Left Termination proof of /tmp/tmpBPdhmi/select-bff.pl

Left Termination of the query pattern select(b,f,f) w.r.t. the given Prolog program could not be shown:

↳ PROLOG

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

select₃(X, .₂(X, Xs), Xs).
select₃(X, .₂(Y, Xs), .₂(Y, Zs)) :- select₃(X, Xs, Zs).

With regard to the inferred argument filtering the predicates were used in the following modes:
select₃: (b,f,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

select_3_in_gaa₃(X, ._2₂(X, Xs), Xs) -> select_3_out_gaa₃(X, ._2₂(X, Xs), Xs)
select_3_in_gaa₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs)) -> if_select_3_in_1_gaa₅(X, Y, Xs, Zs, select_3_in_gaa₃(X, Xs, Zs))
if_select_3_in_1_gaa₅(X, Y, Xs, Zs, select_3_out_gaa₃(X, Xs, Zs)) -> select_3_out_gaa₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs))

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:

select_3_in_gaa₃(X, ._2₂(X, Xs), Xs) -> select_3_out_gaa₃(X, ._2₂(X, Xs), Xs)
select_3_in_gaa₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs)) -> if_select_3_in_1_gaa₅(X, Y, Xs, Zs, select_3_in_gaa₃(X, Xs, Zs))
if_select_3_in_1_gaa₅(X, Y, Xs, Zs, select_3_out_gaa₃(X, Xs, Zs)) -> select_3_out_gaa₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs))

The argument filtering Pi contains the following mapping:
select_3_in_gaa₃(x1, x2, x3) = select_3_in_gaa₁(x1)
._2₂(x1, x2) = ._2
select_3_out_gaa₃(x1, x2, x3) = select_3_out_gaa₁(x2)
if_select_3_in_1_gaa₅(x1, x2, x3, x4, x5) = if_select_3_in_1_gaa₁(x5)

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

SELECT_3_IN_GAA₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs)) -> IF_SELECT_3_IN_1_GAA₅(X, Y, Xs, Zs, select_3_in_gaa₃(X, Xs, Zs))
SELECT_3_IN_GAA₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs)) -> SELECT_3_IN_GAA₃(X, Xs, Zs)

The TRS R consists of the following rules:

select_3_in_gaa₃(X, ._2₂(X, Xs), Xs) -> select_3_out_gaa₃(X, ._2₂(X, Xs), Xs)
select_3_in_gaa₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs)) -> if_select_3_in_1_gaa₅(X, Y, Xs, Zs, select_3_in_gaa₃(X, Xs, Zs))
if_select_3_in_1_gaa₅(X, Y, Xs, Zs, select_3_out_gaa₃(X, Xs, Zs)) -> select_3_out_gaa₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs))

The argument filtering Pi contains the following mapping:
select_3_in_gaa₃(x1, x2, x3) = select_3_in_gaa₁(x1)
._2₂(x1, x2) = ._2
select_3_out_gaa₃(x1, x2, x3) = select_3_out_gaa₁(x2)
if_select_3_in_1_gaa₅(x1, x2, x3, x4, x5) = if_select_3_in_1_gaa₁(x5)
IF_SELECT_3_IN_1_GAA₅(x1, x2, x3, x4, x5) = IF_SELECT_3_IN_1_GAA₁(x5)
SELECT_3_IN_GAA₃(x1, x2, x3) = SELECT_3_IN_GAA₁(x1)

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:

SELECT_3_IN_GAA₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs)) -> IF_SELECT_3_IN_1_GAA₅(X, Y, Xs, Zs, select_3_in_gaa₃(X, Xs, Zs))
SELECT_3_IN_GAA₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs)) -> SELECT_3_IN_GAA₃(X, Xs, Zs)

The TRS R consists of the following rules:

select_3_in_gaa₃(X, ._2₂(X, Xs), Xs) -> select_3_out_gaa₃(X, ._2₂(X, Xs), Xs)
select_3_in_gaa₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs)) -> if_select_3_in_1_gaa₅(X, Y, Xs, Zs, select_3_in_gaa₃(X, Xs, Zs))
if_select_3_in_1_gaa₅(X, Y, Xs, Zs, select_3_out_gaa₃(X, Xs, Zs)) -> select_3_out_gaa₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs))

The argument filtering Pi contains the following mapping:
select_3_in_gaa₃(x1, x2, x3) = select_3_in_gaa₁(x1)
._2₂(x1, x2) = ._2
select_3_out_gaa₃(x1, x2, x3) = select_3_out_gaa₁(x2)
if_select_3_in_1_gaa₅(x1, x2, x3, x4, x5) = if_select_3_in_1_gaa₁(x5)
IF_SELECT_3_IN_1_GAA₅(x1, x2, x3, x4, x5) = IF_SELECT_3_IN_1_GAA₁(x5)
SELECT_3_IN_GAA₃(x1, x2, x3) = SELECT_3_IN_GAA₁(x1)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 1 SCC with 1 less node.

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

  ↳ PrologToPiTRSProof

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

SELECT_3_IN_GAA₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs)) -> SELECT_3_IN_GAA₃(X, Xs, Zs)

The TRS R consists of the following rules:

select_3_in_gaa₃(X, ._2₂(X, Xs), Xs) -> select_3_out_gaa₃(X, ._2₂(X, Xs), Xs)
select_3_in_gaa₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs)) -> if_select_3_in_1_gaa₅(X, Y, Xs, Zs, select_3_in_gaa₃(X, Xs, Zs))
if_select_3_in_1_gaa₅(X, Y, Xs, Zs, select_3_out_gaa₃(X, Xs, Zs)) -> select_3_out_gaa₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs))

The argument filtering Pi contains the following mapping:
select_3_in_gaa₃(x1, x2, x3) = select_3_in_gaa₁(x1)
._2₂(x1, x2) = ._2
select_3_out_gaa₃(x1, x2, x3) = select_3_out_gaa₁(x2)
if_select_3_in_1_gaa₅(x1, x2, x3, x4, x5) = if_select_3_in_1_gaa₁(x5)
SELECT_3_IN_GAA₃(x1, x2, x3) = SELECT_3_IN_GAA₁(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

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

  ↳ PrologToPiTRSProof

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

SELECT_3_IN_GAA₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs)) -> SELECT_3_IN_GAA₃(X, Xs, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2
SELECT_3_IN_GAA₃(x1, x2, x3) = SELECT_3_IN_GAA₁(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

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

                    ↳ QDP

  ↳ PrologToPiTRSProof

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

SELECT_3_IN_GAA₁(X) -> SELECT_3_IN_GAA₁(X)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {SELECT_3_IN_GAA₁}.
With regard to the inferred argument filtering the predicates were used in the following modes:
select₃: (b,f,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

select_3_in_gaa₃(X, ._2₂(X, Xs), Xs) -> select_3_out_gaa₃(X, ._2₂(X, Xs), Xs)
select_3_in_gaa₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs)) -> if_select_3_in_1_gaa₅(X, Y, Xs, Zs, select_3_in_gaa₃(X, Xs, Zs))
if_select_3_in_1_gaa₅(X, Y, Xs, Zs, select_3_out_gaa₃(X, Xs, Zs)) -> select_3_out_gaa₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs))

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:

select_3_in_gaa₃(X, ._2₂(X, Xs), Xs) -> select_3_out_gaa₃(X, ._2₂(X, Xs), Xs)
select_3_in_gaa₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs)) -> if_select_3_in_1_gaa₅(X, Y, Xs, Zs, select_3_in_gaa₃(X, Xs, Zs))
if_select_3_in_1_gaa₅(X, Y, Xs, Zs, select_3_out_gaa₃(X, Xs, Zs)) -> select_3_out_gaa₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs))

SELECT_3_IN_GAA₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs)) -> IF_SELECT_3_IN_1_GAA₅(X, Y, Xs, Zs, select_3_in_gaa₃(X, Xs, Zs))
SELECT_3_IN_GAA₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs)) -> SELECT_3_IN_GAA₃(X, Xs, Zs)

The TRS R consists of the following rules:

select_3_in_gaa₃(X, ._2₂(X, Xs), Xs) -> select_3_out_gaa₃(X, ._2₂(X, Xs), Xs)
select_3_in_gaa₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs)) -> if_select_3_in_1_gaa₅(X, Y, Xs, Zs, select_3_in_gaa₃(X, Xs, Zs))
if_select_3_in_1_gaa₅(X, Y, Xs, Zs, select_3_out_gaa₃(X, Xs, Zs)) -> select_3_out_gaa₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs))

The argument filtering Pi contains the following mapping:
select_3_in_gaa₃(x1, x2, x3) = select_3_in_gaa₁(x1)
._2₂(x1, x2) = ._2
select_3_out_gaa₃(x1, x2, x3) = select_3_out_gaa₂(x1, x2)
if_select_3_in_1_gaa₅(x1, x2, x3, x4, x5) = if_select_3_in_1_gaa₂(x1, x5)
IF_SELECT_3_IN_1_GAA₅(x1, x2, x3, x4, x5) = IF_SELECT_3_IN_1_GAA₂(x1, x5)
SELECT_3_IN_GAA₃(x1, x2, x3) = SELECT_3_IN_GAA₁(x1)

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:

SELECT_3_IN_GAA₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs)) -> IF_SELECT_3_IN_1_GAA₅(X, Y, Xs, Zs, select_3_in_gaa₃(X, Xs, Zs))
SELECT_3_IN_GAA₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs)) -> SELECT_3_IN_GAA₃(X, Xs, Zs)

The TRS R consists of the following rules:

select_3_in_gaa₃(X, ._2₂(X, Xs), Xs) -> select_3_out_gaa₃(X, ._2₂(X, Xs), Xs)
select_3_in_gaa₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs)) -> if_select_3_in_1_gaa₅(X, Y, Xs, Zs, select_3_in_gaa₃(X, Xs, Zs))
if_select_3_in_1_gaa₅(X, Y, Xs, Zs, select_3_out_gaa₃(X, Xs, Zs)) -> select_3_out_gaa₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs))

The argument filtering Pi contains the following mapping:
select_3_in_gaa₃(x1, x2, x3) = select_3_in_gaa₁(x1)
._2₂(x1, x2) = ._2
select_3_out_gaa₃(x1, x2, x3) = select_3_out_gaa₂(x1, x2)
if_select_3_in_1_gaa₅(x1, x2, x3, x4, x5) = if_select_3_in_1_gaa₂(x1, x5)
IF_SELECT_3_IN_1_GAA₅(x1, x2, x3, x4, x5) = IF_SELECT_3_IN_1_GAA₂(x1, x5)
SELECT_3_IN_GAA₃(x1, x2, x3) = SELECT_3_IN_GAA₁(x1)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 1 SCC with 1 less node.

↳ PROLOG

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

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

SELECT_3_IN_GAA₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs)) -> SELECT_3_IN_GAA₃(X, Xs, Zs)

The TRS R consists of the following rules:

select_3_in_gaa₃(X, ._2₂(X, Xs), Xs) -> select_3_out_gaa₃(X, ._2₂(X, Xs), Xs)
select_3_in_gaa₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs)) -> if_select_3_in_1_gaa₅(X, Y, Xs, Zs, select_3_in_gaa₃(X, Xs, Zs))
if_select_3_in_1_gaa₅(X, Y, Xs, Zs, select_3_out_gaa₃(X, Xs, Zs)) -> select_3_out_gaa₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs))

The argument filtering Pi contains the following mapping:
select_3_in_gaa₃(x1, x2, x3) = select_3_in_gaa₁(x1)
._2₂(x1, x2) = ._2
select_3_out_gaa₃(x1, x2, x3) = select_3_out_gaa₂(x1, x2)
if_select_3_in_1_gaa₅(x1, x2, x3, x4, x5) = if_select_3_in_1_gaa₂(x1, x5)
SELECT_3_IN_GAA₃(x1, x2, x3) = SELECT_3_IN_GAA₁(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

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

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

SELECT_3_IN_GAA₃(X, ._2₂(Y, Xs), ._2₂(Y, Zs)) -> SELECT_3_IN_GAA₃(X, Xs, Zs)

↳ PROLOG

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

                    ↳ QDP

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

SELECT_3_IN_GAA₁(X) -> SELECT_3_IN_GAA₁(X)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {SELECT_3_IN_GAA₁}.