Left Termination proof of /tmp/tmpKPWUtX/subset1.pl

Left Termination of the query pattern subset1_in_2(a, g) w.r.t. the given Prolog program could not be shown:

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

Clauses:

member(X, .(Y, Xs)) :- member(X, Xs).
member(X, .(X, Xs)).
subset(.(X, Xs), Ys) :- ','(member(X, Ys), subset(Xs, Ys)).
subset([], Ys).
member1(X, .(Y, Xs)) :- member1(X, Xs).
member1(X, .(X, Xs)).
subset1(.(X, Xs), Ys) :- ','(member1(X, Ys), subset1(Xs, Ys)).
subset1([], Ys).

Queries:

subset1(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:

subset1_in([], Ys) → subset1_out([], Ys)
subset1_in(.(X, Xs), Ys) → U5(X, Xs, Ys, member1_in(X, Ys))
member1_in(X, .(X, Xs)) → member1_out(X, .(X, Xs))
member1_in(X, .(Y, Xs)) → U4(X, Y, Xs, member1_in(X, Xs))
U4(X, Y, Xs, member1_out(X, Xs)) → member1_out(X, .(Y, Xs))
U5(X, Xs, Ys, member1_out(X, Ys)) → U6(X, Xs, Ys, subset1_in(Xs, Ys))
U6(X, Xs, Ys, subset1_out(Xs, Ys)) → subset1_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:

subset1_in([], Ys) → subset1_out([], Ys)
subset1_in(.(X, Xs), Ys) → U5(X, Xs, Ys, member1_in(X, Ys))
member1_in(X, .(X, Xs)) → member1_out(X, .(X, Xs))
member1_in(X, .(Y, Xs)) → U4(X, Y, Xs, member1_in(X, Xs))
U4(X, Y, Xs, member1_out(X, Xs)) → member1_out(X, .(Y, Xs))
U5(X, Xs, Ys, member1_out(X, Ys)) → U6(X, Xs, Ys, subset1_in(Xs, Ys))
U6(X, Xs, Ys, subset1_out(Xs, Ys)) → subset1_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_in(x1, x2) = subset1_in(x2)
[] = []
subset1_out(x1, x2) = subset1_out(x1)
.(x1, x2) = .(x1, x2)
U5(x1, x2, x3, x4) = U5(x3, x4)
member1_in(x1, x2) = member1_in(x2)
member1_out(x1, x2) = member1_out(x1)
U4(x1, x2, x3, x4) = U4(x4)
U6(x1, x2, x3, x4) = U6(x1, x4)

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

SUBSET1_IN(.(X, Xs), Ys) → U5¹(X, Xs, Ys, member1_in(X, Ys))
SUBSET1_IN(.(X, Xs), Ys) → MEMBER1_IN(X, Ys)
MEMBER1_IN(X, .(Y, Xs)) → U4¹(X, Y, Xs, member1_in(X, Xs))
MEMBER1_IN(X, .(Y, Xs)) → MEMBER1_IN(X, Xs)
U5¹(X, Xs, Ys, member1_out(X, Ys)) → U6¹(X, Xs, Ys, subset1_in(Xs, Ys))
U5¹(X, Xs, Ys, member1_out(X, Ys)) → SUBSET1_IN(Xs, Ys)

The TRS R consists of the following rules:

subset1_in([], Ys) → subset1_out([], Ys)
subset1_in(.(X, Xs), Ys) → U5(X, Xs, Ys, member1_in(X, Ys))
member1_in(X, .(X, Xs)) → member1_out(X, .(X, Xs))
member1_in(X, .(Y, Xs)) → U4(X, Y, Xs, member1_in(X, Xs))
U4(X, Y, Xs, member1_out(X, Xs)) → member1_out(X, .(Y, Xs))
U5(X, Xs, Ys, member1_out(X, Ys)) → U6(X, Xs, Ys, subset1_in(Xs, Ys))
U6(X, Xs, Ys, subset1_out(Xs, Ys)) → subset1_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_in(x1, x2) = subset1_in(x2)
[] = []
subset1_out(x1, x2) = subset1_out(x1)
.(x1, x2) = .(x1, x2)
U5(x1, x2, x3, x4) = U5(x3, x4)
member1_in(x1, x2) = member1_in(x2)
member1_out(x1, x2) = member1_out(x1)
U4(x1, x2, x3, x4) = U4(x4)
U6(x1, x2, x3, x4) = U6(x1, x4)
U5¹(x1, x2, x3, x4) = U5¹(x3, x4)
SUBSET1_IN(x1, x2) = SUBSET1_IN(x2)
U4¹(x1, x2, x3, x4) = U4¹(x4)
MEMBER1_IN(x1, x2) = MEMBER1_IN(x2)
U6¹(x1, x2, x3, x4) = U6¹(x1, 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:

