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



↳ PROLOG
  ↳ PrologToPiTRSProof

prefix2(Xs, Ys) :- app3(Xs, underscore, Ys).
app3({}0, X, X).
app3(.2(X, Xs), Ys, .2(X, Zs)) :- app3(Xs, Ys, Zs).


With regard to the inferred argument filtering the predicates were used in the following modes:
prefix2: (b,f)
app3: (b,f,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


prefix_2_in_ga2(Xs, Ys) -> if_prefix_2_in_1_ga3(Xs, Ys, app_3_in_gaa3(Xs, underscore, Ys))
app_3_in_gaa3([]_0, X, X) -> app_3_out_gaa3([]_0, X, X)
app_3_in_gaa3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_gaa5(X, Xs, Ys, Zs, app_3_in_gaa3(Xs, Ys, Zs))
if_app_3_in_1_gaa5(X, Xs, Ys, Zs, app_3_out_gaa3(Xs, Ys, Zs)) -> app_3_out_gaa3(._22(X, Xs), Ys, ._22(X, Zs))
if_prefix_2_in_1_ga3(Xs, Ys, app_3_out_gaa3(Xs, underscore, Ys)) -> prefix_2_out_ga2(Xs, Ys)

The argument filtering Pi contains the following mapping:
prefix_2_in_ga2(x1, x2)  =  prefix_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_prefix_2_in_1_ga3(x1, x2, x3)  =  if_prefix_2_in_1_ga1(x3)
app_3_in_gaa3(x1, x2, x3)  =  app_3_in_gaa1(x1)
app_3_out_gaa3(x1, x2, x3)  =  app_3_out_gaa
if_app_3_in_1_gaa5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_gaa1(x5)
prefix_2_out_ga2(x1, x2)  =  prefix_2_out_ga

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:

prefix_2_in_ga2(Xs, Ys) -> if_prefix_2_in_1_ga3(Xs, Ys, app_3_in_gaa3(Xs, underscore, Ys))
app_3_in_gaa3([]_0, X, X) -> app_3_out_gaa3([]_0, X, X)
app_3_in_gaa3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_gaa5(X, Xs, Ys, Zs, app_3_in_gaa3(Xs, Ys, Zs))
if_app_3_in_1_gaa5(X, Xs, Ys, Zs, app_3_out_gaa3(Xs, Ys, Zs)) -> app_3_out_gaa3(._22(X, Xs), Ys, ._22(X, Zs))
if_prefix_2_in_1_ga3(Xs, Ys, app_3_out_gaa3(Xs, underscore, Ys)) -> prefix_2_out_ga2(Xs, Ys)

The argument filtering Pi contains the following mapping:
prefix_2_in_ga2(x1, x2)  =  prefix_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_prefix_2_in_1_ga3(x1, x2, x3)  =  if_prefix_2_in_1_ga1(x3)
app_3_in_gaa3(x1, x2, x3)  =  app_3_in_gaa1(x1)
app_3_out_gaa3(x1, x2, x3)  =  app_3_out_gaa
if_app_3_in_1_gaa5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_gaa1(x5)
prefix_2_out_ga2(x1, x2)  =  prefix_2_out_ga


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

PREFIX_2_IN_GA2(Xs, Ys) -> IF_PREFIX_2_IN_1_GA3(Xs, Ys, app_3_in_gaa3(Xs, underscore, Ys))
PREFIX_2_IN_GA2(Xs, Ys) -> APP_3_IN_GAA3(Xs, underscore, Ys)
APP_3_IN_GAA3(._22(X, Xs), Ys, ._22(X, Zs)) -> IF_APP_3_IN_1_GAA5(X, Xs, Ys, Zs, app_3_in_gaa3(Xs, Ys, Zs))
APP_3_IN_GAA3(._22(X, Xs), Ys, ._22(X, Zs)) -> APP_3_IN_GAA3(Xs, Ys, Zs)

The TRS R consists of the following rules:

prefix_2_in_ga2(Xs, Ys) -> if_prefix_2_in_1_ga3(Xs, Ys, app_3_in_gaa3(Xs, underscore, Ys))
app_3_in_gaa3([]_0, X, X) -> app_3_out_gaa3([]_0, X, X)
app_3_in_gaa3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_gaa5(X, Xs, Ys, Zs, app_3_in_gaa3(Xs, Ys, Zs))
if_app_3_in_1_gaa5(X, Xs, Ys, Zs, app_3_out_gaa3(Xs, Ys, Zs)) -> app_3_out_gaa3(._22(X, Xs), Ys, ._22(X, Zs))
if_prefix_2_in_1_ga3(Xs, Ys, app_3_out_gaa3(Xs, underscore, Ys)) -> prefix_2_out_ga2(Xs, Ys)

The argument filtering Pi contains the following mapping:
prefix_2_in_ga2(x1, x2)  =  prefix_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_prefix_2_in_1_ga3(x1, x2, x3)  =  if_prefix_2_in_1_ga1(x3)
app_3_in_gaa3(x1, x2, x3)  =  app_3_in_gaa1(x1)
app_3_out_gaa3(x1, x2, x3)  =  app_3_out_gaa
if_app_3_in_1_gaa5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_gaa1(x5)
prefix_2_out_ga2(x1, x2)  =  prefix_2_out_ga
PREFIX_2_IN_GA2(x1, x2)  =  PREFIX_2_IN_GA1(x1)
IF_APP_3_IN_1_GAA5(x1, x2, x3, x4, x5)  =  IF_APP_3_IN_1_GAA1(x5)
IF_PREFIX_2_IN_1_GA3(x1, x2, x3)  =  IF_PREFIX_2_IN_1_GA1(x3)
APP_3_IN_GAA3(x1, x2, x3)  =  APP_3_IN_GAA1(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:

