(0) Obligation:
Clauses:member(X, .(Y, Xs)) :- member(X, Xs).
member(X, .(X, Xs)).
Queries:
member(a,g).
(1) PrologToPrologProblemTransformerProof (SOUND transformation)
Built Prolog problem from termination graph.
(2) Obligation:
Clauses:member1(T25, .(T7, .(T23, T24))) :- member1(T25, T24).
member1(T37, .(T7, .(T37, T38))).
member1(T43, .(T43, T44)).
Queries:
member1(a,g).
(3) PrologToPiTRSProof (SOUND transformation)
We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:member1_in: (f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
member1_in_ag(T25, .(T7, .(T23, T24))) → U1_ag(T25, T7, T23, T24, member1_in_ag(T25, T24))
member1_in_ag(T37, .(T7, .(T37, T38))) → member1_out_ag(T37, .(T7, .(T37, T38)))
member1_in_ag(T43, .(T43, T44)) → member1_out_ag(T43, .(T43, T44))
U1_ag(T25, T7, T23, T24, member1_out_ag(T25, T24)) → member1_out_ag(T25, .(T7, .(T23, T24)))
The argument filtering Pi contains the following mapping:
member1_in_ag(x1, x2) = member1_in_ag(x2)
.(x1, x2) = .(x1, x2)
U1_ag(x1, x2, x3, x4, x5) = U1_ag(x5)
member1_out_ag(x1, x2) = member1_out_ag(x1)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(4) Obligation:
Pi-finite rewrite system:The TRS R consists of the following rules:
member1_in_ag(T25, .(T7, .(T23, T24))) → U1_ag(T25, T7, T23, T24, member1_in_ag(T25, T24))
member1_in_ag(T37, .(T7, .(T37, T38))) → member1_out_ag(T37, .(T7, .(T37, T38)))
member1_in_ag(T43, .(T43, T44)) → member1_out_ag(T43, .(T43, T44))
U1_ag(T25, T7, T23, T24, member1_out_ag(T25, T24)) → member1_out_ag(T25, .(T7, .(T23, T24)))
The argument filtering Pi contains the following mapping:
member1_in_ag(x1, x2) = member1_in_ag(x2)
.(x1, x2) = .(x1, x2)
U1_ag(x1, x2, x3, x4, x5) = U1_ag(x5)
member1_out_ag(x1, x2) = member1_out_ag(x1)
(5) 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:
MEMBER1_IN_AG(T25, .(T7, .(T23, T24))) → U1_AG(T25, T7, T23, T24, member1_in_ag(T25, T24))
MEMBER1_IN_AG(T25, .(T7, .(T23, T24))) → MEMBER1_IN_AG(T25, T24)
The TRS R consists of the following rules:
member1_in_ag(T25, .(T7, .(T23, T24))) → U1_ag(T25, T7, T23, T24, member1_in_ag(T25, T24))
member1_in_ag(T37, .(T7, .(T37, T38))) → member1_out_ag(T37, .(T7, .(T37, T38)))
member1_in_ag(T43, .(T43, T44)) → member1_out_ag(T43, .(T43, T44))
U1_ag(T25, T7, T23, T24, member1_out_ag(T25, T24)) → member1_out_ag(T25, .(T7, .(T23, T24)))
The argument filtering Pi contains the following mapping:
member1_in_ag(x1, x2) = member1_in_ag(x2)
.(x1, x2) = .(x1, x2)
U1_ag(x1, x2, x3, x4, x5) = U1_ag(x5)
member1_out_ag(x1, x2) = member1_out_ag(x1)
MEMBER1_IN_AG(x1, x2) = MEMBER1_IN_AG(x2)
U1_AG(x1, x2, x3, x4, x5) = U1_AG(x5)
We have to consider all (P,R,Pi)-chains
(6) Obligation:
Pi DP problem:The TRS P consists of the following rules:
MEMBER1_IN_AG(T25, .(T7, .(T23, T24))) → U1_AG(T25, T7, T23, T24, member1_in_ag(T25, T24))
MEMBER1_IN_AG(T25, .(T7, .(T23, T24))) → MEMBER1_IN_AG(T25, T24)
The TRS R consists of the following rules:
member1_in_ag(T25, .(T7, .(T23, T24))) → U1_ag(T25, T7, T23, T24, member1_in_ag(T25, T24))
member1_in_ag(T37, .(T7, .(T37, T38))) → member1_out_ag(T37, .(T7, .(T37, T38)))
member1_in_ag(T43, .(T43, T44)) → member1_out_ag(T43, .(T43, T44))
U1_ag(T25, T7, T23, T24, member1_out_ag(T25, T24)) → member1_out_ag(T25, .(T7, .(T23, T24)))
The argument filtering Pi contains the following mapping:
member1_in_ag(x1, x2) = member1_in_ag(x2)
.(x1, x2) = .(x1, x2)
U1_ag(x1, x2, x3, x4, x5) = U1_ag(x5)
member1_out_ag(x1, x2) = member1_out_ag(x1)
MEMBER1_IN_AG(x1, x2) = MEMBER1_IN_AG(x2)
U1_AG(x1, x2, x3, x4, x5) = U1_AG(x5)
We have to consider all (P,R,Pi)-chains
(7) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.(8) Obligation:
Pi DP problem:The TRS P consists of the following rules:
MEMBER1_IN_AG(T25, .(T7, .(T23, T24))) → MEMBER1_IN_AG(T25, T24)
The TRS R consists of the following rules:
member1_in_ag(T25, .(T7, .(T23, T24))) → U1_ag(T25, T7, T23, T24, member1_in_ag(T25, T24))
member1_in_ag(T37, .(T7, .(T37, T38))) → member1_out_ag(T37, .(T7, .(T37, T38)))
member1_in_ag(T43, .(T43, T44)) → member1_out_ag(T43, .(T43, T44))
U1_ag(T25, T7, T23, T24, member1_out_ag(T25, T24)) → member1_out_ag(T25, .(T7, .(T23, T24)))
The argument filtering Pi contains the following mapping:
member1_in_ag(x1, x2) = member1_in_ag(x2)
.(x1, x2) = .(x1, x2)
U1_ag(x1, x2, x3, x4, x5) = U1_ag(x5)
member1_out_ag(x1, x2) = member1_out_ag(x1)
MEMBER1_IN_AG(x1, x2) = MEMBER1_IN_AG(x2)
We have to consider all (P,R,Pi)-chains
(9) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.(10) Obligation:
Pi DP problem:The TRS P consists of the following rules:
MEMBER1_IN_AG(T25, .(T7, .(T23, T24))) → MEMBER1_IN_AG(T25, T24)
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
(11) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.(12) Obligation:
Q DP problem:The TRS P consists of the following rules:
MEMBER1_IN_AG(.(T7, .(T23, T24))) → MEMBER1_IN_AG(T24)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(13) 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(.(T7, .(T23, T24))) → MEMBER1_IN_AG(T24)
The graph contains the following edges 1 > 1