SUBSET1_IN(.(X, Xs), Ys) → U5¹(X, Xs, Ys, member1_in(X, Ys))
SUBSET1_IN(.(X, Xs), Ys) → MEMBER1_IN(X, Ys)
MEMBER1_IN(X, .(Y, Xs)) → U4¹(X, Y, Xs, member1_in(X, Xs))
MEMBER1_IN(X, .(Y, Xs)) → MEMBER1_IN(X, Xs)
U5¹(X, Xs, Ys, member1_out(X, Ys)) → U6¹(X, Xs, Ys, subset1_in(Xs, Ys))
U5¹(X, Xs, Ys, member1_out(X, Ys)) → SUBSET1_IN(Xs, Ys)

The TRS R consists of the following rules:

subset1_in([], Ys) → subset1_out([], Ys)
subset1_in(.(X, Xs), Ys) → U5(X, Xs, Ys, member1_in(X, Ys))
member1_in(X, .(X, Xs)) → member1_out(X, .(X, Xs))
member1_in(X, .(Y, Xs)) → U4(X, Y, Xs, member1_in(X, Xs))
U4(X, Y, Xs, member1_out(X, Xs)) → member1_out(X, .(Y, Xs))
U5(X, Xs, Ys, member1_out(X, Ys)) → U6(X, Xs, Ys, subset1_in(Xs, Ys))
U6(X, Xs, Ys, subset1_out(Xs, Ys)) → subset1_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_in(x1, x2) = subset1_in(x2)
[] = []
subset1_out(x1, x2) = subset1_out(x1)
.(x1, x2) = .(x1, x2)
U5(x1, x2, x3, x4) = U5(x3, x4)
member1_in(x1, x2) = member1_in(x2)
member1_out(x1, x2) = member1_out(x1)
U4(x1, x2, x3, x4) = U4(x4)
U6(x1, x2, x3, x4) = U6(x1, x4)
U5¹(x1, x2, x3, x4) = U5¹(x3, x4)
SUBSET1_IN(x1, x2) = SUBSET1_IN(x2)
U4¹(x1, x2, x3, x4) = U4¹(x4)
MEMBER1_IN(x1, x2) = MEMBER1_IN(x2)
U6¹(x1, x2, x3, x4) = U6¹(x1, 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:

MEMBER1_IN(X, .(Y, Xs)) → MEMBER1_IN(X, Xs)

The TRS R consists of the following rules:

subset1_in([], Ys) → subset1_out([], Ys)
subset1_in(.(X, Xs), Ys) → U5(X, Xs, Ys, member1_in(X, Ys))
member1_in(X, .(X, Xs)) → member1_out(X, .(X, Xs))
member1_in(X, .(Y, Xs)) → U4(X, Y, Xs, member1_in(X, Xs))
U4(X, Y, Xs, member1_out(X, Xs)) → member1_out(X, .(Y, Xs))
U5(X, Xs, Ys, member1_out(X, Ys)) → U6(X, Xs, Ys, subset1_in(Xs, Ys))
U6(X, Xs, Ys, subset1_out(Xs, Ys)) → subset1_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_in(x1, x2) = subset1_in(x2)
[] = []
subset1_out(x1, x2) = subset1_out(x1)
.(x1, x2) = .(x1, x2)
U5(x1, x2, x3, x4) = U5(x3, x4)
member1_in(x1, x2) = member1_in(x2)
member1_out(x1, x2) = member1_out(x1)
U4(x1, x2, x3, x4) = U4(x4)
U6(x1, x2, x3, x4) = U6(x1, x4)
MEMBER1_IN(x1, x2) = MEMBER1_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:

MEMBER1_IN(X, .(Y, Xs)) → MEMBER1_IN(X, Xs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
MEMBER1_IN(x1, x2) = MEMBER1_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:

