Left Termination proof of /tmp/tmp0Jwhw5/subset1.pl

Left Termination of the query pattern subset1(f,b) w.r.t. the given Prolog program could not be shown:

↳ PROLOG

  ↳ UnrequestedClauseRemoverProof

member₂(X, .₂(Y, Xs)) :- member₂(X, Xs).
member₂(X, .₂(X, Xs)).
subset₂(.₂(X, Xs), Ys) :- member₂(X, Ys), subset₂(Xs, Ys).
subset₂({}₀, Ys).
member1₂(X, .₂(Y, Xs)) :- member1₂(X, Xs).
member1₂(X, .₂(X, Xs)).
subset1₂(.₂(X, Xs), Ys) :- member1₂(X, Ys), subset1₂(Xs, Ys).
subset1₂({}₀, Ys).

The clauses

member₂(X, .₂(Y, Xs)) :- member₂(X, Xs).
member₂(X, .₂(X, Xs)).
subset₂(.₂(X, Xs), Ys) :- member₂(X, Ys), subset₂(Xs, Ys).
subset₂({}₀, Ys).

can be ignored, as they are not needed by any of the given querys.

Deleting these clauses results in the following prolog program:

member1₂(X, .₂(Y, Xs)) :- member1₂(X, Xs).
member1₂(X, .₂(X, Xs)).
subset1₂(.₂(X, Xs), Ys) :- member1₂(X, Ys), subset1₂(Xs, Ys).
subset1₂({}₀, Ys).

↳ PROLOG

  ↳ UnrequestedClauseRemoverProof

    ↳ PROLOG

      ↳ PrologToPiTRSProof

      ↳ PrologToPiTRSProof

member1₂(X, .₂(Y, Xs)) :- member1₂(X, Xs).
member1₂(X, .₂(X, Xs)).
subset1₂(.₂(X, Xs), Ys) :- member1₂(X, Ys), subset1₂(Xs, Ys).
subset1₂({}₀, Ys).

With regard to the inferred argument filtering the predicates were used in the following modes:
subset1₂: (f,b)
member1₂: (f,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

subset1_2_in_ag₂(._2₂(X, Xs), Ys) -> if_subset1_2_in_1_ag₄(X, Xs, Ys, member1_2_in_ag₂(X, Ys))
member1_2_in_ag₂(X, ._2₂(Y, Xs)) -> if_member1_2_in_1_ag₄(X, Y, Xs, member1_2_in_ag₂(X, Xs))
member1_2_in_ag₂(X, ._2₂(X, Xs)) -> member1_2_out_ag₂(X, ._2₂(X, Xs))
if_member1_2_in_1_ag₄(X, Y, Xs, member1_2_out_ag₂(X, Xs)) -> member1_2_out_ag₂(X, ._2₂(Y, Xs))
if_subset1_2_in_1_ag₄(X, Xs, Ys, member1_2_out_ag₂(X, Ys)) -> if_subset1_2_in_2_ag₄(X, Xs, Ys, subset1_2_in_ag₂(Xs, Ys))
subset1_2_in_ag₂([]_0, Ys) -> subset1_2_out_ag₂([]_0, Ys)
if_subset1_2_in_2_ag₄(X, Xs, Ys, subset1_2_out_ag₂(Xs, Ys)) -> subset1_2_out_ag₂(._2₂(X, Xs), Ys)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG

↳ PROLOG

  ↳ UnrequestedClauseRemoverProof

    ↳ PROLOG

      ↳ PrologToPiTRSProof

        ↳ PiTRS

          ↳ DependencyPairsProof

      ↳ PrologToPiTRSProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

subset1_2_in_ag₂(._2₂(X, Xs), Ys) -> if_subset1_2_in_1_ag₄(X, Xs, Ys, member1_2_in_ag₂(X, Ys))
member1_2_in_ag₂(X, ._2₂(Y, Xs)) -> if_member1_2_in_1_ag₄(X, Y, Xs, member1_2_in_ag₂(X, Xs))
member1_2_in_ag₂(X, ._2₂(X, Xs)) -> member1_2_out_ag₂(X, ._2₂(X, Xs))
if_member1_2_in_1_ag₄(X, Y, Xs, member1_2_out_ag₂(X, Xs)) -> member1_2_out_ag₂(X, ._2₂(Y, Xs))
if_subset1_2_in_1_ag₄(X, Xs, Ys, member1_2_out_ag₂(X, Ys)) -> if_subset1_2_in_2_ag₄(X, Xs, Ys, subset1_2_in_ag₂(Xs, Ys))
subset1_2_in_ag₂([]_0, Ys) -> subset1_2_out_ag₂([]_0, Ys)
if_subset1_2_in_2_ag₄(X, Xs, Ys, subset1_2_out_ag₂(Xs, Ys)) -> subset1_2_out_ag₂(._2₂(X, Xs), Ys)

SUBSET1_2_IN_AG₂(._2₂(X, Xs), Ys) -> IF_SUBSET1_2_IN_1_AG₄(X, Xs, Ys, member1_2_in_ag₂(X, Ys))
SUBSET1_2_IN_AG₂(._2₂(X, Xs), Ys) -> MEMBER1_2_IN_AG₂(X, Ys)
MEMBER1_2_IN_AG₂(X, ._2₂(Y, Xs)) -> IF_MEMBER1_2_IN_1_AG₄(X, Y, Xs, member1_2_in_ag₂(X, Xs))
MEMBER1_2_IN_AG₂(X, ._2₂(Y, Xs)) -> MEMBER1_2_IN_AG₂(X, Xs)
IF_SUBSET1_2_IN_1_AG₄(X, Xs, Ys, member1_2_out_ag₂(X, Ys)) -> IF_SUBSET1_2_IN_2_AG₄(X, Xs, Ys, subset1_2_in_ag₂(Xs, Ys))
IF_SUBSET1_2_IN_1_AG₄(X, Xs, Ys, member1_2_out_ag₂(X, Ys)) -> SUBSET1_2_IN_AG₂(Xs, Ys)

The TRS R consists of the following rules:

subset1_2_in_ag₂(._2₂(X, Xs), Ys) -> if_subset1_2_in_1_ag₄(X, Xs, Ys, member1_2_in_ag₂(X, Ys))
member1_2_in_ag₂(X, ._2₂(Y, Xs)) -> if_member1_2_in_1_ag₄(X, Y, Xs, member1_2_in_ag₂(X, Xs))
member1_2_in_ag₂(X, ._2₂(X, Xs)) -> member1_2_out_ag₂(X, ._2₂(X, Xs))
if_member1_2_in_1_ag₄(X, Y, Xs, member1_2_out_ag₂(X, Xs)) -> member1_2_out_ag₂(X, ._2₂(Y, Xs))
if_subset1_2_in_1_ag₄(X, Xs, Ys, member1_2_out_ag₂(X, Ys)) -> if_subset1_2_in_2_ag₄(X, Xs, Ys, subset1_2_in_ag₂(Xs, Ys))
subset1_2_in_ag₂([]_0, Ys) -> subset1_2_out_ag₂([]_0, Ys)
if_subset1_2_in_2_ag₄(X, Xs, Ys, subset1_2_out_ag₂(Xs, Ys)) -> subset1_2_out_ag₂(._2₂(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_2_in_ag₂(x1, x2) = subset1_2_in_ag₁(x2)
._2₂(x1, x2) = ._2₂(x1, x2)
[]_0 = []_0
if_subset1_2_in_1_ag₄(x1, x2, x3, x4) = if_subset1_2_in_1_ag₂(x3, x4)
member1_2_in_ag₂(x1, x2) = member1_2_in_ag₁(x2)
if_member1_2_in_1_ag₄(x1, x2, x3, x4) = if_member1_2_in_1_ag₁(x4)
member1_2_out_ag₂(x1, x2) = member1_2_out_ag₁(x1)
if_subset1_2_in_2_ag₄(x1, x2, x3, x4) = if_subset1_2_in_2_ag₂(x1, x4)
subset1_2_out_ag₂(x1, x2) = subset1_2_out_ag₁(x1)
SUBSET1_2_IN_AG₂(x1, x2) = SUBSET1_2_IN_AG₁(x2)
MEMBER1_2_IN_AG₂(x1, x2) = MEMBER1_2_IN_AG₁(x2)
IF_MEMBER1_2_IN_1_AG₄(x1, x2, x3, x4) = IF_MEMBER1_2_IN_1_AG₁(x4)
IF_SUBSET1_2_IN_2_AG₄(x1, x2, x3, x4) = IF_SUBSET1_2_IN_2_AG₂(x1, x4)
IF_SUBSET1_2_IN_1_AG₄(x1, x2, x3, x4) = IF_SUBSET1_2_IN_1_AG₂(x3, x4)

We have to consider all (P,R,Pi)-chains

↳ PROLOG

  ↳ UnrequestedClauseRemoverProof

    ↳ PROLOG

      ↳ PrologToPiTRSProof

        ↳ PiTRS

          ↳ DependencyPairsProof

            ↳ PiDP

              ↳ DependencyGraphProof

      ↳ PrologToPiTRSProof

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

SUBSET1_2_IN_AG₂(._2₂(X, Xs), Ys) -> IF_SUBSET1_2_IN_1_AG₄(X, Xs, Ys, member1_2_in_ag₂(X, Ys))
SUBSET1_2_IN_AG₂(._2₂(X, Xs), Ys) -> MEMBER1_2_IN_AG₂(X, Ys)
MEMBER1_2_IN_AG₂(X, ._2₂(Y, Xs)) -> IF_MEMBER1_2_IN_1_AG₄(X, Y, Xs, member1_2_in_ag₂(X, Xs))
MEMBER1_2_IN_AG₂(X, ._2₂(Y, Xs)) -> MEMBER1_2_IN_AG₂(X, Xs)
IF_SUBSET1_2_IN_1_AG₄(X, Xs, Ys, member1_2_out_ag₂(X, Ys)) -> IF_SUBSET1_2_IN_2_AG₄(X, Xs, Ys, subset1_2_in_ag₂(Xs, Ys))
IF_SUBSET1_2_IN_1_AG₄(X, Xs, Ys, member1_2_out_ag₂(X, Ys)) -> SUBSET1_2_IN_AG₂(Xs, Ys)

The TRS R consists of the following rules:

subset1_2_in_ag₂(._2₂(X, Xs), Ys) -> if_subset1_2_in_1_ag₄(X, Xs, Ys, member1_2_in_ag₂(X, Ys))
member1_2_in_ag₂(X, ._2₂(Y, Xs)) -> if_member1_2_in_1_ag₄(X, Y, Xs, member1_2_in_ag₂(X, Xs))
member1_2_in_ag₂(X, ._2₂(X, Xs)) -> member1_2_out_ag₂(X, ._2₂(X, Xs))
if_member1_2_in_1_ag₄(X, Y, Xs, member1_2_out_ag₂(X, Xs)) -> member1_2_out_ag₂(X, ._2₂(Y, Xs))
if_subset1_2_in_1_ag₄(X, Xs, Ys, member1_2_out_ag₂(X, Ys)) -> if_subset1_2_in_2_ag₄(X, Xs, Ys, subset1_2_in_ag₂(Xs, Ys))
subset1_2_in_ag₂([]_0, Ys) -> subset1_2_out_ag₂([]_0, Ys)
if_subset1_2_in_2_ag₄(X, Xs, Ys, subset1_2_out_ag₂(Xs, Ys)) -> subset1_2_out_ag₂(._2₂(X, Xs), Ys)

↳ PROLOG

  ↳ UnrequestedClauseRemoverProof

    ↳ PROLOG

      ↳ PrologToPiTRSProof

        ↳ PiTRS

          ↳ DependencyPairsProof

            ↳ PiDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ PiDP

                    ↳ UsableRulesProof

                  ↳ PiDP

      ↳ PrologToPiTRSProof

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

MEMBER1_2_IN_AG₂(X, ._2₂(Y, Xs)) -> MEMBER1_2_IN_AG₂(X, Xs)

The TRS R consists of the following rules:

subset1_2_in_ag₂(._2₂(X, Xs), Ys) -> if_subset1_2_in_1_ag₄(X, Xs, Ys, member1_2_in_ag₂(X, Ys))
member1_2_in_ag₂(X, ._2₂(Y, Xs)) -> if_member1_2_in_1_ag₄(X, Y, Xs, member1_2_in_ag₂(X, Xs))
member1_2_in_ag₂(X, ._2₂(X, Xs)) -> member1_2_out_ag₂(X, ._2₂(X, Xs))
if_member1_2_in_1_ag₄(X, Y, Xs, member1_2_out_ag₂(X, Xs)) -> member1_2_out_ag₂(X, ._2₂(Y, Xs))
if_subset1_2_in_1_ag₄(X, Xs, Ys, member1_2_out_ag₂(X, Ys)) -> if_subset1_2_in_2_ag₄(X, Xs, Ys, subset1_2_in_ag₂(Xs, Ys))
subset1_2_in_ag₂([]_0, Ys) -> subset1_2_out_ag₂([]_0, Ys)
if_subset1_2_in_2_ag₄(X, Xs, Ys, subset1_2_out_ag₂(Xs, Ys)) -> subset1_2_out_ag₂(._2₂(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_2_in_ag₂(x1, x2) = subset1_2_in_ag₁(x2)
._2₂(x1, x2) = ._2₂(x1, x2)
[]_0 = []_0
if_subset1_2_in_1_ag₄(x1, x2, x3, x4) = if_subset1_2_in_1_ag₂(x3, x4)
member1_2_in_ag₂(x1, x2) = member1_2_in_ag₁(x2)
if_member1_2_in_1_ag₄(x1, x2, x3, x4) = if_member1_2_in_1_ag₁(x4)
member1_2_out_ag₂(x1, x2) = member1_2_out_ag₁(x1)
if_subset1_2_in_2_ag₄(x1, x2, x3, x4) = if_subset1_2_in_2_ag₂(x1, x4)
subset1_2_out_ag₂(x1, x2) = subset1_2_out_ag₁(x1)
MEMBER1_2_IN_AG₂(x1, x2) = MEMBER1_2_IN_AG₁(x2)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting we can delete all non-usable rules from R.

↳ PROLOG

  ↳ UnrequestedClauseRemoverProof

    ↳ 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_2_IN_AG₂(X, ._2₂(Y, Xs)) -> MEMBER1_2_IN_AG₂(X, Xs)

R is empty.
The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₂(x1, x2)
MEMBER1_2_IN_AG₂(x1, x2) = MEMBER1_2_IN_AG₁(x2)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem by application of Pi.