PREFIX_2_IN_GA2(Xs, Ys) -> IF_PREFIX_2_IN_1_GA3(Xs, Ys, app_3_in_gaa3(Xs, underscore, Ys))
PREFIX_2_IN_GA2(Xs, Ys) -> APP_3_IN_GAA3(Xs, underscore, Ys)
APP_3_IN_GAA3(._22(X, Xs), Ys, ._22(X, Zs)) -> IF_APP_3_IN_1_GAA5(X, Xs, Ys, Zs, app_3_in_gaa3(Xs, Ys, Zs))
APP_3_IN_GAA3(._22(X, Xs), Ys, ._22(X, Zs)) -> APP_3_IN_GAA3(Xs, Ys, Zs)

The TRS R consists of the following rules:

prefix_2_in_ga2(Xs, Ys) -> if_prefix_2_in_1_ga3(Xs, Ys, app_3_in_gaa3(Xs, underscore, Ys))
app_3_in_gaa3([]_0, X, X) -> app_3_out_gaa3([]_0, X, X)
app_3_in_gaa3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_gaa5(X, Xs, Ys, Zs, app_3_in_gaa3(Xs, Ys, Zs))
if_app_3_in_1_gaa5(X, Xs, Ys, Zs, app_3_out_gaa3(Xs, Ys, Zs)) -> app_3_out_gaa3(._22(X, Xs), Ys, ._22(X, Zs))
if_prefix_2_in_1_ga3(Xs, Ys, app_3_out_gaa3(Xs, underscore, Ys)) -> prefix_2_out_ga2(Xs, Ys)

The argument filtering Pi contains the following mapping:
prefix_2_in_ga2(x1, x2)  =  prefix_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_prefix_2_in_1_ga3(x1, x2, x3)  =  if_prefix_2_in_1_ga1(x3)
app_3_in_gaa3(x1, x2, x3)  =  app_3_in_gaa1(x1)
app_3_out_gaa3(x1, x2, x3)  =  app_3_out_gaa
if_app_3_in_1_gaa5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_gaa1(x5)
prefix_2_out_ga2(x1, x2)  =  prefix_2_out_ga
PREFIX_2_IN_GA2(x1, x2)  =  PREFIX_2_IN_GA1(x1)
IF_APP_3_IN_1_GAA5(x1, x2, x3, x4, x5)  =  IF_APP_3_IN_1_GAA1(x5)
IF_PREFIX_2_IN_1_GA3(x1, x2, x3)  =  IF_PREFIX_2_IN_1_GA1(x3)
APP_3_IN_GAA3(x1, x2, x3)  =  APP_3_IN_GAA1(x1)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 1 SCC with 3 less nodes.

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof

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

APP_3_IN_GAA3(._22(X, Xs), Ys, ._22(X, Zs)) -> APP_3_IN_GAA3(Xs, Ys, Zs)

The TRS R consists of the following rules:

prefix_2_in_ga2(Xs, Ys) -> if_prefix_2_in_1_ga3(Xs, Ys, app_3_in_gaa3(Xs, underscore, Ys))
app_3_in_gaa3([]_0, X, X) -> app_3_out_gaa3([]_0, X, X)
app_3_in_gaa3(._22(X, Xs), Ys, ._22(X, Zs)) -> if_app_3_in_1_gaa5(X, Xs, Ys, Zs, app_3_in_gaa3(Xs, Ys, Zs))
if_app_3_in_1_gaa5(X, Xs, Ys, Zs, app_3_out_gaa3(Xs, Ys, Zs)) -> app_3_out_gaa3(._22(X, Xs), Ys, ._22(X, Zs))
if_prefix_2_in_1_ga3(Xs, Ys, app_3_out_gaa3(Xs, underscore, Ys)) -> prefix_2_out_ga2(Xs, Ys)

The argument filtering Pi contains the following mapping:
prefix_2_in_ga2(x1, x2)  =  prefix_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_prefix_2_in_1_ga3(x1, x2, x3)  =  if_prefix_2_in_1_ga1(x3)
app_3_in_gaa3(x1, x2, x3)  =  app_3_in_gaa1(x1)
app_3_out_gaa3(x1, x2, x3)  =  app_3_out_gaa
if_app_3_in_1_gaa5(x1, x2, x3, x4, x5)  =  if_app_3_in_1_gaa1(x5)
prefix_2_out_ga2(x1, x2)  =  prefix_2_out_ga
APP_3_IN_GAA3(x1, x2, x3)  =  APP_3_IN_GAA1(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:

APP_3_IN_GAA3(._22(X, Xs), Ys, ._22(X, Zs)) -> APP_3_IN_GAA3(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._22(x1, x2)
APP_3_IN_GAA3(x1, x2, x3)  =  APP_3_IN_GAA1(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:

APP_3_IN_GAA1(._22(X, Xs)) -> APP_3_IN_GAA1(Xs)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {APP_3_IN_GAA1}.
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: