Left Termination proof of /tmp/tmpfAYjJK/subset.pl

Left Termination of the query pattern subset(b,b) w.r.t. the given Prolog program could successfully be proven:

↳ 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

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).

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

Deleting these clauses results in the following prolog program:

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).

↳ PROLOG

  ↳ UnrequestedClauseRemoverProof

    ↳ PROLOG

      ↳ PrologToPiTRSProof

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).

With regard to the inferred argument filtering the predicates were used in the following modes:
subset₂: (b,b)
member₂: (b,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

subset_2_in_gg₂(._2₂(X, Xs), Ys) -> if_subset_2_in_1_gg₄(X, Xs, Ys, member_2_in_gg₂(X, Ys))
member_2_in_gg₂(X, ._2₂(Y, Xs)) -> if_member_2_in_1_gg₄(X, Y, Xs, member_2_in_gg₂(X, Xs))
member_2_in_gg₂(X, ._2₂(X, Xs)) -> member_2_out_gg₂(X, ._2₂(X, Xs))
if_member_2_in_1_gg₄(X, Y, Xs, member_2_out_gg₂(X, Xs)) -> member_2_out_gg₂(X, ._2₂(Y, Xs))
if_subset_2_in_1_gg₄(X, Xs, Ys, member_2_out_gg₂(X, Ys)) -> if_subset_2_in_2_gg₄(X, Xs, Ys, subset_2_in_gg₂(Xs, Ys))
subset_2_in_gg₂([]_0, Ys) -> subset_2_out_gg₂([]_0, Ys)
if_subset_2_in_2_gg₄(X, Xs, Ys, subset_2_out_gg₂(Xs, Ys)) -> subset_2_out_gg₂(._2₂(X, Xs), Ys)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG

↳ PROLOG

  ↳ UnrequestedClauseRemoverProof

    ↳ PROLOG

      ↳ PrologToPiTRSProof

        ↳ PiTRS

          ↳ DependencyPairsProof

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

subset_2_in_gg₂(._2₂(X, Xs), Ys) -> if_subset_2_in_1_gg₄(X, Xs, Ys, member_2_in_gg₂(X, Ys))
member_2_in_gg₂(X, ._2₂(Y, Xs)) -> if_member_2_in_1_gg₄(X, Y, Xs, member_2_in_gg₂(X, Xs))
member_2_in_gg₂(X, ._2₂(X, Xs)) -> member_2_out_gg₂(X, ._2₂(X, Xs))
if_member_2_in_1_gg₄(X, Y, Xs, member_2_out_gg₂(X, Xs)) -> member_2_out_gg₂(X, ._2₂(Y, Xs))
if_subset_2_in_1_gg₄(X, Xs, Ys, member_2_out_gg₂(X, Ys)) -> if_subset_2_in_2_gg₄(X, Xs, Ys, subset_2_in_gg₂(Xs, Ys))
subset_2_in_gg₂([]_0, Ys) -> subset_2_out_gg₂([]_0, Ys)
if_subset_2_in_2_gg₄(X, Xs, Ys, subset_2_out_gg₂(Xs, Ys)) -> subset_2_out_gg₂(._2₂(X, Xs), Ys)

SUBSET_2_IN_GG₂(._2₂(X, Xs), Ys) -> IF_SUBSET_2_IN_1_GG₄(X, Xs, Ys, member_2_in_gg₂(X, Ys))
SUBSET_2_IN_GG₂(._2₂(X, Xs), Ys) -> MEMBER_2_IN_GG₂(X, Ys)
MEMBER_2_IN_GG₂(X, ._2₂(Y, Xs)) -> IF_MEMBER_2_IN_1_GG₄(X, Y, Xs, member_2_in_gg₂(X, Xs))
MEMBER_2_IN_GG₂(X, ._2₂(Y, Xs)) -> MEMBER_2_IN_GG₂(X, Xs)
IF_SUBSET_2_IN_1_GG₄(X, Xs, Ys, member_2_out_gg₂(X, Ys)) -> IF_SUBSET_2_IN_2_GG₄(X, Xs, Ys, subset_2_in_gg₂(Xs, Ys))
IF_SUBSET_2_IN_1_GG₄(X, Xs, Ys, member_2_out_gg₂(X, Ys)) -> SUBSET_2_IN_GG₂(Xs, Ys)

The TRS R consists of the following rules:

subset_2_in_gg₂(._2₂(X, Xs), Ys) -> if_subset_2_in_1_gg₄(X, Xs, Ys, member_2_in_gg₂(X, Ys))
member_2_in_gg₂(X, ._2₂(Y, Xs)) -> if_member_2_in_1_gg₄(X, Y, Xs, member_2_in_gg₂(X, Xs))
member_2_in_gg₂(X, ._2₂(X, Xs)) -> member_2_out_gg₂(X, ._2₂(X, Xs))
if_member_2_in_1_gg₄(X, Y, Xs, member_2_out_gg₂(X, Xs)) -> member_2_out_gg₂(X, ._2₂(Y, Xs))
if_subset_2_in_1_gg₄(X, Xs, Ys, member_2_out_gg₂(X, Ys)) -> if_subset_2_in_2_gg₄(X, Xs, Ys, subset_2_in_gg₂(Xs, Ys))
subset_2_in_gg₂([]_0, Ys) -> subset_2_out_gg₂([]_0, Ys)
if_subset_2_in_2_gg₄(X, Xs, Ys, subset_2_out_gg₂(Xs, Ys)) -> subset_2_out_gg₂(._2₂(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_2_in_gg₂(x1, x2) = subset_2_in_gg₂(x1, x2)
._2₂(x1, x2) = ._2₂(x1, x2)
[]_0 = []_0
if_subset_2_in_1_gg₄(x1, x2, x3, x4) = if_subset_2_in_1_gg₃(x2, x3, x4)
member_2_in_gg₂(x1, x2) = member_2_in_gg₂(x1, x2)
if_member_2_in_1_gg₄(x1, x2, x3, x4) = if_member_2_in_1_gg₁(x4)
member_2_out_gg₂(x1, x2) = member_2_out_gg
if_subset_2_in_2_gg₄(x1, x2, x3, x4) = if_subset_2_in_2_gg₁(x4)
subset_2_out_gg₂(x1, x2) = subset_2_out_gg
IF_MEMBER_2_IN_1_GG₄(x1, x2, x3, x4) = IF_MEMBER_2_IN_1_GG₁(x4)
IF_SUBSET_2_IN_1_GG₄(x1, x2, x3, x4) = IF_SUBSET_2_IN_1_GG₃(x2, x3, x4)
MEMBER_2_IN_GG₂(x1, x2) = MEMBER_2_IN_GG₂(x1, x2)
SUBSET_2_IN_GG₂(x1, x2) = SUBSET_2_IN_GG₂(x1, x2)
IF_SUBSET_2_IN_2_GG₄(x1, x2, x3, x4) = IF_SUBSET_2_IN_2_GG₁(x4)

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

↳ PROLOG

  ↳ UnrequestedClauseRemoverProof

    ↳ PROLOG

      ↳ PrologToPiTRSProof

        ↳ PiTRS

          ↳ DependencyPairsProof

            ↳ PiDP

              ↳ DependencyGraphProof

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

SUBSET_2_IN_GG₂(._2₂(X, Xs), Ys) -> IF_SUBSET_2_IN_1_GG₄(X, Xs, Ys, member_2_in_gg₂(X, Ys))
SUBSET_2_IN_GG₂(._2₂(X, Xs), Ys) -> MEMBER_2_IN_GG₂(X, Ys)
MEMBER_2_IN_GG₂(X, ._2₂(Y, Xs)) -> IF_MEMBER_2_IN_1_GG₄(X, Y, Xs, member_2_in_gg₂(X, Xs))
MEMBER_2_IN_GG₂(X, ._2₂(Y, Xs)) -> MEMBER_2_IN_GG₂(X, Xs)
IF_SUBSET_2_IN_1_GG₄(X, Xs, Ys, member_2_out_gg₂(X, Ys)) -> IF_SUBSET_2_IN_2_GG₄(X, Xs, Ys, subset_2_in_gg₂(Xs, Ys))
IF_SUBSET_2_IN_1_GG₄(X, Xs, Ys, member_2_out_gg₂(X, Ys)) -> SUBSET_2_IN_GG₂(Xs, Ys)

The TRS R consists of the following rules:

subset_2_in_gg₂(._2₂(X, Xs), Ys) -> if_subset_2_in_1_gg₄(X, Xs, Ys, member_2_in_gg₂(X, Ys))
member_2_in_gg₂(X, ._2₂(Y, Xs)) -> if_member_2_in_1_gg₄(X, Y, Xs, member_2_in_gg₂(X, Xs))
member_2_in_gg₂(X, ._2₂(X, Xs)) -> member_2_out_gg₂(X, ._2₂(X, Xs))
if_member_2_in_1_gg₄(X, Y, Xs, member_2_out_gg₂(X, Xs)) -> member_2_out_gg₂(X, ._2₂(Y, Xs))
if_subset_2_in_1_gg₄(X, Xs, Ys, member_2_out_gg₂(X, Ys)) -> if_subset_2_in_2_gg₄(X, Xs, Ys, subset_2_in_gg₂(Xs, Ys))
subset_2_in_gg₂([]_0, Ys) -> subset_2_out_gg₂([]_0, Ys)
if_subset_2_in_2_gg₄(X, Xs, Ys, subset_2_out_gg₂(Xs, Ys)) -> subset_2_out_gg₂(._2₂(X, Xs), Ys)