MEMBER1_IN(.(Y, Xs)) → MEMBER1_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:

MEMBER1_IN(.(Y, Xs)) → MEMBER1_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:

SUBSET1_IN(.(X, Xs), Ys) → U5¹(X, Xs, Ys, member1_in(X, Ys))
U5¹(X, Xs, Ys, member1_out(X, Ys)) → SUBSET1_IN(Xs, Ys)

The TRS R consists of the following rules:

subset1_in([], Ys) → subset1_out([], Ys)
subset1_in(.(X, Xs), Ys) → U5(X, Xs, Ys, member1_in(X, Ys))
member1_in(X, .(X, Xs)) → member1_out(X, .(X, Xs))
member1_in(X, .(Y, Xs)) → U4(X, Y, Xs, member1_in(X, Xs))
U4(X, Y, Xs, member1_out(X, Xs)) → member1_out(X, .(Y, Xs))
U5(X, Xs, Ys, member1_out(X, Ys)) → U6(X, Xs, Ys, subset1_in(Xs, Ys))
U6(X, Xs, Ys, subset1_out(Xs, Ys)) → subset1_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_in(x1, x2) = subset1_in(x2)
[] = []
subset1_out(x1, x2) = subset1_out(x1)
.(x1, x2) = .(x1, x2)
U5(x1, x2, x3, x4) = U5(x3, x4)
member1_in(x1, x2) = member1_in(x2)
member1_out(x1, x2) = member1_out(x1)
U4(x1, x2, x3, x4) = U4(x4)
U6(x1, x2, x3, x4) = U6(x1, x4)
U5¹(x1, x2, x3, x4) = U5¹(x3, x4)
SUBSET1_IN(x1, x2) = SUBSET1_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

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

  ↳ PrologToPiTRSProof

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

SUBSET1_IN(.(X, Xs), Ys) → U5¹(X, Xs, Ys, member1_in(X, Ys))
U5¹(X, Xs, Ys, member1_out(X, Ys)) → SUBSET1_IN(Xs, Ys)

The TRS R consists of the following rules:

member1_in(X, .(X, Xs)) → member1_out(X, .(X, Xs))
member1_in(X, .(Y, Xs)) → U4(X, Y, Xs, member1_in(X, Xs))
U4(X, Y, Xs, member1_out(X, Xs)) → member1_out(X, .(Y, Xs))

The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
member1_in(x1, x2) = member1_in(x2)
member1_out(x1, x2) = member1_out(x1)
U4(x1, x2, x3, x4) = U4(x4)
U5¹(x1, x2, x3, x4) = U5¹(x3, x4)
SUBSET1_IN(x1, x2) = SUBSET1_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

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Narrowing

  ↳ PrologToPiTRSProof

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

U5¹(Ys, member1_out(X)) → SUBSET1_IN(Ys)
SUBSET1_IN(Ys) → U5¹(Ys, member1_in(Ys))

The TRS R consists of the following rules:

member1_in(.(X, Xs)) → member1_out(X)
member1_in(.(Y, Xs)) → U4(member1_in(Xs))
U4(member1_out(X)) → member1_out(X)

The set Q consists of the following terms:

member1_in(x0)
U4(x0)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule SUBSET1_IN(Ys) → U5¹(Ys, member1_in(Ys)) at position [1] we obtained the following new rules:

SUBSET1_IN(.(x0, x1)) → U5¹(.(x0, x1), U4(member1_in(x1)))
SUBSET1_IN(.(x0, x1)) → U5¹(.(x0, x1), member1_out(x0))

↳ 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:

U5¹(Ys, member1_out(X)) → SUBSET1_IN(Ys)
SUBSET1_IN(.(x0, x1)) → U5¹(.(x0, x1), U4(member1_in(x1)))
SUBSET1_IN(.(x0, x1)) → U5¹(.(x0, x1), member1_out(x0))

The TRS R consists of the following rules:

member1_in(.(X, Xs)) → member1_out(X)
member1_in(.(Y, Xs)) → U4(member1_in(Xs))
U4(member1_out(X)) → member1_out(X)

The set Q consists of the following terms:

member1_in(x0)
U4(x0)

We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule U5¹(Ys, member1_out(X)) → SUBSET1_IN(Ys) we obtained the following new rules:

U5¹(.(z0, z1), member1_out(x1)) → SUBSET1_IN(.(z0, z1))
U5¹(.(z0, z1), member1_out(z0)) → SUBSET1_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:

U5¹(.(z0, z1), member1_out(x1)) → SUBSET1_IN(.(z0, z1))
SUBSET1_IN(.(x0, x1)) → U5¹(.(x0, x1), U4(member1_in(x1)))
SUBSET1_IN(.(x0, x1)) → U5¹(.(x0, x1), member1_out(x0))
U5¹(.(z0, z1), member1_out(z0)) → SUBSET1_IN(.(z0, z1))

The TRS R consists of the following rules:

member1_in(.(X, Xs)) → member1_out(X)
member1_in(.(Y, Xs)) → U4(member1_in(Xs))
U4(member1_out(X)) → member1_out(X)

The set Q consists of the following terms:

member1_in(x0)
U4(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:

U5¹(.(z0, z1), member1_out(x1)) → SUBSET1_IN(.(z0, z1))
SUBSET1_IN(.(x0, x1)) → U5¹(.(x0, x1), U4(member1_in(x1)))
SUBSET1_IN(.(x0, x1)) → U5¹(.(x0, x1), member1_out(x0))
U5¹(.(z0, z1), member1_out(z0)) → SUBSET1_IN(.(z0, z1))

The TRS R consists of the following rules:

member1_in(.(X, Xs)) → member1_out(X)
member1_in(.(Y, Xs)) → U4(member1_in(Xs))
U4(member1_out(X)) → member1_out(X)

s = SUBSET1_IN(.(x0, x1')) evaluates to t =SUBSET1_IN(.(x0, x1'))

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

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

SUBSET1_IN(.(x0, x1')) → U5¹(.(x0, x1'), member1_out(x0))
with rule SUBSET1_IN(.(x0', x1'')) → U5¹(.(x0', x1''), member1_out(x0')) at position [] and matcher [x1'' / x1', x0' / x0]

U5¹(.(x0, x1'), member1_out(x0)) → SUBSET1_IN(.(x0, x1'))
with rule U5¹(.(z0, z1), member1_out(x1)) → SUBSET1_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.

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:

subset1_in([], Ys) → subset1_out([], Ys)
subset1_in(.(X, Xs), Ys) → U5(X, Xs, Ys, member1_in(X, Ys))
member1_in(X, .(X, Xs)) → member1_out(X, .(X, Xs))
member1_in(X, .(Y, Xs)) → U4(X, Y, Xs, member1_in(X, Xs))
U4(X, Y, Xs, member1_out(X, Xs)) → member1_out(X, .(Y, Xs))
U5(X, Xs, Ys, member1_out(X, Ys)) → U6(X, Xs, Ys, subset1_in(Xs, Ys))
U6(X, Xs, Ys, subset1_out(Xs, Ys)) → subset1_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:

subset1_in([], Ys) → subset1_out([], Ys)
subset1_in(.(X, Xs), Ys) → U5(X, Xs, Ys, member1_in(X, Ys))
member1_in(X, .(X, Xs)) → member1_out(X, .(X, Xs))
member1_in(X, .(Y, Xs)) → U4(X, Y, Xs, member1_in(X, Xs))
U4(X, Y, Xs, member1_out(X, Xs)) → member1_out(X, .(Y, Xs))
U5(X, Xs, Ys, member1_out(X, Ys)) → U6(X, Xs, Ys, subset1_in(Xs, Ys))
U6(X, Xs, Ys, subset1_out(Xs, Ys)) → subset1_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_in(x1, x2) = subset1_in(x2)
[] = []
subset1_out(x1, x2) = subset1_out(x1, x2)
.(x1, x2) = .(x1, x2)
U5(x1, x2, x3, x4) = U5(x3, x4)
member1_in(x1, x2) = member1_in(x2)
member1_out(x1, x2) = member1_out(x1, x2)
U4(x1, x2, x3, x4) = U4(x2, x3, x4)
U6(x1, x2, x3, x4) = U6(x1, x3, x4)