↳ PROLOG

  ↳ UnrequestedClauseRemoverProof

    ↳ 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_2_IN_AG₁(._2₂(Y, Xs)) -> MEMBER1_2_IN_AG₁(Xs)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {MEMBER1_2_IN_AG₁}.
By using the subterm criterion together with the size-change analysis 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_2_IN_AG₁(._2₂(Y, Xs)) -> MEMBER1_2_IN_AG₁(Xs)
The graph contains the following edges 1 > 1

↳ PROLOG

  ↳ UnrequestedClauseRemoverProof

    ↳ PROLOG

      ↳ PrologToPiTRSProof

        ↳ PiTRS

          ↳ DependencyPairsProof

            ↳ PiDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ PiDP

                  ↳ PiDP

                    ↳ UsableRulesProof

      ↳ PrologToPiTRSProof

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

SUBSET1_2_IN_AG₂(._2₂(X, Xs), Ys) -> IF_SUBSET1_2_IN_1_AG₄(X, Xs, Ys, member1_2_in_ag₂(X, Ys))
IF_SUBSET1_2_IN_1_AG₄(X, Xs, Ys, member1_2_out_ag₂(X, Ys)) -> SUBSET1_2_IN_AG₂(Xs, Ys)

The TRS R consists of the following rules:

subset1_2_in_ag₂(._2₂(X, Xs), Ys) -> if_subset1_2_in_1_ag₄(X, Xs, Ys, member1_2_in_ag₂(X, Ys))
member1_2_in_ag₂(X, ._2₂(Y, Xs)) -> if_member1_2_in_1_ag₄(X, Y, Xs, member1_2_in_ag₂(X, Xs))
member1_2_in_ag₂(X, ._2₂(X, Xs)) -> member1_2_out_ag₂(X, ._2₂(X, Xs))
if_member1_2_in_1_ag₄(X, Y, Xs, member1_2_out_ag₂(X, Xs)) -> member1_2_out_ag₂(X, ._2₂(Y, Xs))
if_subset1_2_in_1_ag₄(X, Xs, Ys, member1_2_out_ag₂(X, Ys)) -> if_subset1_2_in_2_ag₄(X, Xs, Ys, subset1_2_in_ag₂(Xs, Ys))
subset1_2_in_ag₂([]_0, Ys) -> subset1_2_out_ag₂([]_0, Ys)
if_subset1_2_in_2_ag₄(X, Xs, Ys, subset1_2_out_ag₂(Xs, Ys)) -> subset1_2_out_ag₂(._2₂(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_2_in_ag₂(x1, x2) = subset1_2_in_ag₁(x2)
._2₂(x1, x2) = ._2₂(x1, x2)
[]_0 = []_0
if_subset1_2_in_1_ag₄(x1, x2, x3, x4) = if_subset1_2_in_1_ag₂(x3, x4)
member1_2_in_ag₂(x1, x2) = member1_2_in_ag₁(x2)
if_member1_2_in_1_ag₄(x1, x2, x3, x4) = if_member1_2_in_1_ag₁(x4)
member1_2_out_ag₂(x1, x2) = member1_2_out_ag₁(x1)
if_subset1_2_in_2_ag₄(x1, x2, x3, x4) = if_subset1_2_in_2_ag₂(x1, x4)
subset1_2_out_ag₂(x1, x2) = subset1_2_out_ag₁(x1)
SUBSET1_2_IN_AG₂(x1, x2) = SUBSET1_2_IN_AG₁(x2)
IF_SUBSET1_2_IN_1_AG₄(x1, x2, x3, x4) = IF_SUBSET1_2_IN_1_AG₂(x3, x4)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting we can delete all non-usable rules from R.

↳ PROLOG

  ↳ UnrequestedClauseRemoverProof

    ↳ 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_2_IN_AG₂(._2₂(X, Xs), Ys) -> IF_SUBSET1_2_IN_1_AG₄(X, Xs, Ys, member1_2_in_ag₂(X, Ys))
IF_SUBSET1_2_IN_1_AG₄(X, Xs, Ys, member1_2_out_ag₂(X, Ys)) -> SUBSET1_2_IN_AG₂(Xs, Ys)

The TRS R consists of the following rules:

member1_2_in_ag₂(X, ._2₂(Y, Xs)) -> if_member1_2_in_1_ag₄(X, Y, Xs, member1_2_in_ag₂(X, Xs))
member1_2_in_ag₂(X, ._2₂(X, Xs)) -> member1_2_out_ag₂(X, ._2₂(X, Xs))
if_member1_2_in_1_ag₄(X, Y, Xs, member1_2_out_ag₂(X, Xs)) -> member1_2_out_ag₂(X, ._2₂(Y, Xs))

The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₂(x1, x2)
member1_2_in_ag₂(x1, x2) = member1_2_in_ag₁(x2)
if_member1_2_in_1_ag₄(x1, x2, x3, x4) = if_member1_2_in_1_ag₁(x4)
member1_2_out_ag₂(x1, x2) = member1_2_out_ag₁(x1)
SUBSET1_2_IN_AG₂(x1, x2) = SUBSET1_2_IN_AG₁(x2)
IF_SUBSET1_2_IN_1_AG₄(x1, x2, x3, x4) = IF_SUBSET1_2_IN_1_AG₂(x3, x4)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem by application of Pi.

↳ PROLOG

  ↳ UnrequestedClauseRemoverProof

    ↳ PROLOG

      ↳ PrologToPiTRSProof

        ↳ PiTRS

          ↳ DependencyPairsProof

            ↳ PiDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ PiDP

                  ↳ PiDP

                    ↳ UsableRulesProof

                      ↳ PiDP

                        ↳ PiDPToQDPProof

                          ↳ QDP

      ↳ PrologToPiTRSProof

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

SUBSET1_2_IN_AG₁(Ys) -> IF_SUBSET1_2_IN_1_AG₂(Ys, member1_2_in_ag₁(Ys))
IF_SUBSET1_2_IN_1_AG₂(Ys, member1_2_out_ag₁(X)) -> SUBSET1_2_IN_AG₁(Ys)

The TRS R consists of the following rules:

member1_2_in_ag₁(._2₂(Y, Xs)) -> if_member1_2_in_1_ag₁(member1_2_in_ag₁(Xs))
member1_2_in_ag₁(._2₂(X, Xs)) -> member1_2_out_ag₁(X)
if_member1_2_in_1_ag₁(member1_2_out_ag₁(X)) -> member1_2_out_ag₁(X)

The set Q consists of the following terms:

member1_2_in_ag₁(x0)
if_member1_2_in_1_ag₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_SUBSET1_2_IN_1_AG₂, SUBSET1_2_IN_AG₁}.
With regard to the inferred argument filtering the predicates were used in the following modes:
subset1₂: (f,b)
member1₂: (f,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

subset1_2_in_ag₂(._2₂(X, Xs), Ys) -> if_subset1_2_in_1_ag₄(X, Xs, Ys, member1_2_in_ag₂(X, Ys))
member1_2_in_ag₂(X, ._2₂(Y, Xs)) -> if_member1_2_in_1_ag₄(X, Y, Xs, member1_2_in_ag₂(X, Xs))
member1_2_in_ag₂(X, ._2₂(X, Xs)) -> member1_2_out_ag₂(X, ._2₂(X, Xs))
if_member1_2_in_1_ag₄(X, Y, Xs, member1_2_out_ag₂(X, Xs)) -> member1_2_out_ag₂(X, ._2₂(Y, Xs))
if_subset1_2_in_1_ag₄(X, Xs, Ys, member1_2_out_ag₂(X, Ys)) -> if_subset1_2_in_2_ag₄(X, Xs, Ys, subset1_2_in_ag₂(Xs, Ys))
subset1_2_in_ag₂([]_0, Ys) -> subset1_2_out_ag₂([]_0, Ys)
if_subset1_2_in_2_ag₄(X, Xs, Ys, subset1_2_out_ag₂(Xs, Ys)) -> subset1_2_out_ag₂(._2₂(X, Xs), Ys)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG

↳ PROLOG

  ↳ UnrequestedClauseRemoverProof

    ↳ PROLOG

      ↳ PrologToPiTRSProof

      ↳ PrologToPiTRSProof

        ↳ PiTRS

          ↳ DependencyPairsProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

subset1_2_in_ag₂(._2₂(X, Xs), Ys) -> if_subset1_2_in_1_ag₄(X, Xs, Ys, member1_2_in_ag₂(X, Ys))
member1_2_in_ag₂(X, ._2₂(Y, Xs)) -> if_member1_2_in_1_ag₄(X, Y, Xs, member1_2_in_ag₂(X, Xs))
member1_2_in_ag₂(X, ._2₂(X, Xs)) -> member1_2_out_ag₂(X, ._2₂(X, Xs))
if_member1_2_in_1_ag₄(X, Y, Xs, member1_2_out_ag₂(X, Xs)) -> member1_2_out_ag₂(X, ._2₂(Y, Xs))
if_subset1_2_in_1_ag₄(X, Xs, Ys, member1_2_out_ag₂(X, Ys)) -> if_subset1_2_in_2_ag₄(X, Xs, Ys, subset1_2_in_ag₂(Xs, Ys))
subset1_2_in_ag₂([]_0, Ys) -> subset1_2_out_ag₂([]_0, Ys)
if_subset1_2_in_2_ag₄(X, Xs, Ys, subset1_2_out_ag₂(Xs, Ys)) -> subset1_2_out_ag₂(._2₂(X, Xs), Ys)

SUBSET1_2_IN_AG₂(._2₂(X, Xs), Ys) -> IF_SUBSET1_2_IN_1_AG₄(X, Xs, Ys, member1_2_in_ag₂(X, Ys))
SUBSET1_2_IN_AG₂(._2₂(X, Xs), Ys) -> MEMBER1_2_IN_AG₂(X, Ys)
MEMBER1_2_IN_AG₂(X, ._2₂(Y, Xs)) -> IF_MEMBER1_2_IN_1_AG₄(X, Y, Xs, member1_2_in_ag₂(X, Xs))
MEMBER1_2_IN_AG₂(X, ._2₂(Y, Xs)) -> MEMBER1_2_IN_AG₂(X, Xs)
IF_SUBSET1_2_IN_1_AG₄(X, Xs, Ys, member1_2_out_ag₂(X, Ys)) -> IF_SUBSET1_2_IN_2_AG₄(X, Xs, Ys, subset1_2_in_ag₂(Xs, Ys))
IF_SUBSET1_2_IN_1_AG₄(X, Xs, Ys, member1_2_out_ag₂(X, Ys)) -> SUBSET1_2_IN_AG₂(Xs, Ys)

The TRS R consists of the following rules:

subset1_2_in_ag₂(._2₂(X, Xs), Ys) -> if_subset1_2_in_1_ag₄(X, Xs, Ys, member1_2_in_ag₂(X, Ys))
member1_2_in_ag₂(X, ._2₂(Y, Xs)) -> if_member1_2_in_1_ag₄(X, Y, Xs, member1_2_in_ag₂(X, Xs))
member1_2_in_ag₂(X, ._2₂(X, Xs)) -> member1_2_out_ag₂(X, ._2₂(X, Xs))
if_member1_2_in_1_ag₄(X, Y, Xs, member1_2_out_ag₂(X, Xs)) -> member1_2_out_ag₂(X, ._2₂(Y, Xs))
if_subset1_2_in_1_ag₄(X, Xs, Ys, member1_2_out_ag₂(X, Ys)) -> if_subset1_2_in_2_ag₄(X, Xs, Ys, subset1_2_in_ag₂(Xs, Ys))
subset1_2_in_ag₂([]_0, Ys) -> subset1_2_out_ag₂([]_0, Ys)
if_subset1_2_in_2_ag₄(X, Xs, Ys, subset1_2_out_ag₂(Xs, Ys)) -> subset1_2_out_ag₂(._2₂(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_2_in_ag₂(x1, x2) = subset1_2_in_ag₁(x2)
._2₂(x1, x2) = ._2₂(x1, x2)
[]_0 = []_0
if_subset1_2_in_1_ag₄(x1, x2, x3, x4) = if_subset1_2_in_1_ag₂(x3, x4)
member1_2_in_ag₂(x1, x2) = member1_2_in_ag₁(x2)
if_member1_2_in_1_ag₄(x1, x2, x3, x4) = if_member1_2_in_1_ag₃(x2, x3, x4)
member1_2_out_ag₂(x1, x2) = member1_2_out_ag₂(x1, x2)
if_subset1_2_in_2_ag₄(x1, x2, x3, x4) = if_subset1_2_in_2_ag₃(x1, x3, x4)
subset1_2_out_ag₂(x1, x2) = subset1_2_out_ag₂(x1, x2)
SUBSET1_2_IN_AG₂(x1, x2) = SUBSET1_2_IN_AG₁(x2)
MEMBER1_2_IN_AG₂(x1, x2) = MEMBER1_2_IN_AG₁(x2)
IF_MEMBER1_2_IN_1_AG₄(x1, x2, x3, x4) = IF_MEMBER1_2_IN_1_AG₃(x2, x3, x4)
IF_SUBSET1_2_IN_2_AG₄(x1, x2, x3, x4) = IF_SUBSET1_2_IN_2_AG₃(x1, x3, x4)
IF_SUBSET1_2_IN_1_AG₄(x1, x2, x3, x4) = IF_SUBSET1_2_IN_1_AG₂(x3, x4)

We have to consider all (P,R,Pi)-chains

↳ PROLOG

  ↳ UnrequestedClauseRemoverProof

    ↳ PROLOG

      ↳ PrologToPiTRSProof

      ↳ PrologToPiTRSProof

        ↳ PiTRS

          ↳ DependencyPairsProof

            ↳ PiDP

              ↳ DependencyGraphProof

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

SUBSET1_2_IN_AG₂(._2₂(X, Xs), Ys) -> IF_SUBSET1_2_IN_1_AG₄(X, Xs, Ys, member1_2_in_ag₂(X, Ys))
SUBSET1_2_IN_AG₂(._2₂(X, Xs), Ys) -> MEMBER1_2_IN_AG₂(X, Ys)
MEMBER1_2_IN_AG₂(X, ._2₂(Y, Xs)) -> IF_MEMBER1_2_IN_1_AG₄(X, Y, Xs, member1_2_in_ag₂(X, Xs))
MEMBER1_2_IN_AG₂(X, ._2₂(Y, Xs)) -> MEMBER1_2_IN_AG₂(X, Xs)
IF_SUBSET1_2_IN_1_AG₄(X, Xs, Ys, member1_2_out_ag₂(X, Ys)) -> IF_SUBSET1_2_IN_2_AG₄(X, Xs, Ys, subset1_2_in_ag₂(Xs, Ys))
IF_SUBSET1_2_IN_1_AG₄(X, Xs, Ys, member1_2_out_ag₂(X, Ys)) -> SUBSET1_2_IN_AG₂(Xs, Ys)

