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

subset_in_ga([], X) → subset_out_ga([], X)
subset_in_ga(.(X, Xs), Ys) → U1_ga(X, Xs, Ys, member_in_aa(X, Ys))
member_in_aa(X, .(X, X1)) → member_out_aa(X, .(X, X1))
member_in_aa(X, .(X1, Xs)) → U3_aa(X, X1, Xs, member_in_aa(X, Xs))
U3_aa(X, X1, Xs, member_out_aa(X, Xs)) → member_out_aa(X, .(X1, Xs))
U1_ga(X, Xs, Ys, member_out_aa(X, Ys)) → U2_ga(X, Xs, Ys, subset_in_ag(Xs, Ys))
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_aa(X, Xs))
U3_ag(X, X1, Xs, member_out_aa(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)
U2_ga(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ga(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in_ga(x1, x2)  =  subset_in_ga(x1)
[]  =  []
subset_out_ga(x1, x2)  =  subset_out_ga
.(x1, x2)  =  .
U1_ga(x1, x2, x3, x4)  =  U1_ga(x4)
member_in_aa(x1, x2)  =  member_in_aa
member_out_aa(x1, x2)  =  member_out_aa(x2)
U3_aa(x1, x2, x3, x4)  =  U3_aa(x4)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x4)
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)
member_out_ag(x1, x2)  =  member_out_ag
U3_ag(x1, x2, x3, x4)  =  U3_ag(x4)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x4)

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_ga([], X) → subset_out_ga([], X)
subset_in_ga(.(X, Xs), Ys) → U1_ga(X, Xs, Ys, member_in_aa(X, Ys))
member_in_aa(X, .(X, X1)) → member_out_aa(X, .(X, X1))
member_in_aa(X, .(X1, Xs)) → U3_aa(X, X1, Xs, member_in_aa(X, Xs))
U3_aa(X, X1, Xs, member_out_aa(X, Xs)) → member_out_aa(X, .(X1, Xs))
U1_ga(X, Xs, Ys, member_out_aa(X, Ys)) → U2_ga(X, Xs, Ys, subset_in_ag(Xs, Ys))
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_aa(X, Xs))
U3_ag(X, X1, Xs, member_out_aa(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)
U2_ga(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ga(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in_ga(x1, x2)  =  subset_in_ga(x1)
[]  =  []
subset_out_ga(x1, x2)  =  subset_out_ga
.(x1, x2)  =  .
U1_ga(x1, x2, x3, x4)  =  U1_ga(x4)
member_in_aa(x1, x2)  =  member_in_aa
member_out_aa(x1, x2)  =  member_out_aa(x2)
U3_aa(x1, x2, x3, x4)  =  U3_aa(x4)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x4)
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)
member_out_ag(x1, x2)  =  member_out_ag
U3_ag(x1, x2, x3, x4)  =  U3_ag(x4)
U2_ag(x1, x2, x3, x4)  =  U2_ag(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_GA(.(X, Xs), Ys) → U1_GA(X, Xs, Ys, member_in_aa(X, Ys))
SUBSET_IN_GA(.(X, Xs), Ys) → MEMBER_IN_AA(X, Ys)
MEMBER_IN_AA(X, .(X1, Xs)) → U3_AA(X, X1, Xs, member_in_aa(X, Xs))
MEMBER_IN_AA(X, .(X1, Xs)) → MEMBER_IN_AA(X, Xs)
U1_GA(X, Xs, Ys, member_out_aa(X, Ys)) → U2_GA(X, Xs, Ys, subset_in_ag(Xs, Ys))
U1_GA(X, Xs, Ys, member_out_aa(X, Ys)) → SUBSET_IN_AG(Xs, Ys)
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_aa(X, Xs))
MEMBER_IN_AG(X, .(X1, Xs)) → MEMBER_IN_AA(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_ga([], X) → subset_out_ga([], X)
subset_in_ga(.(X, Xs), Ys) → U1_ga(X, Xs, Ys, member_in_aa(X, Ys))
member_in_aa(X, .(X, X1)) → member_out_aa(X, .(X, X1))
member_in_aa(X, .(X1, Xs)) → U3_aa(X, X1, Xs, member_in_aa(X, Xs))
U3_aa(X, X1, Xs, member_out_aa(X, Xs)) → member_out_aa(X, .(X1, Xs))
U1_ga(X, Xs, Ys, member_out_aa(X, Ys)) → U2_ga(X, Xs, Ys, subset_in_ag(Xs, Ys))
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_aa(X, Xs))
U3_ag(X, X1, Xs, member_out_aa(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)
U2_ga(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ga(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in_ga(x1, x2)  =  subset_in_ga(x1)
[]  =  []
subset_out_ga(x1, x2)  =  subset_out_ga
.(x1, x2)  =  .
U1_ga(x1, x2, x3, x4)  =  U1_ga(x4)
member_in_aa(x1, x2)  =  member_in_aa
member_out_aa(x1, x2)  =  member_out_aa(x2)
U3_aa(x1, x2, x3, x4)  =  U3_aa(x4)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x4)
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)
member_out_ag(x1, x2)  =  member_out_ag
U3_ag(x1, x2, x3, x4)  =  U3_ag(x4)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x4)
MEMBER_IN_AG(x1, x2)  =  MEMBER_IN_AG(x2)
MEMBER_IN_AA(x1, x2)  =  MEMBER_IN_AA
U3_AG(x1, x2, x3, x4)  =  U3_AG(x4)
SUBSET_IN_AG(x1, x2)  =  SUBSET_IN_AG(x2)
SUBSET_IN_GA(x1, x2)  =  SUBSET_IN_GA(x1)
U2_AG(x1, x2, x3, x4)  =  U2_AG(x4)
U2_GA(x1, x2, x3, x4)  =  U2_GA(x4)
U1_GA(x1, x2, x3, x4)  =  U1_GA(x4)
U1_AG(x1, x2, x3, x4)  =  U1_AG(x3, x4)
U3_AA(x1, x2, x3, x4)  =  U3_AA(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_GA(.(X, Xs), Ys) → U1_GA(X, Xs, Ys, member_in_aa(X, Ys))
SUBSET_IN_GA(.(X, Xs), Ys) → MEMBER_IN_AA(X, Ys)
MEMBER_IN_AA(X, .(X1, Xs)) → U3_AA(X, X1, Xs, member_in_aa(X, Xs))
MEMBER_IN_AA(X, .(X1, Xs)) → MEMBER_IN_AA(X, Xs)
U1_GA(X, Xs, Ys, member_out_aa(X, Ys)) → U2_GA(X, Xs, Ys, subset_in_ag(Xs, Ys))
U1_GA(X, Xs, Ys, member_out_aa(X, Ys)) → SUBSET_IN_AG(Xs, Ys)
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_aa(X, Xs))
MEMBER_IN_AG(X, .(X1, Xs)) → MEMBER_IN_AA(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_ga([], X) → subset_out_ga([], X)
subset_in_ga(.(X, Xs), Ys) → U1_ga(X, Xs, Ys, member_in_aa(X, Ys))
member_in_aa(X, .(X, X1)) → member_out_aa(X, .(X, X1))
member_in_aa(X, .(X1, Xs)) → U3_aa(X, X1, Xs, member_in_aa(X, Xs))
U3_aa(X, X1, Xs, member_out_aa(X, Xs)) → member_out_aa(X, .(X1, Xs))
U1_ga(X, Xs, Ys, member_out_aa(X, Ys)) → U2_ga(X, Xs, Ys, subset_in_ag(Xs, Ys))
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_aa(X, Xs))
U3_ag(X, X1, Xs, member_out_aa(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)
U2_ga(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ga(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in_ga(x1, x2)  =  subset_in_ga(x1)
[]  =  []
subset_out_ga(x1, x2)  =  subset_out_ga
.(x1, x2)  =  .
U1_ga(x1, x2, x3, x4)  =  U1_ga(x4)
member_in_aa(x1, x2)  =  member_in_aa
member_out_aa(x1, x2)  =  member_out_aa(x2)
U3_aa(x1, x2, x3, x4)  =  U3_aa(x4)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x4)
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)
member_out_ag(x1, x2)  =  member_out_ag
U3_ag(x1, x2, x3, x4)  =  U3_ag(x4)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x4)
MEMBER_IN_AG(x1, x2)  =  MEMBER_IN_AG(x2)
MEMBER_IN_AA(x1, x2)  =  MEMBER_IN_AA
U3_AG(x1, x2, x3, x4)  =  U3_AG(x4)
SUBSET_IN_AG(x1, x2)  =  SUBSET_IN_AG(x2)
SUBSET_IN_GA(x1, x2)  =  SUBSET_IN_GA(x1)
U2_AG(x1, x2, x3, x4)  =  U2_AG(x4)
U2_GA(x1, x2, x3, x4)  =  U2_GA(x4)
U1_GA(x1, x2, x3, x4)  =  U1_GA(x4)
U1_AG(x1, x2, x3, x4)  =  U1_AG(x3, x4)
U3_AA(x1, x2, x3, x4)  =  U3_AA(x4)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 9 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_AA(X, .(X1, Xs)) → MEMBER_IN_AA(X, Xs)

The TRS R consists of the following rules:

subset_in_ga([], X) → subset_out_ga([], X)
subset_in_ga(.(X, Xs), Ys) → U1_ga(X, Xs, Ys, member_in_aa(X, Ys))
member_in_aa(X, .(X, X1)) → member_out_aa(X, .(X, X1))
member_in_aa(X, .(X1, Xs)) → U3_aa(X, X1, Xs, member_in_aa(X, Xs))
U3_aa(X, X1, Xs, member_out_aa(X, Xs)) → member_out_aa(X, .(X1, Xs))
U1_ga(X, Xs, Ys, member_out_aa(X, Ys)) → U2_ga(X, Xs, Ys, subset_in_ag(Xs, Ys))
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_aa(X, Xs))
U3_ag(X, X1, Xs, member_out_aa(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)
U2_ga(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ga(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in_ga(x1, x2)  =  subset_in_ga(x1)
[]  =  []
subset_out_ga(x1, x2)  =  subset_out_ga
.(x1, x2)  =  .
U1_ga(x1, x2, x3, x4)  =  U1_ga(x4)
member_in_aa(x1, x2)  =  member_in_aa
member_out_aa(x1, x2)  =  member_out_aa(x2)
U3_aa(x1, x2, x3, x4)  =  U3_aa(x4)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x4)
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)
member_out_ag(x1, x2)  =  member_out_ag
U3_ag(x1, x2, x3, x4)  =  U3_ag(x4)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x4)
MEMBER_IN_AA(x1, x2)  =  MEMBER_IN_AA

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_AA(X, .(X1, Xs)) → MEMBER_IN_AA(X, Xs)

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

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_AA → MEMBER_IN_AA

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_AA → MEMBER_IN_AA

The TRS R consists of the following rules:none


s = MEMBER_IN_AA evaluates to t =MEMBER_IN_AA

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_AA to MEMBER_IN_AA.





↳ 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_ga([], X) → subset_out_ga([], X)
subset_in_ga(.(X, Xs), Ys) → U1_ga(X, Xs, Ys, member_in_aa(X, Ys))
member_in_aa(X, .(X, X1)) → member_out_aa(X, .(X, X1))
member_in_aa(X, .(X1, Xs)) → U3_aa(X, X1, Xs, member_in_aa(X, Xs))
U3_aa(X, X1, Xs, member_out_aa(X, Xs)) → member_out_aa(X, .(X1, Xs))
U1_ga(X, Xs, Ys, member_out_aa(X, Ys)) → U2_ga(X, Xs, Ys, subset_in_ag(Xs, Ys))
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_aa(X, Xs))
U3_ag(X, X1, Xs, member_out_aa(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)
U2_ga(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ga(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in_ga(x1, x2)  =  subset_in_ga(x1)
[]  =  []
subset_out_ga(x1, x2)  =  subset_out_ga
.(x1, x2)  =  .
U1_ga(x1, x2, x3, x4)  =  U1_ga(x4)
member_in_aa(x1, x2)  =  member_in_aa
member_out_aa(x1, x2)  =  member_out_aa(x2)
U3_aa(x1, x2, x3, x4)  =  U3_aa(x4)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x4)
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)
member_out_ag(x1, x2)  =  member_out_ag
U3_ag(x1, x2, x3, x4)  =  U3_ag(x4)
U2_ag(x1, x2, x3, x4)  =  U2_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
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_aa(X, Xs))
U3_ag(X, X1, Xs, member_out_aa(X, Xs)) → member_out_ag(X, .(X1, Xs))
member_in_aa(X, .(X, X1)) → member_out_aa(X, .(X, X1))
member_in_aa(X, .(X1, Xs)) → U3_aa(X, X1, Xs, member_in_aa(X, Xs))
U3_aa(X, X1, Xs, member_out_aa(X, Xs)) → member_out_aa(X, .(X1, Xs))

The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .
member_in_aa(x1, x2)  =  member_in_aa
member_out_aa(x1, x2)  =  member_out_aa(x2)
U3_aa(x1, x2, x3, x4)  =  U3_aa(x4)
member_in_ag(x1, x2)  =  member_in_ag(x2)
member_out_ag(x1, x2)  =  member_out_ag
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) → SUBSET_IN_AG(Ys)

The TRS R consists of the following rules:

member_in_ag(.) → member_out_ag
member_in_ag(.) → U3_ag(member_in_aa)
U3_ag(member_out_aa(Xs)) → member_out_ag
member_in_aa → member_out_aa(.)
member_in_aa → U3_aa(member_in_aa)
U3_aa(member_out_aa(Xs)) → member_out_aa(.)

The set Q consists of the following terms:

member_in_ag(x0)
U3_ag(x0)
member_in_aa
U3_aa(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(.) → U1_AG(., member_out_ag)
SUBSET_IN_AG(.) → U1_AG(., U3_ag(member_in_aa))



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ UsableRulesProof
  ↳ PrologToPiTRSProof

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

SUBSET_IN_AG(.) → U1_AG(., member_out_ag)
U1_AG(Ys, member_out_ag) → SUBSET_IN_AG(Ys)
SUBSET_IN_AG(.) → U1_AG(., U3_ag(member_in_aa))

The TRS R consists of the following rules:

member_in_ag(.) → member_out_ag
member_in_ag(.) → U3_ag(member_in_aa)
U3_ag(member_out_aa(Xs)) → member_out_ag
member_in_aa → member_out_aa(.)
member_in_aa → U3_aa(member_in_aa)
U3_aa(member_out_aa(Xs)) → member_out_aa(.)

The set Q consists of the following terms:

member_in_ag(x0)
U3_ag(x0)
member_in_aa
U3_aa(x0)

We have to consider all (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ UsableRulesProof
                              ↳ QDP
                                ↳ QReductionProof
  ↳ PrologToPiTRSProof

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

SUBSET_IN_AG(.) → U1_AG(., member_out_ag)
U1_AG(Ys, member_out_ag) → SUBSET_IN_AG(Ys)
SUBSET_IN_AG(.) → U1_AG(., U3_ag(member_in_aa))

The TRS R consists of the following rules:

member_in_aa → member_out_aa(.)
member_in_aa → U3_aa(member_in_aa)
U3_ag(member_out_aa(Xs)) → member_out_ag
U3_aa(member_out_aa(Xs)) → member_out_aa(.)

The set Q consists of the following terms:

member_in_ag(x0)
U3_ag(x0)
member_in_aa
U3_aa(x0)

We have to consider all (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

member_in_ag(x0)



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ UsableRulesProof
                              ↳ QDP
                                ↳ QReductionProof
                                  ↳ QDP
                                    ↳ Instantiation
  ↳ PrologToPiTRSProof

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

SUBSET_IN_AG(.) → U1_AG(., member_out_ag)
U1_AG(Ys, member_out_ag) → SUBSET_IN_AG(Ys)
SUBSET_IN_AG(.) → U1_AG(., U3_ag(member_in_aa))

The TRS R consists of the following rules:

member_in_aa → member_out_aa(.)
member_in_aa → U3_aa(member_in_aa)
U3_ag(member_out_aa(Xs)) → member_out_ag
U3_aa(member_out_aa(Xs)) → member_out_aa(.)

The set Q consists of the following terms:

U3_ag(x0)
member_in_aa
U3_aa(x0)

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

U1_AG(., member_out_ag) → SUBSET_IN_AG(.)



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ UsableRulesProof
                              ↳ QDP
                                ↳ QReductionProof
                                  ↳ QDP
                                    ↳ Instantiation
                                      ↳ QDP
                                        ↳ NonTerminationProof
  ↳ PrologToPiTRSProof

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

SUBSET_IN_AG(.) → U1_AG(., member_out_ag)
U1_AG(., member_out_ag) → SUBSET_IN_AG(.)
SUBSET_IN_AG(.) → U1_AG(., U3_ag(member_in_aa))

The TRS R consists of the following rules:

member_in_aa → member_out_aa(.)
member_in_aa → U3_aa(member_in_aa)
U3_ag(member_out_aa(Xs)) → member_out_ag
U3_aa(member_out_aa(Xs)) → member_out_aa(.)

The set Q consists of the following terms:

U3_ag(x0)
member_in_aa
U3_aa(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:

SUBSET_IN_AG(.) → U1_AG(., member_out_ag)
U1_AG(., member_out_ag) → SUBSET_IN_AG(.)
SUBSET_IN_AG(.) → U1_AG(., U3_ag(member_in_aa))

The TRS R consists of the following rules:

member_in_aa → member_out_aa(.)
member_in_aa → U3_aa(member_in_aa)
U3_ag(member_out_aa(Xs)) → member_out_ag
U3_aa(member_out_aa(Xs)) → member_out_aa(.)


s = U1_AG(., member_out_ag) evaluates to t =U1_AG(., member_out_ag)

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



Rewriting sequence

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

SUBSET_IN_AG(.) → U1_AG(., member_out_ag)
with rule SUBSET_IN_AG(.) → U1_AG(., member_out_ag)

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: (b,f) (f,b)
member_in: (f,f) (f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

subset_in_ga([], X) → subset_out_ga([], X)
subset_in_ga(.(X, Xs), Ys) → U1_ga(X, Xs, Ys, member_in_aa(X, Ys))
member_in_aa(X, .(X, X1)) → member_out_aa(X, .(X, X1))
member_in_aa(X, .(X1, Xs)) → U3_aa(X, X1, Xs, member_in_aa(X, Xs))
U3_aa(X, X1, Xs, member_out_aa(X, Xs)) → member_out_aa(X, .(X1, Xs))
U1_ga(X, Xs, Ys, member_out_aa(X, Ys)) → U2_ga(X, Xs, Ys, subset_in_ag(Xs, Ys))
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_aa(X, Xs))
U3_ag(X, X1, Xs, member_out_aa(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)
U2_ga(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ga(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in_ga(x1, x2)  =  subset_in_ga(x1)
[]  =  []
subset_out_ga(x1, x2)  =  subset_out_ga(x1)
.(x1, x2)  =  .
U1_ga(x1, x2, x3, x4)  =  U1_ga(x4)
member_in_aa(x1, x2)  =  member_in_aa
member_out_aa(x1, x2)  =  member_out_aa(x2)
U3_aa(x1, x2, x3, x4)  =  U3_aa(x4)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x4)
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)
member_out_ag(x1, x2)  =  member_out_ag(x2)
U3_ag(x1, x2, x3, x4)  =  U3_ag(x4)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x3, x4)

