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



↳ PROLOG
  ↳ PrologToPiTRSProof

append3(.2(H, X), Y, .2(X, Z)) :- append3(X, Y, Z).
append3({}0, Y, Y).


With regard to the inferred argument filtering the predicates were used in the following modes:
append3: (b,b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


append_3_in_gga3(._22(H, X), Y, ._22(X, Z)) -> if_append_3_in_1_gga5(H, X, Y, Z, append_3_in_gga3(X, Y, Z))
append_3_in_gga3([]_0, Y, Y) -> append_3_out_gga3([]_0, Y, Y)
if_append_3_in_1_gga5(H, X, Y, Z, append_3_out_gga3(X, Y, Z)) -> append_3_out_gga3(._22(H, X), Y, ._22(X, Z))

The argument filtering Pi contains the following mapping:
append_3_in_gga3(x1, x2, x3)  =  append_3_in_gga2(x1, x2)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_append_3_in_1_gga5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_gga2(x2, x5)
append_3_out_gga3(x1, x2, x3)  =  append_3_out_gga1(x3)

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:

append_3_in_gga3(._22(H, X), Y, ._22(X, Z)) -> if_append_3_in_1_gga5(H, X, Y, Z, append_3_in_gga3(X, Y, Z))
append_3_in_gga3([]_0, Y, Y) -> append_3_out_gga3([]_0, Y, Y)
if_append_3_in_1_gga5(H, X, Y, Z, append_3_out_gga3(X, Y, Z)) -> append_3_out_gga3(._22(H, X), Y, ._22(X, Z))

The argument filtering Pi contains the following mapping:
append_3_in_gga3(x1, x2, x3)  =  append_3_in_gga2(x1, x2)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_append_3_in_1_gga5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_gga2(x2, x5)
append_3_out_gga3(x1, x2, x3)  =  append_3_out_gga1(x3)


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

APPEND_3_IN_GGA3(._22(H, X), Y, ._22(X, Z)) -> IF_APPEND_3_IN_1_GGA5(H, X, Y, Z, append_3_in_gga3(X, Y, Z))
APPEND_3_IN_GGA3(._22(H, X), Y, ._22(X, Z)) -> APPEND_3_IN_GGA3(X, Y, Z)

The TRS R consists of the following rules:

append_3_in_gga3(._22(H, X), Y, ._22(X, Z)) -> if_append_3_in_1_gga5(H, X, Y, Z, append_3_in_gga3(X, Y, Z))
append_3_in_gga3([]_0, Y, Y) -> append_3_out_gga3([]_0, Y, Y)
if_append_3_in_1_gga5(H, X, Y, Z, append_3_out_gga3(X, Y, Z)) -> append_3_out_gga3(._22(H, X), Y, ._22(X, Z))

The argument filtering Pi contains the following mapping:
append_3_in_gga3(x1, x2, x3)  =  append_3_in_gga2(x1, x2)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_append_3_in_1_gga5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_gga2(x2, x5)
append_3_out_gga3(x1, x2, x3)  =  append_3_out_gga1(x3)
IF_APPEND_3_IN_1_GGA5(x1, x2, x3, x4, x5)  =  IF_APPEND_3_IN_1_GGA2(x2, x5)
APPEND_3_IN_GGA3(x1, x2, x3)  =  APPEND_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:

APPEND_3_IN_GGA3(._22(H, X), Y, ._22(X, Z)) -> IF_APPEND_3_IN_1_GGA5(H, X, Y, Z, append_3_in_gga3(X, Y, Z))
APPEND_3_IN_GGA3(._22(H, X), Y, ._22(X, Z)) -> APPEND_3_IN_GGA3(X, Y, Z)

The TRS R consists of the following rules:

append_3_in_gga3(._22(H, X), Y, ._22(X, Z)) -> if_append_3_in_1_gga5(H, X, Y, Z, append_3_in_gga3(X, Y, Z))
append_3_in_gga3([]_0, Y, Y) -> append_3_out_gga3([]_0, Y, Y)
if_append_3_in_1_gga5(H, X, Y, Z, append_3_out_gga3(X, Y, Z)) -> append_3_out_gga3(._22(H, X), Y, ._22(X, Z))

The argument filtering Pi contains the following mapping:
append_3_in_gga3(x1, x2, x3)  =  append_3_in_gga2(x1, x2)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_append_3_in_1_gga5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_gga2(x2, x5)
append_3_out_gga3(x1, x2, x3)  =  append_3_out_gga1(x3)
IF_APPEND_3_IN_1_GGA5(x1, x2, x3, x4, x5)  =  IF_APPEND_3_IN_1_GGA2(x2, x5)
APPEND_3_IN_GGA3(x1, x2, x3)  =  APPEND_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:

APPEND_3_IN_GGA3(._22(H, X), Y, ._22(X, Z)) -> APPEND_3_IN_GGA3(X, Y, Z)

The TRS R consists of the following rules:

append_3_in_gga3(._22(H, X), Y, ._22(X, Z)) -> if_append_3_in_1_gga5(H, X, Y, Z, append_3_in_gga3(X, Y, Z))
append_3_in_gga3([]_0, Y, Y) -> append_3_out_gga3([]_0, Y, Y)
if_append_3_in_1_gga5(H, X, Y, Z, append_3_out_gga3(X, Y, Z)) -> append_3_out_gga3(._22(H, X), Y, ._22(X, Z))

The argument filtering Pi contains the following mapping:
append_3_in_gga3(x1, x2, x3)  =  append_3_in_gga2(x1, x2)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_append_3_in_1_gga5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_gga2(x2, x5)
append_3_out_gga3(x1, x2, x3)  =  append_3_out_gga1(x3)
APPEND_3_IN_GGA3(x1, x2, x3)  =  APPEND_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:

APPEND_3_IN_GGA3(._22(H, X), Y, ._22(X, Z)) -> APPEND_3_IN_GGA3(X, Y, Z)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._22(x1, x2)
APPEND_3_IN_GGA3(x1, x2, x3)  =  APPEND_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:

APPEND_3_IN_GGA2(._22(H, X), Y) -> APPEND_3_IN_GGA2(X, Y)

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