(0) Obligation:
Clauses:
member(X, .(X, X1)).
member(X, .(X2, Xs)) :- member(X, Xs).
Queries:
member(g,a).
(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:
member_in: (b,f)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
member_in_ga(X, .(X, X1)) → member_out_ga(X, .(X, X1))
member_in_ga(X, .(X2, Xs)) → U1_ga(X, X2, Xs, member_in_ga(X, Xs))
U1_ga(X, X2, Xs, member_out_ga(X, Xs)) → member_out_ga(X, .(X2, Xs))
The argument filtering Pi contains the following mapping:
member_in_ga(
x1,
x2) =
member_in_ga(
x1)
member_out_ga(
x1,
x2) =
member_out_ga(
x1)
U1_ga(
x1,
x2,
x3,
x4) =
U1_ga(
x1,
x4)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(2) Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:
member_in_ga(X, .(X, X1)) → member_out_ga(X, .(X, X1))
member_in_ga(X, .(X2, Xs)) → U1_ga(X, X2, Xs, member_in_ga(X, Xs))
U1_ga(X, X2, Xs, member_out_ga(X, Xs)) → member_out_ga(X, .(X2, Xs))
The argument filtering Pi contains the following mapping:
member_in_ga(
x1,
x2) =
member_in_ga(
x1)
member_out_ga(
x1,
x2) =
member_out_ga(
x1)
U1_ga(
x1,
x2,
x3,
x4) =
U1_ga(
x1,
x4)
(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:
MEMBER_IN_GA(X, .(X2, Xs)) → U1_GA(X, X2, Xs, member_in_ga(X, Xs))
MEMBER_IN_GA(X, .(X2, Xs)) → MEMBER_IN_GA(X, Xs)
The TRS R consists of the following rules:
member_in_ga(X, .(X, X1)) → member_out_ga(X, .(X, X1))
member_in_ga(X, .(X2, Xs)) → U1_ga(X, X2, Xs, member_in_ga(X, Xs))
U1_ga(X, X2, Xs, member_out_ga(X, Xs)) → member_out_ga(X, .(X2, Xs))
The argument filtering Pi contains the following mapping:
member_in_ga(
x1,
x2) =
member_in_ga(
x1)
member_out_ga(
x1,
x2) =
member_out_ga(
x1)
U1_ga(
x1,
x2,
x3,
x4) =
U1_ga(
x1,
x4)
MEMBER_IN_GA(
x1,
x2) =
MEMBER_IN_GA(
x1)
U1_GA(
x1,
x2,
x3,
x4) =
U1_GA(
x1,
x4)
We have to consider all (P,R,Pi)-chains
(4) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
MEMBER_IN_GA(X, .(X2, Xs)) → U1_GA(X, X2, Xs, member_in_ga(X, Xs))
MEMBER_IN_GA(X, .(X2, Xs)) → MEMBER_IN_GA(X, Xs)
The TRS R consists of the following rules:
member_in_ga(X, .(X, X1)) → member_out_ga(X, .(X, X1))
member_in_ga(X, .(X2, Xs)) → U1_ga(X, X2, Xs, member_in_ga(X, Xs))
U1_ga(X, X2, Xs, member_out_ga(X, Xs)) → member_out_ga(X, .(X2, Xs))
The argument filtering Pi contains the following mapping:
member_in_ga(
x1,
x2) =
member_in_ga(
x1)
member_out_ga(
x1,
x2) =
member_out_ga(
x1)
U1_ga(
x1,
x2,
x3,
x4) =
U1_ga(
x1,
x4)
MEMBER_IN_GA(
x1,
x2) =
MEMBER_IN_GA(
x1)
U1_GA(
x1,
x2,
x3,
x4) =
U1_GA(
x1,
x4)
We have to consider all (P,R,Pi)-chains
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.
(6) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
MEMBER_IN_GA(X, .(X2, Xs)) → MEMBER_IN_GA(X, Xs)
The TRS R consists of the following rules:
member_in_ga(X, .(X, X1)) → member_out_ga(X, .(X, X1))
member_in_ga(X, .(X2, Xs)) → U1_ga(X, X2, Xs, member_in_ga(X, Xs))
U1_ga(X, X2, Xs, member_out_ga(X, Xs)) → member_out_ga(X, .(X2, Xs))
The argument filtering Pi contains the following mapping:
member_in_ga(
x1,
x2) =
member_in_ga(
x1)
member_out_ga(
x1,
x2) =
member_out_ga(
x1)
U1_ga(
x1,
x2,
x3,
x4) =
U1_ga(
x1,
x4)
MEMBER_IN_GA(
x1,
x2) =
MEMBER_IN_GA(
x1)
We have to consider all (P,R,Pi)-chains
(7) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
(8) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
MEMBER_IN_GA(X, .(X2, Xs)) → MEMBER_IN_GA(X, Xs)
R is empty.
The argument filtering Pi contains the following mapping:
MEMBER_IN_GA(
x1,
x2) =
MEMBER_IN_GA(
x1)
We have to consider all (P,R,Pi)-chains
(9) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MEMBER_IN_GA(X) → MEMBER_IN_GA(X)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(11) 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 =
MEMBER_IN_GA(
X) evaluates to t =
MEMBER_IN_GA(
X)
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Matcher: [ ]
- Semiunifier: [ ]
Rewriting sequenceThe DP semiunifies directly so there is only one rewrite step from MEMBER_IN_GA(X) to MEMBER_IN_GA(X).
