Left Termination proof of /tmp/tmp7irRD1/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]. With regard to the inferred argument filtering the predicates were used in the following modes:
subset_in: (f,b)
member_in: (f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

subset_in_ag([], X) → subset_out_ag([], X)
subset_in_ag(.(X, Xs), Ys) → U1_ag(X, Xs, Ys, member_in_ag(X, Ys))
member_in_ag(X, .(X, X1)) → member_out_ag(X, .(X, X1))
member_in_ag(X, .(X1, Xs)) → U3_ag(X, X1, Xs, member_in_ag(X, Xs))
U3_ag(X, X1, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(X1, Xs))
U1_ag(X, Xs, Ys, member_out_ag(X, Ys)) → U2_ag(X, Xs, Ys, subset_in_ag(Xs, Ys))
U2_ag(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ag(.(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_ag([], X) → subset_out_ag([], X)
subset_in_ag(.(X, Xs), Ys) → U1_ag(X, Xs, Ys, member_in_ag(X, Ys))
member_in_ag(X, .(X, X1)) → member_out_ag(X, .(X, X1))
member_in_ag(X, .(X1, Xs)) → U3_ag(X, X1, Xs, member_in_ag(X, Xs))
U3_ag(X, X1, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(X1, Xs))
U1_ag(X, Xs, Ys, member_out_ag(X, Ys)) → U2_ag(X, Xs, Ys, subset_in_ag(Xs, Ys))
U2_ag(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in_ag(x1, x2) = subset_in_ag(x2)
subset_out_ag(x1, x2) = subset_out_ag(x1)
U1_ag(x1, x2, x3, x4) = U1_ag(x3, x4)
member_in_ag(x1, x2) = member_in_ag(x2)
.(x1, x2) = .(x1, x2)
member_out_ag(x1, x2) = member_out_ag(x1)
U3_ag(x1, x2, x3, x4) = U3_ag(x4)
U2_ag(x1, x2, x3, x4) = U2_ag(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:

SUBSET_IN_AG(.(X, Xs), Ys) → U1_AG(X, Xs, Ys, member_in_ag(X, Ys))
SUBSET_IN_AG(.(X, Xs), Ys) → MEMBER_IN_AG(X, Ys)
MEMBER_IN_AG(X, .(X1, Xs)) → U3_AG(X, X1, Xs, member_in_ag(X, Xs))
MEMBER_IN_AG(X, .(X1, Xs)) → MEMBER_IN_AG(X, Xs)
U1_AG(X, Xs, Ys, member_out_ag(X, Ys)) → U2_AG(X, Xs, Ys, subset_in_ag(Xs, Ys))
U1_AG(X, Xs, Ys, member_out_ag(X, Ys)) → SUBSET_IN_AG(Xs, Ys)

The TRS R consists of the following rules:

subset_in_ag([], X) → subset_out_ag([], X)
subset_in_ag(.(X, Xs), Ys) → U1_ag(X, Xs, Ys, member_in_ag(X, Ys))
member_in_ag(X, .(X, X1)) → member_out_ag(X, .(X, X1))
member_in_ag(X, .(X1, Xs)) → U3_ag(X, X1, Xs, member_in_ag(X, Xs))
U3_ag(X, X1, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(X1, Xs))
U1_ag(X, Xs, Ys, member_out_ag(X, Ys)) → U2_ag(X, Xs, Ys, subset_in_ag(Xs, Ys))
U2_ag(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in_ag(x1, x2) = subset_in_ag(x2)
subset_out_ag(x1, x2) = subset_out_ag(x1)
U1_ag(x1, x2, x3, x4) = U1_ag(x3, x4)
member_in_ag(x1, x2) = member_in_ag(x2)
.(x1, x2) = .(x1, x2)
member_out_ag(x1, x2) = member_out_ag(x1)
U3_ag(x1, x2, x3, x4) = U3_ag(x4)
U2_ag(x1, x2, x3, x4) = U2_ag(x1, x4)
MEMBER_IN_AG(x1, x2) = MEMBER_IN_AG(x2)
U3_AG(x1, x2, x3, x4) = U3_AG(x4)
SUBSET_IN_AG(x1, x2) = SUBSET_IN_AG(x2)
U2_AG(x1, x2, x3, x4) = U2_AG(x1, x4)
U1_AG(x1, x2, x3, x4) = U1_AG(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_AG(.(X, Xs), Ys) → U1_AG(X, Xs, Ys, member_in_ag(X, Ys))
SUBSET_IN_AG(.(X, Xs), Ys) → MEMBER_IN_AG(X, Ys)
MEMBER_IN_AG(X, .(X1, Xs)) → U3_AG(X, X1, Xs, member_in_ag(X, Xs))
MEMBER_IN_AG(X, .(X1, Xs)) → MEMBER_IN_AG(X, Xs)
U1_AG(X, Xs, Ys, member_out_ag(X, Ys)) → U2_AG(X, Xs, Ys, subset_in_ag(Xs, Ys))
U1_AG(X, Xs, Ys, member_out_ag(X, Ys)) → SUBSET_IN_AG(Xs, Ys)

The TRS R consists of the following rules:

subset_in_ag([], X) → subset_out_ag([], X)
subset_in_ag(.(X, Xs), Ys) → U1_ag(X, Xs, Ys, member_in_ag(X, Ys))
member_in_ag(X, .(X, X1)) → member_out_ag(X, .(X, X1))
member_in_ag(X, .(X1, Xs)) → U3_ag(X, X1, Xs, member_in_ag(X, Xs))
U3_ag(X, X1, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(X1, Xs))
U1_ag(X, Xs, Ys, member_out_ag(X, Ys)) → U2_ag(X, Xs, Ys, subset_in_ag(Xs, Ys))
U2_ag(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in_ag(x1, x2) = subset_in_ag(x2)
subset_out_ag(x1, x2) = subset_out_ag(x1)
U1_ag(x1, x2, x3, x4) = U1_ag(x3, x4)
member_in_ag(x1, x2) = member_in_ag(x2)
.(x1, x2) = .(x1, x2)
member_out_ag(x1, x2) = member_out_ag(x1)
U3_ag(x1, x2, x3, x4) = U3_ag(x4)
U2_ag(x1, x2, x3, x4) = U2_ag(x1, x4)
MEMBER_IN_AG(x1, x2) = MEMBER_IN_AG(x2)
U3_AG(x1, x2, x3, x4) = U3_AG(x4)
SUBSET_IN_AG(x1, x2) = SUBSET_IN_AG(x2)
U2_AG(x1, x2, x3, x4) = U2_AG(x1, x4)
U1_AG(x1, x2, x3, x4) = U1_AG(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_AG(X, .(X1, Xs)) → MEMBER_IN_AG(X, Xs)

The TRS R consists of the following rules:

subset_in_ag([], X) → subset_out_ag([], X)
subset_in_ag(.(X, Xs), Ys) → U1_ag(X, Xs, Ys, member_in_ag(X, Ys))
member_in_ag(X, .(X, X1)) → member_out_ag(X, .(X, X1))
member_in_ag(X, .(X1, Xs)) → U3_ag(X, X1, Xs, member_in_ag(X, Xs))
U3_ag(X, X1, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(X1, Xs))
U1_ag(X, Xs, Ys, member_out_ag(X, Ys)) → U2_ag(X, Xs, Ys, subset_in_ag(Xs, Ys))
U2_ag(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in_ag(x1, x2) = subset_in_ag(x2)
subset_out_ag(x1, x2) = subset_out_ag(x1)
U1_ag(x1, x2, x3, x4) = U1_ag(x3, x4)
member_in_ag(x1, x2) = member_in_ag(x2)
.(x1, x2) = .(x1, x2)
member_out_ag(x1, x2) = member_out_ag(x1)
U3_ag(x1, x2, x3, x4) = U3_ag(x4)
U2_ag(x1, x2, x3, x4) = U2_ag(x1, x4)
MEMBER_IN_AG(x1, x2) = MEMBER_IN_AG(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_AG(X, .(X1, Xs)) → MEMBER_IN_AG(X, Xs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
MEMBER_IN_AG(x1, x2) = MEMBER_IN_AG(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_AG(.(X1, Xs)) → MEMBER_IN_AG(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_AG(.(X1, Xs)) → MEMBER_IN_AG(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_AG(.(X, Xs), Ys) → U1_AG(X, Xs, Ys, member_in_ag(X, Ys))
U1_AG(X, Xs, Ys, member_out_ag(X, Ys)) → SUBSET_IN_AG(Xs, Ys)

The TRS R consists of the following rules:

subset_in_ag([], X) → subset_out_ag([], X)
subset_in_ag(.(X, Xs), Ys) → U1_ag(X, Xs, Ys, member_in_ag(X, Ys))
member_in_ag(X, .(X, X1)) → member_out_ag(X, .(X, X1))
member_in_ag(X, .(X1, Xs)) → U3_ag(X, X1, Xs, member_in_ag(X, Xs))
U3_ag(X, X1, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(X1, Xs))
U1_ag(X, Xs, Ys, member_out_ag(X, Ys)) → U2_ag(X, Xs, Ys, subset_in_ag(Xs, Ys))
U2_ag(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in_ag(x1, x2) = subset_in_ag(x2)
subset_out_ag(x1, x2) = subset_out_ag(x1)
U1_ag(x1, x2, x3, x4) = U1_ag(x3, x4)
member_in_ag(x1, x2) = member_in_ag(x2)
.(x1, x2) = .(x1, x2)
member_out_ag(x1, x2) = member_out_ag(x1)
U3_ag(x1, x2, x3, x4) = U3_ag(x4)
U2_ag(x1, x2, x3, x4) = U2_ag(x1, x4)
SUBSET_IN_AG(x1, x2) = SUBSET_IN_AG(x2)
U1_AG(x1, x2, x3, x4) = U1_AG(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_AG(.(X, Xs), Ys) → U1_AG(X, Xs, Ys, member_in_ag(X, Ys))
U1_AG(X, Xs, Ys, member_out_ag(X, Ys)) → SUBSET_IN_AG(Xs, Ys)

The TRS R consists of the following rules:

member_in_ag(X, .(X, X1)) → member_out_ag(X, .(X, X1))
member_in_ag(X, .(X1, Xs)) → U3_ag(X, X1, Xs, member_in_ag(X, Xs))
U3_ag(X, X1, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(X1, Xs))

The argument filtering Pi contains the following mapping:
member_in_ag(x1, x2) = member_in_ag(x2)
.(x1, x2) = .(x1, x2)
member_out_ag(x1, x2) = member_out_ag(x1)
U3_ag(x1, x2, x3, x4) = U3_ag(x4)
SUBSET_IN_AG(x1, x2) = SUBSET_IN_AG(x2)
U1_AG(x1, x2, x3, x4) = U1_AG(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_AG(Ys) → U1_AG(Ys, member_in_ag(Ys))
U1_AG(Ys, member_out_ag(X)) → SUBSET_IN_AG(Ys)

The TRS R consists of the following rules:

member_in_ag(.(X, X1)) → member_out_ag(X)
member_in_ag(.(X1, Xs)) → U3_ag(member_in_ag(Xs))
U3_ag(member_out_ag(X)) → member_out_ag(X)

The set Q consists of the following terms:

member_in_ag(x0)
U3_ag(x0)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule SUBSET_IN_AG(Ys) → U1_AG(Ys, member_in_ag(Ys)) at position [1] we obtained the following new rules:

SUBSET_IN_AG(.(x0, x1)) → U1_AG(.(x0, x1), member_out_ag(x0))
SUBSET_IN_AG(.(x0, x1)) → U1_AG(.(x0, x1), U3_ag(member_in_ag(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:

U1_AG(Ys, member_out_ag(X)) → SUBSET_IN_AG(Ys)
SUBSET_IN_AG(.(x0, x1)) → U1_AG(.(x0, x1), U3_ag(member_in_ag(x1)))
SUBSET_IN_AG(.(x0, x1)) → U1_AG(.(x0, x1), member_out_ag(x0))

The TRS R consists of the following rules:

member_in_ag(.(X, X1)) → member_out_ag(X)
member_in_ag(.(X1, Xs)) → U3_ag(member_in_ag(Xs))
U3_ag(member_out_ag(X)) → member_out_ag(X)

The set Q consists of the following terms:

member_in_ag(x0)
U3_ag(x0)

We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule U1_AG(Ys, member_out_ag(X)) → SUBSET_IN_AG(Ys) we obtained the following new rules:

U1_AG(.(z0, z1), member_out_ag(x1)) → SUBSET_IN_AG(.(z0, z1))
U1_AG(.(z0, z1), member_out_ag(z0)) → SUBSET_IN_AG(.(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:

U1_AG(.(z0, z1), member_out_ag(x1)) → SUBSET_IN_AG(.(z0, z1))
U1_AG(.(z0, z1), member_out_ag(z0)) → SUBSET_IN_AG(.(z0, z1))
SUBSET_IN_AG(.(x0, x1)) → U1_AG(.(x0, x1), U3_ag(member_in_ag(x1)))
SUBSET_IN_AG(.(x0, x1)) → U1_AG(.(x0, x1), member_out_ag(x0))

The TRS R consists of the following rules:

member_in_ag(.(X, X1)) → member_out_ag(X)
member_in_ag(.(X1, Xs)) → U3_ag(member_in_ag(Xs))
U3_ag(member_out_ag(X)) → member_out_ag(X)

The set Q consists of the following terms:

member_in_ag(x0)
U3_ag(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:

U1_AG(.(z0, z1), member_out_ag(x1)) → SUBSET_IN_AG(.(z0, z1))
U1_AG(.(z0, z1), member_out_ag(z0)) → SUBSET_IN_AG(.(z0, z1))
SUBSET_IN_AG(.(x0, x1)) → U1_AG(.(x0, x1), U3_ag(member_in_ag(x1)))
SUBSET_IN_AG(.(x0, x1)) → U1_AG(.(x0, x1), member_out_ag(x0))

The TRS R consists of the following rules:

member_in_ag(.(X, X1)) → member_out_ag(X)
member_in_ag(.(X1, Xs)) → U3_ag(member_in_ag(Xs))
U3_ag(member_out_ag(X)) → member_out_ag(X)

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

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

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

SUBSET_IN_AG(.(x0, x1')) → U1_AG(.(x0, x1'), member_out_ag(x0))
with rule SUBSET_IN_AG(.(x0', x1'')) → U1_AG(.(x0', x1''), member_out_ag(x0')) at position [] and matcher [x1'' / x1', x0' / x0]

U1_AG(.(x0, x1'), member_out_ag(x0)) → SUBSET_IN_AG(.(x0, x1'))
with rule U1_AG(.(z0, z1), member_out_ag(x1)) → SUBSET_IN_AG(.(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]. With regard to the inferred argument filtering the predicates were used in the following modes:
subset_in: (f,b)
member_in: (f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

subset_in_ag([], X) → subset_out_ag([], X)
subset_in_ag(.(X, Xs), Ys) → U1_ag(X, Xs, Ys, member_in_ag(X, Ys))
member_in_ag(X, .(X, X1)) → member_out_ag(X, .(X, X1))
member_in_ag(X, .(X1, Xs)) → U3_ag(X, X1, Xs, member_in_ag(X, Xs))
U3_ag(X, X1, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(X1, Xs))
U1_ag(X, Xs, Ys, member_out_ag(X, Ys)) → U2_ag(X, Xs, Ys, subset_in_ag(Xs, Ys))
U2_ag(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ag(.(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_ag([], X) → subset_out_ag([], X)
subset_in_ag(.(X, Xs), Ys) → U1_ag(X, Xs, Ys, member_in_ag(X, Ys))
member_in_ag(X, .(X, X1)) → member_out_ag(X, .(X, X1))
member_in_ag(X, .(X1, Xs)) → U3_ag(X, X1, Xs, member_in_ag(X, Xs))
U3_ag(X, X1, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(X1, Xs))
U1_ag(X, Xs, Ys, member_out_ag(X, Ys)) → U2_ag(X, Xs, Ys, subset_in_ag(Xs, Ys))
U2_ag(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in_ag(x1, x2) = subset_in_ag(x2)
subset_out_ag(x1, x2) = subset_out_ag(x1, x2)
U1_ag(x1, x2, x3, x4) = U1_ag(x3, x4)
member_in_ag(x1, x2) = member_in_ag(x2)
.(x1, x2) = .(x1, x2)
member_out_ag(x1, x2) = member_out_ag(x1, x2)
U3_ag(x1, x2, x3, x4) = U3_ag(x2, x3, x4)
U2_ag(x1, x2, x3, x4) = U2_ag(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:

SUBSET_IN_AG(.(X, Xs), Ys) → U1_AG(X, Xs, Ys, member_in_ag(X, Ys))
SUBSET_IN_AG(.(X, Xs), Ys) → MEMBER_IN_AG(X, Ys)
MEMBER_IN_AG(X, .(X1, Xs)) → U3_AG(X, X1, Xs, member_in_ag(X, Xs))
MEMBER_IN_AG(X, .(X1, Xs)) → MEMBER_IN_AG(X, Xs)
U1_AG(X, Xs, Ys, member_out_ag(X, Ys)) → U2_AG(X, Xs, Ys, subset_in_ag(Xs, Ys))
U1_AG(X, Xs, Ys, member_out_ag(X, Ys)) → SUBSET_IN_AG(Xs, Ys)

The TRS R consists of the following rules:

subset_in_ag([], X) → subset_out_ag([], X)
subset_in_ag(.(X, Xs), Ys) → U1_ag(X, Xs, Ys, member_in_ag(X, Ys))
member_in_ag(X, .(X, X1)) → member_out_ag(X, .(X, X1))
member_in_ag(X, .(X1, Xs)) → U3_ag(X, X1, Xs, member_in_ag(X, Xs))
U3_ag(X, X1, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(X1, Xs))
U1_ag(X, Xs, Ys, member_out_ag(X, Ys)) → U2_ag(X, Xs, Ys, subset_in_ag(Xs, Ys))
U2_ag(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in_ag(x1, x2) = subset_in_ag(x2)
subset_out_ag(x1, x2) = subset_out_ag(x1, x2)
U1_ag(x1, x2, x3, x4) = U1_ag(x3, x4)
member_in_ag(x1, x2) = member_in_ag(x2)
.(x1, x2) = .(x1, x2)
member_out_ag(x1, x2) = member_out_ag(x1, x2)
U3_ag(x1, x2, x3, x4) = U3_ag(x2, x3, x4)
U2_ag(x1, x2, x3, x4) = U2_ag(x1, x3, x4)
MEMBER_IN_AG(x1, x2) = MEMBER_IN_AG(x2)
U3_AG(x1, x2, x3, x4) = U3_AG(x2, x3, x4)
SUBSET_IN_AG(x1, x2) = SUBSET_IN_AG(x2)
U2_AG(x1, x2, x3, x4) = U2_AG(x1, x3, x4)
U1_AG(x1, x2, x3, x4) = U1_AG(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_AG(.(X, Xs), Ys) → U1_AG(X, Xs, Ys, member_in_ag(X, Ys))
SUBSET_IN_AG(.(X, Xs), Ys) → MEMBER_IN_AG(X, Ys)
MEMBER_IN_AG(X, .(X1, Xs)) → U3_AG(X, X1, Xs, member_in_ag(X, Xs))
MEMBER_IN_AG(X, .(X1, Xs)) → MEMBER_IN_AG(X, Xs)
U1_AG(X, Xs, Ys, member_out_ag(X, Ys)) → U2_AG(X, Xs, Ys, subset_in_ag(Xs, Ys))
U1_AG(X, Xs, Ys, member_out_ag(X, Ys)) → SUBSET_IN_AG(Xs, Ys)