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_ga([], X) → subset_out_ga([], X)
subset_in_ga(.(X, Xs), Ys) → U1_ga(X, Xs, Ys, member_in_aa(X, Ys))
member_in_aa(X, .(X, X1)) → member_out_aa(X, .(X, X1))
member_in_aa(X, .(X1, Xs)) → U3_aa(X, X1, Xs, member_in_aa(X, Xs))
U3_aa(X, X1, Xs, member_out_aa(X, Xs)) → member_out_aa(X, .(X1, Xs))
U1_ga(X, Xs, Ys, member_out_aa(X, Ys)) → U2_ga(X, Xs, Ys, subset_in_ag(Xs, Ys))
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_aa(X, Xs))
U3_ag(X, X1, Xs, member_out_aa(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)
U2_ga(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ga(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in_ga(x1, x2)  =  subset_in_ga(x1)
[]  =  []
subset_out_ga(x1, x2)  =  subset_out_ga(x1)
.(x1, x2)  =  .
U1_ga(x1, x2, x3, x4)  =  U1_ga(x4)
member_in_aa(x1, x2)  =  member_in_aa
member_out_aa(x1, x2)  =  member_out_aa(x2)
U3_aa(x1, x2, x3, x4)  =  U3_aa(x4)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x4)
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)
member_out_ag(x1, x2)  =  member_out_ag(x2)
U3_ag(x1, x2, x3, x4)  =  U3_ag(x4)
U2_ag(x1, x2, x3, x4)  =  U2_ag(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_GA(.(X, Xs), Ys) → U1_GA(X, Xs, Ys, member_in_aa(X, Ys))
SUBSET_IN_GA(.(X, Xs), Ys) → MEMBER_IN_AA(X, Ys)
MEMBER_IN_AA(X, .(X1, Xs)) → U3_AA(X, X1, Xs, member_in_aa(X, Xs))
MEMBER_IN_AA(X, .(X1, Xs)) → MEMBER_IN_AA(X, Xs)
U1_GA(X, Xs, Ys, member_out_aa(X, Ys)) → U2_GA(X, Xs, Ys, subset_in_ag(Xs, Ys))
U1_GA(X, Xs, Ys, member_out_aa(X, Ys)) → SUBSET_IN_AG(Xs, Ys)
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_aa(X, Xs))
MEMBER_IN_AG(X, .(X1, Xs)) → MEMBER_IN_AA(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_ga([], X) → subset_out_ga([], X)
subset_in_ga(.(X, Xs), Ys) → U1_ga(X, Xs, Ys, member_in_aa(X, Ys))
member_in_aa(X, .(X, X1)) → member_out_aa(X, .(X, X1))
member_in_aa(X, .(X1, Xs)) → U3_aa(X, X1, Xs, member_in_aa(X, Xs))
U3_aa(X, X1, Xs, member_out_aa(X, Xs)) → member_out_aa(X, .(X1, Xs))
U1_ga(X, Xs, Ys, member_out_aa(X, Ys)) → U2_ga(X, Xs, Ys, subset_in_ag(Xs, Ys))
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_aa(X, Xs))
U3_ag(X, X1, Xs, member_out_aa(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)
U2_ga(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ga(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in_ga(x1, x2)  =  subset_in_ga(x1)
[]  =  []
subset_out_ga(x1, x2)  =  subset_out_ga(x1)
.(x1, x2)  =  .
U1_ga(x1, x2, x3, x4)  =  U1_ga(x4)
member_in_aa(x1, x2)  =  member_in_aa
member_out_aa(x1, x2)  =  member_out_aa(x2)
U3_aa(x1, x2, x3, x4)  =  U3_aa(x4)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x4)
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)
member_out_ag(x1, x2)  =  member_out_ag(x2)
U3_ag(x1, x2, x3, x4)  =  U3_ag(x4)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x3, x4)
MEMBER_IN_AG(x1, x2)  =  MEMBER_IN_AG(x2)
MEMBER_IN_AA(x1, x2)  =  MEMBER_IN_AA
U3_AG(x1, x2, x3, x4)  =  U3_AG(x4)
SUBSET_IN_AG(x1, x2)  =  SUBSET_IN_AG(x2)
SUBSET_IN_GA(x1, x2)  =  SUBSET_IN_GA(x1)
U2_AG(x1, x2, x3, x4)  =  U2_AG(x3, x4)
U2_GA(x1, x2, x3, x4)  =  U2_GA(x4)
U1_GA(x1, x2, x3, x4)  =  U1_GA(x4)
U1_AG(x1, x2, x3, x4)  =  U1_AG(x3, x4)
U3_AA(x1, x2, x3, x4)  =  U3_AA(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_GA(.(X, Xs), Ys) → U1_GA(X, Xs, Ys, member_in_aa(X, Ys))
SUBSET_IN_GA(.(X, Xs), Ys) → MEMBER_IN_AA(X, Ys)
MEMBER_IN_AA(X, .(X1, Xs)) → U3_AA(X, X1, Xs, member_in_aa(X, Xs))
MEMBER_IN_AA(X, .(X1, Xs)) → MEMBER_IN_AA(X, Xs)
U1_GA(X, Xs, Ys, member_out_aa(X, Ys)) → U2_GA(X, Xs, Ys, subset_in_ag(Xs, Ys))
U1_GA(X, Xs, Ys, member_out_aa(X, Ys)) → SUBSET_IN_AG(Xs, Ys)
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_aa(X, Xs))
MEMBER_IN_AG(X, .(X1, Xs)) → MEMBER_IN_AA(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_ga([], X) → subset_out_ga([], X)
subset_in_ga(.(X, Xs), Ys) → U1_ga(X, Xs, Ys, member_in_aa(X, Ys))
member_in_aa(X, .(X, X1)) → member_out_aa(X, .(X, X1))
member_in_aa(X, .(X1, Xs)) → U3_aa(X, X1, Xs, member_in_aa(X, Xs))
U3_aa(X, X1, Xs, member_out_aa(X, Xs)) → member_out_aa(X, .(X1, Xs))
U1_ga(X, Xs, Ys, member_out_aa(X, Ys)) → U2_ga(X, Xs, Ys, subset_in_ag(Xs, Ys))
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_aa(X, Xs))
U3_ag(X, X1, Xs, member_out_aa(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)
U2_ga(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ga(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in_ga(x1, x2)  =  subset_in_ga(x1)
[]  =  []
subset_out_ga(x1, x2)  =  subset_out_ga(x1)
.(x1, x2)  =  .
U1_ga(x1, x2, x3, x4)  =  U1_ga(x4)
member_in_aa(x1, x2)  =  member_in_aa
member_out_aa(x1, x2)  =  member_out_aa(x2)
U3_aa(x1, x2, x3, x4)  =  U3_aa(x4)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x4)
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)
member_out_ag(x1, x2)  =  member_out_ag(x2)
U3_ag(x1, x2, x3, x4)  =  U3_ag(x4)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x3, x4)
MEMBER_IN_AG(x1, x2)  =  MEMBER_IN_AG(x2)
MEMBER_IN_AA(x1, x2)  =  MEMBER_IN_AA
U3_AG(x1, x2, x3, x4)  =  U3_AG(x4)
SUBSET_IN_AG(x1, x2)  =  SUBSET_IN_AG(x2)
SUBSET_IN_GA(x1, x2)  =  SUBSET_IN_GA(x1)
U2_AG(x1, x2, x3, x4)  =  U2_AG(x3, x4)
U2_GA(x1, x2, x3, x4)  =  U2_GA(x4)
U1_GA(x1, x2, x3, x4)  =  U1_GA(x4)
U1_AG(x1, x2, x3, x4)  =  U1_AG(x3, x4)
U3_AA(x1, x2, x3, x4)  =  U3_AA(x4)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 9 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_AA(X, .(X1, Xs)) → MEMBER_IN_AA(X, Xs)

The TRS R consists of the following rules:

subset_in_ga([], X) → subset_out_ga([], X)
subset_in_ga(.(X, Xs), Ys) → U1_ga(X, Xs, Ys, member_in_aa(X, Ys))
member_in_aa(X, .(X, X1)) → member_out_aa(X, .(X, X1))
member_in_aa(X, .(X1, Xs)) → U3_aa(X, X1, Xs, member_in_aa(X, Xs))
U3_aa(X, X1, Xs, member_out_aa(X, Xs)) → member_out_aa(X, .(X1, Xs))
U1_ga(X, Xs, Ys, member_out_aa(X, Ys)) → U2_ga(X, Xs, Ys, subset_in_ag(Xs, Ys))
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_aa(X, Xs))
U3_ag(X, X1, Xs, member_out_aa(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)
U2_ga(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ga(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in_ga(x1, x2)  =  subset_in_ga(x1)
[]  =  []
subset_out_ga(x1, x2)  =  subset_out_ga(x1)
.(x1, x2)  =  .
U1_ga(x1, x2, x3, x4)  =  U1_ga(x4)
member_in_aa(x1, x2)  =  member_in_aa
member_out_aa(x1, x2)  =  member_out_aa(x2)
U3_aa(x1, x2, x3, x4)  =  U3_aa(x4)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x4)
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)
member_out_ag(x1, x2)  =  member_out_ag(x2)
U3_ag(x1, x2, x3, x4)  =  U3_ag(x4)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x3, x4)
MEMBER_IN_AA(x1, x2)  =  MEMBER_IN_AA

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_AA(X, .(X1, Xs)) → MEMBER_IN_AA(X, Xs)

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

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_AA → MEMBER_IN_AA

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_AA → MEMBER_IN_AA

The TRS R consists of the following rules:none


s = MEMBER_IN_AA evaluates to t =MEMBER_IN_AA

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_AA to MEMBER_IN_AA.





↳ 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_ga([], X) → subset_out_ga([], X)
subset_in_ga(.(X, Xs), Ys) → U1_ga(X, Xs, Ys, member_in_aa(X, Ys))
member_in_aa(X, .(X, X1)) → member_out_aa(X, .(X, X1))
member_in_aa(X, .(X1, Xs)) → U3_aa(X, X1, Xs, member_in_aa(X, Xs))
U3_aa(X, X1, Xs, member_out_aa(X, Xs)) → member_out_aa(X, .(X1, Xs))
U1_ga(X, Xs, Ys, member_out_aa(X, Ys)) → U2_ga(X, Xs, Ys, subset_in_ag(Xs, Ys))
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_aa(X, Xs))
U3_ag(X, X1, Xs, member_out_aa(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)
U2_ga(X, Xs, Ys, subset_out_ag(Xs, Ys)) → subset_out_ga(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_in_ga(x1, x2)  =  subset_in_ga(x1)
[]  =  []
subset_out_ga(x1, x2)  =  subset_out_ga(x1)
.(x1, x2)  =  .
U1_ga(x1, x2, x3, x4)  =  U1_ga(x4)
member_in_aa(x1, x2)  =  member_in_aa
member_out_aa(x1, x2)  =  member_out_aa(x2)
U3_aa(x1, x2, x3, x4)  =  U3_aa(x4)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x4)
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)
member_out_ag(x1, x2)  =  member_out_ag(x2)
U3_ag(x1, x2, x3, x4)  =  U3_ag(x4)
U2_ag(x1, x2, x3, x4)  =  U2_ag(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_aa(X, Xs))
U3_ag(X, X1, Xs, member_out_aa(X, Xs)) → member_out_ag(X, .(X1, Xs))
member_in_aa(X, .(X, X1)) → member_out_aa(X, .(X, X1))
member_in_aa(X, .(X1, Xs)) → U3_aa(X, X1, Xs, member_in_aa(X, Xs))
U3_aa(X, X1, Xs, member_out_aa(X, Xs)) → member_out_aa(X, .(X1, Xs))

The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .
member_in_aa(x1, x2)  =  member_in_aa
member_out_aa(x1, x2)  =  member_out_aa(x2)
U3_aa(x1, x2, x3, x4)  =  U3_aa(x4)
member_in_ag(x1, x2)  =  member_in_ag(x2)
member_out_ag(x1, x2)  =  member_out_ag(x2)
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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing

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(Ys)) → SUBSET_IN_AG(Ys)

The TRS R consists of the following rules:

member_in_ag(.) → member_out_ag(.)
member_in_ag(.) → U3_ag(member_in_aa)
U3_ag(member_out_aa(Xs)) → member_out_ag(.)
member_in_aa → member_out_aa(.)
member_in_aa → U3_aa(member_in_aa)
U3_aa(member_out_aa(Xs)) → member_out_aa(.)

The set Q consists of the following terms:

member_in_ag(x0)
U3_ag(x0)
member_in_aa
U3_aa(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(.) → U1_AG(., member_out_ag(.))
SUBSET_IN_AG(.) → U1_AG(., U3_ag(member_in_aa))



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ UsableRulesProof

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

SUBSET_IN_AG(.) → U1_AG(., member_out_ag(.))
SUBSET_IN_AG(.) → U1_AG(., U3_ag(member_in_aa))
U1_AG(Ys, member_out_ag(Ys)) → SUBSET_IN_AG(Ys)

The TRS R consists of the following rules:

member_in_ag(.) → member_out_ag(.)
member_in_ag(.) → U3_ag(member_in_aa)
U3_ag(member_out_aa(Xs)) → member_out_ag(.)
member_in_aa → member_out_aa(.)
member_in_aa → U3_aa(member_in_aa)
U3_aa(member_out_aa(Xs)) → member_out_aa(.)

The set Q consists of the following terms:

member_in_ag(x0)
U3_ag(x0)
member_in_aa
U3_aa(x0)

We have to consider all (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ UsableRulesProof
                              ↳ QDP
                                ↳ QReductionProof

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

SUBSET_IN_AG(.) → U1_AG(., member_out_ag(.))
SUBSET_IN_AG(.) → U1_AG(., U3_ag(member_in_aa))
U1_AG(Ys, member_out_ag(Ys)) → SUBSET_IN_AG(Ys)

The TRS R consists of the following rules:

member_in_aa → member_out_aa(.)
member_in_aa → U3_aa(member_in_aa)
U3_ag(member_out_aa(Xs)) → member_out_ag(.)
U3_aa(member_out_aa(Xs)) → member_out_aa(.)

The set Q consists of the following terms:

member_in_ag(x0)
U3_ag(x0)
member_in_aa
U3_aa(x0)

We have to consider all (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

member_in_ag(x0)



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ UsableRulesProof
                              ↳ QDP
                                ↳ QReductionProof
                                  ↳ QDP
                                    ↳ Instantiation

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

SUBSET_IN_AG(.) → U1_AG(., member_out_ag(.))
SUBSET_IN_AG(.) → U1_AG(., U3_ag(member_in_aa))
U1_AG(Ys, member_out_ag(Ys)) → SUBSET_IN_AG(Ys)

The TRS R consists of the following rules:

member_in_aa → member_out_aa(.)
member_in_aa → U3_aa(member_in_aa)
U3_ag(member_out_aa(Xs)) → member_out_ag(.)
U3_aa(member_out_aa(Xs)) → member_out_aa(.)

The set Q consists of the following terms:

U3_ag(x0)
member_in_aa
U3_aa(x0)

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

U1_AG(., member_out_ag(.)) → SUBSET_IN_AG(.)



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ UsableRulesProof
                              ↳ QDP
                                ↳ QReductionProof
                                  ↳ QDP
                                    ↳ Instantiation
                                      ↳ QDP
                                        ↳ NonTerminationProof

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

SUBSET_IN_AG(.) → U1_AG(., member_out_ag(.))
U1_AG(., member_out_ag(.)) → SUBSET_IN_AG(.)
SUBSET_IN_AG(.) → U1_AG(., U3_ag(member_in_aa))

The TRS R consists of the following rules:

member_in_aa → member_out_aa(.)
member_in_aa → U3_aa(member_in_aa)
U3_ag(member_out_aa(Xs)) → member_out_ag(.)
U3_aa(member_out_aa(Xs)) → member_out_aa(.)

The set Q consists of the following terms:

U3_ag(x0)
member_in_aa
U3_aa(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:

SUBSET_IN_AG(.) → U1_AG(., member_out_ag(.))
U1_AG(., member_out_ag(.)) → SUBSET_IN_AG(.)
SUBSET_IN_AG(.) → U1_AG(., U3_ag(member_in_aa))

The TRS R consists of the following rules:

member_in_aa → member_out_aa(.)
member_in_aa → U3_aa(member_in_aa)
U3_ag(member_out_aa(Xs)) → member_out_ag(.)
U3_aa(member_out_aa(Xs)) → member_out_aa(.)


s = U1_AG(., member_out_ag(.)) evaluates to t =U1_AG(., member_out_ag(.))

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



Rewriting sequence

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

SUBSET_IN_AG(.) → U1_AG(., member_out_ag(.))
with rule SUBSET_IN_AG(.) → U1_AG(., member_out_ag(.))

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.