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

SUBSET1_IN(.(X, Xs), Ys) → U5¹(X, Xs, Ys, member1_in(X, Ys))
SUBSET1_IN(.(X, Xs), Ys) → MEMBER1_IN(X, Ys)
MEMBER1_IN(X, .(Y, Xs)) → U4¹(X, Y, Xs, member1_in(X, Xs))
MEMBER1_IN(X, .(Y, Xs)) → MEMBER1_IN(X, Xs)
U5¹(X, Xs, Ys, member1_out(X, Ys)) → U6¹(X, Xs, Ys, subset1_in(Xs, Ys))
U5¹(X, Xs, Ys, member1_out(X, Ys)) → SUBSET1_IN(Xs, Ys)

The TRS R consists of the following rules:

subset1_in([], Ys) → subset1_out([], Ys)
subset1_in(.(X, Xs), Ys) → U5(X, Xs, Ys, member1_in(X, Ys))
member1_in(X, .(X, Xs)) → member1_out(X, .(X, Xs))
member1_in(X, .(Y, Xs)) → U4(X, Y, Xs, member1_in(X, Xs))
U4(X, Y, Xs, member1_out(X, Xs)) → member1_out(X, .(Y, Xs))
U5(X, Xs, Ys, member1_out(X, Ys)) → U6(X, Xs, Ys, subset1_in(Xs, Ys))
U6(X, Xs, Ys, subset1_out(Xs, Ys)) → subset1_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_in(x1, x2) = subset1_in(x2)
[] = []
subset1_out(x1, x2) = subset1_out(x1, x2)
.(x1, x2) = .(x1, x2)
U5(x1, x2, x3, x4) = U5(x3, x4)
member1_in(x1, x2) = member1_in(x2)
member1_out(x1, x2) = member1_out(x1, x2)
U4(x1, x2, x3, x4) = U4(x2, x3, x4)
U6(x1, x2, x3, x4) = U6(x1, x3, x4)
U5¹(x1, x2, x3, x4) = U5¹(x3, x4)
SUBSET1_IN(x1, x2) = SUBSET1_IN(x2)
U4¹(x1, x2, x3, x4) = U4¹(x2, x3, x4)
MEMBER1_IN(x1, x2) = MEMBER1_IN(x2)
U6¹(x1, x2, x3, x4) = U6¹(x1, 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:

SUBSET1_IN(.(X, Xs), Ys) → U5¹(X, Xs, Ys, member1_in(X, Ys))
SUBSET1_IN(.(X, Xs), Ys) → MEMBER1_IN(X, Ys)
MEMBER1_IN(X, .(Y, Xs)) → U4¹(X, Y, Xs, member1_in(X, Xs))
MEMBER1_IN(X, .(Y, Xs)) → MEMBER1_IN(X, Xs)
U5¹(X, Xs, Ys, member1_out(X, Ys)) → U6¹(X, Xs, Ys, subset1_in(Xs, Ys))
U5¹(X, Xs, Ys, member1_out(X, Ys)) → SUBSET1_IN(Xs, Ys)

The TRS R consists of the following rules:

subset1_in([], Ys) → subset1_out([], Ys)
subset1_in(.(X, Xs), Ys) → U5(X, Xs, Ys, member1_in(X, Ys))
member1_in(X, .(X, Xs)) → member1_out(X, .(X, Xs))
member1_in(X, .(Y, Xs)) → U4(X, Y, Xs, member1_in(X, Xs))
U4(X, Y, Xs, member1_out(X, Xs)) → member1_out(X, .(Y, Xs))
U5(X, Xs, Ys, member1_out(X, Ys)) → U6(X, Xs, Ys, subset1_in(Xs, Ys))
U6(X, Xs, Ys, subset1_out(Xs, Ys)) → subset1_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_in(x1, x2) = subset1_in(x2)
[] = []
subset1_out(x1, x2) = subset1_out(x1, x2)
.(x1, x2) = .(x1, x2)
U5(x1, x2, x3, x4) = U5(x3, x4)
member1_in(x1, x2) = member1_in(x2)
member1_out(x1, x2) = member1_out(x1, x2)
U4(x1, x2, x3, x4) = U4(x2, x3, x4)
U6(x1, x2, x3, x4) = U6(x1, x3, x4)
U5¹(x1, x2, x3, x4) = U5¹(x3, x4)
SUBSET1_IN(x1, x2) = SUBSET1_IN(x2)
U4¹(x1, x2, x3, x4) = U4¹(x2, x3, x4)
MEMBER1_IN(x1, x2) = MEMBER1_IN(x2)
U6¹(x1, x2, x3, x4) = U6¹(x1, 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:

