(0) Obligation:

Clauses:

duplicate([], L) :- ','(!, eq(L, [])).
duplicate(X, .(H, .(H, Z))) :- ','(head(X, H), ','(tail(X, T), duplicate(T, Z))).
head([], X1).
head(.(H, X2), H).
tail([], []).
tail(.(X3, T), T).
eq(X, X).

Queries:

duplicate(g,a).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

(2) Obligation:

Clauses:

duplicate1([], []).
duplicate1(.(T14, T15), .(T14, .(T14, T13))) :- duplicate1(T15, T13).

Queries:

duplicate1(g,a).

(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:
duplicate1_in: (b,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

duplicate1_in_ga([], []) → duplicate1_out_ga([], [])
duplicate1_in_ga(.(T14, T15), .(T14, .(T14, T13))) → U1_ga(T14, T15, T13, duplicate1_in_ga(T15, T13))
U1_ga(T14, T15, T13, duplicate1_out_ga(T15, T13)) → duplicate1_out_ga(.(T14, T15), .(T14, .(T14, T13)))

The argument filtering Pi contains the following mapping:
duplicate1_in_ga(x1, x2)  =  duplicate1_in_ga(x1)
[]  =  []
duplicate1_out_ga(x1, x2)  =  duplicate1_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:

duplicate1_in_ga([], []) → duplicate1_out_ga([], [])
duplicate1_in_ga(.(T14, T15), .(T14, .(T14, T13))) → U1_ga(T14, T15, T13, duplicate1_in_ga(T15, T13))
U1_ga(T14, T15, T13, duplicate1_out_ga(T15, T13)) → duplicate1_out_ga(.(T14, T15), .(T14, .(T14, T13)))

The argument filtering Pi contains the following mapping:
duplicate1_in_ga(x1, x2)  =  duplicate1_in_ga(x1)
[]  =  []
duplicate1_out_ga(x1, x2)  =  duplicate1_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:

DUPLICATE1_IN_GA(.(T14, T15), .(T14, .(T14, T13))) → U1_GA(T14, T15, T13, duplicate1_in_ga(T15, T13))
DUPLICATE1_IN_GA(.(T14, T15), .(T14, .(T14, T13))) → DUPLICATE1_IN_GA(T15, T13)

The TRS R consists of the following rules:

duplicate1_in_ga([], []) → duplicate1_out_ga([], [])
duplicate1_in_ga(.(T14, T15), .(T14, .(T14, T13))) → U1_ga(T14, T15, T13, duplicate1_in_ga(T15, T13))
U1_ga(T14, T15, T13, duplicate1_out_ga(T15, T13)) → duplicate1_out_ga(.(T14, T15), .(T14, .(T14, T13)))

The argument filtering Pi contains the following mapping:
duplicate1_in_ga(x1, x2)  =  duplicate1_in_ga(x1)
[]  =  []
duplicate1_out_ga(x1, x2)  =  duplicate1_out_ga(x2)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4)  =  U1_ga(x1, x4)
DUPLICATE1_IN_GA(x1, x2)  =  DUPLICATE1_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:

DUPLICATE1_IN_GA(.(T14, T15), .(T14, .(T14, T13))) → U1_GA(T14, T15, T13, duplicate1_in_ga(T15, T13))
DUPLICATE1_IN_GA(.(T14, T15), .(T14, .(T14, T13))) → DUPLICATE1_IN_GA(T15, T13)

The TRS R consists of the following rules:

duplicate1_in_ga([], []) → duplicate1_out_ga([], [])
duplicate1_in_ga(.(T14, T15), .(T14, .(T14, T13))) → U1_ga(T14, T15, T13, duplicate1_in_ga(T15, T13))
U1_ga(T14, T15, T13, duplicate1_out_ga(T15, T13)) → duplicate1_out_ga(.(T14, T15), .(T14, .(T14, T13)))

The argument filtering Pi contains the following mapping:
duplicate1_in_ga(x1, x2)  =  duplicate1_in_ga(x1)
[]  =  []
duplicate1_out_ga(x1, x2)  =  duplicate1_out_ga(x2)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4)  =  U1_ga(x1, x4)
DUPLICATE1_IN_GA(x1, x2)  =  DUPLICATE1_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:

DUPLICATE1_IN_GA(.(T14, T15), .(T14, .(T14, T13))) → DUPLICATE1_IN_GA(T15, T13)

The TRS R consists of the following rules:

duplicate1_in_ga([], []) → duplicate1_out_ga([], [])
duplicate1_in_ga(.(T14, T15), .(T14, .(T14, T13))) → U1_ga(T14, T15, T13, duplicate1_in_ga(T15, T13))
U1_ga(T14, T15, T13, duplicate1_out_ga(T15, T13)) → duplicate1_out_ga(.(T14, T15), .(T14, .(T14, T13)))

The argument filtering Pi contains the following mapping:
duplicate1_in_ga(x1, x2)  =  duplicate1_in_ga(x1)
[]  =  []
duplicate1_out_ga(x1, x2)  =  duplicate1_out_ga(x2)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3, x4)  =  U1_ga(x1, x4)
DUPLICATE1_IN_GA(x1, x2)  =  DUPLICATE1_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:

DUPLICATE1_IN_GA(.(T14, T15), .(T14, .(T14, T13))) → DUPLICATE1_IN_GA(T15, T13)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
DUPLICATE1_IN_GA(x1, x2)  =  DUPLICATE1_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:

DUPLICATE1_IN_GA(.(T14, T15)) → DUPLICATE1_IN_GA(T15)

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:

  • DUPLICATE1_IN_GA(.(T14, T15)) → DUPLICATE1_IN_GA(T15)
    The graph contains the following edges 1 > 1

(14) TRUE