Left Termination proof of /tmp/tmpAPBtOf/naive

Left Termination of the query pattern reverse_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:

↳ Prolog

  ↳ 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(g,a).

We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
reverse_in: (b,f)
app_in: (b,b,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

reverse_in_ga(.(X, Xs), Ys) → U2_ga(X, Xs, Ys, reverse_in_ga(Xs, Zs))
reverse_in_ga([], []) → reverse_out_ga([], [])
U2_ga(X, Xs, Ys, reverse_out_ga(Xs, Zs)) → U3_ga(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
app_in_gga(.(X, Xs), Ys, .(X, Zs)) → U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
app_in_gga([], Ys, Ys) → app_out_gga([], Ys, Ys)
U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) → app_out_gga(.(X, Xs), Ys, .(X, Zs))
U3_ga(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) → reverse_out_ga(.(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_in_ga(.(X, Xs), Ys) → U2_ga(X, Xs, Ys, reverse_in_ga(Xs, Zs))
reverse_in_ga([], []) → reverse_out_ga([], [])
U2_ga(X, Xs, Ys, reverse_out_ga(Xs, Zs)) → U3_ga(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
app_in_gga(.(X, Xs), Ys, .(X, Zs)) → U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
app_in_gga([], Ys, Ys) → app_out_gga([], Ys, Ys)
U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) → app_out_gga(.(X, Xs), Ys, .(X, Zs))
U3_ga(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) → reverse_out_ga(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
reverse_in_ga(x1, x2) = reverse_in_ga(x1)
.(x1, x2) = .(x1, x2)
U2_ga(x1, x2, x3, x4) = U2_ga(x1, x4)
[] = []
reverse_out_ga(x1, x2) = reverse_out_ga(x2)
U3_ga(x1, x2, x3, x4) = U3_ga(x4)
app_in_gga(x1, x2, x3) = app_in_gga(x1, x2)
U1_gga(x1, x2, x3, x4, x5) = U1_gga(x1, x5)
app_out_gga(x1, x2, x3) = app_out_gga(x3)

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_GA(.(X, Xs), Ys) → U2_GA(X, Xs, Ys, reverse_in_ga(Xs, Zs))
REVERSE_IN_GA(.(X, Xs), Ys) → REVERSE_IN_GA(Xs, Zs)
U2_GA(X, Xs, Ys, reverse_out_ga(Xs, Zs)) → U3_GA(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
U2_GA(X, Xs, Ys, reverse_out_ga(Xs, Zs)) → APP_IN_GGA(Zs, .(X, []), Ys)
APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) → U1_GGA(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GGA(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_in_ga(.(X, Xs), Ys) → U2_ga(X, Xs, Ys, reverse_in_ga(Xs, Zs))
reverse_in_ga([], []) → reverse_out_ga([], [])
U2_ga(X, Xs, Ys, reverse_out_ga(Xs, Zs)) → U3_ga(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
app_in_gga(.(X, Xs), Ys, .(X, Zs)) → U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
app_in_gga([], Ys, Ys) → app_out_gga([], Ys, Ys)
U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) → app_out_gga(.(X, Xs), Ys, .(X, Zs))
U3_ga(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) → reverse_out_ga(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
reverse_in_ga(x1, x2) = reverse_in_ga(x1)
.(x1, x2) = .(x1, x2)
U2_ga(x1, x2, x3, x4) = U2_ga(x1, x4)
[] = []
reverse_out_ga(x1, x2) = reverse_out_ga(x2)
U3_ga(x1, x2, x3, x4) = U3_ga(x4)
app_in_gga(x1, x2, x3) = app_in_gga(x1, x2)
U1_gga(x1, x2, x3, x4, x5) = U1_gga(x1, x5)
app_out_gga(x1, x2, x3) = app_out_gga(x3)
U1_GGA(x1, x2, x3, x4, x5) = U1_GGA(x1, x5)
APP_IN_GGA(x1, x2, x3) = APP_IN_GGA(x1, x2)
REVERSE_IN_GA(x1, x2) = REVERSE_IN_GA(x1)
U3_GA(x1, x2, x3, x4) = U3_GA(x4)
U2_GA(x1, x2, x3, x4) = U2_GA(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_IN_GA(.(X, Xs), Ys) → U2_GA(X, Xs, Ys, reverse_in_ga(Xs, Zs))
REVERSE_IN_GA(.(X, Xs), Ys) → REVERSE_IN_GA(Xs, Zs)
U2_GA(X, Xs, Ys, reverse_out_ga(Xs, Zs)) → U3_GA(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
U2_GA(X, Xs, Ys, reverse_out_ga(Xs, Zs)) → APP_IN_GGA(Zs, .(X, []), Ys)
APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) → U1_GGA(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GGA(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_in_ga(.(X, Xs), Ys) → U2_ga(X, Xs, Ys, reverse_in_ga(Xs, Zs))
reverse_in_ga([], []) → reverse_out_ga([], [])
U2_ga(X, Xs, Ys, reverse_out_ga(Xs, Zs)) → U3_ga(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
app_in_gga(.(X, Xs), Ys, .(X, Zs)) → U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
app_in_gga([], Ys, Ys) → app_out_gga([], Ys, Ys)
U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) → app_out_gga(.(X, Xs), Ys, .(X, Zs))
U3_ga(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) → reverse_out_ga(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
reverse_in_ga(x1, x2) = reverse_in_ga(x1)
.(x1, x2) = .(x1, x2)
U2_ga(x1, x2, x3, x4) = U2_ga(x1, x4)
[] = []
reverse_out_ga(x1, x2) = reverse_out_ga(x2)
U3_ga(x1, x2, x3, x4) = U3_ga(x4)
app_in_gga(x1, x2, x3) = app_in_gga(x1, x2)
U1_gga(x1, x2, x3, x4, x5) = U1_gga(x1, x5)
app_out_gga(x1, x2, x3) = app_out_gga(x3)
U1_GGA(x1, x2, x3, x4, x5) = U1_GGA(x1, x5)
APP_IN_GGA(x1, x2, x3) = APP_IN_GGA(x1, x2)
REVERSE_IN_GA(x1, x2) = REVERSE_IN_GA(x1)
U3_GA(x1, x2, x3, x4) = U3_GA(x4)
U2_GA(x1, x2, x3, x4) = U2_GA(x1, x4)

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

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

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

The TRS R consists of the following rules:

reverse_in_ga(.(X, Xs), Ys) → U2_ga(X, Xs, Ys, reverse_in_ga(Xs, Zs))
reverse_in_ga([], []) → reverse_out_ga([], [])
U2_ga(X, Xs, Ys, reverse_out_ga(Xs, Zs)) → U3_ga(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
app_in_gga(.(X, Xs), Ys, .(X, Zs)) → U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
app_in_gga([], Ys, Ys) → app_out_gga([], Ys, Ys)
U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) → app_out_gga(.(X, Xs), Ys, .(X, Zs))
U3_ga(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) → reverse_out_ga(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
reverse_in_ga(x1, x2) = reverse_in_ga(x1)
.(x1, x2) = .(x1, x2)
U2_ga(x1, x2, x3, x4) = U2_ga(x1, x4)
[] = []
reverse_out_ga(x1, x2) = reverse_out_ga(x2)
U3_ga(x1, x2, x3, x4) = U3_ga(x4)
app_in_gga(x1, x2, x3) = app_in_gga(x1, x2)
U1_gga(x1, x2, x3, x4, x5) = U1_gga(x1, x5)
app_out_gga(x1, x2, x3) = app_out_gga(x3)
APP_IN_GGA(x1, x2, x3) = APP_IN_GGA(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

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

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

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

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

APP_IN_GGA(.(X, Xs), Ys) → APP_IN_GGA(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_GGA(.(X, Xs), Ys) → APP_IN_GGA(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_IN_GA(.(X, Xs), Ys) → REVERSE_IN_GA(Xs, Zs)

The TRS R consists of the following rules:

reverse_in_ga(.(X, Xs), Ys) → U2_ga(X, Xs, Ys, reverse_in_ga(Xs, Zs))
reverse_in_ga([], []) → reverse_out_ga([], [])
U2_ga(X, Xs, Ys, reverse_out_ga(Xs, Zs)) → U3_ga(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
app_in_gga(.(X, Xs), Ys, .(X, Zs)) → U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
app_in_gga([], Ys, Ys) → app_out_gga([], Ys, Ys)
U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) → app_out_gga(.(X, Xs), Ys, .(X, Zs))
U3_ga(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) → reverse_out_ga(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
reverse_in_ga(x1, x2) = reverse_in_ga(x1)
.(x1, x2) = .(x1, x2)
U2_ga(x1, x2, x3, x4) = U2_ga(x1, x4)
[] = []
reverse_out_ga(x1, x2) = reverse_out_ga(x2)
U3_ga(x1, x2, x3, x4) = U3_ga(x4)
app_in_gga(x1, x2, x3) = app_in_gga(x1, x2)
U1_gga(x1, x2, x3, x4, x5) = U1_gga(x1, x5)
app_out_gga(x1, x2, x3) = app_out_gga(x3)
REVERSE_IN_GA(x1, x2) = REVERSE_IN_GA(x1)

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

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

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

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

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

                        ↳ QDPSizeChangeProof

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

REVERSE_IN_GA(.(X, Xs)) → REVERSE_IN_GA(Xs)

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

REVERSE_IN_GA(.(X, Xs)) → REVERSE_IN_GA(Xs)
The graph contains the following edges 1 > 1