Left Termination proof of /tmp/tmp2ksPQd/select-bff.pl

Left Termination of the query pattern select_in_3(g, a, a) w.r.t. the given Prolog program could not be shown:

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

Clauses:

select(X, .(X, Xs), Xs).
select(X, .(Y, Xs), .(Y, Zs)) :- select(X, Xs, Zs).

Queries:

select(g,a,a).

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

select_in_gaa(X, .(X, Xs), Xs) → select_out_gaa(X, .(X, Xs), Xs)
select_in_gaa(X, .(Y, Xs), .(Y, Zs)) → U1_gaa(X, Y, Xs, Zs, select_in_gaa(X, Xs, Zs))
U1_gaa(X, Y, Xs, Zs, select_out_gaa(X, Xs, Zs)) → select_out_gaa(X, .(Y, Xs), .(Y, Zs))

The argument filtering Pi contains the following mapping:
select_in_gaa(x1, x2, x3) = select_in_gaa(x1)
select_out_gaa(x1, x2, x3) = select_out_gaa
U1_gaa(x1, x2, x3, x4, x5) = U1_gaa(x5)

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_in_gaa(X, .(X, Xs), Xs) → select_out_gaa(X, .(X, Xs), Xs)
select_in_gaa(X, .(Y, Xs), .(Y, Zs)) → U1_gaa(X, Y, Xs, Zs, select_in_gaa(X, Xs, Zs))
U1_gaa(X, Y, Xs, Zs, select_out_gaa(X, Xs, Zs)) → select_out_gaa(X, .(Y, Xs), .(Y, Zs))

The argument filtering Pi contains the following mapping:
select_in_gaa(x1, x2, x3) = select_in_gaa(x1)
select_out_gaa(x1, x2, x3) = select_out_gaa
U1_gaa(x1, x2, x3, x4, x5) = U1_gaa(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:

SELECT_IN_GAA(X, .(Y, Xs), .(Y, Zs)) → U1_GAA(X, Y, Xs, Zs, select_in_gaa(X, Xs, Zs))
SELECT_IN_GAA(X, .(Y, Xs), .(Y, Zs)) → SELECT_IN_GAA(X, Xs, Zs)

The TRS R consists of the following rules:

select_in_gaa(X, .(X, Xs), Xs) → select_out_gaa(X, .(X, Xs), Xs)
select_in_gaa(X, .(Y, Xs), .(Y, Zs)) → U1_gaa(X, Y, Xs, Zs, select_in_gaa(X, Xs, Zs))
U1_gaa(X, Y, Xs, Zs, select_out_gaa(X, Xs, Zs)) → select_out_gaa(X, .(Y, Xs), .(Y, Zs))

The argument filtering Pi contains the following mapping:
select_in_gaa(x1, x2, x3) = select_in_gaa(x1)
select_out_gaa(x1, x2, x3) = select_out_gaa
U1_gaa(x1, x2, x3, x4, x5) = U1_gaa(x5)
SELECT_IN_GAA(x1, x2, x3) = SELECT_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5) = U1_GAA(x5)

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_IN_GAA(X, .(Y, Xs), .(Y, Zs)) → U1_GAA(X, Y, Xs, Zs, select_in_gaa(X, Xs, Zs))
SELECT_IN_GAA(X, .(Y, Xs), .(Y, Zs)) → SELECT_IN_GAA(X, Xs, Zs)

The TRS R consists of the following rules:

select_in_gaa(X, .(X, Xs), Xs) → select_out_gaa(X, .(X, Xs), Xs)
select_in_gaa(X, .(Y, Xs), .(Y, Zs)) → U1_gaa(X, Y, Xs, Zs, select_in_gaa(X, Xs, Zs))
U1_gaa(X, Y, Xs, Zs, select_out_gaa(X, Xs, Zs)) → select_out_gaa(X, .(Y, Xs), .(Y, Zs))

The argument filtering Pi contains the following mapping:
select_in_gaa(x1, x2, x3) = select_in_gaa(x1)
select_out_gaa(x1, x2, x3) = select_out_gaa
U1_gaa(x1, x2, x3, x4, x5) = U1_gaa(x5)
SELECT_IN_GAA(x1, x2, x3) = SELECT_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5) = U1_GAA(x5)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] 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_IN_GAA(X, .(Y, Xs), .(Y, Zs)) → SELECT_IN_GAA(X, Xs, Zs)

The TRS R consists of the following rules:

select_in_gaa(X, .(X, Xs), Xs) → select_out_gaa(X, .(X, Xs), Xs)
select_in_gaa(X, .(Y, Xs), .(Y, Zs)) → U1_gaa(X, Y, Xs, Zs, select_in_gaa(X, Xs, Zs))
U1_gaa(X, Y, Xs, Zs, select_out_gaa(X, Xs, Zs)) → select_out_gaa(X, .(Y, Xs), .(Y, Zs))

The argument filtering Pi contains the following mapping:
select_in_gaa(x1, x2, x3) = select_in_gaa(x1)
select_out_gaa(x1, x2, x3) = select_out_gaa
U1_gaa(x1, x2, x3, x4, x5) = U1_gaa(x5)
SELECT_IN_GAA(x1, x2, x3) = SELECT_IN_GAA(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

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

  ↳ PrologToPiTRSProof

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

SELECT_IN_GAA(X, .(Y, Xs), .(Y, Zs)) → SELECT_IN_GAA(X, Xs, Zs)

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

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

                    ↳ QDP

                      ↳ NonTerminationProof

  ↳ PrologToPiTRSProof

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

SELECT_IN_GAA(X) → SELECT_IN_GAA(X)

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:

SELECT_IN_GAA(X) → SELECT_IN_GAA(X)

The TRS R consists of the following rules:none

s = SELECT_IN_GAA(X) evaluates to t =SELECT_IN_GAA(X)

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 SELECT_IN_GAA(X) to SELECT_IN_GAA(X).

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

select_in_gaa(X, .(X, Xs), Xs) → select_out_gaa(X, .(X, Xs), Xs)
select_in_gaa(X, .(Y, Xs), .(Y, Zs)) → U1_gaa(X, Y, Xs, Zs, select_in_gaa(X, Xs, Zs))
U1_gaa(X, Y, Xs, Zs, select_out_gaa(X, Xs, Zs)) → select_out_gaa(X, .(Y, Xs), .(Y, Zs))

The argument filtering Pi contains the following mapping:
select_in_gaa(x1, x2, x3) = select_in_gaa(x1)
select_out_gaa(x1, x2, x3) = select_out_gaa(x1)
U1_gaa(x1, x2, x3, x4, x5) = U1_gaa(x1, x5)

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_in_gaa(X, .(X, Xs), Xs) → select_out_gaa(X, .(X, Xs), Xs)
select_in_gaa(X, .(Y, Xs), .(Y, Zs)) → U1_gaa(X, Y, Xs, Zs, select_in_gaa(X, Xs, Zs))
U1_gaa(X, Y, Xs, Zs, select_out_gaa(X, Xs, Zs)) → select_out_gaa(X, .(Y, Xs), .(Y, Zs))

The argument filtering Pi contains the following mapping:
select_in_gaa(x1, x2, x3) = select_in_gaa(x1)
select_out_gaa(x1, x2, x3) = select_out_gaa(x1)
U1_gaa(x1, x2, x3, x4, x5) = U1_gaa(x1, 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:

SELECT_IN_GAA(X, .(Y, Xs), .(Y, Zs)) → U1_GAA(X, Y, Xs, Zs, select_in_gaa(X, Xs, Zs))
SELECT_IN_GAA(X, .(Y, Xs), .(Y, Zs)) → SELECT_IN_GAA(X, Xs, Zs)

The TRS R consists of the following rules:

select_in_gaa(X, .(X, Xs), Xs) → select_out_gaa(X, .(X, Xs), Xs)
select_in_gaa(X, .(Y, Xs), .(Y, Zs)) → U1_gaa(X, Y, Xs, Zs, select_in_gaa(X, Xs, Zs))
U1_gaa(X, Y, Xs, Zs, select_out_gaa(X, Xs, Zs)) → select_out_gaa(X, .(Y, Xs), .(Y, Zs))

The argument filtering Pi contains the following mapping:
select_in_gaa(x1, x2, x3) = select_in_gaa(x1)
select_out_gaa(x1, x2, x3) = select_out_gaa(x1)
U1_gaa(x1, x2, x3, x4, x5) = U1_gaa(x1, x5)
SELECT_IN_GAA(x1, x2, x3) = SELECT_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5) = U1_GAA(x1, x5)

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_IN_GAA(X, .(Y, Xs), .(Y, Zs)) → U1_GAA(X, Y, Xs, Zs, select_in_gaa(X, Xs, Zs))
SELECT_IN_GAA(X, .(Y, Xs), .(Y, Zs)) → SELECT_IN_GAA(X, Xs, Zs)

The TRS R consists of the following rules:

select_in_gaa(X, .(X, Xs), Xs) → select_out_gaa(X, .(X, Xs), Xs)
select_in_gaa(X, .(Y, Xs), .(Y, Zs)) → U1_gaa(X, Y, Xs, Zs, select_in_gaa(X, Xs, Zs))
U1_gaa(X, Y, Xs, Zs, select_out_gaa(X, Xs, Zs)) → select_out_gaa(X, .(Y, Xs), .(Y, Zs))

The argument filtering Pi contains the following mapping:
select_in_gaa(x1, x2, x3) = select_in_gaa(x1)
select_out_gaa(x1, x2, x3) = select_out_gaa(x1)
U1_gaa(x1, x2, x3, x4, x5) = U1_gaa(x1, x5)
SELECT_IN_GAA(x1, x2, x3) = SELECT_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5) = U1_GAA(x1, x5)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] 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_IN_GAA(X, .(Y, Xs), .(Y, Zs)) → SELECT_IN_GAA(X, Xs, Zs)

The TRS R consists of the following rules:

select_in_gaa(X, .(X, Xs), Xs) → select_out_gaa(X, .(X, Xs), Xs)
select_in_gaa(X, .(Y, Xs), .(Y, Zs)) → U1_gaa(X, Y, Xs, Zs, select_in_gaa(X, Xs, Zs))
U1_gaa(X, Y, Xs, Zs, select_out_gaa(X, Xs, Zs)) → select_out_gaa(X, .(Y, Xs), .(Y, Zs))

The argument filtering Pi contains the following mapping:
select_in_gaa(x1, x2, x3) = select_in_gaa(x1)
select_out_gaa(x1, x2, x3) = select_out_gaa(x1)
U1_gaa(x1, x2, x3, x4, x5) = U1_gaa(x1, x5)
SELECT_IN_GAA(x1, x2, x3) = SELECT_IN_GAA(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

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

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

SELECT_IN_GAA(X, .(Y, Xs), .(Y, Zs)) → SELECT_IN_GAA(X, Xs, Zs)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

                    ↳ QDP

                      ↳ NonTerminationProof

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

SELECT_IN_GAA(X) → SELECT_IN_GAA(X)

The TRS R consists of the following rules:none

s = SELECT_IN_GAA(X) evaluates to t =SELECT_IN_GAA(X)

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 SELECT_IN_GAA(X) to SELECT_IN_GAA(X).