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



↳ PROLOG
  ↳ PrologToPiTRSProof

interleave3({}0, Xs, Xs).
interleave3(.2(X, Xs), Ys, .2(X, Zs)) :- interleave3(Ys, Xs, Zs).


With regard to the inferred argument filtering the predicates were used in the following modes:
interleave3: (b,b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


interleave_3_in_gga3([]_0, Xs, Xs) -> interleave_3_out_gga3([]_0, Xs, Xs)
interleave_3_in_gga3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_interleave_3_in_1_gga5(X, Xs, Ys, Zs, interleave_3_in_gga3(Ys, Xs, Zs))
if_interleave_3_in_1_gga5(X, Xs, Ys, Zs, interleave_3_out_gga3(Ys, Xs, Zs)) -> interleave_3_out_gga3(._22(X, Xs), Ys, ._22(X, Zs))

The argument filtering Pi contains the following mapping:
interleave_3_in_gga3(x1, x2, x3)  =  interleave_3_in_gga2(x1, x2)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
interleave_3_out_gga3(x1, x2, x3)  =  interleave_3_out_gga1(x3)
if_interleave_3_in_1_gga5(x1, x2, x3, x4, x5)  =  if_interleave_3_in_1_gga2(x1, x5)

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:

interleave_3_in_gga3([]_0, Xs, Xs) -> interleave_3_out_gga3([]_0, Xs, Xs)
interleave_3_in_gga3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_interleave_3_in_1_gga5(X, Xs, Ys, Zs, interleave_3_in_gga3(Ys, Xs, Zs))
if_interleave_3_in_1_gga5(X, Xs, Ys, Zs, interleave_3_out_gga3(Ys, Xs, Zs)) -> interleave_3_out_gga3(._22(X, Xs), Ys, ._22(X, Zs))

The argument filtering Pi contains the following mapping:
interleave_3_in_gga3(x1, x2, x3)  =  interleave_3_in_gga2(x1, x2)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
interleave_3_out_gga3(x1, x2, x3)  =  interleave_3_out_gga1(x3)
if_interleave_3_in_1_gga5(x1, x2, x3, x4, x5)  =  if_interleave_3_in_1_gga2(x1, x5)


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

INTERLEAVE_3_IN_GGA3(._22(X, Xs), Ys, ._22(X, Zs)) -> IF_INTERLEAVE_3_IN_1_GGA5(X, Xs, Ys, Zs, interleave_3_in_gga3(Ys, Xs, Zs))
INTERLEAVE_3_IN_GGA3(._22(X, Xs), Ys, ._22(X, Zs)) -> INTERLEAVE_3_IN_GGA3(Ys, Xs, Zs)

The TRS R consists of the following rules:

interleave_3_in_gga3([]_0, Xs, Xs) -> interleave_3_out_gga3([]_0, Xs, Xs)
interleave_3_in_gga3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_interleave_3_in_1_gga5(X, Xs, Ys, Zs, interleave_3_in_gga3(Ys, Xs, Zs))
if_interleave_3_in_1_gga5(X, Xs, Ys, Zs, interleave_3_out_gga3(Ys, Xs, Zs)) -> interleave_3_out_gga3(._22(X, Xs), Ys, ._22(X, Zs))

The argument filtering Pi contains the following mapping:
interleave_3_in_gga3(x1, x2, x3)  =  interleave_3_in_gga2(x1, x2)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
interleave_3_out_gga3(x1, x2, x3)  =  interleave_3_out_gga1(x3)
if_interleave_3_in_1_gga5(x1, x2, x3, x4, x5)  =  if_interleave_3_in_1_gga2(x1, x5)
IF_INTERLEAVE_3_IN_1_GGA5(x1, x2, x3, x4, x5)  =  IF_INTERLEAVE_3_IN_1_GGA2(x1, x5)
INTERLEAVE_3_IN_GGA3(x1, x2, x3)  =  INTERLEAVE_3_IN_GGA2(x1, x2)

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:

INTERLEAVE_3_IN_GGA3(._22(X, Xs), Ys, ._22(X, Zs)) -> IF_INTERLEAVE_3_IN_1_GGA5(X, Xs, Ys, Zs, interleave_3_in_gga3(Ys, Xs, Zs))
INTERLEAVE_3_IN_GGA3(._22(X, Xs), Ys, ._22(X, Zs)) -> INTERLEAVE_3_IN_GGA3(Ys, Xs, Zs)

The TRS R consists of the following rules:

interleave_3_in_gga3([]_0, Xs, Xs) -> interleave_3_out_gga3([]_0, Xs, Xs)
interleave_3_in_gga3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_interleave_3_in_1_gga5(X, Xs, Ys, Zs, interleave_3_in_gga3(Ys, Xs, Zs))
if_interleave_3_in_1_gga5(X, Xs, Ys, Zs, interleave_3_out_gga3(Ys, Xs, Zs)) -> interleave_3_out_gga3(._22(X, Xs), Ys, ._22(X, Zs))

The argument filtering Pi contains the following mapping:
interleave_3_in_gga3(x1, x2, x3)  =  interleave_3_in_gga2(x1, x2)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
interleave_3_out_gga3(x1, x2, x3)  =  interleave_3_out_gga1(x3)
if_interleave_3_in_1_gga5(x1, x2, x3, x4, x5)  =  if_interleave_3_in_1_gga2(x1, x5)
IF_INTERLEAVE_3_IN_1_GGA5(x1, x2, x3, x4, x5)  =  IF_INTERLEAVE_3_IN_1_GGA2(x1, x5)
INTERLEAVE_3_IN_GGA3(x1, x2, x3)  =  INTERLEAVE_3_IN_GGA2(x1, x2)

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:

INTERLEAVE_3_IN_GGA3(._22(X, Xs), Ys, ._22(X, Zs)) -> INTERLEAVE_3_IN_GGA3(Ys, Xs, Zs)

The TRS R consists of the following rules:

interleave_3_in_gga3([]_0, Xs, Xs) -> interleave_3_out_gga3([]_0, Xs, Xs)
interleave_3_in_gga3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_interleave_3_in_1_gga5(X, Xs, Ys, Zs, interleave_3_in_gga3(Ys, Xs, Zs))
if_interleave_3_in_1_gga5(X, Xs, Ys, Zs, interleave_3_out_gga3(Ys, Xs, Zs)) -> interleave_3_out_gga3(._22(X, Xs), Ys, ._22(X, Zs))

The argument filtering Pi contains the following mapping:
interleave_3_in_gga3(x1, x2, x3)  =  interleave_3_in_gga2(x1, x2)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
interleave_3_out_gga3(x1, x2, x3)  =  interleave_3_out_gga1(x3)
if_interleave_3_in_1_gga5(x1, x2, x3, x4, x5)  =  if_interleave_3_in_1_gga2(x1, x5)
INTERLEAVE_3_IN_GGA3(x1, x2, x3)  =  INTERLEAVE_3_IN_GGA2(x1, x2)

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:

INTERLEAVE_3_IN_GGA3(._22(X, Xs), Ys, ._22(X, Zs)) -> INTERLEAVE_3_IN_GGA3(Ys, Xs, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._22(x1, x2)
INTERLEAVE_3_IN_GGA3(x1, x2, x3)  =  INTERLEAVE_3_IN_GGA2(x1, x2)

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:

INTERLEAVE_3_IN_GGA2(._22(X, Xs), Ys) -> INTERLEAVE_3_IN_GGA2(Ys, Xs)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {INTERLEAVE_3_IN_GGA2}.
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: