(0) Obligation:
Clauses:
duplicate([], []).
duplicate(X, .(H, .(H, Z))) :- ','(no(empty(X)), ','(head(X, H), ','(tail(X, T), duplicate(T, Z)))).
head([], X1).
head(.(H, X2), H).
tail([], []).
tail(.(X3, T), T).
empty([]).
no(X) :- ','(X, ','(!, failure(a))).
no(X4).
failure(b).
Query: duplicate(g,a)
(1) PrologToPrologProblemTransformerProof (SOUND transformation)
Built Prolog problem from termination graph ICLP10.
(2) Obligation:
Clauses:
duplicateA([], []).
duplicateA(.(T35, T36), .(T35, .(T35, T30))) :- duplicateA(T36, T30).
Query: duplicateA(g,a)
(3) PrologToPiTRSProof (SOUND transformation)
We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
duplicateA_in: (b,f)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
duplicateA_in_ga([], []) → duplicateA_out_ga([], [])
duplicateA_in_ga(.(T35, T36), .(T35, .(T35, T30))) → U1_ga(T35, T36, T30, duplicateA_in_ga(T36, T30))
U1_ga(T35, T36, T30, duplicateA_out_ga(T36, T30)) → duplicateA_out_ga(.(T35, T36), .(T35, .(T35, T30)))
The argument filtering Pi contains the following mapping:
duplicateA_in_ga(
x1,
x2) =
duplicateA_in_ga(
x1)
[] =
[]
duplicateA_out_ga(
x1,
x2) =
duplicateA_out_ga(
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ga(
x1,
x2,
x3,
x4) =
U1_ga(
x1,
x4)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(4) Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:
duplicateA_in_ga([], []) → duplicateA_out_ga([], [])
duplicateA_in_ga(.(T35, T36), .(T35, .(T35, T30))) → U1_ga(T35, T36, T30, duplicateA_in_ga(T36, T30))
U1_ga(T35, T36, T30, duplicateA_out_ga(T36, T30)) → duplicateA_out_ga(.(T35, T36), .(T35, .(T35, T30)))
The argument filtering Pi contains the following mapping:
duplicateA_in_ga(
x1,
x2) =
duplicateA_in_ga(
x1)
[] =
[]
duplicateA_out_ga(
x1,
x2) =
duplicateA_out_ga(
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ga(
x1,
x2,
x3,
x4) =
U1_ga(
x1,
x4)
(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:
DUPLICATEA_IN_GA(.(T35, T36), .(T35, .(T35, T30))) → U1_GA(T35, T36, T30, duplicateA_in_ga(T36, T30))
DUPLICATEA_IN_GA(.(T35, T36), .(T35, .(T35, T30))) → DUPLICATEA_IN_GA(T36, T30)
The TRS R consists of the following rules:
duplicateA_in_ga([], []) → duplicateA_out_ga([], [])
duplicateA_in_ga(.(T35, T36), .(T35, .(T35, T30))) → U1_ga(T35, T36, T30, duplicateA_in_ga(T36, T30))
U1_ga(T35, T36, T30, duplicateA_out_ga(T36, T30)) → duplicateA_out_ga(.(T35, T36), .(T35, .(T35, T30)))
The argument filtering Pi contains the following mapping:
duplicateA_in_ga(
x1,
x2) =
duplicateA_in_ga(
x1)
[] =
[]
duplicateA_out_ga(
x1,
x2) =
duplicateA_out_ga(
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ga(
x1,
x2,
x3,
x4) =
U1_ga(
x1,
x4)
DUPLICATEA_IN_GA(
x1,
x2) =
DUPLICATEA_IN_GA(
x1)
U1_GA(
x1,
x2,
x3,
x4) =
U1_GA(
x1,
x4)
We have to consider all (P,R,Pi)-chains
(6) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
DUPLICATEA_IN_GA(.(T35, T36), .(T35, .(T35, T30))) → U1_GA(T35, T36, T30, duplicateA_in_ga(T36, T30))
DUPLICATEA_IN_GA(.(T35, T36), .(T35, .(T35, T30))) → DUPLICATEA_IN_GA(T36, T30)
The TRS R consists of the following rules:
duplicateA_in_ga([], []) → duplicateA_out_ga([], [])
duplicateA_in_ga(.(T35, T36), .(T35, .(T35, T30))) → U1_ga(T35, T36, T30, duplicateA_in_ga(T36, T30))
U1_ga(T35, T36, T30, duplicateA_out_ga(T36, T30)) → duplicateA_out_ga(.(T35, T36), .(T35, .(T35, T30)))
The argument filtering Pi contains the following mapping:
duplicateA_in_ga(
x1,
x2) =
duplicateA_in_ga(
x1)
[] =
[]
duplicateA_out_ga(
x1,
x2) =
duplicateA_out_ga(
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ga(
x1,
x2,
x3,
x4) =
U1_ga(
x1,
x4)
DUPLICATEA_IN_GA(
x1,
x2) =
DUPLICATEA_IN_GA(
x1)
U1_GA(
x1,
x2,
x3,
x4) =
U1_GA(
x1,
x4)
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:
DUPLICATEA_IN_GA(.(T35, T36), .(T35, .(T35, T30))) → DUPLICATEA_IN_GA(T36, T30)
The TRS R consists of the following rules:
duplicateA_in_ga([], []) → duplicateA_out_ga([], [])
duplicateA_in_ga(.(T35, T36), .(T35, .(T35, T30))) → U1_ga(T35, T36, T30, duplicateA_in_ga(T36, T30))
U1_ga(T35, T36, T30, duplicateA_out_ga(T36, T30)) → duplicateA_out_ga(.(T35, T36), .(T35, .(T35, T30)))
The argument filtering Pi contains the following mapping:
duplicateA_in_ga(
x1,
x2) =
duplicateA_in_ga(
x1)
[] =
[]
duplicateA_out_ga(
x1,
x2) =
duplicateA_out_ga(
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ga(
x1,
x2,
x3,
x4) =
U1_ga(
x1,
x4)
DUPLICATEA_IN_GA(
x1,
x2) =
DUPLICATEA_IN_GA(
x1)
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:
DUPLICATEA_IN_GA(.(T35, T36), .(T35, .(T35, T30))) → DUPLICATEA_IN_GA(T36, T30)
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x1,
x2)
DUPLICATEA_IN_GA(
x1,
x2) =
DUPLICATEA_IN_GA(
x1)
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:
DUPLICATEA_IN_GA(.(T35, T36)) → DUPLICATEA_IN_GA(T36)
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:
- DUPLICATEA_IN_GA(.(T35, T36)) → DUPLICATEA_IN_GA(T36)
The graph contains the following edges 1 > 1
(14) YES