The TRS R consists of the following rules:

subset_in_ag([], X) → subset_out_ag([], X)
subset_in_ag(.(X, Xs), Ys) → U1_ag(X, Xs, Ys, member_in_ag(X, Ys))
member_in_ag(X, .(X, X1)) → member_out_ag(X, .(X, X1))
member_in_ag(X, .(X1, Xs)) → U3_ag(X, X1, Xs, member_in_ag(X, Xs))
U3_ag(X, X1, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(X1, Xs))
U1_ag(X, Xs, Ys, member_out_ag(X, Ys)) → U2_ag(X, Xs, Ys, subset_in_ag(Xs, Ys))
U2_ag(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in_ag(x1, x2) = subset_in_ag(x2)
subset_out_ag(x1, x2) = subset_out_ag(x1, x2)
U1_ag(x1, x2, x3, x4) = U1_ag(x3, x4)
member_in_ag(x1, x2) = member_in_ag(x2)
.(x1, x2) = .(x1, x2)
member_out_ag(x1, x2) = member_out_ag(x1, x2)
U3_ag(x1, x2, x3, x4) = U3_ag(x2, x3, x4)
U2_ag(x1, x2, x3, x4) = U2_ag(x1, x3, x4)
MEMBER_IN_AG(x1, x2) = MEMBER_IN_AG(x2)
U3_AG(x1, x2, x3, x4) = U3_AG(x2, x3, x4)
SUBSET_IN_AG(x1, x2) = SUBSET_IN_AG(x2)
U2_AG(x1, x2, x3, x4) = U2_AG(x1, x3, x4)
U1_AG(x1, x2, x3, x4) = U1_AG(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_AG(X, .(X1, Xs)) → MEMBER_IN_AG(X, Xs)

The TRS R consists of the following rules:

subset_in_ag([], X) → subset_out_ag([], X)
subset_in_ag(.(X, Xs), Ys) → U1_ag(X, Xs, Ys, member_in_ag(X, Ys))
member_in_ag(X, .(X, X1)) → member_out_ag(X, .(X, X1))
member_in_ag(X, .(X1, Xs)) → U3_ag(X, X1, Xs, member_in_ag(X, Xs))
U3_ag(X, X1, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(X1, Xs))
U1_ag(X, Xs, Ys, member_out_ag(X, Ys)) → U2_ag(X, Xs, Ys, subset_in_ag(Xs, Ys))
U2_ag(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in_ag(x1, x2) = subset_in_ag(x2)
subset_out_ag(x1, x2) = subset_out_ag(x1, x2)
U1_ag(x1, x2, x3, x4) = U1_ag(x3, x4)
member_in_ag(x1, x2) = member_in_ag(x2)
.(x1, x2) = .(x1, x2)
member_out_ag(x1, x2) = member_out_ag(x1, x2)
U3_ag(x1, x2, x3, x4) = U3_ag(x2, x3, x4)
U2_ag(x1, x2, x3, x4) = U2_ag(x1, x3, x4)
MEMBER_IN_AG(x1, x2) = MEMBER_IN_AG(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_AG(X, .(X1, Xs)) → MEMBER_IN_AG(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_AG(.(X1, Xs)) → MEMBER_IN_AG(Xs)

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

MEMBER_IN_AG(.(X1, Xs)) → MEMBER_IN_AG(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_AG(.(X, Xs), Ys) → U1_AG(X, Xs, Ys, member_in_ag(X, Ys))
U1_AG(X, Xs, Ys, member_out_ag(X, Ys)) → SUBSET_IN_AG(Xs, Ys)

The TRS R consists of the following rules:

subset_in_ag([], X) → subset_out_ag([], X)
subset_in_ag(.(X, Xs), Ys) → U1_ag(X, Xs, Ys, member_in_ag(X, Ys))
member_in_ag(X, .(X, X1)) → member_out_ag(X, .(X, X1))
member_in_ag(X, .(X1, Xs)) → U3_ag(X, X1, Xs, member_in_ag(X, Xs))
U3_ag(X, X1, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(X1, Xs))
U1_ag(X, Xs, Ys, member_out_ag(X, Ys)) → U2_ag(X, Xs, Ys, subset_in_ag(Xs, Ys))
U2_ag(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in_ag(x1, x2) = subset_in_ag(x2)
subset_out_ag(x1, x2) = subset_out_ag(x1, x2)
U1_ag(x1, x2, x3, x4) = U1_ag(x3, x4)
member_in_ag(x1, x2) = member_in_ag(x2)
.(x1, x2) = .(x1, x2)
member_out_ag(x1, x2) = member_out_ag(x1, x2)
U3_ag(x1, x2, x3, x4) = U3_ag(x2, x3, x4)
U2_ag(x1, x2, x3, x4) = U2_ag(x1, x3, x4)
SUBSET_IN_AG(x1, x2) = SUBSET_IN_AG(x2)
U1_AG(x1, x2, x3, x4) = U1_AG(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_AG(.(X, Xs), Ys) → U1_AG(X, Xs, Ys, member_in_ag(X, Ys))
U1_AG(X, Xs, Ys, member_out_ag(X, Ys)) → SUBSET_IN_AG(Xs, Ys)

The TRS R consists of the following rules:

member_in_ag(X, .(X, X1)) → member_out_ag(X, .(X, X1))
member_in_ag(X, .(X1, Xs)) → U3_ag(X, X1, Xs, member_in_ag(X, Xs))
U3_ag(X, X1, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(X1, Xs))

The argument filtering Pi contains the following mapping:
member_in_ag(x1, x2) = member_in_ag(x2)
.(x1, x2) = .(x1, x2)
member_out_ag(x1, x2) = member_out_ag(x1, x2)
U3_ag(x1, x2, x3, x4) = U3_ag(x2, x3, x4)
SUBSET_IN_AG(x1, x2) = SUBSET_IN_AG(x2)
U1_AG(x1, x2, x3, x4) = U1_AG(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:

U1_AG(Ys, member_out_ag(X, Ys)) → SUBSET_IN_AG(Ys)
SUBSET_IN_AG(Ys) → U1_AG(Ys, member_in_ag(Ys))

The TRS R consists of the following rules:

member_in_ag(.(X, X1)) → member_out_ag(X, .(X, X1))
member_in_ag(.(X1, Xs)) → U3_ag(X1, Xs, member_in_ag(Xs))
U3_ag(X1, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(X1, Xs))

The set Q consists of the following terms:

member_in_ag(x0)
U3_ag(x0, x1, x2)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule SUBSET_IN_AG(Ys) → U1_AG(Ys, member_in_ag(Ys)) at position [1] we obtained the following new rules:

SUBSET_IN_AG(.(x0, x1)) → U1_AG(.(x0, x1), U3_ag(x0, x1, member_in_ag(x1)))
SUBSET_IN_AG(.(x0, x1)) → U1_AG(.(x0, x1), member_out_ag(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:

U1_AG(Ys, member_out_ag(X, Ys)) → SUBSET_IN_AG(Ys)
SUBSET_IN_AG(.(x0, x1)) → U1_AG(.(x0, x1), U3_ag(x0, x1, member_in_ag(x1)))
SUBSET_IN_AG(.(x0, x1)) → U1_AG(.(x0, x1), member_out_ag(x0, .(x0, x1)))

The TRS R consists of the following rules:

member_in_ag(.(X, X1)) → member_out_ag(X, .(X, X1))
member_in_ag(.(X1, Xs)) → U3_ag(X1, Xs, member_in_ag(Xs))
U3_ag(X1, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(X1, Xs))

The set Q consists of the following terms:

member_in_ag(x0)
U3_ag(x0, x1, x2)

We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule U1_AG(Ys, member_out_ag(X, Ys)) → SUBSET_IN_AG(Ys) we obtained the following new rules:

U1_AG(.(z0, z1), member_out_ag(x1, .(z0, z1))) → SUBSET_IN_AG(.(z0, z1))
U1_AG(.(z0, z1), member_out_ag(z0, .(z0, z1))) → SUBSET_IN_AG(.(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:

SUBSET_IN_AG(.(x0, x1)) → U1_AG(.(x0, x1), U3_ag(x0, x1, member_in_ag(x1)))
SUBSET_IN_AG(.(x0, x1)) → U1_AG(.(x0, x1), member_out_ag(x0, .(x0, x1)))
U1_AG(.(z0, z1), member_out_ag(x1, .(z0, z1))) → SUBSET_IN_AG(.(z0, z1))
U1_AG(.(z0, z1), member_out_ag(z0, .(z0, z1))) → SUBSET_IN_AG(.(z0, z1))

The TRS R consists of the following rules:

member_in_ag(.(X, X1)) → member_out_ag(X, .(X, X1))
member_in_ag(.(X1, Xs)) → U3_ag(X1, Xs, member_in_ag(Xs))
U3_ag(X1, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(X1, Xs))

The set Q consists of the following terms:

member_in_ag(x0)
U3_ag(x0, x1, x2)

SUBSET_IN_AG(.(x0, x1)) → U1_AG(.(x0, x1), U3_ag(x0, x1, member_in_ag(x1)))
SUBSET_IN_AG(.(x0, x1)) → U1_AG(.(x0, x1), member_out_ag(x0, .(x0, x1)))
U1_AG(.(z0, z1), member_out_ag(x1, .(z0, z1))) → SUBSET_IN_AG(.(z0, z1))
U1_AG(.(z0, z1), member_out_ag(z0, .(z0, z1))) → SUBSET_IN_AG(.(z0, z1))

The TRS R consists of the following rules:

member_in_ag(.(X, X1)) → member_out_ag(X, .(X, X1))
member_in_ag(.(X1, Xs)) → U3_ag(X1, Xs, member_in_ag(Xs))
U3_ag(X1, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(X1, Xs))

s = U1_AG(.(z0, z1), member_out_ag(x1', .(z0, z1))) evaluates to t =U1_AG(.(z0, z1), member_out_ag(z0, .(z0, z1)))

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

Matcher: [x1' / z0]
Semiunifier: [ ]

Rewriting sequence

U1_AG(.(z0, z1), member_out_ag(x1', .(z0, z1))) → SUBSET_IN_AG(.(z0, z1))
with rule U1_AG(.(z0', z1'), member_out_ag(x1'', .(z0', z1'))) → SUBSET_IN_AG(.(z0', z1')) at position [] and matcher [x1'' / x1', z1' / z1, z0' / z0]

SUBSET_IN_AG(.(z0, z1)) → U1_AG(.(z0, z1), member_out_ag(z0, .(z0, z1)))
with rule SUBSET_IN_AG(.(x0, x1)) → U1_AG(.(x0, x1), member_out_ag(x0, .(x0, x1)))

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.