(12) FALSE
(13) PrologToPiTRSProof (SOUND transformation)
We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
member_in: (b,f)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
member_in_ga(X, .(X, X1)) → member_out_ga(X, .(X, X1))
member_in_ga(X, .(X2, Xs)) → U1_ga(X, X2, Xs, member_in_ga(X, Xs))
U1_ga(X, X2, Xs, member_out_ga(X, Xs)) → member_out_ga(X, .(X2, Xs))
The argument filtering Pi contains the following mapping:
member_in_ga(
x1,
x2) =
member_in_ga(
x1)
member_out_ga(
x1,
x2) =
member_out_ga
U1_ga(
x1,
x2,
x3,
x4) =
U1_ga(
x4)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(14) Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:
member_in_ga(X, .(X, X1)) → member_out_ga(X, .(X, X1))
member_in_ga(X, .(X2, Xs)) → U1_ga(X, X2, Xs, member_in_ga(X, Xs))
U1_ga(X, X2, Xs, member_out_ga(X, Xs)) → member_out_ga(X, .(X2, Xs))
The argument filtering Pi contains the following mapping:
member_in_ga(
x1,
x2) =
member_in_ga(
x1)
member_out_ga(
x1,
x2) =
member_out_ga
U1_ga(
x1,
x2,
x3,
x4) =
U1_ga(
x4)
(15) 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:
MEMBER_IN_GA(X, .(X2, Xs)) → U1_GA(X, X2, Xs, member_in_ga(X, Xs))
MEMBER_IN_GA(X, .(X2, Xs)) → MEMBER_IN_GA(X, Xs)
The TRS R consists of the following rules:
member_in_ga(X, .(X, X1)) → member_out_ga(X, .(X, X1))
member_in_ga(X, .(X2, Xs)) → U1_ga(X, X2, Xs, member_in_ga(X, Xs))
U1_ga(X, X2, Xs, member_out_ga(X, Xs)) → member_out_ga(X, .(X2, Xs))
The argument filtering Pi contains the following mapping:
member_in_ga(
x1,
x2) =
member_in_ga(
x1)
member_out_ga(
x1,
x2) =
member_out_ga
U1_ga(
x1,
x2,
x3,
x4) =
U1_ga(
x4)
MEMBER_IN_GA(
x1,
x2) =
MEMBER_IN_GA(
x1)
U1_GA(
x1,
x2,
x3,
x4) =
U1_GA(
x4)
We have to consider all (P,R,Pi)-chains
(16) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
MEMBER_IN_GA(X, .(X2, Xs)) → U1_GA(X, X2, Xs, member_in_ga(X, Xs))
MEMBER_IN_GA(X, .(X2, Xs)) → MEMBER_IN_GA(X, Xs)
The TRS R consists of the following rules:
member_in_ga(X, .(X, X1)) → member_out_ga(X, .(X, X1))
member_in_ga(X, .(X2, Xs)) → U1_ga(X, X2, Xs, member_in_ga(X, Xs))
U1_ga(X, X2, Xs, member_out_ga(X, Xs)) → member_out_ga(X, .(X2, Xs))
The argument filtering Pi contains the following mapping:
member_in_ga(
x1,
x2) =
member_in_ga(
x1)
member_out_ga(
x1,
x2) =
member_out_ga
U1_ga(
x1,
x2,
x3,
x4) =
U1_ga(
x4)
MEMBER_IN_GA(
x1,
x2) =
MEMBER_IN_GA(
x1)
U1_GA(
x1,
x2,
x3,
x4) =
U1_GA(
x4)
We have to consider all (P,R,Pi)-chains
(17) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.
(18) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
MEMBER_IN_GA(X, .(X2, Xs)) → MEMBER_IN_GA(X, Xs)
The TRS R consists of the following rules:
member_in_ga(X, .(X, X1)) → member_out_ga(X, .(X, X1))
member_in_ga(X, .(X2, Xs)) → U1_ga(X, X2, Xs, member_in_ga(X, Xs))
U1_ga(X, X2, Xs, member_out_ga(X, Xs)) → member_out_ga(X, .(X2, Xs))
The argument filtering Pi contains the following mapping:
member_in_ga(
x1,
x2) =
member_in_ga(
x1)
member_out_ga(
x1,
x2) =
member_out_ga
U1_ga(
x1,
x2,
x3,
x4) =
U1_ga(
x4)
MEMBER_IN_GA(
x1,
x2) =
MEMBER_IN_GA(
x1)
We have to consider all (P,R,Pi)-chains
(19) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
(20) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
MEMBER_IN_GA(X, .(X2, Xs)) → MEMBER_IN_GA(X, Xs)
R is empty.
The argument filtering Pi contains the following mapping:
MEMBER_IN_GA(
x1,
x2) =
MEMBER_IN_GA(
x1)
We have to consider all (P,R,Pi)-chains
(21) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(22) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MEMBER_IN_GA(X) → MEMBER_IN_GA(X)
R is empty.
Q is empty.
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 semiunifying a rule from P directly.
s =
MEMBER_IN_GA(
X) evaluates to t =
MEMBER_IN_GA(
X)
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [ ]
- Matcher: [ ]
Rewriting sequenceThe DP semiunifies directly so there is only one rewrite step from MEMBER_IN_GA(X) to MEMBER_IN_GA(X).
(24) FALSE