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

Left Termination of the query pattern subset_in_2(g, g) w.r.t. the given Prolog program could successfully be proven:

↳ Prolog

  ↳ 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:

subset(g,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:

subset_in([], Ys) → subset_out([], Ys)
subset_in(.(X, Xs), Ys) → U2(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X, Xs)) → member_out(X, .(X, Xs))
member_in(X, .(Y, Xs)) → U1(X, Y, Xs, member_in(X, Xs))
U1(X, Y, Xs, member_out(X, Xs)) → member_out(X, .(Y, Xs))
U2(X, Xs, Ys, member_out(X, Ys)) → U3(X, Xs, Ys, subset_in(Xs, Ys))
U3(X, Xs, Ys, subset_out(Xs, Ys)) → subset_out(.(X, Xs), Ys)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

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

subset_in([], Ys) → subset_out([], Ys)
subset_in(.(X, Xs), Ys) → U2(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X, Xs)) → member_out(X, .(X, Xs))
member_in(X, .(Y, Xs)) → U1(X, Y, Xs, member_in(X, Xs))
U1(X, Y, Xs, member_out(X, Xs)) → member_out(X, .(Y, Xs))
U2(X, Xs, Ys, member_out(X, Ys)) → U3(X, Xs, Ys, subset_in(Xs, Ys))
U3(X, Xs, Ys, subset_out(Xs, Ys)) → subset_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in(x1, x2) = subset_in(x1, x2)
[] = []
subset_out(x1, x2) = subset_out
.(x1, x2) = .(x1, x2)
U2(x1, x2, x3, x4) = U2(x2, x3, x4)
member_in(x1, x2) = member_in(x1, x2)
member_out(x1, x2) = member_out
U1(x1, x2, x3, x4) = U1(x4)
U3(x1, x2, x3, x4) = U3(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(.(X, Xs), Ys) → U2¹(X, Xs, Ys, member_in(X, Ys))
SUBSET_IN(.(X, Xs), Ys) → MEMBER_IN(X, Ys)
MEMBER_IN(X, .(Y, Xs)) → U1¹(X, Y, Xs, member_in(X, Xs))
MEMBER_IN(X, .(Y, Xs)) → MEMBER_IN(X, Xs)
U2¹(X, Xs, Ys, member_out(X, Ys)) → U3¹(X, Xs, Ys, subset_in(Xs, Ys))
U2¹(X, Xs, Ys, member_out(X, Ys)) → SUBSET_IN(Xs, Ys)

The TRS R consists of the following rules:

subset_in([], Ys) → subset_out([], Ys)
subset_in(.(X, Xs), Ys) → U2(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X, Xs)) → member_out(X, .(X, Xs))
member_in(X, .(Y, Xs)) → U1(X, Y, Xs, member_in(X, Xs))
U1(X, Y, Xs, member_out(X, Xs)) → member_out(X, .(Y, Xs))
U2(X, Xs, Ys, member_out(X, Ys)) → U3(X, Xs, Ys, subset_in(Xs, Ys))
U3(X, Xs, Ys, subset_out(Xs, Ys)) → subset_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in(x1, x2) = subset_in(x1, x2)
[] = []
subset_out(x1, x2) = subset_out
.(x1, x2) = .(x1, x2)
U2(x1, x2, x3, x4) = U2(x2, x3, x4)
member_in(x1, x2) = member_in(x1, x2)
member_out(x1, x2) = member_out
U1(x1, x2, x3, x4) = U1(x4)
U3(x1, x2, x3, x4) = U3(x4)
U3¹(x1, x2, x3, x4) = U3¹(x4)
MEMBER_IN(x1, x2) = MEMBER_IN(x1, x2)
SUBSET_IN(x1, x2) = SUBSET_IN(x1, x2)
U2¹(x1, x2, x3, x4) = U2¹(x2, x3, x4)
U1¹(x1, x2, x3, x4) = U1¹(x4)

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

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

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

SUBSET_IN(.(X, Xs), Ys) → U2¹(X, Xs, Ys, member_in(X, Ys))
SUBSET_IN(.(X, Xs), Ys) → MEMBER_IN(X, Ys)
MEMBER_IN(X, .(Y, Xs)) → U1¹(X, Y, Xs, member_in(X, Xs))
MEMBER_IN(X, .(Y, Xs)) → MEMBER_IN(X, Xs)
U2¹(X, Xs, Ys, member_out(X, Ys)) → U3¹(X, Xs, Ys, subset_in(Xs, Ys))
U2¹(X, Xs, Ys, member_out(X, Ys)) → SUBSET_IN(Xs, Ys)

The TRS R consists of the following rules:

subset_in([], Ys) → subset_out([], Ys)
subset_in(.(X, Xs), Ys) → U2(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X, Xs)) → member_out(X, .(X, Xs))
member_in(X, .(Y, Xs)) → U1(X, Y, Xs, member_in(X, Xs))
U1(X, Y, Xs, member_out(X, Xs)) → member_out(X, .(Y, Xs))
U2(X, Xs, Ys, member_out(X, Ys)) → U3(X, Xs, Ys, subset_in(Xs, Ys))
U3(X, Xs, Ys, subset_out(Xs, Ys)) → subset_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in(x1, x2) = subset_in(x1, x2)
[] = []
subset_out(x1, x2) = subset_out
.(x1, x2) = .(x1, x2)
U2(x1, x2, x3, x4) = U2(x2, x3, x4)
member_in(x1, x2) = member_in(x1, x2)
member_out(x1, x2) = member_out
U1(x1, x2, x3, x4) = U1(x4)
U3(x1, x2, x3, x4) = U3(x4)
U3¹(x1, x2, x3, x4) = U3¹(x4)
MEMBER_IN(x1, x2) = MEMBER_IN(x1, x2)
SUBSET_IN(x1, x2) = SUBSET_IN(x1, x2)
U2¹(x1, x2, x3, x4) = U2¹(x2, x3, x4)
U1¹(x1, x2, x3, x4) = U1¹(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

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

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

The TRS R consists of the following rules:

subset_in([], Ys) → subset_out([], Ys)
subset_in(.(X, Xs), Ys) → U2(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X, Xs)) → member_out(X, .(X, Xs))
member_in(X, .(Y, Xs)) → U1(X, Y, Xs, member_in(X, Xs))
U1(X, Y, Xs, member_out(X, Xs)) → member_out(X, .(Y, Xs))
U2(X, Xs, Ys, member_out(X, Ys)) → U3(X, Xs, Ys, subset_in(Xs, Ys))
U3(X, Xs, Ys, subset_out(Xs, Ys)) → subset_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in(x1, x2) = subset_in(x1, x2)
[] = []
subset_out(x1, x2) = subset_out
.(x1, x2) = .(x1, x2)
U2(x1, x2, x3, x4) = U2(x2, x3, x4)
member_in(x1, x2) = member_in(x1, x2)
member_out(x1, x2) = member_out
U1(x1, x2, x3, x4) = U1(x4)
U3(x1, x2, x3, x4) = U3(x4)
MEMBER_IN(x1, x2) = MEMBER_IN(x1, 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

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

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

R is empty.
Pi is empty.
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

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

MEMBER_IN(X, .(Y, Xs)) → MEMBER_IN(X, 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(X, .(Y, Xs)) → MEMBER_IN(X, Xs)
The graph contains the following edges 1 >= 1, 2 > 2

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

SUBSET_IN(.(X, Xs), Ys) → U2¹(X, Xs, Ys, member_in(X, Ys))
U2¹(X, Xs, Ys, member_out(X, Ys)) → SUBSET_IN(Xs, Ys)

The TRS R consists of the following rules:

subset_in([], Ys) → subset_out([], Ys)
subset_in(.(X, Xs), Ys) → U2(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X, Xs)) → member_out(X, .(X, Xs))
member_in(X, .(Y, Xs)) → U1(X, Y, Xs, member_in(X, Xs))
U1(X, Y, Xs, member_out(X, Xs)) → member_out(X, .(Y, Xs))
U2(X, Xs, Ys, member_out(X, Ys)) → U3(X, Xs, Ys, subset_in(Xs, Ys))
U3(X, Xs, Ys, subset_out(Xs, Ys)) → subset_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in(x1, x2) = subset_in(x1, x2)
[] = []
subset_out(x1, x2) = subset_out
.(x1, x2) = .(x1, x2)
U2(x1, x2, x3, x4) = U2(x2, x3, x4)
member_in(x1, x2) = member_in(x1, x2)
member_out(x1, x2) = member_out
U1(x1, x2, x3, x4) = U1(x4)
U3(x1, x2, x3, x4) = U3(x4)
SUBSET_IN(x1, x2) = SUBSET_IN(x1, x2)
U2¹(x1, x2, x3, x4) = U2¹(x2, 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

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

SUBSET_IN(.(X, Xs), Ys) → U2¹(X, Xs, Ys, member_in(X, Ys))
U2¹(X, Xs, Ys, member_out(X, Ys)) → SUBSET_IN(Xs, Ys)

The TRS R consists of the following rules:

member_in(X, .(X, Xs)) → member_out(X, .(X, Xs))
member_in(X, .(Y, Xs)) → U1(X, Y, Xs, member_in(X, Xs))
U1(X, Y, Xs, member_out(X, Xs)) → member_out(X, .(Y, Xs))

The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
member_in(x1, x2) = member_in(x1, x2)
member_out(x1, x2) = member_out
U1(x1, x2, x3, x4) = U1(x4)
SUBSET_IN(x1, x2) = SUBSET_IN(x1, x2)
U2¹(x1, x2, x3, x4) = U2¹(x2, 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

                        ↳ QDPSizeChangeProof

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

U2¹(Xs, Ys, member_out) → SUBSET_IN(Xs, Ys)
SUBSET_IN(.(X, Xs), Ys) → U2¹(Xs, Ys, member_in(X, Ys))

The TRS R consists of the following rules:

member_in(X, .(X, Xs)) → member_out
member_in(X, .(Y, Xs)) → U1(member_in(X, Xs))
U1(member_out) → member_out

The set Q consists of the following terms:

member_in(x0, x1)
U1(x0)

We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

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

U2¹(Xs, Ys, member_out) → SUBSET_IN(Xs, Ys)
The graph contains the following edges 1 >= 1, 2 >= 2
SUBSET_IN(.(X, Xs), Ys) → U2¹(Xs, Ys, member_in(X, Ys))
The graph contains the following edges 1 > 1, 2 >= 2