4) = U3(x1, x4)
member_out(x1, x2) = member_out(x1)
U2(x1, x2, x3, x4) = U2(x1, x2, x4)
[] = []
subset_out(x1, x2) = subset_out(x1)
U3¹(x1, x2, x3, x4) = U3¹(x1, x4)
MEMBER_IN(x1, x2) = MEMBER_IN(x1)
SUBSET_IN(x1, x2) = SUBSET_IN(x1)
U2¹(x1, x2, x3, x4) = U2¹(x1, x2, x4)
U1¹(x1, x2, x3, x4) = U1¹(x1, x2, 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(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x1, x2, x4)
member_in(x1, x2) = member_in(x1)
U3(x1, x2, x3, x4) = U3(x1, x4)
member_out(x1, x2) = member_out(x1)
U2(x1, x2, x3, x4) = U2(x1, x2, x4)
[] = []
subset_out(x1, x2) = subset_out(x1)
MEMBER_IN(x1, x2) = MEMBER_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

  ↳ 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

                        ↳ NonTerminationProof

              ↳ PiDP

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

MEMBER_IN(X) → MEMBER_IN(X)

The TRS R consists of the following rules:none

s = MEMBER_IN(X) evaluates to t =MEMBER_IN(X)

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

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from MEMBER_IN(X) to MEMBER_IN(X).

↳ 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(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x1, x2, x4)
member_in(x1, x2) = member_in(x1)
U3(x1, x2, x3, x4) = U3(x1, x4)
member_out(x1, x2) = member_out(x1)
U2(x1, x2, x3, x4) = U2(x1, x2, x4)
[] = []
subset_out(x1, x2) = subset_out(x1)
SUBSET_IN(x1, x2) = SUBSET_IN(x1)
U1¹(x1, x2, x3, x4) = U1¹(x1, x2, 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(x1)
U3(x1, x2, x3, x4) = U3(x1, x4)
member_out(x1, x2) = member_out(x1)
SUBSET_IN(x1, x2) = SUBSET_IN(x1)
U1¹(x1, x2, x3, x4) = U1¹(x1, x2, 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

                        ↳ QDPSizeChangeProof

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

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

The TRS R consists of the following rules:

member_in(X) → U3(X, member_in(X))
member_in(X) → member_out(X)
U3(X, member_out(X)) → member_out(X)

The set Q consists of the following terms:

member_in(x0)
U3(x0, x1)

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:

U1¹(X, Xs, member_out(X)) → SUBSET_IN(Xs)
The graph contains the following edges 2 >= 1
SUBSET_IN(.(X, Xs)) → U1¹(X, Xs, member_in(X))
The graph contains the following edges 1 > 1, 1 > 2