Left Termination proof of /tmp/tmpGT4VKF/naive

Left Termination of the query pattern reverse_in_2(a, g) w.r.t. the given Prolog program could not be shown:

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

Clauses:

app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs).
app([], Ys, Ys).
reverse(.(X, Xs), Ys) :- ','(reverse(Xs, Zs), app(Zs, .(X, []), Ys)).
reverse([], []).

Queries:

reverse(a,g).

We use the technique of [30].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

reverse_in([], []) → reverse_out([], [])
reverse_in(.(X, Xs), Ys) → U2(X, Xs, Ys, reverse_in(Xs, Zs))
U2(X, Xs, Ys, reverse_out(Xs, Zs)) → U3(X, Xs, Ys, app_in(Zs, .(X, []), Ys))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
app_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Xs, Ys, app_out(Zs, .(X, []), Ys)) → reverse_out(.(X, Xs), Ys)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

  ↳ PrologToPiTRSProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

reverse_in([], []) → reverse_out([], [])
reverse_in(.(X, Xs), Ys) → U2(X, Xs, Ys, reverse_in(Xs, Zs))
U2(X, Xs, Ys, reverse_out(Xs, Zs)) → U3(X, Xs, Ys, app_in(Zs, .(X, []), Ys))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
app_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Xs, Ys, app_out(Zs, .(X, []), Ys)) → reverse_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
reverse_in(x1, x2) = reverse_in
[] = []
reverse_out(x1, x2) = reverse_out(x1, x2)
.(x1, x2) = .(x2)
U2(x1, x2, x3, x4) = U2(x4)
U3(x1, x2, x3, x4) = U3(x2, x4)
app_in(x1, x2, x3) = app_in(x1, x2)
app_out(x1, x2, x3) = app_out(x3)
U1(x1, x2, x3, x4, x5) = U1(x5)

Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

REVERSE_IN(.(X, Xs), Ys) → U2¹(X, Xs, Ys, reverse_in(Xs, Zs))
REVERSE_IN(.(X, Xs), Ys) → REVERSE_IN(Xs, Zs)
U2¹(X, Xs, Ys, reverse_out(Xs, Zs)) → U3¹(X, Xs, Ys, app_in(Zs, .(X, []), Ys))
U2¹(X, Xs, Ys, reverse_out(Xs, Zs)) → APP_IN(Zs, .(X, []), Ys)
APP_IN(.(X, Xs), Ys, .(X, Zs)) → U1¹(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_in([], []) → reverse_out([], [])
reverse_in(.(X, Xs), Ys) → U2(X, Xs, Ys, reverse_in(Xs, Zs))
U2(X, Xs, Ys, reverse_out(Xs, Zs)) → U3(X, Xs, Ys, app_in(Zs, .(X, []), Ys))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
app_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Xs, Ys, app_out(Zs, .(X, []), Ys)) → reverse_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
reverse_in(x1, x2) = reverse_in
[] = []
reverse_out(x1, x2) = reverse_out(x1, x2)
.(x1, x2) = .(x2)
U2(x1, x2, x3, x4) = U2(x4)
U3(x1, x2, x3, x4) = U3(x2, x4)
app_in(x1, x2, x3) = app_in(x1, x2)
app_out(x1, x2, x3) = app_out(x3)
U1(x1, x2, x3, x4, x5) = U1(x5)
REVERSE_IN(x1, x2) = REVERSE_IN
U3¹(x1, x2, x3, x4) = U3¹(x2, x4)
U2¹(x1, x2, x3, x4) = U2¹(x4)
U1¹(x1, x2, x3, x4, x5) = U1¹(x5)
APP_IN(x1, x2, x3) = APP_IN(x1, x2)

We have to consider all (P,R,Pi)-chains

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

  ↳ PrologToPiTRSProof

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

REVERSE_IN(.(X, Xs), Ys) → U2¹(X, Xs, Ys, reverse_in(Xs, Zs))
REVERSE_IN(.(X, Xs), Ys) → REVERSE_IN(Xs, Zs)
U2¹(X, Xs, Ys, reverse_out(Xs, Zs)) → U3¹(X, Xs, Ys, app_in(Zs, .(X, []), Ys))
U2¹(X, Xs, Ys, reverse_out(Xs, Zs)) → APP_IN(Zs, .(X, []), Ys)
APP_IN(.(X, Xs), Ys, .(X, Zs)) → U1¹(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_in([], []) → reverse_out([], [])
reverse_in(.(X, Xs), Ys) → U2(X, Xs, Ys, reverse_in(Xs, Zs))
U2(X, Xs, Ys, reverse_out(Xs, Zs)) → U3(X, Xs, Ys, app_in(Zs, .(X, []), Ys))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
app_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Xs, Ys, app_out(Zs, .(X, []), Ys)) → reverse_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
reverse_in(x1, x2) = reverse_in
[] = []
reverse_out(x1, x2) = reverse_out(x1, x2)
.(x1, x2) = .(x2)
U2(x1, x2, x3, x4) = U2(x4)
U3(x1, x2, x3, x4) = U3(x2, x4)
app_in(x1, x2, x3) = app_in(x1, x2)
app_out(x1, x2, x3) = app_out(x3)
U1(x1, x2, x3, x4, x5) = U1(x5)
REVERSE_IN(x1, x2) = REVERSE_IN
U3¹(x1, x2, x3, x4) = U3¹(x2, x4)
U2¹(x1, x2, x3, x4) = U2¹(x4)
U1¹(x1, x2, x3, x4, x5) = U1¹(x5)
APP_IN(x1, x2, x3) = APP_IN(x1, x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 4 less nodes.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_in([], []) → reverse_out([], [])
reverse_in(.(X, Xs), Ys) → U2(X, Xs, Ys, reverse_in(Xs, Zs))
U2(X, Xs, Ys, reverse_out(Xs, Zs)) → U3(X, Xs, Ys, app_in(Zs, .(X, []), Ys))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
app_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Xs, Ys, app_out(Zs, .(X, []), Ys)) → reverse_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
reverse_in(x1, x2) = reverse_in
[] = []
reverse_out(x1, x2) = reverse_out(x1, x2)
.(x1, x2) = .(x2)
U2(x1, x2, x3, x4) = U2(x4)
U3(x1, x2, x3, x4) = U3(x2, x4)
app_in(x1, x2, x3) = app_in(x1, x2)
app_out(x1, x2, x3) = app_out(x3)
U1(x1, x2, x3, x4, x5) = U1(x5)
APP_IN(x1, x2, x3) = APP_IN(x1, x2)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

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

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

APP_IN(.(Xs), Ys) → APP_IN(Xs, Ys)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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_IN(.(Xs), Ys) → APP_IN(Xs, Ys)
The graph contains the following edges 1 > 1, 2 >= 2

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

  ↳ PrologToPiTRSProof

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

REVERSE_IN(.(X, Xs), Ys) → REVERSE_IN(Xs, Zs)

The TRS R consists of the following rules:

reverse_in([], []) → reverse_out([], [])
reverse_in(.(X, Xs), Ys) → U2(X, Xs, Ys, reverse_in(Xs, Zs))
U2(X, Xs, Ys, reverse_out(Xs, Zs)) → U3(X, Xs, Ys, app_in(Zs, .(X, []), Ys))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
app_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Xs, Ys, app_out(Zs, .(X, []), Ys)) → reverse_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
reverse_in(x1, x2) = reverse_in
[] = []
reverse_out(x1, x2) = reverse_out(x1, x2)
.(x1, x2) = .(x2)
U2(x1, x2, x3, x4) = U2(x4)
U3(x1, x2, x3, x4) = U3(x2, x4)
app_in(x1, x2, x3) = app_in(x1, x2)
app_out(x1, x2, x3) = app_out(x3)
U1(x1, x2, x3, x4, x5) = U1(x5)
REVERSE_IN(x1, x2) = REVERSE_IN

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

  ↳ PrologToPiTRSProof

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

REVERSE_IN(.(X, Xs), Ys) → REVERSE_IN(Xs, Zs)

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

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ NonTerminationProof

  ↳ PrologToPiTRSProof

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

REVERSE_IN → REVERSE_IN

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

REVERSE_IN → REVERSE_IN

The TRS R consists of the following rules:none

s = REVERSE_IN evaluates to t =REVERSE_IN

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from REVERSE_IN to REVERSE_IN.

We use the technique of [30].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

reverse_in([], []) → reverse_out([], [])
reverse_in(.(X, Xs), Ys) → U2(X, Xs, Ys, reverse_in(Xs, Zs))
U2(X, Xs, Ys, reverse_out(Xs, Zs)) → U3(X, Xs, Ys, app_in(Zs, .(X, []), Ys))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
app_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Xs, Ys, app_out(Zs, .(X, []), Ys)) → reverse_out(.(X, Xs), Ys)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

reverse_in([], []) → reverse_out([], [])
reverse_in(.(X, Xs), Ys) → U2(X, Xs, Ys, reverse_in(Xs, Zs))
U2(X, Xs, Ys, reverse_out(Xs, Zs)) → U3(X, Xs, Ys, app_in(Zs, .(X, []), Ys))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
app_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Xs, Ys, app_out(Zs, .(X, []), Ys)) → reverse_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
reverse_in(x1, x2) = reverse_in
[] = []
reverse_out(x1, x2) = reverse_out(x1, x2)
.(x1, x2) = .(x2)
U2(x1, x2, x3, x4) = U2(x4)
U3(x1, x2, x3, x4) = U3(x2, x4)
app_in(x1, x2, x3) = app_in(x1, x2)
app_out(x1, x2, x3) = app_out(x1, x2, x3)
U1(x1, x2, x3, x4, x5) = U1(x2, x3, x5)

Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

REVERSE_IN(.(X, Xs), Ys) → U2¹(X, Xs, Ys, reverse_in(Xs, Zs))
REVERSE_IN(.(X, Xs), Ys) → REVERSE_IN(Xs, Zs)
U2¹(X, Xs, Ys, reverse_out(Xs, Zs)) → U3¹(X, Xs, Ys, app_in(Zs, .(X, []), Ys))
U2¹(X, Xs, Ys, reverse_out(Xs, Zs)) → APP_IN(Zs, .(X, []), Ys)
APP_IN(.(X, Xs), Ys, .(X, Zs)) → U1¹(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_in([], []) → reverse_out([], [])
reverse_in(.(X, Xs), Ys) → U2(X, Xs, Ys, reverse_in(Xs, Zs))
U2(X, Xs, Ys, reverse_out(Xs, Zs)) → U3(X, Xs, Ys, app_in(Zs, .(X, []), Ys))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
app_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Xs, Ys, app_out(Zs, .(X, []), Ys)) → reverse_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
reverse_in(x1, x2) = reverse_in
[] = []
reverse_out(x1, x2) = reverse_out(x1, x2)
.(x1, x2) = .(x2)
U2(x1, x2, x3, x4) = U2(x4)
U3(x1, x2, x3, x4) = U3(x2, x4)
app_in(x1, x2, x3) = app_in(x1, x2)
app_out(x1, x2, x3) = app_out(x1, x2, x3)
U1(x1, x2, x3, x4, x5) = U1(x2, x3, x5)
REVERSE_IN(x1, x2) = REVERSE_IN
U3¹(x1, x2, x3, x4) = U3¹(x2, x4)
U2¹(x1, x2, x3, x4) = U2¹(x4)
U1¹(x1, x2, x3, x4, x5) = U1¹(x2, x3, x5)
APP_IN(x1, x2, x3) = APP_IN(x1, x2)

We have to consider all (P,R,Pi)-chains

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

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

REVERSE_IN(.(X, Xs), Ys) → U2¹(X, Xs, Ys, reverse_in(Xs, Zs))
REVERSE_IN(.(X, Xs), Ys) → REVERSE_IN(Xs, Zs)
U2¹(X, Xs, Ys, reverse_out(Xs, Zs)) → U3¹(X, Xs, Ys, app_in(Zs, .(X, []), Ys))
U2¹(X, Xs, Ys, reverse_out(Xs, Zs)) → APP_IN(Zs, .(X, []), Ys)
APP_IN(.(X, Xs), Ys, .(X, Zs)) → U1¹(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_in([], []) → reverse_out([], [])
reverse_in(.(X, Xs), Ys) → U2(X, Xs, Ys, reverse_in(Xs, Zs))
U2(X, Xs, Ys, reverse_out(Xs, Zs)) → U3(X, Xs, Ys, app_in(Zs, .(X, []), Ys))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
app_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Xs, Ys, app_out(Zs, .(X, []), Ys)) → reverse_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
reverse_in(x1, x2) = reverse_in
[] = []
reverse_out(x1, x2) = reverse_out(x1, x2)
.(x1, x2) = .(x2)
U2(x1, x2, x3, x4) = U2(x4)
U3(x1, x2, x3, x4) = U3(x2, x4)
app_in(x1, x2, x3) = app_in(x1, x2)
app_out(x1, x2, x3) = app_out(x1, x2, x3)
U1(x1, x2, x3, x4, x5) = U1(x2, x3, x5)
REVERSE_IN(x1, x2) = REVERSE_IN
U3¹(x1, x2, x3, x4) = U3¹(x2, x4)
U2¹(x1, x2, x3, x4) = U2¹(x4)
U1¹(x1, x2, x3, x4, x5) = U1¹(x2, x3, x5)
APP_IN(x1, x2, x3) = APP_IN(x1, x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 4 less nodes.

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_in([], []) → reverse_out([], [])
reverse_in(.(X, Xs), Ys) → U2(X, Xs, Ys, reverse_in(Xs, Zs))
U2(X, Xs, Ys, reverse_out(Xs, Zs)) → U3(X, Xs, Ys, app_in(Zs, .(X, []), Ys))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
app_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Xs, Ys, app_out(Zs, .(X, []), Ys)) → reverse_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
reverse_in(x1, x2) = reverse_in
[] = []
reverse_out(x1, x2) = reverse_out(x1, x2)
.(x1, x2) = .(x2)
U2(x1, x2, x3, x4) = U2(x4)
U3(x1, x2, x3, x4) = U3(x2, x4)
app_in(x1, x2, x3) = app_in(x1, x2)
app_out(x1, x2, x3) = app_out(x1, x2, x3)
U1(x1, x2, x3, x4, x5) = U1(x2, x3, x5)
APP_IN(x1, x2, x3) = APP_IN(x1, x2)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ 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_IN(.(Xs), Ys) → APP_IN(Xs, Ys)

From the DPs we obtained the following set of size-change graphs:

APP_IN(.(Xs), Ys) → APP_IN(Xs, Ys)
The graph contains the following edges 1 > 1, 2 >= 2

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

REVERSE_IN(.(X, Xs), Ys) → REVERSE_IN(Xs, Zs)

The TRS R consists of the following rules:

reverse_in([], []) → reverse_out([], [])
reverse_in(.(X, Xs), Ys) → U2(X, Xs, Ys, reverse_in(Xs, Zs))
U2(X, Xs, Ys, reverse_out(Xs, Zs)) → U3(X, Xs, Ys, app_in(Zs, .(X, []), Ys))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
app_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Xs, Ys, app_out(Zs, .(X, []), Ys)) → reverse_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
reverse_in(x1, x2) = reverse_in
[] = []
reverse_out(x1, x2) = reverse_out(x1, x2)
.(x1, x2) = .(x2)
U2(x1, x2, x3, x4) = U2(x4)
U3(x1, x2, x3, x4) = U3(x2, x4)
app_in(x1, x2, x3) = app_in(x1, x2)
app_out(x1, x2, x3) = app_out(x1, x2, x3)
U1(x1, x2, x3, x4, x5) = U1(x2, x3, x5)
REVERSE_IN(x1, x2) = REVERSE_IN

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

REVERSE_IN(.(X, Xs), Ys) → REVERSE_IN(Xs, Zs)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ NonTerminationProof

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

REVERSE_IN → REVERSE_IN

The TRS R consists of the following rules:none

s = REVERSE_IN evaluates to t =REVERSE_IN

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from REVERSE_IN to REVERSE_IN.