Left Termination of the query pattern duplicate(b,f) w.r.t. the given Prolog program could successfully be proven:



↳ PROLOG
  ↳ PrologToPiTRSProof

duplicate2({}0, {}0).
duplicate2(.2(X, Y), .2(X, .2(X, Z))) :- duplicate2(Y, Z).


With regard to the inferred argument filtering the predicates were used in the following modes:
duplicate2: (b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


duplicate_2_in_ga2([]_0, []_0) -> duplicate_2_out_ga2([]_0, []_0)
duplicate_2_in_ga2(._22(X, Y), ._22(X, ._22(X, Z))) -> if_duplicate_2_in_1_ga4(X, Y, Z, duplicate_2_in_ga2(Y, Z))
if_duplicate_2_in_1_ga4(X, Y, Z, duplicate_2_out_ga2(Y, Z)) -> duplicate_2_out_ga2(._22(X, Y), ._22(X, ._22(X, Z)))

The argument filtering Pi contains the following mapping:
duplicate_2_in_ga2(x1, x2)  =  duplicate_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
duplicate_2_out_ga2(x1, x2)  =  duplicate_2_out_ga1(x2)
if_duplicate_2_in_1_ga4(x1, x2, x3, x4)  =  if_duplicate_2_in_1_ga2(x1, x4)

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_2_in_ga2([]_0, []_0) -> duplicate_2_out_ga2([]_0, []_0)
duplicate_2_in_ga2(._22(X, Y), ._22(X, ._22(X, Z))) -> if_duplicate_2_in_1_ga4(X, Y, Z, duplicate_2_in_ga2(Y, Z))
if_duplicate_2_in_1_ga4(X, Y, Z, duplicate_2_out_ga2(Y, Z)) -> duplicate_2_out_ga2(._22(X, Y), ._22(X, ._22(X, Z)))

The argument filtering Pi contains the following mapping:
duplicate_2_in_ga2(x1, x2)  =  duplicate_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
duplicate_2_out_ga2(x1, x2)  =  duplicate_2_out_ga1(x2)
if_duplicate_2_in_1_ga4(x1, x2, x3, x4)  =  if_duplicate_2_in_1_ga2(x1, x4)


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

DUPLICATE_2_IN_GA2(._22(X, Y), ._22(X, ._22(X, Z))) -> IF_DUPLICATE_2_IN_1_GA4(X, Y, Z, duplicate_2_in_ga2(Y, Z))
DUPLICATE_2_IN_GA2(._22(X, Y), ._22(X, ._22(X, Z))) -> DUPLICATE_2_IN_GA2(Y, Z)

The TRS R consists of the following rules:

duplicate_2_in_ga2([]_0, []_0) -> duplicate_2_out_ga2([]_0, []_0)
duplicate_2_in_ga2(._22(X, Y), ._22(X, ._22(X, Z))) -> if_duplicate_2_in_1_ga4(X, Y, Z, duplicate_2_in_ga2(Y, Z))
if_duplicate_2_in_1_ga4(X, Y, Z, duplicate_2_out_ga2(Y, Z)) -> duplicate_2_out_ga2(._22(X, Y), ._22(X, ._22(X, Z)))

The argument filtering Pi contains the following mapping:
duplicate_2_in_ga2(x1, x2)  =  duplicate_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
duplicate_2_out_ga2(x1, x2)  =  duplicate_2_out_ga1(x2)
if_duplicate_2_in_1_ga4(x1, x2, x3, x4)  =  if_duplicate_2_in_1_ga2(x1, x4)
IF_DUPLICATE_2_IN_1_GA4(x1, x2, x3, x4)  =  IF_DUPLICATE_2_IN_1_GA2(x1, x4)
DUPLICATE_2_IN_GA2(x1, x2)  =  DUPLICATE_2_IN_GA1(x1)

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_2_IN_GA2(._22(X, Y), ._22(X, ._22(X, Z))) -> IF_DUPLICATE_2_IN_1_GA4(X, Y, Z, duplicate_2_in_ga2(Y, Z))
DUPLICATE_2_IN_GA2(._22(X, Y), ._22(X, ._22(X, Z))) -> DUPLICATE_2_IN_GA2(Y, Z)

The TRS R consists of the following rules:

duplicate_2_in_ga2([]_0, []_0) -> duplicate_2_out_ga2([]_0, []_0)
duplicate_2_in_ga2(._22(X, Y), ._22(X, ._22(X, Z))) -> if_duplicate_2_in_1_ga4(X, Y, Z, duplicate_2_in_ga2(Y, Z))
if_duplicate_2_in_1_ga4(X, Y, Z, duplicate_2_out_ga2(Y, Z)) -> duplicate_2_out_ga2(._22(X, Y), ._22(X, ._22(X, Z)))

The argument filtering Pi contains the following mapping:
duplicate_2_in_ga2(x1, x2)  =  duplicate_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
duplicate_2_out_ga2(x1, x2)  =  duplicate_2_out_ga1(x2)
if_duplicate_2_in_1_ga4(x1, x2, x3, x4)  =  if_duplicate_2_in_1_ga2(x1, x4)
IF_DUPLICATE_2_IN_1_GA4(x1, x2, x3, x4)  =  IF_DUPLICATE_2_IN_1_GA2(x1, x4)
DUPLICATE_2_IN_GA2(x1, x2)  =  DUPLICATE_2_IN_GA1(x1)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph 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_2_IN_GA2(._22(X, Y), ._22(X, ._22(X, Z))) -> DUPLICATE_2_IN_GA2(Y, Z)

The TRS R consists of the following rules:

duplicate_2_in_ga2([]_0, []_0) -> duplicate_2_out_ga2([]_0, []_0)
duplicate_2_in_ga2(._22(X, Y), ._22(X, ._22(X, Z))) -> if_duplicate_2_in_1_ga4(X, Y, Z, duplicate_2_in_ga2(Y, Z))
if_duplicate_2_in_1_ga4(X, Y, Z, duplicate_2_out_ga2(Y, Z)) -> duplicate_2_out_ga2(._22(X, Y), ._22(X, ._22(X, Z)))

The argument filtering Pi contains the following mapping:
duplicate_2_in_ga2(x1, x2)  =  duplicate_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
duplicate_2_out_ga2(x1, x2)  =  duplicate_2_out_ga1(x2)
if_duplicate_2_in_1_ga4(x1, x2, x3, x4)  =  if_duplicate_2_in_1_ga2(x1, x4)
DUPLICATE_2_IN_GA2(x1, x2)  =  DUPLICATE_2_IN_GA1(x1)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting 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_2_IN_GA2(._22(X, Y), ._22(X, ._22(X, Z))) -> DUPLICATE_2_IN_GA2(Y, Z)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._22(x1, x2)
DUPLICATE_2_IN_GA2(x1, x2)  =  DUPLICATE_2_IN_GA1(x1)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem 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_2_IN_GA1(._22(X, Y)) -> DUPLICATE_2_IN_GA1(Y)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {DUPLICATE_2_IN_GA1}.
By using the subterm criterion together with the size-change analysis 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: