Left Termination of the query pattern subset1_in_2(a, g) w.r.t. the given Prolog program could not be shown:

0 Prolog
  ↳1 PrologToPiTRSProof (⇐)
    ↳2 PiTRS
    ↳3 DependencyPairsProof (⇔)
    ↳4 PiDP
    ↳5 DependencyGraphProof (⇔)
    ↳6 AND
      ↳7 PiDP
        ↳8 UsableRulesProof (⇔)
        ↳9 PiDP
        ↳10 PiDPToQDPProof (⇐)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔)
        ↳13 TRUE
      ↳14 PiDP
        ↳15 UsableRulesProof (⇔)
        ↳16 PiDP
        ↳17 PiDPToQDPProof (⇐)
        ↳18 QDP
        ↳19 Narrowing (⇐)
        ↳20 QDP
        ↳21 Instantiation (⇔)
        ↳22 QDP
        ↳23 NonTerminationProof (⇔)
        ↳24 FALSE
  ↳25 PrologToPiTRSProof (⇐)
    ↳26 PiTRS
    ↳27 DependencyPairsProof (⇔)
    ↳28 PiDP
    ↳29 DependencyGraphProof (⇔)
    ↳30 AND
      ↳31 PiDP
        ↳32 UsableRulesProof (⇔)
        ↳33 PiDP
        ↳34 PiDPToQDPProof (⇐)
        ↳35 QDP
        ↳36 QDPSizeChangeProof (⇔)
        ↳37 TRUE
      ↳38 PiDP
        ↳39 UsableRulesProof (⇔)
        ↳40 PiDP
        ↳41 PiDPToQDPProof (⇐)
        ↳42 QDP
        ↳43 Narrowing (⇐)
        ↳44 QDP
        ↳45 Instantiation (⇔)
        ↳46 QDP
        ↳47 NonTerminationProof (⇔)
        ↳48 FALSE

(0) Obligation:

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:

subset1(a,g).

(1) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
subset1_in: (f,b)
member1_in: (f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

subset1_in_ag(.(X, Xs), Ys) → U5_ag(X, Xs, Ys, member1_in_ag(X, Ys))
member1_in_ag(X, .(Y, Xs)) → U4_ag(X, Y, Xs, member1_in_ag(X, Xs))
member1_in_ag(X, .(X, Xs)) → member1_out_ag(X, .(X, Xs))
U4_ag(X, Y, Xs, member1_out_ag(X, Xs)) → member1_out_ag(X, .(Y, Xs))
U5_ag(X, Xs, Ys, member1_out_ag(X, Ys)) → U6_ag(X, Xs, Ys, subset1_in_ag(Xs, Ys))
subset1_in_ag([], Ys) → subset1_out_ag([], Ys)
U6_ag(X, Xs, Ys, subset1_out_ag(Xs, Ys)) → subset1_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_in_ag(x1, x2)  =  subset1_in_ag(x2)
U5_ag(x1, x2, x3, x4)  =  U5_ag(x3, x4)
member1_in_ag(x1, x2)  =  member1_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
U4_ag(x1, x2, x3, x4)  =  U4_ag(x4)
member1_out_ag(x1, x2)  =  member1_out_ag(x1)
U6_ag(x1, x2, x3, x4)  =  U6_ag(x1, x4)
subset1_out_ag(x1, x2)  =  subset1_out_ag(x1)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(2) Obligation:

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

subset1_in_ag(.(X, Xs), Ys) → U5_ag(X, Xs, Ys, member1_in_ag(X, Ys))
member1_in_ag(X, .(Y, Xs)) → U4_ag(X, Y, Xs, member1_in_ag(X, Xs))
member1_in_ag(X, .(X, Xs)) → member1_out_ag(X, .(X, Xs))
U4_ag(X, Y, Xs, member1_out_ag(X, Xs)) → member1_out_ag(X, .(Y, Xs))
U5_ag(X, Xs, Ys, member1_out_ag(X, Ys)) → U6_ag(X, Xs, Ys, subset1_in_ag(Xs, Ys))
subset1_in_ag([], Ys) → subset1_out_ag([], Ys)
U6_ag(X, Xs, Ys, subset1_out_ag(Xs, Ys)) → subset1_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_in_ag(x1, x2)  =  subset1_in_ag(x2)
U5_ag(x1, x2, x3, x4)  =  U5_ag(x3, x4)
member1_in_ag(x1, x2)  =  member1_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
U4_ag(x1, x2, x3, x4)  =  U4_ag(x4)
member1_out_ag(x1, x2)  =  member1_out_ag(x1)
U6_ag(x1, x2, x3, x4)  =  U6_ag(x1, x4)
subset1_out_ag(x1, x2)  =  subset1_out_ag(x1)

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

SUBSET1_IN_AG(.(X, Xs), Ys) → U5_AG(X, Xs, Ys, member1_in_ag(X, Ys))
SUBSET1_IN_AG(.(X, Xs), Ys) → MEMBER1_IN_AG(X, Ys)
MEMBER1_IN_AG(X, .(Y, Xs)) → U4_AG(X, Y, Xs, member1_in_ag(X, Xs))
MEMBER1_IN_AG(X, .(Y, Xs)) → MEMBER1_IN_AG(X, Xs)
U5_AG(X, Xs, Ys, member1_out_ag(X, Ys)) → U6_AG(X, Xs, Ys, subset1_in_ag(Xs, Ys))
U5_AG(X, Xs, Ys, member1_out_ag(X, Ys)) → SUBSET1_IN_AG(Xs, Ys)

The TRS R consists of the following rules:

subset1_in_ag(.(X, Xs), Ys) → U5_ag(X, Xs, Ys, member1_in_ag(X, Ys))
member1_in_ag(X, .(Y, Xs)) → U4_ag(X, Y, Xs, member1_in_ag(X, Xs))
member1_in_ag(X, .(X, Xs)) → member1_out_ag(X, .(X, Xs))
U4_ag(X, Y, Xs, member1_out_ag(X, Xs)) → member1_out_ag(X, .(Y, Xs))
U5_ag(X, Xs, Ys, member1_out_ag(X, Ys)) → U6_ag(X, Xs, Ys, subset1_in_ag(Xs, Ys))
subset1_in_ag([], Ys) → subset1_out_ag([], Ys)
U6_ag(X, Xs, Ys, subset1_out_ag(Xs, Ys)) → subset1_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_in_ag(x1, x2)  =  subset1_in_ag(x2)
U5_ag(x1, x2, x3, x4)  =  U5_ag(x3, x4)
member1_in_ag(x1, x2)  =  member1_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
U4_ag(x1, x2, x3, x4)  =  U4_ag(x4)
member1_out_ag(x1, x2)  =  member1_out_ag(x1)
U6_ag(x1, x2, x3, x4)  =  U6_ag(x1, x4)
subset1_out_ag(x1, x2)  =  subset1_out_ag(x1)
SUBSET1_IN_AG(x1, x2)  =  SUBSET1_IN_AG(x2)
U5_AG(x1, x2, x3, x4)  =  U5_AG(x3, x4)
MEMBER1_IN_AG(x1, x2)  =  MEMBER1_IN_AG(x2)
U4_AG(x1, x2, x3, x4)  =  U4_AG(x4)
U6_AG(x1, x2, x3, x4)  =  U6_AG(x1, x4)

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

(4) Obligation:

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

SUBSET1_IN_AG(.(X, Xs), Ys) → U5_AG(X, Xs, Ys, member1_in_ag(X, Ys))
SUBSET1_IN_AG(.(X, Xs), Ys) → MEMBER1_IN_AG(X, Ys)
MEMBER1_IN_AG(X, .(Y, Xs)) → U4_AG(X, Y, Xs, member1_in_ag(X, Xs))
MEMBER1_IN_AG(X, .(Y, Xs)) → MEMBER1_IN_AG(X, Xs)
U5_AG(X, Xs, Ys, member1_out_ag(X, Ys)) → U6_AG(X, Xs, Ys, subset1_in_ag(Xs, Ys))
U5_AG(X, Xs, Ys, member1_out_ag(X, Ys)) → SUBSET1_IN_AG(Xs, Ys)

The TRS R consists of the following rules:

subset1_in_ag(.(X, Xs), Ys) → U5_ag(X, Xs, Ys, member1_in_ag(X, Ys))
member1_in_ag(X, .(Y, Xs)) → U4_ag(X, Y, Xs, member1_in_ag(X, Xs))
member1_in_ag(X, .(X, Xs)) → member1_out_ag(X, .(X, Xs))
U4_ag(X, Y, Xs, member1_out_ag(X, Xs)) → member1_out_ag(X, .(Y, Xs))
U5_ag(X, Xs, Ys, member1_out_ag(X, Ys)) → U6_ag(X, Xs, Ys, subset1_in_ag(Xs, Ys))
subset1_in_ag([], Ys) → subset1_out_ag([], Ys)
U6_ag(X, Xs, Ys, subset1_out_ag(Xs, Ys)) → subset1_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_in_ag(x1, x2)  =  subset1_in_ag(x2)
U5_ag(x1, x2, x3, x4)  =  U5_ag(x3, x4)
member1_in_ag(x1, x2)  =  member1_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
U4_ag(x1, x2, x3, x4)  =  U4_ag(x4)
member1_out_ag(x1, x2)  =  member1_out_ag(x1)
U6_ag(x1, x2, x3, x4)  =  U6_ag(x1, x4)
subset1_out_ag(x1, x2)  =  subset1_out_ag(x1)
SUBSET1_IN_AG(x1, x2)  =  SUBSET1_IN_AG(x2)
U5_AG(x1, x2, x3, x4)  =  U5_AG(x3, x4)
MEMBER1_IN_AG(x1, x2)  =  MEMBER1_IN_AG(x2)
U4_AG(x1, x2, x3, x4)  =  U4_AG(x4)
U6_AG(x1, x2, x3, x4)  =  U6_AG(x1, x4)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

MEMBER1_IN_AG(X, .(Y, Xs)) → MEMBER1_IN_AG(X, Xs)

The TRS R consists of the following rules:

subset1_in_ag(.(X, Xs), Ys) → U5_ag(X, Xs, Ys, member1_in_ag(X, Ys))
member1_in_ag(X, .(Y, Xs)) → U4_ag(X, Y, Xs, member1_in_ag(X, Xs))
member1_in_ag(X, .(X, Xs)) → member1_out_ag(X, .(X, Xs))
U4_ag(X, Y, Xs, member1_out_ag(X, Xs)) → member1_out_ag(X, .(Y, Xs))
U5_ag(X, Xs, Ys, member1_out_ag(X, Ys)) → U6_ag(X, Xs, Ys, subset1_in_ag(Xs, Ys))
subset1_in_ag([], Ys) → subset1_out_ag([], Ys)
U6_ag(X, Xs, Ys, subset1_out_ag(Xs, Ys)) → subset1_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_in_ag(x1, x2)  =  subset1_in_ag(x2)
U5_ag(x1, x2, x3, x4)  =  U5_ag(x3, x4)
member1_in_ag(x1, x2)  =  member1_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
U4_ag(x1, x2, x3, x4)  =  U4_ag(x4)
member1_out_ag(x1, x2)  =  member1_out_ag(x1)
U6_ag(x1, x2, x3, x4)  =  U6_ag(x1, x4)
subset1_out_ag(x1, x2)  =  subset1_out_ag(x1)
MEMBER1_IN_AG(x1, x2)  =  MEMBER1_IN_AG(x2)

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

(8) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(9) Obligation:

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

MEMBER1_IN_AG(X, .(Y, Xs)) → MEMBER1_IN_AG(X, Xs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
MEMBER1_IN_AG(x1, x2)  =  MEMBER1_IN_AG(x2)

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

(10) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(11) Obligation:

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

MEMBER1_IN_AG(.(Y, Xs)) → MEMBER1_IN_AG(Xs)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

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

  • MEMBER1_IN_AG(.(Y, Xs)) → MEMBER1_IN_AG(Xs)
    The graph contains the following edges 1 > 1

(13) TRUE

(14) Obligation:

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

U5_AG(X, Xs, Ys, member1_out_ag(X, Ys)) → SUBSET1_IN_AG(Xs, Ys)
SUBSET1_IN_AG(.(X, Xs), Ys) → U5_AG(X, Xs, Ys, member1_in_ag(X, Ys))

The TRS R consists of the following rules:

subset1_in_ag(.(X, Xs), Ys) → U5_ag(X, Xs, Ys, member1_in_ag(X, Ys))
member1_in_ag(X, .(Y, Xs)) → U4_ag(X, Y, Xs, member1_in_ag(X, Xs))
member1_in_ag(X, .(X, Xs)) → member1_out_ag(X, .(X, Xs))
U4_ag(X, Y, Xs, member1_out_ag(X, Xs)) → member1_out_ag(X, .(Y, Xs))
U5_ag(X, Xs, Ys, member1_out_ag(X, Ys)) → U6_ag(X, Xs, Ys, subset1_in_ag(Xs, Ys))
subset1_in_ag([], Ys) → subset1_out_ag([], Ys)
U6_ag(X, Xs, Ys, subset1_out_ag(Xs, Ys)) → subset1_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_in_ag(x1, x2)  =  subset1_in_ag(x2)
U5_ag(x1, x2, x3, x4)  =  U5_ag(x3, x4)
member1_in_ag(x1, x2)  =  member1_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
U4_ag(x1, x2, x3, x4)  =  U4_ag(x4)
member1_out_ag(x1, x2)  =  member1_out_ag(x1)
U6_ag(x1, x2, x3, x4)  =  U6_ag(x1, x4)
subset1_out_ag(x1, x2)  =  subset1_out_ag(x1)
SUBSET1_IN_AG(x1, x2)  =  SUBSET1_IN_AG(x2)
U5_AG(x1, x2, x3, x4)  =  U5_AG(x3, x4)

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

(15) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(16) Obligation:

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

U5_AG(X, Xs, Ys, member1_out_ag(X, Ys)) → SUBSET1_IN_AG(Xs, Ys)
SUBSET1_IN_AG(.(X, Xs), Ys) → U5_AG(X, Xs, Ys, member1_in_ag(X, Ys))

The TRS R consists of the following rules:

member1_in_ag(X, .(Y, Xs)) → U4_ag(X, Y, Xs, member1_in_ag(X, Xs))
member1_in_ag(X, .(X, Xs)) → member1_out_ag(X, .(X, Xs))
U4_ag(X, Y, Xs, member1_out_ag(X, Xs)) → member1_out_ag(X, .(Y, Xs))

The argument filtering Pi contains the following mapping:
member1_in_ag(x1, x2)  =  member1_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
U4_ag(x1, x2, x3, x4)  =  U4_ag(x4)
member1_out_ag(x1, x2)  =  member1_out_ag(x1)
SUBSET1_IN_AG(x1, x2)  =  SUBSET1_IN_AG(x2)
U5_AG(x1, x2, x3, x4)  =  U5_AG(x3, x4)

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

(17) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(18) Obligation:

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

U5_AG(Ys, member1_out_ag(X)) → SUBSET1_IN_AG(Ys)
SUBSET1_IN_AG(Ys) → U5_AG(Ys, member1_in_ag(Ys))

The TRS R consists of the following rules:

member1_in_ag(.(Y, Xs)) → U4_ag(member1_in_ag(Xs))
member1_in_ag(.(X, Xs)) → member1_out_ag(X)
U4_ag(member1_out_ag(X)) → member1_out_ag(X)

The set Q consists of the following terms:

member1_in_ag(x0)
U4_ag(x0)

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

(19) Narrowing (SOUND transformation)

By narrowing [LPAR04] the rule SUBSET1_IN_AG(Ys) → U5_AG(Ys, member1_in_ag(Ys)) at position [1] we obtained the following new rules [LPAR04]:

SUBSET1_IN_AG(.(x0, x1)) → U5_AG(.(x0, x1), U4_ag(member1_in_ag(x1)))
SUBSET1_IN_AG(.(x0, x1)) → U5_AG(.(x0, x1), member1_out_ag(x0))

(20) Obligation:

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

U5_AG(Ys, member1_out_ag(X)) → SUBSET1_IN_AG(Ys)
SUBSET1_IN_AG(.(x0, x1)) → U5_AG(.(x0, x1), U4_ag(member1_in_ag(x1)))
SUBSET1_IN_AG(.(x0, x1)) → U5_AG(.(x0, x1), member1_out_ag(x0))

The TRS R consists of the following rules:

member1_in_ag(.(Y, Xs)) → U4_ag(member1_in_ag(Xs))
member1_in_ag(.(X, Xs)) → member1_out_ag(X)
U4_ag(member1_out_ag(X)) → member1_out_ag(X)

The set Q consists of the following terms:

member1_in_ag(x0)
U4_ag(x0)

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

(21) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule U5_AG(Ys, member1_out_ag(X)) → SUBSET1_IN_AG(Ys) we obtained the following new rules [LPAR04]:

U5_AG(.(z0, z1), member1_out_ag(x1)) → SUBSET1_IN_AG(.(z0, z1))
U5_AG(.(z0, z1), member1_out_ag(z0)) → SUBSET1_IN_AG(.(z0, z1))

(22) Obligation:

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

SUBSET1_IN_AG(.(x0, x1)) → U5_AG(.(x0, x1), U4_ag(member1_in_ag(x1)))
SUBSET1_IN_AG(.(x0, x1)) → U5_AG(.(x0, x1), member1_out_ag(x0))
U5_AG(.(z0, z1), member1_out_ag(x1)) → SUBSET1_IN_AG(.(z0, z1))
U5_AG(.(z0, z1), member1_out_ag(z0)) → SUBSET1_IN_AG(.(z0, z1))

The TRS R consists of the following rules:

member1_in_ag(.(Y, Xs)) → U4_ag(member1_in_ag(Xs))
member1_in_ag(.(X, Xs)) → member1_out_ag(X)
U4_ag(member1_out_ag(X)) → member1_out_ag(X)

The set Q consists of the following terms:

member1_in_ag(x0)
U4_ag(x0)

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

(23) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = U5_AG(.(z0, z1), member1_out_ag(x1')) evaluates to t =U5_AG(.(z0, z1), member1_out_ag(z0))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Semiunifier: [ ]
  • Matcher: [x1' / z0]



Rewriting sequence

U5_AG(.(z0, z1), member1_out_ag(x1')) → SUBSET1_IN_AG(.(z0, z1))
with rule U5_AG(.(z0', z1'), member1_out_ag(x1'')) → SUBSET1_IN_AG(.(z0', z1')) at position [] and matcher [z0' / z0, z1' / z1, x1'' / x1']

SUBSET1_IN_AG(.(z0, z1)) → U5_AG(.(z0, z1), member1_out_ag(z0))
with rule SUBSET1_IN_AG(.(x0, x1)) → U5_AG(.(x0, x1), member1_out_ag(x0))

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.



(24) FALSE

(25) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
subset1_in: (f,b)
member1_in: (f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

subset1_in_ag(.(X, Xs), Ys) → U5_ag(X, Xs, Ys, member1_in_ag(X, Ys))
member1_in_ag(X, .(Y, Xs)) → U4_ag(X, Y, Xs, member1_in_ag(X, Xs))
member1_in_ag(X, .(X, Xs)) → member1_out_ag(X, .(X, Xs))
U4_ag(X, Y, Xs, member1_out_ag(X, Xs)) → member1_out_ag(X, .(Y, Xs))
U5_ag(X, Xs, Ys, member1_out_ag(X, Ys)) → U6_ag(X, Xs, Ys, subset1_in_ag(Xs, Ys))
subset1_in_ag([], Ys) → subset1_out_ag([], Ys)
U6_ag(X, Xs, Ys, subset1_out_ag(Xs, Ys)) → subset1_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_in_ag(x1, x2)  =  subset1_in_ag(x2)
U5_ag(x1, x2, x3, x4)  =  U5_ag(x3, x4)
member1_in_ag(x1, x2)  =  member1_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
U4_ag(x1, x2, x3, x4)  =  U4_ag(x2, x3, x4)
member1_out_ag(x1, x2)  =  member1_out_ag(x1, x2)
U6_ag(x1, x2, x3, x4)  =  U6_ag(x1, x3, x4)
subset1_out_ag(x1, x2)  =  subset1_out_ag(x1, x2)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(26) Obligation:

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

subset1_in_ag(.(X, Xs), Ys) → U5_ag(X, Xs, Ys, member1_in_ag(X, Ys))
member1_in_ag(X, .(Y, Xs)) → U4_ag(X, Y, Xs, member1_in_ag(X, Xs))
member1_in_ag(X, .(X, Xs)) → member1_out_ag(X, .(X, Xs))
U4_ag(X, Y, Xs, member1_out_ag(X, Xs)) → member1_out_ag(X, .(Y, Xs))
U5_ag(X, Xs, Ys, member1_out_ag(X, Ys)) → U6_ag(X, Xs, Ys, subset1_in_ag(Xs, Ys))
subset1_in_ag([], Ys) → subset1_out_ag([], Ys)
U6_ag(X, Xs, Ys, subset1_out_ag(Xs, Ys)) → subset1_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_in_ag(x1, x2)  =  subset1_in_ag(x2)
U5_ag(x1, x2, x3, x4)  =  U5_ag(x3, x4)
member1_in_ag(x1, x2)  =  member1_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
U4_ag(x1, x2, x3, x4)  =  U4_ag(x2, x3, x4)
member1_out_ag(x1, x2)  =  member1_out_ag(x1, x2)
U6_ag(x1, x2, x3, x4)  =  U6_ag(x1, x3, x4)
subset1_out_ag(x1, x2)  =  subset1_out_ag(x1, x2)

(27) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

SUBSET1_IN_AG(.(X, Xs), Ys) → U5_AG(X, Xs, Ys, member1_in_ag(X, Ys))
SUBSET1_IN_AG(.(X, Xs), Ys) → MEMBER1_IN_AG(X, Ys)
MEMBER1_IN_AG(X, .(Y, Xs)) → U4_AG(X, Y, Xs, member1_in_ag(X, Xs))
MEMBER1_IN_AG(X, .(Y, Xs)) → MEMBER1_IN_AG(X, Xs)
U5_AG(X, Xs, Ys, member1_out_ag(X, Ys)) → U6_AG(X, Xs, Ys, subset1_in_ag(Xs, Ys))
U5_AG(X, Xs, Ys, member1_out_ag(X, Ys)) → SUBSET1_IN_AG(Xs, Ys)

The TRS R consists of the following rules:

subset1_in_ag(.(X, Xs), Ys) → U5_ag(X, Xs, Ys, member1_in_ag(X, Ys))
member1_in_ag(X, .(Y, Xs)) → U4_ag(X, Y, Xs, member1_in_ag(X, Xs))
member1_in_ag(X, .(X, Xs)) → member1_out_ag(X, .(X, Xs))
U4_ag(X, Y, Xs, member1_out_ag(X, Xs)) → member1_out_ag(X, .(Y, Xs))
U5_ag(X, Xs, Ys, member1_out_ag(X, Ys)) → U6_ag(X, Xs, Ys, subset1_in_ag(Xs, Ys))
subset1_in_ag([], Ys) → subset1_out_ag([], Ys)
U6_ag(X, Xs, Ys, subset1_out_ag(Xs, Ys)) → subset1_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_in_ag(x1, x2)  =  subset1_in_ag(x2)
U5_ag(x1, x2, x3, x4)  =  U5_ag(x3, x4)
member1_in_ag(x1, x2)  =  member1_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
U4_ag(x1, x2, x3, x4)  =  U4_ag(x2, x3, x4)
member1_out_ag(x1, x2)  =  member1_out_ag(x1, x2)
U6_ag(x1, x2, x3, x4)  =  U6_ag(x1, x3, x4)
subset1_out_ag(x1, x2)  =  subset1_out_ag(x1, x2)
SUBSET1_IN_AG(x1, x2)  =  SUBSET1_IN_AG(x2)
U5_AG(x1, x2, x3, x4)  =  U5_AG(x3, x4)
MEMBER1_IN_AG(x1, x2)  =  MEMBER1_IN_AG(x2)
U4_AG(x1, x2, x3, x4)  =  U4_AG(x2, x3, x4)
U6_AG(x1, x2, x3, x4)  =  U6_AG(x1, x3, x4)

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

(28) Obligation:

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

SUBSET1_IN_AG(.(X, Xs), Ys) → U5_AG(X, Xs, Ys, member1_in_ag(X, Ys))
SUBSET1_IN_AG(.(X, Xs), Ys) → MEMBER1_IN_AG(X, Ys)
MEMBER1_IN_AG(X, .(Y, Xs)) → U4_AG(X, Y, Xs, member1_in_ag(X, Xs))
MEMBER1_IN_AG(X, .(Y, Xs)) → MEMBER1_IN_AG(X, Xs)
U5_AG(X, Xs, Ys, member1_out_ag(X, Ys)) → U6_AG(X, Xs, Ys, subset1_in_ag(Xs, Ys))
U5_AG(X, Xs, Ys, member1_out_ag(X, Ys)) → SUBSET1_IN_AG(Xs, Ys)

The TRS R consists of the following rules:

subset1_in_ag(.(X, Xs), Ys) → U5_ag(X, Xs, Ys, member1_in_ag(X, Ys))
member1_in_ag(X, .(Y, Xs)) → U4_ag(X, Y, Xs, member1_in_ag(X, Xs))
member1_in_ag(X, .(X, Xs)) → member1_out_ag(X, .(X, Xs))
U4_ag(X, Y, Xs, member1_out_ag(X, Xs)) → member1_out_ag(X, .(Y, Xs))
U5_ag(X, Xs, Ys, member1_out_ag(X, Ys)) → U6_ag(X, Xs, Ys, subset1_in_ag(Xs, Ys))
subset1_in_ag([], Ys) → subset1_out_ag([], Ys)
U6_ag(X, Xs, Ys, subset1_out_ag(Xs, Ys)) → subset1_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_in_ag(x1, x2)  =  subset1_in_ag(x2)
U5_ag(x1, x2, x3, x4)  =  U5_ag(x3, x4)
member1_in_ag(x1, x2)  =  member1_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
U4_ag(x1, x2, x3, x4)  =  U4_ag(x2, x3, x4)
member1_out_ag(x1, x2)  =  member1_out_ag(x1, x2)
U6_ag(x1, x2, x3, x4)  =  U6_ag(x1, x3, x4)
subset1_out_ag(x1, x2)  =  subset1_out_ag(x1, x2)
SUBSET1_IN_AG(x1, x2)  =  SUBSET1_IN_AG(x2)
U5_AG(x1, x2, x3, x4)  =  U5_AG(x3, x4)
MEMBER1_IN_AG(x1, x2)  =  MEMBER1_IN_AG(x2)
U4_AG(x1, x2, x3, x4)  =  U4_AG(x2, x3, x4)
U6_AG(x1, x2, x3, x4)  =  U6_AG(x1, x3, x4)

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

(29) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 3 less nodes.

(30) Complex Obligation (AND)

(31) Obligation:

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

MEMBER1_IN_AG(X, .(Y, Xs)) → MEMBER1_IN_AG(X, Xs)

The TRS R consists of the following rules:

subset1_in_ag(.(X, Xs), Ys) → U5_ag(X, Xs, Ys, member1_in_ag(X, Ys))
member1_in_ag(X, .(Y, Xs)) → U4_ag(X, Y, Xs, member1_in_ag(X, Xs))
member1_in_ag(X, .(X, Xs)) → member1_out_ag(X, .(X, Xs))
U4_ag(X, Y, Xs, member1_out_ag(X, Xs)) → member1_out_ag(X, .(Y, Xs))
U5_ag(X, Xs, Ys, member1_out_ag(X, Ys)) → U6_ag(X, Xs, Ys, subset1_in_ag(Xs, Ys))
subset1_in_ag([], Ys) → subset1_out_ag([], Ys)
U6_ag(X, Xs, Ys, subset1_out_ag(Xs, Ys)) → subset1_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_in_ag(x1, x2)  =  subset1_in_ag(x2)
U5_ag(x1, x2, x3, x4)  =  U5_ag(x3, x4)
member1_in_ag(x1, x2)  =  member1_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
U4_ag(x1, x2, x3, x4)  =  U4_ag(x2, x3, x4)
member1_out_ag(x1, x2)  =  member1_out_ag(x1, x2)
U6_ag(x1, x2, x3, x4)  =  U6_ag(x1, x3, x4)
subset1_out_ag(x1, x2)  =  subset1_out_ag(x1, x2)
MEMBER1_IN_AG(x1, x2)  =  MEMBER1_IN_AG(x2)

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

(32) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(33) Obligation:

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

MEMBER1_IN_AG(X, .(Y, Xs)) → MEMBER1_IN_AG(X, Xs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
MEMBER1_IN_AG(x1, x2)  =  MEMBER1_IN_AG(x2)

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

(34) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(35) Obligation:

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

MEMBER1_IN_AG(.(Y, Xs)) → MEMBER1_IN_AG(Xs)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(36) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

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

  • MEMBER1_IN_AG(.(Y, Xs)) → MEMBER1_IN_AG(Xs)
    The graph contains the following edges 1 > 1

(37) TRUE

(38) Obligation:

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

U5_AG(X, Xs, Ys, member1_out_ag(X, Ys)) → SUBSET1_IN_AG(Xs, Ys)
SUBSET1_IN_AG(.(X, Xs), Ys) → U5_AG(X, Xs, Ys, member1_in_ag(X, Ys))

The TRS R consists of the following rules:

subset1_in_ag(.(X, Xs), Ys) → U5_ag(X, Xs, Ys, member1_in_ag(X, Ys))
member1_in_ag(X, .(Y, Xs)) → U4_ag(X, Y, Xs, member1_in_ag(X, Xs))
member1_in_ag(X, .(X, Xs)) → member1_out_ag(X, .(X, Xs))
U4_ag(X, Y, Xs, member1_out_ag(X, Xs)) → member1_out_ag(X, .(Y, Xs))
U5_ag(X, Xs, Ys, member1_out_ag(X, Ys)) → U6_ag(X, Xs, Ys, subset1_in_ag(Xs, Ys))
subset1_in_ag([], Ys) → subset1_out_ag([], Ys)
U6_ag(X, Xs, Ys, subset1_out_ag(Xs, Ys)) → subset1_out_ag(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset1_in_ag(x1, x2)  =  subset1_in_ag(x2)
U5_ag(x1, x2, x3, x4)  =  U5_ag(x3, x4)
member1_in_ag(x1, x2)  =  member1_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
U4_ag(x1, x2, x3, x4)  =  U4_ag(x2, x3, x4)
member1_out_ag(x1, x2)  =  member1_out_ag(x1, x2)
U6_ag(x1, x2, x3, x4)  =  U6_ag(x1, x3, x4)
subset1_out_ag(x1, x2)  =  subset1_out_ag(x1, x2)
SUBSET1_IN_AG(x1, x2)  =  SUBSET1_IN_AG(x2)
U5_AG(x1, x2, x3, x4)  =  U5_AG(x3, x4)

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

(39) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(40) Obligation:

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

U5_AG(X, Xs, Ys, member1_out_ag(X, Ys)) → SUBSET1_IN_AG(Xs, Ys)
SUBSET1_IN_AG(.(X, Xs), Ys) → U5_AG(X, Xs, Ys, member1_in_ag(X, Ys))

The TRS R consists of the following rules:

member1_in_ag(X, .(Y, Xs)) → U4_ag(X, Y, Xs, member1_in_ag(X, Xs))
member1_in_ag(X, .(X, Xs)) → member1_out_ag(X, .(X, Xs))
U4_ag(X, Y, Xs, member1_out_ag(X, Xs)) → member1_out_ag(X, .(Y, Xs))

The argument filtering Pi contains the following mapping:
member1_in_ag(x1, x2)  =  member1_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
U4_ag(x1, x2, x3, x4)  =  U4_ag(x2, x3, x4)
member1_out_ag(x1, x2)  =  member1_out_ag(x1, x2)
SUBSET1_IN_AG(x1, x2)  =  SUBSET1_IN_AG(x2)
U5_AG(x1, x2, x3, x4)  =  U5_AG(x3, x4)

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

(41) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(42) Obligation:

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

U5_AG(Ys, member1_out_ag(X, Ys)) → SUBSET1_IN_AG(Ys)
SUBSET1_IN_AG(Ys) → U5_AG(Ys, member1_in_ag(Ys))

The TRS R consists of the following rules:

member1_in_ag(.(Y, Xs)) → U4_ag(Y, Xs, member1_in_ag(Xs))
member1_in_ag(.(X, Xs)) → member1_out_ag(X, .(X, Xs))
U4_ag(Y, Xs, member1_out_ag(X, Xs)) → member1_out_ag(X, .(Y, Xs))

The set Q consists of the following terms:

member1_in_ag(x0)
U4_ag(x0, x1, x2)

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

(43) Narrowing (SOUND transformation)

By narrowing [LPAR04] the rule SUBSET1_IN_AG(Ys) → U5_AG(Ys, member1_in_ag(Ys)) at position [1] we obtained the following new rules [LPAR04]:

SUBSET1_IN_AG(.(x0, x1)) → U5_AG(.(x0, x1), U4_ag(x0, x1, member1_in_ag(x1)))
SUBSET1_IN_AG(.(x0, x1)) → U5_AG(.(x0, x1), member1_out_ag(x0, .(x0, x1)))

(44) Obligation:

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

U5_AG(Ys, member1_out_ag(X, Ys)) → SUBSET1_IN_AG(Ys)
SUBSET1_IN_AG(.(x0, x1)) → U5_AG(.(x0, x1), U4_ag(x0, x1, member1_in_ag(x1)))
SUBSET1_IN_AG(.(x0, x1)) → U5_AG(.(x0, x1), member1_out_ag(x0, .(x0, x1)))

The TRS R consists of the following rules:

member1_in_ag(.(Y, Xs)) → U4_ag(Y, Xs, member1_in_ag(Xs))
member1_in_ag(.(X, Xs)) → member1_out_ag(X, .(X, Xs))
U4_ag(Y, Xs, member1_out_ag(X, Xs)) → member1_out_ag(X, .(Y, Xs))

The set Q consists of the following terms:

member1_in_ag(x0)
U4_ag(x0, x1, x2)

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

(45) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule U5_AG(Ys, member1_out_ag(X, Ys)) → SUBSET1_IN_AG(Ys) we obtained the following new rules [LPAR04]:

U5_AG(.(z0, z1), member1_out_ag(x1, .(z0, z1))) → SUBSET1_IN_AG(.(z0, z1))
U5_AG(.(z0, z1), member1_out_ag(z0, .(z0, z1))) → SUBSET1_IN_AG(.(z0, z1))

(46) Obligation:

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

SUBSET1_IN_AG(.(x0, x1)) → U5_AG(.(x0, x1), U4_ag(x0, x1, member1_in_ag(x1)))
SUBSET1_IN_AG(.(x0, x1)) → U5_AG(.(x0, x1), member1_out_ag(x0, .(x0, x1)))
U5_AG(.(z0, z1), member1_out_ag(x1, .(z0, z1))) → SUBSET1_IN_AG(.(z0, z1))
U5_AG(.(z0, z1), member1_out_ag(z0, .(z0, z1))) → SUBSET1_IN_AG(.(z0, z1))

The TRS R consists of the following rules:

member1_in_ag(.(Y, Xs)) → U4_ag(Y, Xs, member1_in_ag(Xs))
member1_in_ag(.(X, Xs)) → member1_out_ag(X, .(X, Xs))
U4_ag(Y, Xs, member1_out_ag(X, Xs)) → member1_out_ag(X, .(Y, Xs))

The set Q consists of the following terms:

member1_in_ag(x0)
U4_ag(x0, x1, x2)

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

(47) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = U5_AG(.(z0, z1), member1_out_ag(x1', .(z0, z1))) evaluates to t =U5_AG(.(z0, z1), member1_out_ag(z0, .(z0, z1)))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [x1' / z0]
  • Semiunifier: [ ]



Rewriting sequence

U5_AG(.(z0, z1), member1_out_ag(x1', .(z0, z1))) → SUBSET1_IN_AG(.(z0, z1))
with rule U5_AG(.(z0', z1'), member1_out_ag(x1'', .(z0', z1'))) → SUBSET1_IN_AG(.(z0', z1')) at position [] and matcher [z0' / z0, z1' / z1, x1'' / x1']

SUBSET1_IN_AG(.(z0, z1)) → U5_AG(.(z0, z1), member1_out_ag(z0, .(z0, z1)))
with rule SUBSET1_IN_AG(.(x0, x1)) → U5_AG(.(x0, x1), member1_out_ag(x0, .(x0, x1)))

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.



(48) FALSE