The TRS R consists of the following rules:

subset1_2_in_ag₂(._2₂(X, Xs), Ys) -> if_subset1_2_in_1_ag₄(X, Xs, Ys, member1_2_in_ag₂(X, Ys))
member1_2_in_ag₂(X, ._2₂(Y, Xs)) -> if_member1_2_in_1_ag₄(X, Y, Xs, member1_2_in_ag₂(X, Xs))
member1_2_in_ag₂(X, ._2₂(X, Xs)) -> member1_2_out_ag₂(X, ._2₂(X, Xs))
if_member1_2_in_1_ag₄(X, Y, Xs, member1_2_out_ag₂(X, Xs)) -> member1_2_out_ag₂(X, ._2₂(Y, Xs))
if_subset1_2_in_1_ag₄(X, Xs, Ys, member1_2_out_ag₂(X, Ys)) -> if_subset1_2_in_2_ag₄(X, Xs, Ys, subset1_2_in_ag₂(Xs, Ys))
subset1_2_in_ag₂([]_0, Ys) -> subset1_2_out_ag₂([]_0, Ys)
if_subset1_2_in_2_ag₄(X, Xs, Ys, subset1_2_out_ag₂(Xs, Ys)) -> subset1_2_out_ag₂(._2₂(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_2_in_ag₂(x1, x2) = subset1_2_in_ag₁(x2)
._2₂(x1, x2) = ._2₂(x1, x2)
[]_0 = []_0
if_subset1_2_in_1_ag₄(x1, x2, x3, x4) = if_subset1_2_in_1_ag₂(x3, x4)
member1_2_in_ag₂(x1, x2) = member1_2_in_ag₁(x2)
if_member1_2_in_1_ag₄(x1, x2, x3, x4) = if_member1_2_in_1_ag₃(x2, x3, x4)
member1_2_out_ag₂(x1, x2) = member1_2_out_ag₂(x1, x2)
if_subset1_2_in_2_ag₄(x1, x2, x3, x4) = if_subset1_2_in_2_ag₃(x1, x3, x4)
subset1_2_out_ag₂(x1, x2) = subset1_2_out_ag₂(x1, x2)
SUBSET1_2_IN_AG₂(x1, x2) = SUBSET1_2_IN_AG₁(x2)
MEMBER1_2_IN_AG₂(x1, x2) = MEMBER1_2_IN_AG₁(x2)
IF_MEMBER1_2_IN_1_AG₄(x1, x2, x3, x4) = IF_MEMBER1_2_IN_1_AG₃(x2, x3, x4)
IF_SUBSET1_2_IN_2_AG₄(x1, x2, x3, x4) = IF_SUBSET1_2_IN_2_AG₃(x1, x3, x4)
IF_SUBSET1_2_IN_1_AG₄(x1, x2, x3, x4) = IF_SUBSET1_2_IN_1_AG₂(x3, x4)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 2 SCCs with 3 less nodes.

↳ PROLOG

  ↳ UnrequestedClauseRemoverProof

    ↳ PROLOG

      ↳ PrologToPiTRSProof

      ↳ PrologToPiTRSProof

        ↳ PiTRS

          ↳ DependencyPairsProof

            ↳ PiDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ PiDP

                    ↳ UsableRulesProof

                  ↳ PiDP

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

MEMBER1_2_IN_AG₂(X, ._2₂(Y, Xs)) -> MEMBER1_2_IN_AG₂(X, Xs)

The TRS R consists of the following rules:

subset1_2_in_ag₂(._2₂(X, Xs), Ys) -> if_subset1_2_in_1_ag₄(X, Xs, Ys, member1_2_in_ag₂(X, Ys))
member1_2_in_ag₂(X, ._2₂(Y, Xs)) -> if_member1_2_in_1_ag₄(X, Y, Xs, member1_2_in_ag₂(X, Xs))
member1_2_in_ag₂(X, ._2₂(X, Xs)) -> member1_2_out_ag₂(X, ._2₂(X, Xs))
if_member1_2_in_1_ag₄(X, Y, Xs, member1_2_out_ag₂(X, Xs)) -> member1_2_out_ag₂(X, ._2₂(Y, Xs))
if_subset1_2_in_1_ag₄(X, Xs, Ys, member1_2_out_ag₂(X, Ys)) -> if_subset1_2_in_2_ag₄(X, Xs, Ys, subset1_2_in_ag₂(Xs, Ys))
subset1_2_in_ag₂([]_0, Ys) -> subset1_2_out_ag₂([]_0, Ys)
if_subset1_2_in_2_ag₄(X, Xs, Ys, subset1_2_out_ag₂(Xs, Ys)) -> subset1_2_out_ag₂(._2₂(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_2_in_ag₂(x1, x2) = subset1_2_in_ag₁(x2)
._2₂(x1, x2) = ._2₂(x1, x2)
[]_0 = []_0
if_subset1_2_in_1_ag₄(x1, x2, x3, x4) = if_subset1_2_in_1_ag₂(x3, x4)
member1_2_in_ag₂(x1, x2) = member1_2_in_ag₁(x2)
if_member1_2_in_1_ag₄(x1, x2, x3, x4) = if_member1_2_in_1_ag₃(x2, x3, x4)
member1_2_out_ag₂(x1, x2) = member1_2_out_ag₂(x1, x2)
if_subset1_2_in_2_ag₄(x1, x2, x3, x4) = if_subset1_2_in_2_ag₃(x1, x3, x4)
subset1_2_out_ag₂(x1, x2) = subset1_2_out_ag₂(x1, x2)
MEMBER1_2_IN_AG₂(x1, x2) = MEMBER1_2_IN_AG₁(x2)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting we can delete all non-usable rules from R.

↳ PROLOG

  ↳ UnrequestedClauseRemoverProof

    ↳ PROLOG

      ↳ PrologToPiTRSProof

      ↳ PrologToPiTRSProof

        ↳ PiTRS

          ↳ DependencyPairsProof

            ↳ PiDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ PiDP

                    ↳ UsableRulesProof

                      ↳ PiDP

                  ↳ PiDP

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

MEMBER1_2_IN_AG₂(X, ._2₂(Y, Xs)) -> MEMBER1_2_IN_AG₂(X, Xs)

R is empty.
The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₂(x1, x2)
MEMBER1_2_IN_AG₂(x1, x2) = MEMBER1_2_IN_AG₁(x2)

We have to consider all (P,R,Pi)-chains

↳ PROLOG

  ↳ UnrequestedClauseRemoverProof

    ↳ PROLOG

      ↳ PrologToPiTRSProof

      ↳ PrologToPiTRSProof

        ↳ PiTRS

          ↳ DependencyPairsProof

            ↳ PiDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ PiDP

                  ↳ PiDP

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

SUBSET1_2_IN_AG₂(._2₂(X, Xs), Ys) -> IF_SUBSET1_2_IN_1_AG₄(X, Xs, Ys, member1_2_in_ag₂(X, Ys))
IF_SUBSET1_2_IN_1_AG₄(X, Xs, Ys, member1_2_out_ag₂(X, Ys)) -> SUBSET1_2_IN_AG₂(Xs, Ys)

The TRS R consists of the following rules:

subset1_2_in_ag₂(._2₂(X, Xs), Ys) -> if_subset1_2_in_1_ag₄(X, Xs, Ys, member1_2_in_ag₂(X, Ys))
member1_2_in_ag₂(X, ._2₂(Y, Xs)) -> if_member1_2_in_1_ag₄(X, Y, Xs, member1_2_in_ag₂(X, Xs))
member1_2_in_ag₂(X, ._2₂(X, Xs)) -> member1_2_out_ag₂(X, ._2₂(X, Xs))
if_member1_2_in_1_ag₄(X, Y, Xs, member1_2_out_ag₂(X, Xs)) -> member1_2_out_ag₂(X, ._2₂(Y, Xs))
if_subset1_2_in_1_ag₄(X, Xs, Ys, member1_2_out_ag₂(X, Ys)) -> if_subset1_2_in_2_ag₄(X, Xs, Ys, subset1_2_in_ag₂(Xs, Ys))
subset1_2_in_ag₂([]_0, Ys) -> subset1_2_out_ag₂([]_0, Ys)
if_subset1_2_in_2_ag₄(X, Xs, Ys, subset1_2_out_ag₂(Xs, Ys)) -> subset1_2_out_ag₂(._2₂(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_2_in_ag₂(x1, x2) = subset1_2_in_ag₁(x2)
._2₂(x1, x2) = ._2₂(x1, x2)
[]_0 = []_0
if_subset1_2_in_1_ag₄(x1, x2, x3, x4) = if_subset1_2_in_1_ag₂(x3, x4)
member1_2_in_ag₂(x1, x2) = member1_2_in_ag₁(x2)
if_member1_2_in_1_ag₄(x1, x2, x3, x4) = if_member1_2_in_1_ag₃(x2, x3, x4)
member1_2_out_ag₂(x1, x2) = member1_2_out_ag₂(x1, x2)
if_subset1_2_in_2_ag₄(x1, x2, x3, x4) = if_subset1_2_in_2_ag₃(x1, x3, x4)
subset1_2_out_ag₂(x1, x2) = subset1_2_out_ag₂(x1, x2)
SUBSET1_2_IN_AG₂(x1, x2) = SUBSET1_2_IN_AG₁(x2)
IF_SUBSET1_2_IN_1_AG₄(x1, x2, x3, x4) = IF_SUBSET1_2_IN_1_AG₂(x3, x4)

We have to consider all (P,R,Pi)-chains