Left Termination proof of /tmp/tmpu4qz-P/inorder-bf.pl

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

↳ PROLOG

  ↳ PrologToPiTRSProof

in_order₂(void₀, {}₀).
in_order₂(tree₃(X, Left, Right), Xs) :- in_order₂(Left, Ls), in_order₂(Right, Rs), app₃(Ls, .₂(X, Rs), Xs).
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:
in_order₂: (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:

in_order_2_in_ga₂(void_0, []_0) -> in_order_2_out_ga₂(void_0, []_0)
in_order_2_in_ga₂(tree_3₃(X, Left, Right), Xs) -> if_in_order_2_in_1_ga₅(X, Left, Right, Xs, in_order_2_in_ga₂(Left, Ls))
if_in_order_2_in_1_ga₅(X, Left, Right, Xs, in_order_2_out_ga₂(Left, Ls)) -> if_in_order_2_in_2_ga₆(X, Left, Right, Xs, Ls, in_order_2_in_ga₂(Right, Rs))
if_in_order_2_in_2_ga₆(X, Left, Right, Xs, Ls, in_order_2_out_ga₂(Right, Rs)) -> if_in_order_2_in_3_ga₇(X, Left, Right, Xs, Ls, Rs, app_3_in_gga₃(Ls, ._2₂(X, Rs), Xs))
app_3_in_gga₃([]_0, X, X) -> app_3_out_gga₃([]_0, X, X)
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))
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_in_order_2_in_3_ga₇(X, Left, Right, Xs, Ls, Rs, app_3_out_gga₃(Ls, ._2₂(X, Rs), Xs)) -> in_order_2_out_ga₂(tree_3₃(X, Left, Right), Xs)

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:

in_order_2_in_ga₂(void_0, []_0) -> in_order_2_out_ga₂(void_0, []_0)
in_order_2_in_ga₂(tree_3₃(X, Left, Right), Xs) -> if_in_order_2_in_1_ga₅(X, Left, Right, Xs, in_order_2_in_ga₂(Left, Ls))
if_in_order_2_in_1_ga₅(X, Left, Right, Xs, in_order_2_out_ga₂(Left, Ls)) -> if_in_order_2_in_2_ga₆(X, Left, Right, Xs, Ls, in_order_2_in_ga₂(Right, Rs))
if_in_order_2_in_2_ga₆(X, Left, Right, Xs, Ls, in_order_2_out_ga₂(Right, Rs)) -> if_in_order_2_in_3_ga₇(X, Left, Right, Xs, Ls, Rs, app_3_in_gga₃(Ls, ._2₂(X, Rs), Xs))
app_3_in_gga₃([]_0, X, X) -> app_3_out_gga₃([]_0, X, X)
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))
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_in_order_2_in_3_ga₇(X, Left, Right, Xs, Ls, Rs, app_3_out_gga₃(Ls, ._2₂(X, Rs), Xs)) -> in_order_2_out_ga₂(tree_3₃(X, Left, Right), Xs)

IN_ORDER_2_IN_GA₂(tree_3₃(X, Left, Right), Xs) -> IF_IN_ORDER_2_IN_1_GA₅(X, Left, Right, Xs, in_order_2_in_ga₂(Left, Ls))
IN_ORDER_2_IN_GA₂(tree_3₃(X, Left, Right), Xs) -> IN_ORDER_2_IN_GA₂(Left, Ls)
IF_IN_ORDER_2_IN_1_GA₅(X, Left, Right, Xs, in_order_2_out_ga₂(Left, Ls)) -> IF_IN_ORDER_2_IN_2_GA₆(X, Left, Right, Xs, Ls, in_order_2_in_ga₂(Right, Rs))
IF_IN_ORDER_2_IN_1_GA₅(X, Left, Right, Xs, in_order_2_out_ga₂(Left, Ls)) -> IN_ORDER_2_IN_GA₂(Right, Rs)
IF_IN_ORDER_2_IN_2_GA₆(X, Left, Right, Xs, Ls, in_order_2_out_ga₂(Right, Rs)) -> IF_IN_ORDER_2_IN_3_GA₇(X, Left, Right, Xs, Ls, Rs, app_3_in_gga₃(Ls, ._2₂(X, Rs), Xs))
IF_IN_ORDER_2_IN_2_GA₆(X, Left, Right, Xs, Ls, in_order_2_out_ga₂(Right, Rs)) -> APP_3_IN_GGA₃(Ls, ._2₂(X, Rs), Xs)
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:

in_order_2_in_ga₂(void_0, []_0) -> in_order_2_out_ga₂(void_0, []_0)
in_order_2_in_ga₂(tree_3₃(X, Left, Right), Xs) -> if_in_order_2_in_1_ga₅(X, Left, Right, Xs, in_order_2_in_ga₂(Left, Ls))
if_in_order_2_in_1_ga₅(X, Left, Right, Xs, in_order_2_out_ga₂(Left, Ls)) -> if_in_order_2_in_2_ga₆(X, Left, Right, Xs, Ls, in_order_2_in_ga₂(Right, Rs))
if_in_order_2_in_2_ga₆(X, Left, Right, Xs, Ls, in_order_2_out_ga₂(Right, Rs)) -> if_in_order_2_in_3_ga₇(X, Left, Right, Xs, Ls, Rs, app_3_in_gga₃(Ls, ._2₂(X, Rs), Xs))
app_3_in_gga₃([]_0, X, X) -> app_3_out_gga₃([]_0, X, X)
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))
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_in_order_2_in_3_ga₇(X, Left, Right, Xs, Ls, Rs, app_3_out_gga₃(Ls, ._2₂(X, Rs), Xs)) -> in_order_2_out_ga₂(tree_3₃(X, Left, Right), Xs)

The argument filtering Pi contains the following mapping:
in_order_2_in_ga₂(x1, x2) = in_order_2_in_ga₁(x1)
void_0 = void_0
[]_0 = []_0
tree_3₃(x1, x2, x3) = tree_3₃(x1, x2, x3)
._2₂(x1, x2) = ._2₂(x1, x2)
in_order_2_out_ga₂(x1, x2) = in_order_2_out_ga₁(x2)
if_in_order_2_in_1_ga₅(x1, x2, x3, x4, x5) = if_in_order_2_in_1_ga₃(x1, x3, x5)
if_in_order_2_in_2_ga₆(x1, x2, x3, x4, x5, x6) = if_in_order_2_in_2_ga₃(x1, x5, x6)
if_in_order_2_in_3_ga₇(x1, x2, x3, x4, x5, x6, x7) = if_in_order_2_in_3_ga₁(x7)
app_3_in_gga₃(x1, x2, x3) = app_3_in_gga₂(x1, x2)
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)
IF_IN_ORDER_2_IN_3_GA₇(x1, x2, x3, x4, x5, x6, x7) = IF_IN_ORDER_2_IN_3_GA₁(x7)
IF_IN_ORDER_2_IN_1_GA₅(x1, x2, x3, x4, x5) = IF_IN_ORDER_2_IN_1_GA₃(x1, x3, x5)
IF_APP_3_IN_1_GGA₅(x1, x2, x3, x4, x5) = IF_APP_3_IN_1_GGA₂(x1, x5)
IN_ORDER_2_IN_GA₂(x1, x2) = IN_ORDER_2_IN_GA₁(x1)
APP_3_IN_GGA₃(x1, x2, x3) = APP_3_IN_GGA₂(x1, x2)
IF_IN_ORDER_2_IN_2_GA₆(x1, x2, x3, x4, x5, x6) = IF_IN_ORDER_2_IN_2_GA₃(x1, x5, x6)

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:

IN_ORDER_2_IN_GA₂(tree_3₃(X, Left, Right), Xs) -> IF_IN_ORDER_2_IN_1_GA₅(X, Left, Right, Xs, in_order_2_in_ga₂(Left, Ls))
IN_ORDER_2_IN_GA₂(tree_3₃(X, Left, Right), Xs) -> IN_ORDER_2_IN_GA₂(Left, Ls)
IF_IN_ORDER_2_IN_1_GA₅(X, Left, Right, Xs, in_order_2_out_ga₂(Left, Ls)) -> IF_IN_ORDER_2_IN_2_GA₆(X, Left, Right, Xs, Ls, in_order_2_in_ga₂(Right, Rs))
IF_IN_ORDER_2_IN_1_GA₅(X, Left, Right, Xs, in_order_2_out_ga₂(Left, Ls)) -> IN_ORDER_2_IN_GA₂(Right, Rs)
IF_IN_ORDER_2_IN_2_GA₆(X, Left, Right, Xs, Ls, in_order_2_out_ga₂(Right, Rs)) -> IF_IN_ORDER_2_IN_3_GA₇(X, Left, Right, Xs, Ls, Rs, app_3_in_gga₃(Ls, ._2₂(X, Rs), Xs))
IF_IN_ORDER_2_IN_2_GA₆(X, Left, Right, Xs, Ls, in_order_2_out_ga₂(Right, Rs)) -> APP_3_IN_GGA₃(Ls, ._2₂(X, Rs), Xs)
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:

in_order_2_in_ga₂(void_0, []_0) -> in_order_2_out_ga₂(void_0, []_0)
in_order_2_in_ga₂(tree_3₃(X, Left, Right), Xs) -> if_in_order_2_in_1_ga₅(X, Left, Right, Xs, in_order_2_in_ga₂(Left, Ls))
if_in_order_2_in_1_ga₅(X, Left, Right, Xs, in_order_2_out_ga₂(Left, Ls)) -> if_in_order_2_in_2_ga₆(X, Left, Right, Xs, Ls, in_order_2_in_ga₂(Right, Rs))
if_in_order_2_in_2_ga₆(X, Left, Right, Xs, Ls, in_order_2_out_ga₂(Right, Rs)) -> if_in_order_2_in_3_ga₇(X, Left, Right, Xs, Ls, Rs, app_3_in_gga₃(Ls, ._2₂(X, Rs), Xs))
app_3_in_gga₃([]_0, X, X) -> app_3_out_gga₃([]_0, X, X)
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))
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_in_order_2_in_3_ga₇(X, Left, Right, Xs, Ls, Rs, app_3_out_gga₃(Ls, ._2₂(X, Rs), Xs)) -> in_order_2_out_ga₂(tree_3₃(X, Left, Right), Xs)

↳ 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:

in_order_2_in_ga₂(void_0, []_0) -> in_order_2_out_ga₂(void_0, []_0)
in_order_2_in_ga₂(tree_3₃(X, Left, Right), Xs) -> if_in_order_2_in_1_ga₅(X, Left, Right, Xs, in_order_2_in_ga₂(Left, Ls))
if_in_order_2_in_1_ga₅(X, Left, Right, Xs, in_order_2_out_ga₂(Left, Ls)) -> if_in_order_2_in_2_ga₆(X, Left, Right, Xs, Ls, in_order_2_in_ga₂(Right, Rs))
if_in_order_2_in_2_ga₆(X, Left, Right, Xs, Ls, in_order_2_out_ga₂(Right, Rs)) -> if_in_order_2_in_3_ga₇(X, Left, Right, Xs, Ls, Rs, app_3_in_gga₃(Ls, ._2₂(X, Rs), Xs))
app_3_in_gga₃([]_0, X, X) -> app_3_out_gga₃([]_0, X, X)
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))
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_in_order_2_in_3_ga₇(X, Left, Right, Xs, Ls, Rs, app_3_out_gga₃(Ls, ._2₂(X, Rs), Xs)) -> in_order_2_out_ga₂(tree_3₃(X, Left, Right), Xs)

The argument filtering Pi contains the following mapping:
in_order_2_in_ga₂(x1, x2) = in_order_2_in_ga₁(x1)
void_0 = void_0
[]_0 = []_0
tree_3₃(x1, x2, x3) = tree_3₃(x1, x2, x3)
._2₂(x1, x2) = ._2₂(x1, x2)
in_order_2_out_ga₂(x1, x2) = in_order_2_out_ga₁(x2)
if_in_order_2_in_1_ga₅(x1, x2, x3, x4, x5) = if_in_order_2_in_1_ga₃(x1, x3, x5)
if_in_order_2_in_2_ga₆(x1, x2, x3, x4, x5, x6) = if_in_order_2_in_2_ga₃(x1, x5, x6)
if_in_order_2_in_3_ga₇(x1, x2, x3, x4, x5, x6, x7) = if_in_order_2_in_3_ga₁(x7)
app_3_in_gga₃(x1, x2, x3) = app_3_in_gga₂(x1, x2)
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)

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

                ↳ PiDPToQDPProof

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

IN_ORDER_2_IN_GA₂(tree_3₃(X, Left, Right), Xs) -> IN_ORDER_2_IN_GA₂(Left, Ls)
IN_ORDER_2_IN_GA₂(tree_3₃(X, Left, Right), Xs) -> IF_IN_ORDER_2_IN_1_GA₅(X, Left, Right, Xs, in_order_2_in_ga₂(Left, Ls))
IF_IN_ORDER_2_IN_1_GA₅(X, Left, Right, Xs, in_order_2_out_ga₂(Left, Ls)) -> IN_ORDER_2_IN_GA₂(Right, Rs)

The TRS R consists of the following rules:

in_order_2_in_ga₂(void_0, []_0) -> in_order_2_out_ga₂(void_0, []_0)
in_order_2_in_ga₂(tree_3₃(X, Left, Right), Xs) -> if_in_order_2_in_1_ga₅(X, Left, Right, Xs, in_order_2_in_ga₂(Left, Ls))
if_in_order_2_in_1_ga₅(X, Left, Right, Xs, in_order_2_out_ga₂(Left, Ls)) -> if_in_order_2_in_2_ga₆(X, Left, Right, Xs, Ls, in_order_2_in_ga₂(Right, Rs))
if_in_order_2_in_2_ga₆(X, Left, Right, Xs, Ls, in_order_2_out_ga₂(Right, Rs)) -> if_in_order_2_in_3_ga₇(X, Left, Right, Xs, Ls, Rs, app_3_in_gga₃(Ls, ._2₂(X, Rs), Xs))
app_3_in_gga₃([]_0, X, X) -> app_3_out_gga₃([]_0, X, X)
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))
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_in_order_2_in_3_ga₇(X, Left, Right, Xs, Ls, Rs, app_3_out_gga₃(Ls, ._2₂(X, Rs), Xs)) -> in_order_2_out_ga₂(tree_3₃(X, Left, Right), Xs)

The argument filtering Pi contains the following mapping:
in_order_2_in_ga₂(x1, x2) = in_order_2_in_ga₁(x1)
void_0 = void_0
[]_0 = []_0
tree_3₃(x1, x2, x3) = tree_3₃(x1, x2, x3)
._2₂(x1, x2) = ._2₂(x1, x2)
in_order_2_out_ga₂(x1, x2) = in_order_2_out_ga₁(x2)
if_in_order_2_in_1_ga₅(x1, x2, x3, x4, x5) = if_in_order_2_in_1_ga₃(x1, x3, x5)
if_in_order_2_in_2_ga₆(x1, x2, x3, x4, x5, x6) = if_in_order_2_in_2_ga₃(x1, x5, x6)
if_in_order_2_in_3_ga₇(x1, x2, x3, x4, x5, x6, x7) = if_in_order_2_in_3_ga₁(x7)
app_3_in_gga₃(x1, x2, x3) = app_3_in_gga₂(x1, x2)
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)
IF_IN_ORDER_2_IN_1_GA₅(x1, x2, x3, x4, x5) = IF_IN_ORDER_2_IN_1_GA₃(x1, x3, x5)
IN_ORDER_2_IN_GA₂(x1, x2) = IN_ORDER_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

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

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

IN_ORDER_2_IN_GA₁(tree_3₃(X, Left, Right)) -> IN_ORDER_2_IN_GA₁(Left)
IN_ORDER_2_IN_GA₁(tree_3₃(X, Left, Right)) -> IF_IN_ORDER_2_IN_1_GA₃(X, Right, in_order_2_in_ga₁(Left))
IF_IN_ORDER_2_IN_1_GA₃(X, Right, in_order_2_out_ga₁(Ls)) -> IN_ORDER_2_IN_GA₁(Right)

The TRS R consists of the following rules:

in_order_2_in_ga₁(void_0) -> in_order_2_out_ga₁([]_0)
in_order_2_in_ga₁(tree_3₃(X, Left, Right)) -> if_in_order_2_in_1_ga₃(X, Right, in_order_2_in_ga₁(Left))
if_in_order_2_in_1_ga₃(X, Right, in_order_2_out_ga₁(Ls)) -> if_in_order_2_in_2_ga₃(X, Ls, in_order_2_in_ga₁(Right))
if_in_order_2_in_2_ga₃(X, Ls, in_order_2_out_ga₁(Rs)) -> if_in_order_2_in_3_ga₁(app_3_in_gga₂(Ls, ._2₂(X, Rs)))
app_3_in_gga₂([]_0, X) -> app_3_out_gga₁(X)
app_3_in_gga₂(._2₂(X, Xs), Ys) -> if_app_3_in_1_gga₂(X, app_3_in_gga₂(Xs, Ys))
if_app_3_in_1_gga₂(X, app_3_out_gga₁(Zs)) -> app_3_out_gga₁(._2₂(X, Zs))
if_in_order_2_in_3_ga₁(app_3_out_gga₁(Xs)) -> in_order_2_out_ga₁(Xs)

The set Q consists of the following terms:

in_order_2_in_ga₁(x0)
if_in_order_2_in_1_ga₃(x0, x1, x2)
if_in_order_2_in_2_ga₃(x0, x1, x2)
app_3_in_gga₂(x0, x1)
if_app_3_in_1_gga₂(x0, x1)
if_in_order_2_in_3_ga₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IN_ORDER_2_IN_GA₁, IF_IN_ORDER_2_IN_1_GA₃}.
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:

IN_ORDER_2_IN_GA₁(tree_3₃(X, Left, Right)) -> IN_ORDER_2_IN_GA₁(Left)
The graph contains the following edges 1 > 1
IN_ORDER_2_IN_GA₁(tree_3₃(X, Left, Right)) -> IF_IN_ORDER_2_IN_1_GA₃(X, Right, in_order_2_in_ga₁(Left))
The graph contains the following edges 1 > 1, 1 > 2
IF_IN_ORDER_2_IN_1_GA₃(X, Right, in_order_2_out_ga₁(Ls)) -> IN_ORDER_2_IN_GA₁(Right)
The graph contains the following edges 2 >= 1