↳ PROLOG

  ↳ UnrequestedClauseRemoverProof

    ↳ PROLOG

      ↳ PrologToPiTRSProof

        ↳ PiTRS

          ↳ DependencyPairsProof

            ↳ PiDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ PiDP

                    ↳ UsableRulesProof

                  ↳ PiDP

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

MEMBER_2_IN_GG₂(X, ._2₂(Y, Xs)) -> MEMBER_2_IN_GG₂(X, Xs)

The TRS R consists of the following rules:

subset_2_in_gg₂(._2₂(X, Xs), Ys) -> if_subset_2_in_1_gg₄(X, Xs, Ys, member_2_in_gg₂(X, Ys))
member_2_in_gg₂(X, ._2₂(Y, Xs)) -> if_member_2_in_1_gg₄(X, Y, Xs, member_2_in_gg₂(X, Xs))
member_2_in_gg₂(X, ._2₂(X, Xs)) -> member_2_out_gg₂(X, ._2₂(X, Xs))
if_member_2_in_1_gg₄(X, Y, Xs, member_2_out_gg₂(X, Xs)) -> member_2_out_gg₂(X, ._2₂(Y, Xs))
if_subset_2_in_1_gg₄(X, Xs, Ys, member_2_out_gg₂(X, Ys)) -> if_subset_2_in_2_gg₄(X, Xs, Ys, subset_2_in_gg₂(Xs, Ys))
subset_2_in_gg₂([]_0, Ys) -> subset_2_out_gg₂([]_0, Ys)
if_subset_2_in_2_gg₄(X, Xs, Ys, subset_2_out_gg₂(Xs, Ys)) -> subset_2_out_gg₂(._2₂(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_2_in_gg₂(x1, x2) = subset_2_in_gg₂(x1, x2)
._2₂(x1, x2) = ._2₂(x1, x2)
[]_0 = []_0
if_subset_2_in_1_gg₄(x1, x2, x3, x4) = if_subset_2_in_1_gg₃(x2, x3, x4)
member_2_in_gg₂(x1, x2) = member_2_in_gg₂(x1, x2)
if_member_2_in_1_gg₄(x1, x2, x3, x4) = if_member_2_in_1_gg₁(x4)
member_2_out_gg₂(x1, x2) = member_2_out_gg
if_subset_2_in_2_gg₄(x1, x2, x3, x4) = if_subset_2_in_2_gg₁(x4)
subset_2_out_gg₂(x1, x2) = subset_2_out_gg
MEMBER_2_IN_GG₂(x1, x2) = MEMBER_2_IN_GG₂(x1, 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

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

MEMBER_2_IN_GG₂(X, ._2₂(Y, Xs)) -> MEMBER_2_IN_GG₂(X, Xs)

R is empty.
Pi is empty.
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

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

MEMBER_2_IN_GG₂(X, ._2₂(Y, Xs)) -> MEMBER_2_IN_GG₂(X, Xs)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {MEMBER_2_IN_GG₂}.
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:

MEMBER_2_IN_GG₂(X, ._2₂(Y, Xs)) -> MEMBER_2_IN_GG₂(X, Xs)
The graph contains the following edges 1 >= 1, 2 > 2

↳ PROLOG

  ↳ UnrequestedClauseRemoverProof

    ↳ PROLOG

      ↳ PrologToPiTRSProof

        ↳ PiTRS

          ↳ DependencyPairsProof

            ↳ PiDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ PiDP

                  ↳ PiDP

                    ↳ UsableRulesProof

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

IF_SUBSET_2_IN_1_GG₄(X, Xs, Ys, member_2_out_gg₂(X, Ys)) -> SUBSET_2_IN_GG₂(Xs, Ys)
SUBSET_2_IN_GG₂(._2₂(X, Xs), Ys) -> IF_SUBSET_2_IN_1_GG₄(X, Xs, Ys, member_2_in_gg₂(X, Ys))

The TRS R consists of the following rules:

subset_2_in_gg₂(._2₂(X, Xs), Ys) -> if_subset_2_in_1_gg₄(X, Xs, Ys, member_2_in_gg₂(X, Ys))
member_2_in_gg₂(X, ._2₂(Y, Xs)) -> if_member_2_in_1_gg₄(X, Y, Xs, member_2_in_gg₂(X, Xs))
member_2_in_gg₂(X, ._2₂(X, Xs)) -> member_2_out_gg₂(X, ._2₂(X, Xs))
if_member_2_in_1_gg₄(X, Y, Xs, member_2_out_gg₂(X, Xs)) -> member_2_out_gg₂(X, ._2₂(Y, Xs))
if_subset_2_in_1_gg₄(X, Xs, Ys, member_2_out_gg₂(X, Ys)) -> if_subset_2_in_2_gg₄(X, Xs, Ys, subset_2_in_gg₂(Xs, Ys))
subset_2_in_gg₂([]_0, Ys) -> subset_2_out_gg₂([]_0, Ys)
if_subset_2_in_2_gg₄(X, Xs, Ys, subset_2_out_gg₂(Xs, Ys)) -> subset_2_out_gg₂(._2₂(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_2_in_gg₂(x1, x2) = subset_2_in_gg₂(x1, x2)
._2₂(x1, x2) = ._2₂(x1, x2)
[]_0 = []_0
if_subset_2_in_1_gg₄(x1, x2, x3, x4) = if_subset_2_in_1_gg₃(x2, x3, x4)
member_2_in_gg₂(x1, x2) = member_2_in_gg₂(x1, x2)
if_member_2_in_1_gg₄(x1, x2, x3, x4) = if_member_2_in_1_gg₁(x4)
member_2_out_gg₂(x1, x2) = member_2_out_gg
if_subset_2_in_2_gg₄(x1, x2, x3, x4) = if_subset_2_in_2_gg₁(x4)
subset_2_out_gg₂(x1, x2) = subset_2_out_gg
IF_SUBSET_2_IN_1_GG₄(x1, x2, x3, x4) = IF_SUBSET_2_IN_1_GG₃(x2, x3, x4)
SUBSET_2_IN_GG₂(x1, x2) = SUBSET_2_IN_GG₂(x1, 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

                  ↳ PiDP

                    ↳ UsableRulesProof

                      ↳ PiDP

                        ↳ PiDPToQDPProof

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

IF_SUBSET_2_IN_1_GG₄(X, Xs, Ys, member_2_out_gg₂(X, Ys)) -> SUBSET_2_IN_GG₂(Xs, Ys)
SUBSET_2_IN_GG₂(._2₂(X, Xs), Ys) -> IF_SUBSET_2_IN_1_GG₄(X, Xs, Ys, member_2_in_gg₂(X, Ys))

The TRS R consists of the following rules:

member_2_in_gg₂(X, ._2₂(Y, Xs)) -> if_member_2_in_1_gg₄(X, Y, Xs, member_2_in_gg₂(X, Xs))
member_2_in_gg₂(X, ._2₂(X, Xs)) -> member_2_out_gg₂(X, ._2₂(X, Xs))
if_member_2_in_1_gg₄(X, Y, Xs, member_2_out_gg₂(X, Xs)) -> member_2_out_gg₂(X, ._2₂(Y, Xs))

The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₂(x1, x2)
member_2_in_gg₂(x1, x2) = member_2_in_gg₂(x1, x2)
if_member_2_in_1_gg₄(x1, x2, x3, x4) = if_member_2_in_1_gg₁(x4)
member_2_out_gg₂(x1, x2) = member_2_out_gg
IF_SUBSET_2_IN_1_GG₄(x1, x2, x3, x4) = IF_SUBSET_2_IN_1_GG₃(x2, x3, x4)
SUBSET_2_IN_GG₂(x1, x2) = SUBSET_2_IN_GG₂(x1, 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

                  ↳ PiDP

                    ↳ UsableRulesProof

                      ↳ PiDP

                        ↳ PiDPToQDPProof

                          ↳ QDP

                            ↳ QDPSizeChangeProof

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

IF_SUBSET_2_IN_1_GG₃(Xs, Ys, member_2_out_gg) -> SUBSET_2_IN_GG₂(Xs, Ys)
SUBSET_2_IN_GG₂(._2₂(X, Xs), Ys) -> IF_SUBSET_2_IN_1_GG₃(Xs, Ys, member_2_in_gg₂(X, Ys))

The TRS R consists of the following rules:

member_2_in_gg₂(X, ._2₂(Y, Xs)) -> if_member_2_in_1_gg₁(member_2_in_gg₂(X, Xs))
member_2_in_gg₂(X, ._2₂(X, Xs)) -> member_2_out_gg
if_member_2_in_1_gg₁(member_2_out_gg) -> member_2_out_gg

The set Q consists of the following terms:

member_2_in_gg₂(x0, x1)
if_member_2_in_1_gg₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {SUBSET_2_IN_GG₂, IF_SUBSET_2_IN_1_GG₃}.
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:

SUBSET_2_IN_GG₂(._2₂(X, Xs), Ys) -> IF_SUBSET_2_IN_1_GG₃(Xs, Ys, member_2_in_gg₂(X, Ys))
The graph contains the following edges 1 > 1, 2 >= 2
IF_SUBSET_2_IN_1_GG₃(Xs, Ys, member_2_out_gg) -> SUBSET_2_IN_GG₂(Xs, Ys)
The graph contains the following edges 1 >= 1, 2 >= 2