Left Termination proof of /tmp/tmpedCIzg/lessleaves.pl

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

↳ PROLOG

  ↳ PrologToPiTRSProof

append₃(nil₀, Y, Y).
append₃(cons₂(U, V), Y, cons₂(U, Z)) :- append₃(V, Y, Z).
lessleaves₂(nil₀, cons₂(W, Z)).
lessleaves₂(cons₂(U, V), cons₂(W, Z)) :- append₃(U, V, U1), append₃(W, Z, W1), lessleaves₂(U1, W1).

With regard to the inferred argument filtering the predicates were used in the following modes:
lessleaves₂: (b,b)
append₃: (b,b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

lessleaves_2_in_gg₂(nil_0, cons_2₂(W, Z)) -> lessleaves_2_out_gg₂(nil_0, cons_2₂(W, Z))
lessleaves_2_in_gg₂(cons_2₂(U, V), cons_2₂(W, Z)) -> if_lessleaves_2_in_1_gg₅(U, V, W, Z, append_3_in_gga₃(U, V, U1))
append_3_in_gga₃(nil_0, Y, Y) -> append_3_out_gga₃(nil_0, Y, Y)
append_3_in_gga₃(cons_2₂(U, V), Y, cons_2₂(U, Z)) -> if_append_3_in_1_gga₅(U, V, Y, Z, append_3_in_gga₃(V, Y, Z))
if_append_3_in_1_gga₅(U, V, Y, Z, append_3_out_gga₃(V, Y, Z)) -> append_3_out_gga₃(cons_2₂(U, V), Y, cons_2₂(U, Z))
if_lessleaves_2_in_1_gg₅(U, V, W, Z, append_3_out_gga₃(U, V, U1)) -> if_lessleaves_2_in_2_gg₆(U, V, W, Z, U1, append_3_in_gga₃(W, Z, W1))
if_lessleaves_2_in_2_gg₆(U, V, W, Z, U1, append_3_out_gga₃(W, Z, W1)) -> if_lessleaves_2_in_3_gg₇(U, V, W, Z, U1, W1, lessleaves_2_in_gg₂(U1, W1))
if_lessleaves_2_in_3_gg₇(U, V, W, Z, U1, W1, lessleaves_2_out_gg₂(U1, W1)) -> lessleaves_2_out_gg₂(cons_2₂(U, V), cons_2₂(W, Z))

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:

lessleaves_2_in_gg₂(nil_0, cons_2₂(W, Z)) -> lessleaves_2_out_gg₂(nil_0, cons_2₂(W, Z))
lessleaves_2_in_gg₂(cons_2₂(U, V), cons_2₂(W, Z)) -> if_lessleaves_2_in_1_gg₅(U, V, W, Z, append_3_in_gga₃(U, V, U1))
append_3_in_gga₃(nil_0, Y, Y) -> append_3_out_gga₃(nil_0, Y, Y)
append_3_in_gga₃(cons_2₂(U, V), Y, cons_2₂(U, Z)) -> if_append_3_in_1_gga₅(U, V, Y, Z, append_3_in_gga₃(V, Y, Z))
if_append_3_in_1_gga₅(U, V, Y, Z, append_3_out_gga₃(V, Y, Z)) -> append_3_out_gga₃(cons_2₂(U, V), Y, cons_2₂(U, Z))
if_lessleaves_2_in_1_gg₅(U, V, W, Z, append_3_out_gga₃(U, V, U1)) -> if_lessleaves_2_in_2_gg₆(U, V, W, Z, U1, append_3_in_gga₃(W, Z, W1))
if_lessleaves_2_in_2_gg₆(U, V, W, Z, U1, append_3_out_gga₃(W, Z, W1)) -> if_lessleaves_2_in_3_gg₇(U, V, W, Z, U1, W1, lessleaves_2_in_gg₂(U1, W1))
if_lessleaves_2_in_3_gg₇(U, V, W, Z, U1, W1, lessleaves_2_out_gg₂(U1, W1)) -> lessleaves_2_out_gg₂(cons_2₂(U, V), cons_2₂(W, Z))

LESSLEAVES_2_IN_GG₂(cons_2₂(U, V), cons_2₂(W, Z)) -> IF_LESSLEAVES_2_IN_1_GG₅(U, V, W, Z, append_3_in_gga₃(U, V, U1))
LESSLEAVES_2_IN_GG₂(cons_2₂(U, V), cons_2₂(W, Z)) -> APPEND_3_IN_GGA₃(U, V, U1)
APPEND_3_IN_GGA₃(cons_2₂(U, V), Y, cons_2₂(U, Z)) -> IF_APPEND_3_IN_1_GGA₅(U, V, Y, Z, append_3_in_gga₃(V, Y, Z))
APPEND_3_IN_GGA₃(cons_2₂(U, V), Y, cons_2₂(U, Z)) -> APPEND_3_IN_GGA₃(V, Y, Z)
IF_LESSLEAVES_2_IN_1_GG₅(U, V, W, Z, append_3_out_gga₃(U, V, U1)) -> IF_LESSLEAVES_2_IN_2_GG₆(U, V, W, Z, U1, append_3_in_gga₃(W, Z, W1))
IF_LESSLEAVES_2_IN_1_GG₅(U, V, W, Z, append_3_out_gga₃(U, V, U1)) -> APPEND_3_IN_GGA₃(W, Z, W1)
IF_LESSLEAVES_2_IN_2_GG₆(U, V, W, Z, U1, append_3_out_gga₃(W, Z, W1)) -> IF_LESSLEAVES_2_IN_3_GG₇(U, V, W, Z, U1, W1, lessleaves_2_in_gg₂(U1, W1))
IF_LESSLEAVES_2_IN_2_GG₆(U, V, W, Z, U1, append_3_out_gga₃(W, Z, W1)) -> LESSLEAVES_2_IN_GG₂(U1, W1)

The TRS R consists of the following rules:

lessleaves_2_in_gg₂(nil_0, cons_2₂(W, Z)) -> lessleaves_2_out_gg₂(nil_0, cons_2₂(W, Z))
lessleaves_2_in_gg₂(cons_2₂(U, V), cons_2₂(W, Z)) -> if_lessleaves_2_in_1_gg₅(U, V, W, Z, append_3_in_gga₃(U, V, U1))
append_3_in_gga₃(nil_0, Y, Y) -> append_3_out_gga₃(nil_0, Y, Y)
append_3_in_gga₃(cons_2₂(U, V), Y, cons_2₂(U, Z)) -> if_append_3_in_1_gga₅(U, V, Y, Z, append_3_in_gga₃(V, Y, Z))
if_append_3_in_1_gga₅(U, V, Y, Z, append_3_out_gga₃(V, Y, Z)) -> append_3_out_gga₃(cons_2₂(U, V), Y, cons_2₂(U, Z))
if_lessleaves_2_in_1_gg₅(U, V, W, Z, append_3_out_gga₃(U, V, U1)) -> if_lessleaves_2_in_2_gg₆(U, V, W, Z, U1, append_3_in_gga₃(W, Z, W1))
if_lessleaves_2_in_2_gg₆(U, V, W, Z, U1, append_3_out_gga₃(W, Z, W1)) -> if_lessleaves_2_in_3_gg₇(U, V, W, Z, U1, W1, lessleaves_2_in_gg₂(U1, W1))
if_lessleaves_2_in_3_gg₇(U, V, W, Z, U1, W1, lessleaves_2_out_gg₂(U1, W1)) -> lessleaves_2_out_gg₂(cons_2₂(U, V), cons_2₂(W, Z))

The argument filtering Pi contains the following mapping:
lessleaves_2_in_gg₂(x1, x2) = lessleaves_2_in_gg₂(x1, x2)
nil_0 = nil_0
cons_2₂(x1, x2) = cons_2₂(x1, x2)
lessleaves_2_out_gg₂(x1, x2) = lessleaves_2_out_gg
if_lessleaves_2_in_1_gg₅(x1, x2, x3, x4, x5) = if_lessleaves_2_in_1_gg₃(x3, x4, x5)
append_3_in_gga₃(x1, x2, x3) = append_3_in_gga₂(x1, x2)
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(x1, x5)
if_lessleaves_2_in_2_gg₆(x1, x2, x3, x4, x5, x6) = if_lessleaves_2_in_2_gg₂(x5, x6)
if_lessleaves_2_in_3_gg₇(x1, x2, x3, x4, x5, x6, x7) = if_lessleaves_2_in_3_gg₁(x7)
IF_APPEND_3_IN_1_GGA₅(x1, x2, x3, x4, x5) = IF_APPEND_3_IN_1_GGA₂(x1, x5)
IF_LESSLEAVES_2_IN_3_GG₇(x1, x2, x3, x4, x5, x6, x7) = IF_LESSLEAVES_2_IN_3_GG₁(x7)
IF_LESSLEAVES_2_IN_2_GG₆(x1, x2, x3, x4, x5, x6) = IF_LESSLEAVES_2_IN_2_GG₂(x5, x6)
IF_LESSLEAVES_2_IN_1_GG₅(x1, x2, x3, x4, x5) = IF_LESSLEAVES_2_IN_1_GG₃(x3, x4, x5)
LESSLEAVES_2_IN_GG₂(x1, x2) = LESSLEAVES_2_IN_GG₂(x1, x2)
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:

LESSLEAVES_2_IN_GG₂(cons_2₂(U, V), cons_2₂(W, Z)) -> IF_LESSLEAVES_2_IN_1_GG₅(U, V, W, Z, append_3_in_gga₃(U, V, U1))
LESSLEAVES_2_IN_GG₂(cons_2₂(U, V), cons_2₂(W, Z)) -> APPEND_3_IN_GGA₃(U, V, U1)
APPEND_3_IN_GGA₃(cons_2₂(U, V), Y, cons_2₂(U, Z)) -> IF_APPEND_3_IN_1_GGA₅(U, V, Y, Z, append_3_in_gga₃(V, Y, Z))
APPEND_3_IN_GGA₃(cons_2₂(U, V), Y, cons_2₂(U, Z)) -> APPEND_3_IN_GGA₃(V, Y, Z)
IF_LESSLEAVES_2_IN_1_GG₅(U, V, W, Z, append_3_out_gga₃(U, V, U1)) -> IF_LESSLEAVES_2_IN_2_GG₆(U, V, W, Z, U1, append_3_in_gga₃(W, Z, W1))
IF_LESSLEAVES_2_IN_1_GG₅(U, V, W, Z, append_3_out_gga₃(U, V, U1)) -> APPEND_3_IN_GGA₃(W, Z, W1)
IF_LESSLEAVES_2_IN_2_GG₆(U, V, W, Z, U1, append_3_out_gga₃(W, Z, W1)) -> IF_LESSLEAVES_2_IN_3_GG₇(U, V, W, Z, U1, W1, lessleaves_2_in_gg₂(U1, W1))
IF_LESSLEAVES_2_IN_2_GG₆(U, V, W, Z, U1, append_3_out_gga₃(W, Z, W1)) -> LESSLEAVES_2_IN_GG₂(U1, W1)

The TRS R consists of the following rules:

lessleaves_2_in_gg₂(nil_0, cons_2₂(W, Z)) -> lessleaves_2_out_gg₂(nil_0, cons_2₂(W, Z))
lessleaves_2_in_gg₂(cons_2₂(U, V), cons_2₂(W, Z)) -> if_lessleaves_2_in_1_gg₅(U, V, W, Z, append_3_in_gga₃(U, V, U1))
append_3_in_gga₃(nil_0, Y, Y) -> append_3_out_gga₃(nil_0, Y, Y)
append_3_in_gga₃(cons_2₂(U, V), Y, cons_2₂(U, Z)) -> if_append_3_in_1_gga₅(U, V, Y, Z, append_3_in_gga₃(V, Y, Z))
if_append_3_in_1_gga₅(U, V, Y, Z, append_3_out_gga₃(V, Y, Z)) -> append_3_out_gga₃(cons_2₂(U, V), Y, cons_2₂(U, Z))
if_lessleaves_2_in_1_gg₅(U, V, W, Z, append_3_out_gga₃(U, V, U1)) -> if_lessleaves_2_in_2_gg₆(U, V, W, Z, U1, append_3_in_gga₃(W, Z, W1))
if_lessleaves_2_in_2_gg₆(U, V, W, Z, U1, append_3_out_gga₃(W, Z, W1)) -> if_lessleaves_2_in_3_gg₇(U, V, W, Z, U1, W1, lessleaves_2_in_gg₂(U1, W1))
if_lessleaves_2_in_3_gg₇(U, V, W, Z, U1, W1, lessleaves_2_out_gg₂(U1, W1)) -> lessleaves_2_out_gg₂(cons_2₂(U, V), cons_2₂(W, Z))

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

APPEND_3_IN_GGA₃(cons_2₂(U, V), Y, cons_2₂(U, Z)) -> APPEND_3_IN_GGA₃(V, Y, Z)

The TRS R consists of the following rules:

lessleaves_2_in_gg₂(nil_0, cons_2₂(W, Z)) -> lessleaves_2_out_gg₂(nil_0, cons_2₂(W, Z))
lessleaves_2_in_gg₂(cons_2₂(U, V), cons_2₂(W, Z)) -> if_lessleaves_2_in_1_gg₅(U, V, W, Z, append_3_in_gga₃(U, V, U1))
append_3_in_gga₃(nil_0, Y, Y) -> append_3_out_gga₃(nil_0, Y, Y)
append_3_in_gga₃(cons_2₂(U, V), Y, cons_2₂(U, Z)) -> if_append_3_in_1_gga₅(U, V, Y, Z, append_3_in_gga₃(V, Y, Z))
if_append_3_in_1_gga₅(U, V, Y, Z, append_3_out_gga₃(V, Y, Z)) -> append_3_out_gga₃(cons_2₂(U, V), Y, cons_2₂(U, Z))
if_lessleaves_2_in_1_gg₅(U, V, W, Z, append_3_out_gga₃(U, V, U1)) -> if_lessleaves_2_in_2_gg₆(U, V, W, Z, U1, append_3_in_gga₃(W, Z, W1))
if_lessleaves_2_in_2_gg₆(U, V, W, Z, U1, append_3_out_gga₃(W, Z, W1)) -> if_lessleaves_2_in_3_gg₇(U, V, W, Z, U1, W1, lessleaves_2_in_gg₂(U1, W1))
if_lessleaves_2_in_3_gg₇(U, V, W, Z, U1, W1, lessleaves_2_out_gg₂(U1, W1)) -> lessleaves_2_out_gg₂(cons_2₂(U, V), cons_2₂(W, Z))

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

APPEND_3_IN_GGA₃(cons_2₂(U, V), Y, cons_2₂(U, Z)) -> APPEND_3_IN_GGA₃(V, Y, Z)

R is empty.
The argument filtering Pi contains the following mapping:
cons_2₂(x1, x2) = cons_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

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

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

APPEND_3_IN_GGA₂(cons_2₂(U, V), Y) -> APPEND_3_IN_GGA₂(V, 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:

APPEND_3_IN_GGA₂(cons_2₂(U, V), Y) -> APPEND_3_IN_GGA₂(V, Y)
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:

LESSLEAVES_2_IN_GG₂(cons_2₂(U, V), cons_2₂(W, Z)) -> IF_LESSLEAVES_2_IN_1_GG₅(U, V, W, Z, append_3_in_gga₃(U, V, U1))
IF_LESSLEAVES_2_IN_1_GG₅(U, V, W, Z, append_3_out_gga₃(U, V, U1)) -> IF_LESSLEAVES_2_IN_2_GG₆(U, V, W, Z, U1, append_3_in_gga₃(W, Z, W1))
IF_LESSLEAVES_2_IN_2_GG₆(U, V, W, Z, U1, append_3_out_gga₃(W, Z, W1)) -> LESSLEAVES_2_IN_GG₂(U1, W1)

The TRS R consists of the following rules:

lessleaves_2_in_gg₂(nil_0, cons_2₂(W, Z)) -> lessleaves_2_out_gg₂(nil_0, cons_2₂(W, Z))
lessleaves_2_in_gg₂(cons_2₂(U, V), cons_2₂(W, Z)) -> if_lessleaves_2_in_1_gg₅(U, V, W, Z, append_3_in_gga₃(U, V, U1))
append_3_in_gga₃(nil_0, Y, Y) -> append_3_out_gga₃(nil_0, Y, Y)
append_3_in_gga₃(cons_2₂(U, V), Y, cons_2₂(U, Z)) -> if_append_3_in_1_gga₅(U, V, Y, Z, append_3_in_gga₃(V, Y, Z))
if_append_3_in_1_gga₅(U, V, Y, Z, append_3_out_gga₃(V, Y, Z)) -> append_3_out_gga₃(cons_2₂(U, V), Y, cons_2₂(U, Z))
if_lessleaves_2_in_1_gg₅(U, V, W, Z, append_3_out_gga₃(U, V, U1)) -> if_lessleaves_2_in_2_gg₆(U, V, W, Z, U1, append_3_in_gga₃(W, Z, W1))
if_lessleaves_2_in_2_gg₆(U, V, W, Z, U1, append_3_out_gga₃(W, Z, W1)) -> if_lessleaves_2_in_3_gg₇(U, V, W, Z, U1, W1, lessleaves_2_in_gg₂(U1, W1))
if_lessleaves_2_in_3_gg₇(U, V, W, Z, U1, W1, lessleaves_2_out_gg₂(U1, W1)) -> lessleaves_2_out_gg₂(cons_2₂(U, V), cons_2₂(W, Z))

The argument filtering Pi contains the following mapping:
lessleaves_2_in_gg₂(x1, x2) = lessleaves_2_in_gg₂(x1, x2)
nil_0 = nil_0
cons_2₂(x1, x2) = cons_2₂(x1, x2)
lessleaves_2_out_gg₂(x1, x2) = lessleaves_2_out_gg
if_lessleaves_2_in_1_gg₅(x1, x2, x3, x4, x5) = if_lessleaves_2_in_1_gg₃(x3, x4, x5)
append_3_in_gga₃(x1, x2, x3) = append_3_in_gga₂(x1, x2)
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(x1, x5)
if_lessleaves_2_in_2_gg₆(x1, x2, x3, x4, x5, x6) = if_lessleaves_2_in_2_gg₂(x5, x6)
if_lessleaves_2_in_3_gg₇(x1, x2, x3, x4, x5, x6, x7) = if_lessleaves_2_in_3_gg₁(x7)
IF_LESSLEAVES_2_IN_2_GG₆(x1, x2, x3, x4, x5, x6) = IF_LESSLEAVES_2_IN_2_GG₂(x5, x6)
IF_LESSLEAVES_2_IN_1_GG₅(x1, x2, x3, x4, x5) = IF_LESSLEAVES_2_IN_1_GG₃(x3, x4, x5)
LESSLEAVES_2_IN_GG₂(x1, x2) = LESSLEAVES_2_IN_GG₂(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

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

LESSLEAVES_2_IN_GG₂(cons_2₂(U, V), cons_2₂(W, Z)) -> IF_LESSLEAVES_2_IN_1_GG₅(U, V, W, Z, append_3_in_gga₃(U, V, U1))
IF_LESSLEAVES_2_IN_1_GG₅(U, V, W, Z, append_3_out_gga₃(U, V, U1)) -> IF_LESSLEAVES_2_IN_2_GG₆(U, V, W, Z, U1, append_3_in_gga₃(W, Z, W1))
IF_LESSLEAVES_2_IN_2_GG₆(U, V, W, Z, U1, append_3_out_gga₃(W, Z, W1)) -> LESSLEAVES_2_IN_GG₂(U1, W1)

The TRS R consists of the following rules:

append_3_in_gga₃(nil_0, Y, Y) -> append_3_out_gga₃(nil_0, Y, Y)
append_3_in_gga₃(cons_2₂(U, V), Y, cons_2₂(U, Z)) -> if_append_3_in_1_gga₅(U, V, Y, Z, append_3_in_gga₃(V, Y, Z))
if_append_3_in_1_gga₅(U, V, Y, Z, append_3_out_gga₃(V, Y, Z)) -> append_3_out_gga₃(cons_2₂(U, V), Y, cons_2₂(U, Z))

The argument filtering Pi contains the following mapping:
nil_0 = nil_0
cons_2₂(x1, x2) = cons_2₂(x1, x2)
append_3_in_gga₃(x1, x2, x3) = append_3_in_gga₂(x1, x2)
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(x1, x5)
IF_LESSLEAVES_2_IN_2_GG₆(x1, x2, x3, x4, x5, x6) = IF_LESSLEAVES_2_IN_2_GG₂(x5, x6)
IF_LESSLEAVES_2_IN_1_GG₅(x1, x2, x3, x4, x5) = IF_LESSLEAVES_2_IN_1_GG₃(x3, x4, x5)
LESSLEAVES_2_IN_GG₂(x1, x2) = LESSLEAVES_2_IN_GG₂(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

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

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

LESSLEAVES_2_IN_GG₂(cons_2₂(U, V), cons_2₂(W, Z)) -> IF_LESSLEAVES_2_IN_1_GG₃(W, Z, append_3_in_gga₂(U, V))
IF_LESSLEAVES_2_IN_1_GG₃(W, Z, append_3_out_gga₁(U1)) -> IF_LESSLEAVES_2_IN_2_GG₂(U1, append_3_in_gga₂(W, Z))
IF_LESSLEAVES_2_IN_2_GG₂(U1, append_3_out_gga₁(W1)) -> LESSLEAVES_2_IN_GG₂(U1, W1)

The TRS R consists of the following rules:

append_3_in_gga₂(nil_0, Y) -> append_3_out_gga₁(Y)
append_3_in_gga₂(cons_2₂(U, V), Y) -> if_append_3_in_1_gga₂(U, append_3_in_gga₂(V, Y))
if_append_3_in_1_gga₂(U, append_3_out_gga₁(Z)) -> append_3_out_gga₁(cons_2₂(U, Z))

The set Q consists of the following terms:

append_3_in_gga₂(x0, x1)
if_append_3_in_1_gga₂(x0, x1)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_LESSLEAVES_2_IN_1_GG₃, LESSLEAVES_2_IN_GG₂, IF_LESSLEAVES_2_IN_2_GG₂}.
By using a polynomial ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

IF_LESSLEAVES_2_IN_1_GG₃(W, Z, append_3_out_gga₁(U1)) -> IF_LESSLEAVES_2_IN_2_GG₂(U1, append_3_in_gga₂(W, Z))
IF_LESSLEAVES_2_IN_2_GG₂(U1, append_3_out_gga₁(W1)) -> LESSLEAVES_2_IN_GG₂(U1, W1)

Strictly oriented rules of the TRS R:

append_3_in_gga₂(nil_0, Y) -> append_3_out_gga₁(Y)

Used ordering: POLO with Polynomial interpretation:

POL(nil_0) = 2
POL(append_3_out_gga₁(x₁)) = 1 + x₁
POL(LESSLEAVES_2_IN_GG₂(x₁, x₂)) = x₁ + x₂
POL(if_append_3_in_1_gga₂(x₁, x₂)) = x₁ + x₂
POL(IF_LESSLEAVES_2_IN_2_GG₂(x₁, x₂)) = x₁ + x₂
POL(IF_LESSLEAVES_2_IN_1_GG₃(x₁, x₂, x₃)) = x₁ + x₂ + x₃
POL(append_3_in_gga₂(x₁, x₂)) = x₁ + x₂
POL(cons_2₂(x₁, x₂)) = x₁ + x₂

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

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

LESSLEAVES_2_IN_GG₂(cons_2₂(U, V), cons_2₂(W, Z)) -> IF_LESSLEAVES_2_IN_1_GG₃(W, Z, append_3_in_gga₂(U, V))

The TRS R consists of the following rules:

append_3_in_gga₂(cons_2₂(U, V), Y) -> if_append_3_in_1_gga₂(U, append_3_in_gga₂(V, Y))
if_append_3_in_1_gga₂(U, append_3_out_gga₁(Z)) -> append_3_out_gga₁(cons_2₂(U, Z))

The set Q consists of the following terms:

append_3_in_gga₂(x0, x1)
if_append_3_in_1_gga₂(x0, x1)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_LESSLEAVES_2_IN_1_GG₃, LESSLEAVES_2_IN_GG₂}.
By using a polynomial ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

LESSLEAVES_2_IN_GG₂(cons_2₂(U, V), cons_2₂(W, Z)) -> IF_LESSLEAVES_2_IN_1_GG₃(W, Z, append_3_in_gga₂(U, V))

Strictly oriented rules of the TRS R:

append_3_in_gga₂(cons_2₂(U, V), Y) -> if_append_3_in_1_gga₂(U, append_3_in_gga₂(V, Y))

Used ordering: POLO with Polynomial interpretation:

POL(append_3_out_gga₁(x₁)) = x₁
POL(LESSLEAVES_2_IN_GG₂(x₁, x₂)) = x₁ + x₂
POL(if_append_3_in_1_gga₂(x₁, x₂)) = 1 + 2·x₁ + x₂
POL(IF_LESSLEAVES_2_IN_1_GG₃(x₁, x₂, x₃)) = x₁ + x₂ + x₃
POL(append_3_in_gga₂(x₁, x₂)) = 1 + 2·x₁ + x₂
POL(cons_2₂(x₁, x₂)) = 1 + 2·x₁ + x₂

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

                              ↳ QDP

                                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

if_append_3_in_1_gga₂(U, append_3_out_gga₁(Z)) -> append_3_out_gga₁(cons_2₂(U, Z))

The set Q consists of the following terms:

append_3_in_gga₂(x0, x1)
if_append_3_in_1_gga₂(x0, x1)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are empty set.
The TRS P is empty. Hence, there is no (P,Q,R) chain.