MEMBER1_IN(X, .(Y, Xs)) → MEMBER1_IN(X, Xs)

The TRS R consists of the following rules:

subset1_in([], Ys) → subset1_out([], Ys)
subset1_in(.(X, Xs), Ys) → U5(X, Xs, Ys, member1_in(X, Ys))
member1_in(X, .(X, Xs)) → member1_out(X, .(X, Xs))
member1_in(X, .(Y, Xs)) → U4(X, Y, Xs, member1_in(X, Xs))
U4(X, Y, Xs, member1_out(X, Xs)) → member1_out(X, .(Y, Xs))
U5(X, Xs, Ys, member1_out(X, Ys)) → U6(X, Xs, Ys, subset1_in(Xs, Ys))
U6(X, Xs, Ys, subset1_out(Xs, Ys)) → subset1_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_in(x1, x2) = subset1_in(x2)
[] = []
subset1_out(x1, x2) = subset1_out(x1, x2)
.(x1, x2) = .(x1, x2)
U5(x1, x2, x3, x4) = U5(x3, x4)
member1_in(x1, x2) = member1_in(x2)
member1_out(x1, x2) = member1_out(x1, x2)
U4(x1, x2, x3, x4) = U4(x2, x3, x4)
U6(x1, x2, x3, x4) = U6(x1, x3, x4)
MEMBER1_IN(x1, x2) = MEMBER1_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:

MEMBER1_IN(X, .(Y, Xs)) → MEMBER1_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:

MEMBER1_IN(.(Y, Xs)) → MEMBER1_IN(Xs)

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

MEMBER1_IN(.(Y, Xs)) → MEMBER1_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:

SUBSET1_IN(.(X, Xs), Ys) → U5¹(X, Xs, Ys, member1_in(X, Ys))
U5¹(X, Xs, Ys, member1_out(X, Ys)) → SUBSET1_IN(Xs, Ys)

The TRS R consists of the following rules:

subset1_in([], Ys) → subset1_out([], Ys)
subset1_in(.(X, Xs), Ys) → U5(X, Xs, Ys, member1_in(X, Ys))
member1_in(X, .(X, Xs)) → member1_out(X, .(X, Xs))
member1_in(X, .(Y, Xs)) → U4(X, Y, Xs, member1_in(X, Xs))
U4(X, Y, Xs, member1_out(X, Xs)) → member1_out(X, .(Y, Xs))
U5(X, Xs, Ys, member1_out(X, Ys)) → U6(X, Xs, Ys, subset1_in(Xs, Ys))
U6(X, Xs, Ys, subset1_out(Xs, Ys)) → subset1_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_in(x1, x2) = subset1_in(x2)
[] = []
subset1_out(x1, x2) = subset1_out(x1, x2)
.(x1, x2) = .(x1, x2)
U5(x1, x2, x3, x4) = U5(x3, x4)
member1_in(x1, x2) = member1_in(x2)
member1_out(x1, x2) = member1_out(x1, x2)
U4(x1, x2, x3, x4) = U4(x2, x3, x4)
U6(x1, x2, x3, x4) = U6(x1, x3, x4)
U5¹(x1, x2, x3, x4) = U5¹(x3, x4)
SUBSET1_IN(x1, x2) = SUBSET1_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

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

SUBSET1_IN(.(X, Xs), Ys) → U5¹(X, Xs, Ys, member1_in(X, Ys))
U5¹(X, Xs, Ys, member1_out(X, Ys)) → SUBSET1_IN(Xs, Ys)

The TRS R consists of the following rules:

member1_in(X, .(X, Xs)) → member1_out(X, .(X, Xs))
member1_in(X, .(Y, Xs)) → U4(X, Y, Xs, member1_in(X, Xs))
U4(X, Y, Xs, member1_out(X, Xs)) → member1_out(X, .(Y, Xs))

The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
member1_in(x1, x2) = member1_in(x2)
member1_out(x1, x2) = member1_out(x1, x2)
U4(x1, x2, x3, x4) = U4(x2, x3, x4)
U5¹(x1, x2, x3, x4) = U5¹(x3, x4)
SUBSET1_IN(x1, x2) = SUBSET1_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

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Narrowing

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

SUBSET1_IN(Ys) → U5¹(Ys, member1_in(Ys))
U5¹(Ys, member1_out(X, Ys)) → SUBSET1_IN(Ys)

The TRS R consists of the following rules:

member1_in(.(X, Xs)) → member1_out(X, .(X, Xs))
member1_in(.(Y, Xs)) → U4(Y, Xs, member1_in(Xs))
U4(Y, Xs, member1_out(X, Xs)) → member1_out(X, .(Y, Xs))

The set Q consists of the following terms:

member1_in(x0)
U4(x0, x1, x2)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule SUBSET1_IN(Ys) → U5¹(Ys, member1_in(Ys)) at position [1] we obtained the following new rules:

SUBSET1_IN(.(x0, x1)) → U5¹(.(x0, x1), U4(x0, x1, member1_in(x1)))
SUBSET1_IN(.(x0, x1)) → U5¹(.(x0, x1), member1_out(x0, .(x0, 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:

SUBSET1_IN(.(x0, x1)) → U5¹(.(x0, x1), U4(x0, x1, member1_in(x1)))
SUBSET1_IN(.(x0, x1)) → U5¹(.(x0, x1), member1_out(x0, .(x0, x1)))
U5¹(Ys, member1_out(X, Ys)) → SUBSET1_IN(Ys)

The TRS R consists of the following rules:

member1_in(.(X, Xs)) → member1_out(X, .(X, Xs))
member1_in(.(Y, Xs)) → U4(Y, Xs, member1_in(Xs))
U4(Y, Xs, member1_out(X, Xs)) → member1_out(X, .(Y, Xs))

The set Q consists of the following terms:

member1_in(x0)
U4(x0, x1, x2)

We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule U5¹(Ys, member1_out(X, Ys)) → SUBSET1_IN(Ys) we obtained the following new rules:

U5¹(.(z0, z1), member1_out(z0, .(z0, z1))) → SUBSET1_IN(.(z0, z1))
U5¹(.(z0, z1), member1_out(x1, .(z0, z1))) → SUBSET1_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:

U5¹(.(z0, z1), member1_out(z0, .(z0, z1))) → SUBSET1_IN(.(z0, z1))
SUBSET1_IN(.(x0, x1)) → U5¹(.(x0, x1), U4(x0, x1, member1_in(x1)))
U5¹(.(z0, z1), member1_out(x1, .(z0, z1))) → SUBSET1_IN(.(z0, z1))
SUBSET1_IN(.(x0, x1)) → U5¹(.(x0, x1), member1_out(x0, .(x0, x1)))

The TRS R consists of the following rules:

member1_in(.(X, Xs)) → member1_out(X, .(X, Xs))
member1_in(.(Y, Xs)) → U4(Y, Xs, member1_in(Xs))
U4(Y, Xs, member1_out(X, Xs)) → member1_out(X, .(Y, Xs))

The set Q consists of the following terms:

member1_in(x0)
U4(x0, x1, x2)

U5¹(.(z0, z1), member1_out(z0, .(z0, z1))) → SUBSET1_IN(.(z0, z1))
SUBSET1_IN(.(x0, x1)) → U5¹(.(x0, x1), U4(x0, x1, member1_in(x1)))
U5¹(.(z0, z1), member1_out(x1, .(z0, z1))) → SUBSET1_IN(.(z0, z1))
SUBSET1_IN(.(x0, x1)) → U5¹(.(x0, x1), member1_out(x0, .(x0, x1)))

The TRS R consists of the following rules:

member1_in(.(X, Xs)) → member1_out(X, .(X, Xs))
member1_in(.(Y, Xs)) → U4(Y, Xs, member1_in(Xs))
U4(Y, Xs, member1_out(X, Xs)) → member1_out(X, .(Y, Xs))

s = SUBSET1_IN(.(x0, x1)) evaluates to t =SUBSET1_IN(.(x0, x1))

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

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

SUBSET1_IN(.(x0, x1)) → U5¹(.(x0, x1), member1_out(x0, .(x0, x1)))
with rule SUBSET1_IN(.(x0', x1')) → U5¹(.(x0', x1'), member1_out(x0', .(x0', x1'))) at position [] and matcher [x1' / x1, x0' / x0]

U5¹(.(x0, x1), member1_out(x0, .(x0, x1))) → SUBSET1_IN(.(x0, x1))
with rule U5¹(.(z0, z1), member1_out(z0, .(z0, z1))) → SUBSET1_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.