Left Termination of the query pattern duplicate_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

duplicate([], []).
duplicate(.(X, Y), .(X, .(X, Z))) :- duplicate(Y, Z).

Queries:

duplicate(g,a).

We use the technique of [30].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

duplicate_in(.(X, Y), .(X, .(X, Z))) → U1(X, Y, Z, duplicate_in(Y, Z))
duplicate_in([], []) → duplicate_out([], [])
U1(X, Y, Z, duplicate_out(Y, Z)) → duplicate_out(.(X, Y), .(X, .(X, Z)))

The argument filtering Pi contains the following mapping:
duplicate_in(x1, x2)  =  duplicate_in(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x1, x4)
[]  =  []
duplicate_out(x1, x2)  =  duplicate_out(x2)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

duplicate_in(.(X, Y), .(X, .(X, Z))) → U1(X, Y, Z, duplicate_in(Y, Z))
duplicate_in([], []) → duplicate_out([], [])
U1(X, Y, Z, duplicate_out(Y, Z)) → duplicate_out(.(X, Y), .(X, .(X, Z)))

The argument filtering Pi contains the following mapping:
duplicate_in(x1, x2)  =  duplicate_in(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x1, x4)
[]  =  []
duplicate_out(x1, x2)  =  duplicate_out(x2)


Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

DUPLICATE_IN(.(X, Y), .(X, .(X, Z))) → U11(X, Y, Z, duplicate_in(Y, Z))
DUPLICATE_IN(.(X, Y), .(X, .(X, Z))) → DUPLICATE_IN(Y, Z)

The TRS R consists of the following rules:

duplicate_in(.(X, Y), .(X, .(X, Z))) → U1(X, Y, Z, duplicate_in(Y, Z))
duplicate_in([], []) → duplicate_out([], [])
U1(X, Y, Z, duplicate_out(Y, Z)) → duplicate_out(.(X, Y), .(X, .(X, Z)))

The argument filtering Pi contains the following mapping:
duplicate_in(x1, x2)  =  duplicate_in(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x1, x4)
[]  =  []
duplicate_out(x1, x2)  =  duplicate_out(x2)
DUPLICATE_IN(x1, x2)  =  DUPLICATE_IN(x1)
U11(x1, x2, x3, x4)  =  U11(x1, x4)

We have to consider all (P,R,Pi)-chains

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof

Pi DP problem:
The TRS P consists of the following rules:

DUPLICATE_IN(.(X, Y), .(X, .(X, Z))) → U11(X, Y, Z, duplicate_in(Y, Z))
DUPLICATE_IN(.(X, Y), .(X, .(X, Z))) → DUPLICATE_IN(Y, Z)

The TRS R consists of the following rules:

duplicate_in(.(X, Y), .(X, .(X, Z))) → U1(X, Y, Z, duplicate_in(Y, Z))
duplicate_in([], []) → duplicate_out([], [])
U1(X, Y, Z, duplicate_out(Y, Z)) → duplicate_out(.(X, Y), .(X, .(X, Z)))

The argument filtering Pi contains the following mapping:
duplicate_in(x1, x2)  =  duplicate_in(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x1, x4)
[]  =  []
duplicate_out(x1, x2)  =  duplicate_out(x2)
DUPLICATE_IN(x1, x2)  =  DUPLICATE_IN(x1)
U11(x1, x2, x3, x4)  =  U11(x1, x4)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 1 less node.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof

Pi DP problem:
The TRS P consists of the following rules:

DUPLICATE_IN(.(X, Y), .(X, .(X, Z))) → DUPLICATE_IN(Y, Z)

The TRS R consists of the following rules:

duplicate_in(.(X, Y), .(X, .(X, Z))) → U1(X, Y, Z, duplicate_in(Y, Z))
duplicate_in([], []) → duplicate_out([], [])
U1(X, Y, Z, duplicate_out(Y, Z)) → duplicate_out(.(X, Y), .(X, .(X, Z)))

The argument filtering Pi contains the following mapping:
duplicate_in(x1, x2)  =  duplicate_in(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x1, x4)
[]  =  []
duplicate_out(x1, x2)  =  duplicate_out(x2)
DUPLICATE_IN(x1, x2)  =  DUPLICATE_IN(x1)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof

Pi DP problem:
The TRS P consists of the following rules:

DUPLICATE_IN(.(X, Y), .(X, .(X, Z))) → DUPLICATE_IN(Y, Z)

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

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ QDPSizeChangeProof

Q DP problem:
The TRS P consists of the following rules:

DUPLICATE_IN(.(X, Y)) → DUPLICATE_IN(Y)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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: