Left Termination proof of /tmp/tmpnuU9Dn/naive

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

↳ PROLOG

  ↳ PrologToPiTRSProof

app₃(.₂(X, Xs), Ys, .₂(X, Zs)) :- app₃(Xs, Ys, Zs).
app₃({}₀, Ys, Ys).
reverse₂(.₂(X, Xs), Ys) :- reverse₂(Xs, Zs), app₃(Zs, .₂(X, {}₀), Ys).
reverse₂({}₀, {}₀).

With regard to the inferred argument filtering the predicates were used in the following modes:
reverse₂: (b,f)
app₃: (b,b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

reverse_2_in_ga₂(._2₂(X, Xs), Ys) -> if_reverse_2_in_1_ga₄(X, Xs, Ys, reverse_2_in_ga₂(Xs, Zs))
reverse_2_in_ga₂([]_0, []_0) -> reverse_2_out_ga₂([]_0, []_0)
if_reverse_2_in_1_ga₄(X, Xs, Ys, reverse_2_out_ga₂(Xs, Zs)) -> if_reverse_2_in_2_ga₅(X, Xs, Ys, Zs, app_3_in_gga₃(Zs, ._2₂(X, []_0), Ys))
app_3_in_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_in_gga₃(Xs, Ys, Zs))
app_3_in_gga₃([]_0, Ys, Ys) -> app_3_out_gga₃([]_0, Ys, Ys)
if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_out_gga₃(Xs, Ys, Zs)) -> app_3_out_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_reverse_2_in_2_ga₅(X, Xs, Ys, Zs, app_3_out_gga₃(Zs, ._2₂(X, []_0), Ys)) -> reverse_2_out_ga₂(._2₂(X, Xs), Ys)

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:

reverse_2_in_ga₂(._2₂(X, Xs), Ys) -> if_reverse_2_in_1_ga₄(X, Xs, Ys, reverse_2_in_ga₂(Xs, Zs))
reverse_2_in_ga₂([]_0, []_0) -> reverse_2_out_ga₂([]_0, []_0)
if_reverse_2_in_1_ga₄(X, Xs, Ys, reverse_2_out_ga₂(Xs, Zs)) -> if_reverse_2_in_2_ga₅(X, Xs, Ys, Zs, app_3_in_gga₃(Zs, ._2₂(X, []_0), Ys))
app_3_in_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_in_gga₃(Xs, Ys, Zs))
app_3_in_gga₃([]_0, Ys, Ys) -> app_3_out_gga₃([]_0, Ys, Ys)
if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_out_gga₃(Xs, Ys, Zs)) -> app_3_out_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_reverse_2_in_2_ga₅(X, Xs, Ys, Zs, app_3_out_gga₃(Zs, ._2₂(X, []_0), Ys)) -> reverse_2_out_ga₂(._2₂(X, Xs), Ys)

REVERSE_2_IN_GA₂(._2₂(X, Xs), Ys) -> IF_REVERSE_2_IN_1_GA₄(X, Xs, Ys, reverse_2_in_ga₂(Xs, Zs))
REVERSE_2_IN_GA₂(._2₂(X, Xs), Ys) -> REVERSE_2_IN_GA₂(Xs, Zs)
IF_REVERSE_2_IN_1_GA₄(X, Xs, Ys, reverse_2_out_ga₂(Xs, Zs)) -> IF_REVERSE_2_IN_2_GA₅(X, Xs, Ys, Zs, app_3_in_gga₃(Zs, ._2₂(X, []_0), Ys))
IF_REVERSE_2_IN_1_GA₄(X, Xs, Ys, reverse_2_out_ga₂(Xs, Zs)) -> APP_3_IN_GGA₃(Zs, ._2₂(X, []_0), Ys)
APP_3_IN_GGA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> IF_APP_3_IN_1_GGA₅(X, Xs, Ys, Zs, app_3_in_gga₃(Xs, Ys, Zs))
APP_3_IN_GGA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_GGA₃(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_2_in_ga₂(._2₂(X, Xs), Ys) -> if_reverse_2_in_1_ga₄(X, Xs, Ys, reverse_2_in_ga₂(Xs, Zs))
reverse_2_in_ga₂([]_0, []_0) -> reverse_2_out_ga₂([]_0, []_0)
if_reverse_2_in_1_ga₄(X, Xs, Ys, reverse_2_out_ga₂(Xs, Zs)) -> if_reverse_2_in_2_ga₅(X, Xs, Ys, Zs, app_3_in_gga₃(Zs, ._2₂(X, []_0), Ys))
app_3_in_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_in_gga₃(Xs, Ys, Zs))
app_3_in_gga₃([]_0, Ys, Ys) -> app_3_out_gga₃([]_0, Ys, Ys)
if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_out_gga₃(Xs, Ys, Zs)) -> app_3_out_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_reverse_2_in_2_ga₅(X, Xs, Ys, Zs, app_3_out_gga₃(Zs, ._2₂(X, []_0), Ys)) -> reverse_2_out_ga₂(._2₂(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
reverse_2_in_ga₂(x1, x2) = reverse_2_in_ga₁(x1)
._2₂(x1, x2) = ._2₂(x1, x2)
[]_0 = []_0
if_reverse_2_in_1_ga₄(x1, x2, x3, x4) = if_reverse_2_in_1_ga₂(x1, x4)
reverse_2_out_ga₂(x1, x2) = reverse_2_out_ga₁(x2)
if_reverse_2_in_2_ga₅(x1, x2, x3, x4, x5) = if_reverse_2_in_2_ga₁(x5)
app_3_in_gga₃(x1, x2, x3) = app_3_in_gga₂(x1, x2)
if_app_3_in_1_gga₅(x1, x2, x3, x4, x5) = if_app_3_in_1_gga₂(x1, x5)
app_3_out_gga₃(x1, x2, x3) = app_3_out_gga₁(x3)
IF_APP_3_IN_1_GGA₅(x1, x2, x3, x4, x5) = IF_APP_3_IN_1_GGA₂(x1, x5)
APP_3_IN_GGA₃(x1, x2, x3) = APP_3_IN_GGA₂(x1, x2)
REVERSE_2_IN_GA₂(x1, x2) = REVERSE_2_IN_GA₁(x1)
IF_REVERSE_2_IN_2_GA₅(x1, x2, x3, x4, x5) = IF_REVERSE_2_IN_2_GA₁(x5)
IF_REVERSE_2_IN_1_GA₄(x1, x2, x3, x4) = IF_REVERSE_2_IN_1_GA₂(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:

REVERSE_2_IN_GA₂(._2₂(X, Xs), Ys) -> IF_REVERSE_2_IN_1_GA₄(X, Xs, Ys, reverse_2_in_ga₂(Xs, Zs))
REVERSE_2_IN_GA₂(._2₂(X, Xs), Ys) -> REVERSE_2_IN_GA₂(Xs, Zs)
IF_REVERSE_2_IN_1_GA₄(X, Xs, Ys, reverse_2_out_ga₂(Xs, Zs)) -> IF_REVERSE_2_IN_2_GA₅(X, Xs, Ys, Zs, app_3_in_gga₃(Zs, ._2₂(X, []_0), Ys))
IF_REVERSE_2_IN_1_GA₄(X, Xs, Ys, reverse_2_out_ga₂(Xs, Zs)) -> APP_3_IN_GGA₃(Zs, ._2₂(X, []_0), Ys)
APP_3_IN_GGA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> IF_APP_3_IN_1_GGA₅(X, Xs, Ys, Zs, app_3_in_gga₃(Xs, Ys, Zs))
APP_3_IN_GGA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_GGA₃(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_2_in_ga₂(._2₂(X, Xs), Ys) -> if_reverse_2_in_1_ga₄(X, Xs, Ys, reverse_2_in_ga₂(Xs, Zs))
reverse_2_in_ga₂([]_0, []_0) -> reverse_2_out_ga₂([]_0, []_0)
if_reverse_2_in_1_ga₄(X, Xs, Ys, reverse_2_out_ga₂(Xs, Zs)) -> if_reverse_2_in_2_ga₅(X, Xs, Ys, Zs, app_3_in_gga₃(Zs, ._2₂(X, []_0), Ys))
app_3_in_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_in_gga₃(Xs, Ys, Zs))
app_3_in_gga₃([]_0, Ys, Ys) -> app_3_out_gga₃([]_0, Ys, Ys)
if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_out_gga₃(Xs, Ys, Zs)) -> app_3_out_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_reverse_2_in_2_ga₅(X, Xs, Ys, Zs, app_3_out_gga₃(Zs, ._2₂(X, []_0), Ys)) -> reverse_2_out_ga₂(._2₂(X, Xs), Ys)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

APP_3_IN_GGA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_GGA₃(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_2_in_ga₂(._2₂(X, Xs), Ys) -> if_reverse_2_in_1_ga₄(X, Xs, Ys, reverse_2_in_ga₂(Xs, Zs))
reverse_2_in_ga₂([]_0, []_0) -> reverse_2_out_ga₂([]_0, []_0)
if_reverse_2_in_1_ga₄(X, Xs, Ys, reverse_2_out_ga₂(Xs, Zs)) -> if_reverse_2_in_2_ga₅(X, Xs, Ys, Zs, app_3_in_gga₃(Zs, ._2₂(X, []_0), Ys))
app_3_in_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_in_gga₃(Xs, Ys, Zs))
app_3_in_gga₃([]_0, Ys, Ys) -> app_3_out_gga₃([]_0, Ys, Ys)
if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_out_gga₃(Xs, Ys, Zs)) -> app_3_out_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_reverse_2_in_2_ga₅(X, Xs, Ys, Zs, app_3_out_gga₃(Zs, ._2₂(X, []_0), Ys)) -> reverse_2_out_ga₂(._2₂(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
reverse_2_in_ga₂(x1, x2) = reverse_2_in_ga₁(x1)
._2₂(x1, x2) = ._2₂(x1, x2)
[]_0 = []_0
if_reverse_2_in_1_ga₄(x1, x2, x3, x4) = if_reverse_2_in_1_ga₂(x1, x4)
reverse_2_out_ga₂(x1, x2) = reverse_2_out_ga₁(x2)
if_reverse_2_in_2_ga₅(x1, x2, x3, x4, x5) = if_reverse_2_in_2_ga₁(x5)
app_3_in_gga₃(x1, x2, x3) = app_3_in_gga₂(x1, x2)
if_app_3_in_1_gga₅(x1, x2, x3, x4, x5) = if_app_3_in_1_gga₂(x1, x5)
app_3_out_gga₃(x1, x2, x3) = app_3_out_gga₁(x3)
APP_3_IN_GGA₃(x1, x2, x3) = APP_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

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

APP_3_IN_GGA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_GGA₃(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₂(x1, x2)
APP_3_IN_GGA₃(x1, x2, x3) = APP_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

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

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

APP_3_IN_GGA₂(._2₂(X, Xs), Ys) -> APP_3_IN_GGA₂(Xs, Ys)

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_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:

APP_3_IN_GGA₂(._2₂(X, Xs), Ys) -> APP_3_IN_GGA₂(Xs, Ys)
The graph contains the following edges 1 > 1, 2 >= 2

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

REVERSE_2_IN_GA₂(._2₂(X, Xs), Ys) -> REVERSE_2_IN_GA₂(Xs, Zs)

The TRS R consists of the following rules:

reverse_2_in_ga₂(._2₂(X, Xs), Ys) -> if_reverse_2_in_1_ga₄(X, Xs, Ys, reverse_2_in_ga₂(Xs, Zs))
reverse_2_in_ga₂([]_0, []_0) -> reverse_2_out_ga₂([]_0, []_0)
if_reverse_2_in_1_ga₄(X, Xs, Ys, reverse_2_out_ga₂(Xs, Zs)) -> if_reverse_2_in_2_ga₅(X, Xs, Ys, Zs, app_3_in_gga₃(Zs, ._2₂(X, []_0), Ys))
app_3_in_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_in_gga₃(Xs, Ys, Zs))
app_3_in_gga₃([]_0, Ys, Ys) -> app_3_out_gga₃([]_0, Ys, Ys)
if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_out_gga₃(Xs, Ys, Zs)) -> app_3_out_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_reverse_2_in_2_ga₅(X, Xs, Ys, Zs, app_3_out_gga₃(Zs, ._2₂(X, []_0), Ys)) -> reverse_2_out_ga₂(._2₂(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
reverse_2_in_ga₂(x1, x2) = reverse_2_in_ga₁(x1)
._2₂(x1, x2) = ._2₂(x1, x2)
[]_0 = []_0
if_reverse_2_in_1_ga₄(x1, x2, x3, x4) = if_reverse_2_in_1_ga₂(x1, x4)
reverse_2_out_ga₂(x1, x2) = reverse_2_out_ga₁(x2)
if_reverse_2_in_2_ga₅(x1, x2, x3, x4, x5) = if_reverse_2_in_2_ga₁(x5)
app_3_in_gga₃(x1, x2, x3) = app_3_in_gga₂(x1, x2)
if_app_3_in_1_gga₅(x1, x2, x3, x4, x5) = if_app_3_in_1_gga₂(x1, x5)
app_3_out_gga₃(x1, x2, x3) = app_3_out_gga₁(x3)
REVERSE_2_IN_GA₂(x1, x2) = REVERSE_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

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

REVERSE_2_IN_GA₂(._2₂(X, Xs), Ys) -> REVERSE_2_IN_GA₂(Xs, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₂(x1, x2)
REVERSE_2_IN_GA₂(x1, x2) = REVERSE_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

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

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

REVERSE_2_IN_GA₁(._2₂(X, Xs)) -> REVERSE_2_IN_GA₁(Xs)

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

REVERSE_2_IN_GA₁(._2₂(X, Xs)) -> REVERSE_2_IN_GA₁(Xs)
The graph contains the following edges 1 > 1