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 PrologToDTProblemTransformerProof (⇐)
↳2 TRIPLES
↳3 TriplesToPiDPProof (⇐)
↳4 PiDP
↳5 DependencyGraphProof (⇔)
↳6 AND
    ↳7 PiDP
        ↳8 UsableRulesProof (⇔)
        ↳9 PiDP
        ↳10 PiDPToQDPProof (⇐)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔)
        ↳13 YES
    ↳14 PiDP
        ↳15 PiDPToQDPProof (⇐)
        ↳16 QDP
        ↳17 Narrowing (⇐)
        ↳18 QDP
        ↳19 Instantiation (⇔)
        ↳20 QDP
        ↳21 NonTerminationProof (⇔)
        ↳22 NO

(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) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph.

(2) Obligation:

Triples:

member110(T50, .(T48, T49)) :- member110(T50, T49).
subset11(.(T26, T27), .(T24, T25)) :- member110(T26, T25).
subset11(.(T26, T31), .(T24, T25)) :- ','(member1c10(T26, T25), subset11(T31, .(T24, T25))).
subset11(.(T72, T74), .(T72, T73)) :- subset11(T74, .(T72, T73)).

Clauses:

subset1c1(.(T26, T31), .(T24, T25)) :- ','(member1c10(T26, T25), subset1c1(T31, .(T24, T25))).
subset1c1(.(T72, T74), .(T72, T73)) :- subset1c1(T74, .(T72, T73)).
subset1c1([], T81).
subset1c1([], T83).
member1c10(T50, .(T48, T49)) :- member1c10(T50, T49).
member1c10(T58, .(T58, T59)).

Afs:

subset11(x1, x2)  =  subset11(x2)

(3) TriplesToPiDPProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
subset11_in: (f,b)
member110_in: (f,b)
member1c10_in: (f,b)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

SUBSET11_IN_AG(.(T26, T27), .(T24, T25)) → U2_AG(T26, T27, T24, T25, member110_in_ag(T26, T25))
SUBSET11_IN_AG(.(T26, T27), .(T24, T25)) → MEMBER110_IN_AG(T26, T25)
MEMBER110_IN_AG(T50, .(T48, T49)) → U1_AG(T50, T48, T49, member110_in_ag(T50, T49))
MEMBER110_IN_AG(T50, .(T48, T49)) → MEMBER110_IN_AG(T50, T49)
SUBSET11_IN_AG(.(T26, T31), .(T24, T25)) → U3_AG(T26, T31, T24, T25, member1c10_in_ag(T26, T25))
U3_AG(T26, T31, T24, T25, member1c10_out_ag(T26, T25)) → U4_AG(T26, T31, T24, T25, subset11_in_ag(T31, .(T24, T25)))
U3_AG(T26, T31, T24, T25, member1c10_out_ag(T26, T25)) → SUBSET11_IN_AG(T31, .(T24, T25))
SUBSET11_IN_AG(.(T72, T74), .(T72, T73)) → U5_AG(T72, T74, T73, subset11_in_ag(T74, .(T72, T73)))
SUBSET11_IN_AG(.(T72, T74), .(T72, T73)) → SUBSET11_IN_AG(T74, .(T72, T73))

The TRS R consists of the following rules:

member1c10_in_ag(T50, .(T48, T49)) → U10_ag(T50, T48, T49, member1c10_in_ag(T50, T49))
member1c10_in_ag(T58, .(T58, T59)) → member1c10_out_ag(T58, .(T58, T59))
U10_ag(T50, T48, T49, member1c10_out_ag(T50, T49)) → member1c10_out_ag(T50, .(T48, T49))

The argument filtering Pi contains the following mapping:
subset11_in_ag(x1, x2)  =  subset11_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
member110_in_ag(x1, x2)  =  member110_in_ag(x2)
member1c10_in_ag(x1, x2)  =  member1c10_in_ag(x2)
U10_ag(x1, x2, x3, x4)  =  U10_ag(x2, x3, x4)
member1c10_out_ag(x1, x2)  =  member1c10_out_ag(x1, x2)
SUBSET11_IN_AG(x1, x2)  =  SUBSET11_IN_AG(x2)
U2_AG(x1, x2, x3, x4, x5)  =  U2_AG(x3, x4, x5)
MEMBER110_IN_AG(x1, x2)  =  MEMBER110_IN_AG(x2)
U1_AG(x1, x2, x3, x4)  =  U1_AG(x2, x3, x4)
U3_AG(x1, x2, x3, x4, x5)  =  U3_AG(x3, x4, x5)
U4_AG(x1, x2, x3, x4, x5)  =  U4_AG(x3, x4, x5)
U5_AG(x1, x2, x3, x4)  =  U5_AG(x1, x3, x4)

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

Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES

(4) Obligation:

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

SUBSET11_IN_AG(.(T26, T27), .(T24, T25)) → U2_AG(T26, T27, T24, T25, member110_in_ag(T26, T25))
SUBSET11_IN_AG(.(T26, T27), .(T24, T25)) → MEMBER110_IN_AG(T26, T25)
MEMBER110_IN_AG(T50, .(T48, T49)) → U1_AG(T50, T48, T49, member110_in_ag(T50, T49))
MEMBER110_IN_AG(T50, .(T48, T49)) → MEMBER110_IN_AG(T50, T49)
SUBSET11_IN_AG(.(T26, T31), .(T24, T25)) → U3_AG(T26, T31, T24, T25, member1c10_in_ag(T26, T25))
U3_AG(T26, T31, T24, T25, member1c10_out_ag(T26, T25)) → U4_AG(T26, T31, T24, T25, subset11_in_ag(T31, .(T24, T25)))
U3_AG(T26, T31, T24, T25, member1c10_out_ag(T26, T25)) → SUBSET11_IN_AG(T31, .(T24, T25))
SUBSET11_IN_AG(.(T72, T74), .(T72, T73)) → U5_AG(T72, T74, T73, subset11_in_ag(T74, .(T72, T73)))
SUBSET11_IN_AG(.(T72, T74), .(T72, T73)) → SUBSET11_IN_AG(T74, .(T72, T73))

The TRS R consists of the following rules:

member1c10_in_ag(T50, .(T48, T49)) → U10_ag(T50, T48, T49, member1c10_in_ag(T50, T49))
member1c10_in_ag(T58, .(T58, T59)) → member1c10_out_ag(T58, .(T58, T59))
U10_ag(T50, T48, T49, member1c10_out_ag(T50, T49)) → member1c10_out_ag(T50, .(T48, T49))

The argument filtering Pi contains the following mapping:
subset11_in_ag(x1, x2)  =  subset11_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
member110_in_ag(x1, x2)  =  member110_in_ag(x2)
member1c10_in_ag(x1, x2)  =  member1c10_in_ag(x2)
U10_ag(x1, x2, x3, x4)  =  U10_ag(x2, x3, x4)
member1c10_out_ag(x1, x2)  =  member1c10_out_ag(x1, x2)
SUBSET11_IN_AG(x1, x2)  =  SUBSET11_IN_AG(x2)
U2_AG(x1, x2, x3, x4, x5)  =  U2_AG(x3, x4, x5)
MEMBER110_IN_AG(x1, x2)  =  MEMBER110_IN_AG(x2)
U1_AG(x1, x2, x3, x4)  =  U1_AG(x2, x3, x4)
U3_AG(x1, x2, x3, x4, x5)  =  U3_AG(x3, x4, x5)
U4_AG(x1, x2, x3, x4, x5)  =  U4_AG(x3, x4, x5)
U5_AG(x1, x2, x3, x4)  =  U5_AG(x1, x3, 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 5 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

MEMBER110_IN_AG(T50, .(T48, T49)) → MEMBER110_IN_AG(T50, T49)

The TRS R consists of the following rules:

member1c10_in_ag(T50, .(T48, T49)) → U10_ag(T50, T48, T49, member1c10_in_ag(T50, T49))
member1c10_in_ag(T58, .(T58, T59)) → member1c10_out_ag(T58, .(T58, T59))
U10_ag(T50, T48, T49, member1c10_out_ag(T50, T49)) → member1c10_out_ag(T50, .(T48, T49))

The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
member1c10_in_ag(x1, x2)  =  member1c10_in_ag(x2)
U10_ag(x1, x2, x3, x4)  =  U10_ag(x2, x3, x4)
member1c10_out_ag(x1, x2)  =  member1c10_out_ag(x1, x2)
MEMBER110_IN_AG(x1, x2)  =  MEMBER110_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:

MEMBER110_IN_AG(T50, .(T48, T49)) → MEMBER110_IN_AG(T50, T49)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
MEMBER110_IN_AG(x1, x2)  =  MEMBER110_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:

MEMBER110_IN_AG(.(T48, T49)) → MEMBER110_IN_AG(T49)

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:

  • MEMBER110_IN_AG(.(T48, T49)) → MEMBER110_IN_AG(T49)
    The graph contains the following edges 1 > 1

(13) YES

(14) Obligation:

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

SUBSET11_IN_AG(.(T26, T31), .(T24, T25)) → U3_AG(T26, T31, T24, T25, member1c10_in_ag(T26, T25))
U3_AG(T26, T31, T24, T25, member1c10_out_ag(T26, T25)) → SUBSET11_IN_AG(T31, .(T24, T25))
SUBSET11_IN_AG(.(T72, T74), .(T72, T73)) → SUBSET11_IN_AG(T74, .(T72, T73))

The TRS R consists of the following rules:

member1c10_in_ag(T50, .(T48, T49)) → U10_ag(T50, T48, T49, member1c10_in_ag(T50, T49))
member1c10_in_ag(T58, .(T58, T59)) → member1c10_out_ag(T58, .(T58, T59))
U10_ag(T50, T48, T49, member1c10_out_ag(T50, T49)) → member1c10_out_ag(T50, .(T48, T49))

The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
member1c10_in_ag(x1, x2)  =  member1c10_in_ag(x2)
U10_ag(x1, x2, x3, x4)  =  U10_ag(x2, x3, x4)
member1c10_out_ag(x1, x2)  =  member1c10_out_ag(x1, x2)
SUBSET11_IN_AG(x1, x2)  =  SUBSET11_IN_AG(x2)
U3_AG(x1, x2, x3, x4, x5)  =  U3_AG(x3, x4, x5)

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

(15) PiDPToQDPProof (SOUND transformation)

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

(16) Obligation:

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

SUBSET11_IN_AG(.(T24, T25)) → U3_AG(T24, T25, member1c10_in_ag(T25))
U3_AG(T24, T25, member1c10_out_ag(T26, T25)) → SUBSET11_IN_AG(.(T24, T25))
SUBSET11_IN_AG(.(T72, T73)) → SUBSET11_IN_AG(.(T72, T73))

The TRS R consists of the following rules:

member1c10_in_ag(.(T48, T49)) → U10_ag(T48, T49, member1c10_in_ag(T49))
member1c10_in_ag(.(T58, T59)) → member1c10_out_ag(T58, .(T58, T59))
U10_ag(T48, T49, member1c10_out_ag(T50, T49)) → member1c10_out_ag(T50, .(T48, T49))

The set Q consists of the following terms:

member1c10_in_ag(x0)
U10_ag(x0, x1, x2)

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

(17) Narrowing (SOUND transformation)

By narrowing [LPAR04] the rule SUBSET11_IN_AG(.(T24, T25)) → U3_AG(T24, T25, member1c10_in_ag(T25)) at position [2] we obtained the following new rules [LPAR04]:

SUBSET11_IN_AG(.(y0, .(x0, x1))) → U3_AG(y0, .(x0, x1), U10_ag(x0, x1, member1c10_in_ag(x1)))
SUBSET11_IN_AG(.(y0, .(x0, x1))) → U3_AG(y0, .(x0, x1), member1c10_out_ag(x0, .(x0, x1)))

(18) Obligation:

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

U3_AG(T24, T25, member1c10_out_ag(T26, T25)) → SUBSET11_IN_AG(.(T24, T25))
SUBSET11_IN_AG(.(T72, T73)) → SUBSET11_IN_AG(.(T72, T73))
SUBSET11_IN_AG(.(y0, .(x0, x1))) → U3_AG(y0, .(x0, x1), U10_ag(x0, x1, member1c10_in_ag(x1)))
SUBSET11_IN_AG(.(y0, .(x0, x1))) → U3_AG(y0, .(x0, x1), member1c10_out_ag(x0, .(x0, x1)))

The TRS R consists of the following rules:

member1c10_in_ag(.(T48, T49)) → U10_ag(T48, T49, member1c10_in_ag(T49))
member1c10_in_ag(.(T58, T59)) → member1c10_out_ag(T58, .(T58, T59))
U10_ag(T48, T49, member1c10_out_ag(T50, T49)) → member1c10_out_ag(T50, .(T48, T49))

The set Q consists of the following terms:

member1c10_in_ag(x0)
U10_ag(x0, x1, x2)

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

(19) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule U3_AG(T24, T25, member1c10_out_ag(T26, T25)) → SUBSET11_IN_AG(.(T24, T25)) we obtained the following new rules [LPAR04]:

U3_AG(z0, .(z1, z2), member1c10_out_ag(x2, .(z1, z2))) → SUBSET11_IN_AG(.(z0, .(z1, z2)))
U3_AG(z0, .(z1, z2), member1c10_out_ag(z1, .(z1, z2))) → SUBSET11_IN_AG(.(z0, .(z1, z2)))

(20) Obligation:

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

SUBSET11_IN_AG(.(T72, T73)) → SUBSET11_IN_AG(.(T72, T73))
SUBSET11_IN_AG(.(y0, .(x0, x1))) → U3_AG(y0, .(x0, x1), U10_ag(x0, x1, member1c10_in_ag(x1)))
SUBSET11_IN_AG(.(y0, .(x0, x1))) → U3_AG(y0, .(x0, x1), member1c10_out_ag(x0, .(x0, x1)))
U3_AG(z0, .(z1, z2), member1c10_out_ag(x2, .(z1, z2))) → SUBSET11_IN_AG(.(z0, .(z1, z2)))
U3_AG(z0, .(z1, z2), member1c10_out_ag(z1, .(z1, z2))) → SUBSET11_IN_AG(.(z0, .(z1, z2)))

The TRS R consists of the following rules:

member1c10_in_ag(.(T48, T49)) → U10_ag(T48, T49, member1c10_in_ag(T49))
member1c10_in_ag(.(T58, T59)) → member1c10_out_ag(T58, .(T58, T59))
U10_ag(T48, T49, member1c10_out_ag(T50, T49)) → member1c10_out_ag(T50, .(T48, T49))

The set Q consists of the following terms:

member1c10_in_ag(x0)
U10_ag(x0, x1, x2)

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

(21) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = SUBSET11_IN_AG(.(T72, T73)) evaluates to t =SUBSET11_IN_AG(.(T72, T73))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [ ]
  • Semiunifier: [ ]



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from SUBSET11_IN_AG(.(T72, T73)) to SUBSET11_IN_AG(.(T72, T73)).



(22) NO