(0) Obligation:
Clauses:overlap(Xs, Ys) :- ','(member2(X, Xs), member1(X, Ys)).
has_a_or_b(Xs) :- overlap(Xs, .(a, .(b, []))).
member1(X, .(Y, Xs)) :- member1(X, Xs).
member1(X, .(X, Xs)).
member2(X, .(Y, Xs)) :- member2(X, Xs).
member2(X, .(X, Xs)).
Queries:
overlap(g,g).
(1) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph.
(2) Obligation:
Triples:p3(X25, .(T15, T16), T6) :- p3(X25, T16, T6).
p3(X41, .(X41, T21), T6) :- member19(X41, T6).
member19(X67, .(T32, T33)) :- member19(X67, T33).
overlap1(T5, T6) :- p3(X5, T5, T6).
Clauses:
qc3(X25, .(T15, T16), T6) :- qc3(X25, T16, T6).
qc3(X41, .(X41, T21), T6) :- member1c9(X41, T6).
member1c9(X67, .(T32, T33)) :- member1c9(X67, T33).
member1c9(X83, .(X83, T38)).
Afs:
overlap1(x1, x2) = overlap1(x1, 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:overlap1_in: (b,b)
p3_in: (f,b,b)
member19_in: (b,b)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:
OVERLAP1_IN_GG(T5, T6) → U4_GG(T5, T6, p3_in_agg(X5, T5, T6))
OVERLAP1_IN_GG(T5, T6) → P3_IN_AGG(X5, T5, T6)
P3_IN_AGG(X25, .(T15, T16), T6) → U1_AGG(X25, T15, T16, T6, p3_in_agg(X25, T16, T6))
P3_IN_AGG(X25, .(T15, T16), T6) → P3_IN_AGG(X25, T16, T6)
P3_IN_AGG(X41, .(X41, T21), T6) → U2_AGG(X41, T21, T6, member19_in_gg(X41, T6))
P3_IN_AGG(X41, .(X41, T21), T6) → MEMBER19_IN_GG(X41, T6)
MEMBER19_IN_GG(X67, .(T32, T33)) → U3_GG(X67, T32, T33, member19_in_gg(X67, T33))
MEMBER19_IN_GG(X67, .(T32, T33)) → MEMBER19_IN_GG(X67, T33)
R is empty.
The argument filtering Pi contains the following mapping:
p3_in_agg(x1, x2, x3) = p3_in_agg(x2, x3)
.(x1, x2) = .(x1, x2)
member19_in_gg(x1, x2) = member19_in_gg(x1, x2)
OVERLAP1_IN_GG(x1, x2) = OVERLAP1_IN_GG(x1, x2)
U4_GG(x1, x2, x3) = U4_GG(x1, x2, x3)
P3_IN_AGG(x1, x2, x3) = P3_IN_AGG(x2, x3)
U1_AGG(x1, x2, x3, x4, x5) = U1_AGG(x2, x3, x4, x5)
U2_AGG(x1, x2, x3, x4) = U2_AGG(x1, x2, x3, x4)
MEMBER19_IN_GG(x1, x2) = MEMBER19_IN_GG(x1, x2)
U3_GG(x1, x2, x3, x4) = U3_GG(x1, x2, 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:
OVERLAP1_IN_GG(T5, T6) → U4_GG(T5, T6, p3_in_agg(X5, T5, T6))
OVERLAP1_IN_GG(T5, T6) → P3_IN_AGG(X5, T5, T6)
P3_IN_AGG(X25, .(T15, T16), T6) → U1_AGG(X25, T15, T16, T6, p3_in_agg(X25, T16, T6))
P3_IN_AGG(X25, .(T15, T16), T6) → P3_IN_AGG(X25, T16, T6)
P3_IN_AGG(X41, .(X41, T21), T6) → U2_AGG(X41, T21, T6, member19_in_gg(X41, T6))
P3_IN_AGG(X41, .(X41, T21), T6) → MEMBER19_IN_GG(X41, T6)
MEMBER19_IN_GG(X67, .(T32, T33)) → U3_GG(X67, T32, T33, member19_in_gg(X67, T33))
MEMBER19_IN_GG(X67, .(T32, T33)) → MEMBER19_IN_GG(X67, T33)
R is empty.
The argument filtering Pi contains the following mapping:
p3_in_agg(x1, x2, x3) = p3_in_agg(x2, x3)
.(x1, x2) = .(x1, x2)
member19_in_gg(x1, x2) = member19_in_gg(x1, x2)
OVERLAP1_IN_GG(x1, x2) = OVERLAP1_IN_GG(x1, x2)
U4_GG(x1, x2, x3) = U4_GG(x1, x2, x3)
P3_IN_AGG(x1, x2, x3) = P3_IN_AGG(x2, x3)
U1_AGG(x1, x2, x3, x4, x5) = U1_AGG(x2, x3, x4, x5)
U2_AGG(x1, x2, x3, x4) = U2_AGG(x1, x2, x3, x4)
MEMBER19_IN_GG(x1, x2) = MEMBER19_IN_GG(x1, x2)
U3_GG(x1, x2, x3, x4) = U3_GG(x1, x2, 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 6 less nodes.(6) Complex Obligation (AND)
(7) Obligation:
Pi DP problem:The TRS P consists of the following rules:
MEMBER19_IN_GG(X67, .(T32, T33)) → MEMBER19_IN_GG(X67, T33)
R is empty.
Pi is empty.
We have to consider all (P,R,Pi)-chains
(8) PiDPToQDPProof (EQUIVALENT transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.(9) Obligation:
Q DP problem:The TRS P consists of the following rules:
MEMBER19_IN_GG(X67, .(T32, T33)) → MEMBER19_IN_GG(X67, T33)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(10) 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:
- MEMBER19_IN_GG(X67, .(T32, T33)) → MEMBER19_IN_GG(X67, T33)
The graph contains the following edges 1 >= 1, 2 > 2
(11) YES
(12) Obligation:
Pi DP problem:The TRS P consists of the following rules:
P3_IN_AGG(X25, .(T15, T16), T6) → P3_IN_AGG(X25, T16, T6)
R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
P3_IN_AGG(x1, x2, x3) = P3_IN_AGG(x2, x3)
We have to consider all (P,R,Pi)-chains
(13) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.(14) Obligation:
Q DP problem:The TRS P consists of the following rules:
P3_IN_AGG(.(T15, T16), T6) → P3_IN_AGG(T16, T6)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(15) 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:
- P3_IN_AGG(.(T15, T16), T6) → P3_IN_AGG(T16, T6)
The graph contains the following edges 1 > 1, 2 >= 2