Left Termination of the query pattern subset_in_2(g, a) 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(g,a).

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(.(X, Xs), Ys) → U1(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X1, Xs)) → U3(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U3(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))
U1(X, Xs, Ys, member_out(X, Ys)) → U2(X, Xs, Ys, subset_in(Xs, Ys))
subset_in([], X) → subset_out([], X)
U2(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)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x2, x4)
member_in(x1, x2)  =  member_in(x1)
U3(x1, x2, x3, x4)  =  U3(x4)
member_out(x1, x2)  =  member_out
U2(x1, x2, x3, x4)  =  U2(x4)
[]  =  []
subset_out(x1, x2)  =  subset_out

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(.(X, Xs), Ys) → U1(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X1, Xs)) → U3(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U3(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))
U1(X, Xs, Ys, member_out(X, Ys)) → U2(X, Xs, Ys, subset_in(Xs, Ys))
subset_in([], X) → subset_out([], X)
U2(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)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x2, x4)
member_in(x1, x2)  =  member_in(x1)
U3(x1, x2, x3, x4)  =  U3(x4)
member_out(x1, x2)  =  member_out
U2(x1, x2, x3, x4)  =  U2(x4)
[]  =  []
subset_out(x1, x2)  =  subset_out


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) → U11(X, Xs, Ys, member_in(X, Ys))
SUBSET_IN(.(X, Xs), Ys) → MEMBER_IN(X, Ys)
MEMBER_IN(X, .(X1, Xs)) → U31(X, X1, Xs, member_in(X, Xs))
MEMBER_IN(X, .(X1, Xs)) → MEMBER_IN(X, Xs)
U11(X, Xs, Ys, member_out(X, Ys)) → U21(X, Xs, Ys, subset_in(Xs, Ys))
U11(X, Xs, Ys, member_out(X, Ys)) → SUBSET_IN(Xs, Ys)

The TRS R consists of the following rules:

subset_in(.(X, Xs), Ys) → U1(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X1, Xs)) → U3(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U3(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))
U1(X, Xs, Ys, member_out(X, Ys)) → U2(X, Xs, Ys, subset_in(Xs, Ys))
subset_in([], X) → subset_out([], X)
U2(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)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x2, x4)
member_in(x1, x2)  =  member_in(x1)
U3(x1, x2, x3, x4)  =  U3(x4)
member_out(x1, x2)  =  member_out
U2(x1, x2, x3, x4)  =  U2(x4)
[]  =  []
subset_out(x1, x2)  =  subset_out
U31(x1, x2, x3, x4)  =  U31(x4)
MEMBER_IN(x1, x2)  =  MEMBER_IN(x1)
SUBSET_IN(x1, x2)  =  SUBSET_IN(x1)
U21(x1, x2, x3, x4)  =  U21(x4)
U11(x1, x2, x3, x4)  =  U11(x2, 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(.(X, Xs), Ys) → U11(X, Xs, Ys, member_in(X, Ys))
SUBSET_IN(.(X, Xs), Ys) → MEMBER_IN(X, Ys)
MEMBER_IN(X, .(X1, Xs)) → U31(X, X1, Xs, member_in(X, Xs))
MEMBER_IN(X, .(X1, Xs)) → MEMBER_IN(X, Xs)
U11(X, Xs, Ys, member_out(X, Ys)) → U21(X, Xs, Ys, subset_in(Xs, Ys))
U11(X, Xs, Ys, member_out(X, Ys)) → SUBSET_IN(Xs, Ys)

The TRS R consists of the following rules:

subset_in(.(X, Xs), Ys) → U1(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X1, Xs)) → U3(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U3(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))
U1(X, Xs, Ys, member_out(X, Ys)) → U2(X, Xs, Ys, subset_in(Xs, Ys))
subset_in([], X) → subset_out([], X)
U2(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)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x2, x4)
member_in(x1, x2)  =  member_in(x1)
U3(x1, x2, x3, x4)  =  U3(x4)
member_out(x1, x2)  =  member_out
U2(x1, x2, x3, x4)  =  U2(x4)
[]  =  []
subset_out(x1, x2)  =  subset_out
U31(x1, x2, x3, x4)  =  U31(x4)
MEMBER_IN(x1, x2)  =  MEMBER_IN(x1)
SUBSET_IN(x1, x2)  =  SUBSET_IN(x1)
U21(x1, x2, x3, x4)  =  U21(x4)
U11(x1, x2, x3, x4)  =  U11(x2, 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(X, .(X1, Xs)) → MEMBER_IN(X, Xs)

The TRS R consists of the following rules:

subset_in(.(X, Xs), Ys) → U1(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X1, Xs)) → U3(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U3(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))
U1(X, Xs, Ys, member_out(X, Ys)) → U2(X, Xs, Ys, subset_in(Xs, Ys))
subset_in([], X) → subset_out([], X)
U2(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)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x2, x4)
member_in(x1, x2)  =  member_in(x1)
U3(x1, x2, x3, x4)  =  U3(x4)
member_out(x1, x2)  =  member_out
U2(x1, x2, x3, x4)  =  U2(x4)
[]  =  []
subset_out(x1, x2)  =  subset_out
MEMBER_IN(x1, x2)  =  MEMBER_IN(x1)

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(X, .(X1, Xs)) → MEMBER_IN(X, Xs)

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

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
                        ↳ NonTerminationProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

MEMBER_IN(X) → MEMBER_IN(X)

R is empty.
Q is empty.
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 semiunifying a rule from P directly.

The TRS P consists of the following rules:

MEMBER_IN(X) → MEMBER_IN(X)

The TRS R consists of the following rules:none


s = MEMBER_IN(X) evaluates to t =MEMBER_IN(X)

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




Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from MEMBER_IN(X) to MEMBER_IN(X).





↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
PiDP
                ↳ UsableRulesProof
  ↳ PrologToPiTRSProof

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

SUBSET_IN(.(X, Xs), Ys) → U11(X, Xs, Ys, member_in(X, Ys))
U11(X, Xs, Ys, member_out(X, Ys)) → SUBSET_IN(Xs, Ys)

The TRS R consists of the following rules:

subset_in(.(X, Xs), Ys) → U1(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X1, Xs)) → U3(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U3(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))
U1(X, Xs, Ys, member_out(X, Ys)) → U2(X, Xs, Ys, subset_in(Xs, Ys))
subset_in([], X) → subset_out([], X)
U2(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)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x2, x4)
member_in(x1, x2)  =  member_in(x1)
U3(x1, x2, x3, x4)  =  U3(x4)
member_out(x1, x2)  =  member_out
U2(x1, x2, x3, x4)  =  U2(x4)
[]  =  []
subset_out(x1, x2)  =  subset_out
SUBSET_IN(x1, x2)  =  SUBSET_IN(x1)
U11(x1, x2, x3, x4)  =  U11(x2, 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(.(X, Xs), Ys) → U11(X, Xs, Ys, member_in(X, Ys))
U11(X, Xs, Ys, member_out(X, Ys)) → SUBSET_IN(Xs, Ys)

The TRS R consists of the following rules:

member_in(X, .(X1, Xs)) → U3(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U3(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))

The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
member_in(x1, x2)  =  member_in(x1)
U3(x1, x2, x3, x4)  =  U3(x4)
member_out(x1, x2)  =  member_out
SUBSET_IN(x1, x2)  =  SUBSET_IN(x1)
U11(x1, x2, x3, x4)  =  U11(x2, 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
  ↳ PrologToPiTRSProof

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

SUBSET_IN(.(X, Xs)) → U11(Xs, member_in(X))
U11(Xs, member_out) → SUBSET_IN(Xs)

The TRS R consists of the following rules:

member_in(X) → U3(member_in(X))
member_in(X) → member_out
U3(member_out) → member_out

The set Q consists of the following terms:

member_in(x0)
U3(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:


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(.(X, Xs), Ys) → U1(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X1, Xs)) → U3(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U3(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))
U1(X, Xs, Ys, member_out(X, Ys)) → U2(X, Xs, Ys, subset_in(Xs, Ys))
subset_in([], X) → subset_out([], X)
U2(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)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x1, x2, x4)
member_in(x1, x2)  =  member_in(x1)
U3(x1, x2, x3, x4)  =  U3(x1, x4)
member_out(x1, x2)  =  member_out(x1)
U2(x1, x2, x3, x4)  =  U2(x1, x2, x4)
[]  =  []
subset_out(x1, x2)  =  subset_out(x1)

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(.(X, Xs), Ys) → U1(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X1, Xs)) → U3(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U3(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))
U1(X, Xs, Ys, member_out(X, Ys)) → U2(X, Xs, Ys, subset_in(Xs, Ys))
subset_in([], X) → subset_out([], X)
U2(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)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x1, x2, x4)
member_in(x1, x2)  =  member_in(x1)
U3(x1, x2, x3, x4)  =  U3(x1, x4)
member_out(x1, x2)  =  member_out(x1)
U2(x1, x2, x3, x4)  =  U2(x1, x2, x4)
[]  =  []
subset_out(x1, x2)  =  subset_out(x1)


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) → U11(X, Xs, Ys, member_in(X, Ys))
SUBSET_IN(.(X, Xs), Ys) → MEMBER_IN(X, Ys)
MEMBER_IN(X, .(X1, Xs)) → U31(X, X1, Xs, member_in(X, Xs))
MEMBER_IN(X, .(X1, Xs)) → MEMBER_IN(X, Xs)
U11(X, Xs, Ys, member_out(X, Ys)) → U21(X, Xs, Ys, subset_in(Xs, Ys))
U11(X, Xs, Ys, member_out(X, Ys)) → SUBSET_IN(Xs, Ys)

The TRS R consists of the following rules:

subset_in(.(X, Xs), Ys) → U1(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X1, Xs)) → U3(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U3(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))
U1(X, Xs, Ys, member_out(X, Ys)) → U2(X, Xs, Ys, subset_in(Xs, Ys))
subset_in([], X) → subset_out([], X)
U2(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)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x1, x2, x4)
member_in(x1, x2)  =  member_in(x1)
U3(x1, x2, x3, x4)  =  U3(x1, x4)
member_out(x1, x2)  =  member_out(x1)
U2(x1, x2, x3, x4)  =  U2(x1, x2, x4)
[]  =  []
subset_out(x1, x2)  =  subset_out(x1)
U31(x1, x2, x3, x4)  =  U31(x1, x4)
MEMBER_IN(x1, x2)  =  MEMBER_IN(x1)
SUBSET_IN(x1, x2)  =  SUBSET_IN(x1)
U21(x1, x2, x3, x4)  =  U21(x1, x2, x4)
U11(x1, x2, x3, x4)  =  U11(x1, x2, 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(.(X, Xs), Ys) → U11(X, Xs, Ys, member_in(X, Ys))
SUBSET_IN(.(X, Xs), Ys) → MEMBER_IN(X, Ys)
MEMBER_IN(X, .(X1, Xs)) → U31(X, X1, Xs, member_in(X, Xs))
MEMBER_IN(X, .(X1, Xs)) → MEMBER_IN(X, Xs)
U11(X, Xs, Ys, member_out(X, Ys)) → U21(X, Xs, Ys, subset_in(Xs, Ys))
U11(X, Xs, Ys, member_out(X, Ys)) → SUBSET_IN(Xs, Ys)

The TRS R consists of the following rules:

subset_in(.(X, Xs), Ys) → U1(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X1, Xs)) → U3(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U3(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))
U1(X, Xs, Ys, member_out(X, Ys)) → U2(X, Xs, Ys, subset_in(Xs, Ys))
subset_in([], X) → subset_out([], X)
U2(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)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x1, x2, x4)
member_in(x1, x2)  =  member_in(x1)
U3(x1, x2, x3, x4)  =  U3(x1, x4)
member_out(x1, x2)  =  member_out(x1)
U2(x1, x2, x3, x4)  =  U2(x1, x2, x4)
[]  =  []
subset_out(x1, x2)  =  subset_out(x1)
U31(x1, x2, x3, x4)  =  U31(x1, x4)
MEMBER_IN(x1, x2)  =  MEMBER_IN(x1)
SUBSET_IN(x1, x2)  =  SUBSET_IN(x1)
U21(x1, x2, x3, x4)  =  U21(x1, x2, x4)
U11(x1, x2, x3, x4)  =  U11(x1, x2, 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(X, .(X1, Xs)) → MEMBER_IN(X, Xs)

The TRS R consists of the following rules:

subset_in(.(X, Xs), Ys) → U1(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X1, Xs)) → U3(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U3(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))
U1(X, Xs, Ys, member_out(X, Ys)) → U2(X, Xs, Ys, subset_in(Xs, Ys))
subset_in([], X) → subset_out([], X)
U2(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)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x1, x2, x4)
member_in(x1, x2)  =  member_in(x1)
U3(x1, x2, x3, x4)  =  U3(x1, x4)
member_out(x1, x2)  =  member_out(x1)
U2(x1, x2, x3, x4)  =  U2(x1, x2, x4)
[]  =  []
subset_out(x1, x2)  =  subset_out(x1)
MEMBER_IN(x1, x2)  =  MEMBER_IN(x1)

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(X, .(X1, Xs)) → MEMBER_IN(X, Xs)

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

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
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
QDP
                        ↳ NonTerminationProof
              ↳ PiDP

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

MEMBER_IN(X) → MEMBER_IN(X)

R is empty.
Q is empty.
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 semiunifying a rule from P directly.

The TRS P consists of the following rules:

MEMBER_IN(X) → MEMBER_IN(X)

The TRS R consists of the following rules:none


s = MEMBER_IN(X) evaluates to t =MEMBER_IN(X)

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




Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from MEMBER_IN(X) to MEMBER_IN(X).





↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ 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) → U11(X, Xs, Ys, member_in(X, Ys))
U11(X, Xs, Ys, member_out(X, Ys)) → SUBSET_IN(Xs, Ys)

The TRS R consists of the following rules:

subset_in(.(X, Xs), Ys) → U1(X, Xs, Ys, member_in(X, Ys))
member_in(X, .(X1, Xs)) → U3(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U3(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))
U1(X, Xs, Ys, member_out(X, Ys)) → U2(X, Xs, Ys, subset_in(Xs, Ys))
subset_in([], X) → subset_out([], X)
U2(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)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x1, x2, x4)
member_in(x1, x2)  =  member_in(x1)
U3(x1, x2, x3, x4)  =  U3(x1, x4)
member_out(x1, x2)  =  member_out(x1)
U2(x1, x2, x3, x4)  =  U2(x1, x2, x4)
[]  =  []
subset_out(x1, x2)  =  subset_out(x1)
SUBSET_IN(x1, x2)  =  SUBSET_IN(x1)
U11(x1, x2, x3, x4)  =  U11(x1, x2, 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(.(X, Xs), Ys) → U11(X, Xs, Ys, member_in(X, Ys))
U11(X, Xs, Ys, member_out(X, Ys)) → SUBSET_IN(Xs, Ys)

The TRS R consists of the following rules:

member_in(X, .(X1, Xs)) → U3(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U3(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))

The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
member_in(x1, x2)  =  member_in(x1)
U3(x1, x2, x3, x4)  =  U3(x1, x4)
member_out(x1, x2)  =  member_out(x1)
SUBSET_IN(x1, x2)  =  SUBSET_IN(x1)
U11(x1, x2, x3, x4)  =  U11(x1, x2, 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
                        ↳ QDPSizeChangeProof

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

U11(X, Xs, member_out(X)) → SUBSET_IN(Xs)
SUBSET_IN(.(X, Xs)) → U11(X, Xs, member_in(X))

The TRS R consists of the following rules:

member_in(X) → U3(X, member_in(X))
member_in(X) → member_out(X)
U3(X, member_out(X)) → member_out(X)

The set Q consists of the following terms:

member_in(x0)
U3(x0, x1)

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: