Left Termination proof of /tmp/tmpMEJF9O/flat-oi.pl

Left Termination of the query pattern flat_in_2(a, g) w.r.t. the given Prolog program could not be shown:

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

Clauses:

right(tree(X, XS1, XS2), XS2).
flat(niltree, nil).
flat(tree(X, niltree, XS), cons(X, YS)) :- ','(right(tree(X, niltree, XS), ZS), flat(ZS, YS)).
flat(tree(X, tree(Y, YS1, YS2), XS), ZS) :- flat(tree(Y, YS1, tree(X, YS2, XS)), ZS).

Queries:

flat(a,g).

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

flat_in_ag(niltree, nil) → flat_out_ag(niltree, nil)
flat_in_ag(tree(X, niltree, XS), cons(X, YS)) → U1_ag(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
right_in_aa(tree(X, XS1, XS2), XS2) → right_out_aa(tree(X, XS1, XS2), XS2)
U1_ag(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → U2_ag(X, XS, YS, flat_in_ag(ZS, YS))
flat_in_ag(tree(X, tree(Y, YS1, YS2), XS), ZS) → U3_ag(X, Y, YS1, YS2, XS, ZS, flat_in_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS))
U3_ag(X, Y, YS1, YS2, XS, ZS, flat_out_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS)) → flat_out_ag(tree(X, tree(Y, YS1, YS2), XS), ZS)
U2_ag(X, XS, YS, flat_out_ag(ZS, YS)) → flat_out_ag(tree(X, niltree, XS), cons(X, YS))

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:

flat_in_ag(niltree, nil) → flat_out_ag(niltree, nil)
flat_in_ag(tree(X, niltree, XS), cons(X, YS)) → U1_ag(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
right_in_aa(tree(X, XS1, XS2), XS2) → right_out_aa(tree(X, XS1, XS2), XS2)
U1_ag(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → U2_ag(X, XS, YS, flat_in_ag(ZS, YS))
flat_in_ag(tree(X, tree(Y, YS1, YS2), XS), ZS) → U3_ag(X, Y, YS1, YS2, XS, ZS, flat_in_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS))
U3_ag(X, Y, YS1, YS2, XS, ZS, flat_out_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS)) → flat_out_ag(tree(X, tree(Y, YS1, YS2), XS), ZS)
U2_ag(X, XS, YS, flat_out_ag(ZS, YS)) → flat_out_ag(tree(X, niltree, XS), cons(X, YS))

The argument filtering Pi contains the following mapping:
flat_in_ag(x1, x2) = flat_in_ag(x2)
nil = nil
flat_out_ag(x1, x2) = flat_out_ag
cons(x1, x2) = cons(x1, x2)
U1_ag(x1, x2, x3, x4) = U1_ag(x3, x4)
right_in_aa(x1, x2) = right_in_aa
right_out_aa(x1, x2) = right_out_aa
U2_ag(x1, x2, x3, x4) = U2_ag(x4)
U3_ag(x1, x2, x3, x4, x5, x6, x7) = U3_ag(x7)
tree(x1, x2, x3) = tree(x1, x2, 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:

FLAT_IN_AG(tree(X, niltree, XS), cons(X, YS)) → U1_AG(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
FLAT_IN_AG(tree(X, niltree, XS), cons(X, YS)) → RIGHT_IN_AA(tree(X, niltree, XS), ZS)
U1_AG(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → U2_AG(X, XS, YS, flat_in_ag(ZS, YS))
U1_AG(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → FLAT_IN_AG(ZS, YS)
FLAT_IN_AG(tree(X, tree(Y, YS1, YS2), XS), ZS) → U3_AG(X, Y, YS1, YS2, XS, ZS, flat_in_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS))
FLAT_IN_AG(tree(X, tree(Y, YS1, YS2), XS), ZS) → FLAT_IN_AG(tree(Y, YS1, tree(X, YS2, XS)), ZS)

The TRS R consists of the following rules:

flat_in_ag(niltree, nil) → flat_out_ag(niltree, nil)
flat_in_ag(tree(X, niltree, XS), cons(X, YS)) → U1_ag(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
right_in_aa(tree(X, XS1, XS2), XS2) → right_out_aa(tree(X, XS1, XS2), XS2)
U1_ag(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → U2_ag(X, XS, YS, flat_in_ag(ZS, YS))
flat_in_ag(tree(X, tree(Y, YS1, YS2), XS), ZS) → U3_ag(X, Y, YS1, YS2, XS, ZS, flat_in_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS))
U3_ag(X, Y, YS1, YS2, XS, ZS, flat_out_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS)) → flat_out_ag(tree(X, tree(Y, YS1, YS2), XS), ZS)
U2_ag(X, XS, YS, flat_out_ag(ZS, YS)) → flat_out_ag(tree(X, niltree, XS), cons(X, YS))

The argument filtering Pi contains the following mapping:
flat_in_ag(x1, x2) = flat_in_ag(x2)
nil = nil
flat_out_ag(x1, x2) = flat_out_ag
cons(x1, x2) = cons(x1, x2)
U1_ag(x1, x2, x3, x4) = U1_ag(x3, x4)
right_in_aa(x1, x2) = right_in_aa
right_out_aa(x1, x2) = right_out_aa
U2_ag(x1, x2, x3, x4) = U2_ag(x4)
U3_ag(x1, x2, x3, x4, x5, x6, x7) = U3_ag(x7)
tree(x1, x2, x3) = tree(x1, x2, x3)
U3_AG(x1, x2, x3, x4, x5, x6, x7) = U3_AG(x7)
FLAT_IN_AG(x1, x2) = FLAT_IN_AG(x2)
RIGHT_IN_AA(x1, x2) = RIGHT_IN_AA
U2_AG(x1, x2, x3, x4) = U2_AG(x4)
U1_AG(x1, x2, x3, x4) = U1_AG(x3, x4)

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:

FLAT_IN_AG(tree(X, niltree, XS), cons(X, YS)) → U1_AG(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
FLAT_IN_AG(tree(X, niltree, XS), cons(X, YS)) → RIGHT_IN_AA(tree(X, niltree, XS), ZS)
U1_AG(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → U2_AG(X, XS, YS, flat_in_ag(ZS, YS))
U1_AG(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → FLAT_IN_AG(ZS, YS)
FLAT_IN_AG(tree(X, tree(Y, YS1, YS2), XS), ZS) → U3_AG(X, Y, YS1, YS2, XS, ZS, flat_in_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS))
FLAT_IN_AG(tree(X, tree(Y, YS1, YS2), XS), ZS) → FLAT_IN_AG(tree(Y, YS1, tree(X, YS2, XS)), ZS)

The TRS R consists of the following rules:

flat_in_ag(niltree, nil) → flat_out_ag(niltree, nil)
flat_in_ag(tree(X, niltree, XS), cons(X, YS)) → U1_ag(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
right_in_aa(tree(X, XS1, XS2), XS2) → right_out_aa(tree(X, XS1, XS2), XS2)
U1_ag(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → U2_ag(X, XS, YS, flat_in_ag(ZS, YS))
flat_in_ag(tree(X, tree(Y, YS1, YS2), XS), ZS) → U3_ag(X, Y, YS1, YS2, XS, ZS, flat_in_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS))
U3_ag(X, Y, YS1, YS2, XS, ZS, flat_out_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS)) → flat_out_ag(tree(X, tree(Y, YS1, YS2), XS), ZS)
U2_ag(X, XS, YS, flat_out_ag(ZS, YS)) → flat_out_ag(tree(X, niltree, XS), cons(X, YS))

The argument filtering Pi contains the following mapping:
flat_in_ag(x1, x2) = flat_in_ag(x2)
nil = nil
flat_out_ag(x1, x2) = flat_out_ag
cons(x1, x2) = cons(x1, x2)
U1_ag(x1, x2, x3, x4) = U1_ag(x3, x4)
right_in_aa(x1, x2) = right_in_aa
right_out_aa(x1, x2) = right_out_aa
U2_ag(x1, x2, x3, x4) = U2_ag(x4)
U3_ag(x1, x2, x3, x4, x5, x6, x7) = U3_ag(x7)
tree(x1, x2, x3) = tree(x1, x2, x3)
U3_AG(x1, x2, x3, x4, x5, x6, x7) = U3_AG(x7)
FLAT_IN_AG(x1, x2) = FLAT_IN_AG(x2)
RIGHT_IN_AA(x1, x2) = RIGHT_IN_AA
U2_AG(x1, x2, x3, x4) = U2_AG(x4)
U1_AG(x1, x2, x3, x4) = U1_AG(x3, x4)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 3 less nodes.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

  ↳ PrologToPiTRSProof

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

FLAT_IN_AG(tree(X, niltree, XS), cons(X, YS)) → U1_AG(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
U1_AG(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → FLAT_IN_AG(ZS, YS)
FLAT_IN_AG(tree(X, tree(Y, YS1, YS2), XS), ZS) → FLAT_IN_AG(tree(Y, YS1, tree(X, YS2, XS)), ZS)

The TRS R consists of the following rules:

flat_in_ag(niltree, nil) → flat_out_ag(niltree, nil)
flat_in_ag(tree(X, niltree, XS), cons(X, YS)) → U1_ag(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
right_in_aa(tree(X, XS1, XS2), XS2) → right_out_aa(tree(X, XS1, XS2), XS2)
U1_ag(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → U2_ag(X, XS, YS, flat_in_ag(ZS, YS))
flat_in_ag(tree(X, tree(Y, YS1, YS2), XS), ZS) → U3_ag(X, Y, YS1, YS2, XS, ZS, flat_in_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS))
U3_ag(X, Y, YS1, YS2, XS, ZS, flat_out_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS)) → flat_out_ag(tree(X, tree(Y, YS1, YS2), XS), ZS)
U2_ag(X, XS, YS, flat_out_ag(ZS, YS)) → flat_out_ag(tree(X, niltree, XS), cons(X, YS))

The argument filtering Pi contains the following mapping:
flat_in_ag(x1, x2) = flat_in_ag(x2)
nil = nil
flat_out_ag(x1, x2) = flat_out_ag
cons(x1, x2) = cons(x1, x2)
U1_ag(x1, x2, x3, x4) = U1_ag(x3, x4)
right_in_aa(x1, x2) = right_in_aa
right_out_aa(x1, x2) = right_out_aa
U2_ag(x1, x2, x3, x4) = U2_ag(x4)
U3_ag(x1, x2, x3, x4, x5, x6, x7) = U3_ag(x7)
tree(x1, x2, x3) = tree(x1, x2, x3)
FLAT_IN_AG(x1, x2) = FLAT_IN_AG(x2)
U1_AG(x1, x2, x3, x4) = U1_AG(x3, x4)

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:

FLAT_IN_AG(tree(X, niltree, XS), cons(X, YS)) → U1_AG(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
U1_AG(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → FLAT_IN_AG(ZS, YS)
FLAT_IN_AG(tree(X, tree(Y, YS1, YS2), XS), ZS) → FLAT_IN_AG(tree(Y, YS1, tree(X, YS2, XS)), ZS)

The TRS R consists of the following rules:

right_in_aa(tree(X, XS1, XS2), XS2) → right_out_aa(tree(X, XS1, XS2), XS2)

The argument filtering Pi contains the following mapping:
cons(x1, x2) = cons(x1, x2)
right_in_aa(x1, x2) = right_in_aa
right_out_aa(x1, x2) = right_out_aa
tree(x1, x2, x3) = tree(x1, x2, x3)
FLAT_IN_AG(x1, x2) = FLAT_IN_AG(x2)
U1_AG(x1, x2, x3, x4) = U1_AG(x3, x4)

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

                      ↳ UsableRulesReductionPairsProof

  ↳ PrologToPiTRSProof

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

U1_AG(YS, right_out_aa) → FLAT_IN_AG(YS)
FLAT_IN_AG(cons(X, YS)) → U1_AG(YS, right_in_aa)
FLAT_IN_AG(ZS) → FLAT_IN_AG(ZS)

The TRS R consists of the following rules:

right_in_aa → right_out_aa

The set Q consists of the following terms:

right_in_aa

We have to consider all (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

U1_AG(YS, right_out_aa) → FLAT_IN_AG(YS)
FLAT_IN_AG(cons(X, YS)) → U1_AG(YS, right_in_aa)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(FLAT_IN_AG(x₁)) = 2 + x₁
POL(U1_AG(x₁, x₂)) = 1 + x₁ + x₂
POL(cons(x₁, x₂)) = 2 + x₁ + 2·x₂
POL(right_in_aa) = 2
POL(right_out_aa) = 2

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

                    ↳ QDP

                      ↳ UsableRulesReductionPairsProof

                        ↳ QDP

                          ↳ UsableRulesProof

  ↳ PrologToPiTRSProof

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

FLAT_IN_AG(ZS) → FLAT_IN_AG(ZS)

The TRS R consists of the following rules:

right_in_aa → right_out_aa

The set Q consists of the following terms:

right_in_aa

We have to consider all (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

                    ↳ QDP

                      ↳ UsableRulesReductionPairsProof

                        ↳ QDP

                          ↳ UsableRulesProof

                            ↳ QDP

                              ↳ QReductionProof

  ↳ PrologToPiTRSProof

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

FLAT_IN_AG(ZS) → FLAT_IN_AG(ZS)

R is empty.
The set Q consists of the following terms:

right_in_aa

We have to consider all (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

right_in_aa

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

                    ↳ QDP

                      ↳ UsableRulesReductionPairsProof

                        ↳ QDP

                          ↳ UsableRulesProof

                            ↳ QDP

                              ↳ QReductionProof

                                ↳ QDP

                                  ↳ NonTerminationProof

  ↳ PrologToPiTRSProof

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

FLAT_IN_AG(ZS) → FLAT_IN_AG(ZS)

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:

FLAT_IN_AG(ZS) → FLAT_IN_AG(ZS)

The TRS R consists of the following rules:none

s = FLAT_IN_AG(ZS) evaluates to t =FLAT_IN_AG(ZS)

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 FLAT_IN_AG(ZS) to FLAT_IN_AG(ZS).

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

flat_in_ag(niltree, nil) → flat_out_ag(niltree, nil)
flat_in_ag(tree(X, niltree, XS), cons(X, YS)) → U1_ag(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
right_in_aa(tree(X, XS1, XS2), XS2) → right_out_aa(tree(X, XS1, XS2), XS2)
U1_ag(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → U2_ag(X, XS, YS, flat_in_ag(ZS, YS))
flat_in_ag(tree(X, tree(Y, YS1, YS2), XS), ZS) → U3_ag(X, Y, YS1, YS2, XS, ZS, flat_in_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS))
U3_ag(X, Y, YS1, YS2, XS, ZS, flat_out_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS)) → flat_out_ag(tree(X, tree(Y, YS1, YS2), XS), ZS)
U2_ag(X, XS, YS, flat_out_ag(ZS, YS)) → flat_out_ag(tree(X, niltree, XS), cons(X, YS))

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:

flat_in_ag(niltree, nil) → flat_out_ag(niltree, nil)
flat_in_ag(tree(X, niltree, XS), cons(X, YS)) → U1_ag(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
right_in_aa(tree(X, XS1, XS2), XS2) → right_out_aa(tree(X, XS1, XS2), XS2)
U1_ag(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → U2_ag(X, XS, YS, flat_in_ag(ZS, YS))
flat_in_ag(tree(X, tree(Y, YS1, YS2), XS), ZS) → U3_ag(X, Y, YS1, YS2, XS, ZS, flat_in_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS))
U3_ag(X, Y, YS1, YS2, XS, ZS, flat_out_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS)) → flat_out_ag(tree(X, tree(Y, YS1, YS2), XS), ZS)
U2_ag(X, XS, YS, flat_out_ag(ZS, YS)) → flat_out_ag(tree(X, niltree, XS), cons(X, YS))

The argument filtering Pi contains the following mapping:
flat_in_ag(x1, x2) = flat_in_ag(x2)
nil = nil
flat_out_ag(x1, x2) = flat_out_ag(x2)
cons(x1, x2) = cons(x1, x2)
U1_ag(x1, x2, x3, x4) = U1_ag(x1, x3, x4)
right_in_aa(x1, x2) = right_in_aa
right_out_aa(x1, x2) = right_out_aa
U2_ag(x1, x2, x3, x4) = U2_ag(x1, x3, x4)
U3_ag(x1, x2, x3, x4, x5, x6, x7) = U3_ag(x6, x7)
tree(x1, x2, x3) = tree(x1, x2, 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:

FLAT_IN_AG(tree(X, niltree, XS), cons(X, YS)) → U1_AG(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
FLAT_IN_AG(tree(X, niltree, XS), cons(X, YS)) → RIGHT_IN_AA(tree(X, niltree, XS), ZS)
U1_AG(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → U2_AG(X, XS, YS, flat_in_ag(ZS, YS))
U1_AG(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → FLAT_IN_AG(ZS, YS)
FLAT_IN_AG(tree(X, tree(Y, YS1, YS2), XS), ZS) → U3_AG(X, Y, YS1, YS2, XS, ZS, flat_in_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS))
FLAT_IN_AG(tree(X, tree(Y, YS1, YS2), XS), ZS) → FLAT_IN_AG(tree(Y, YS1, tree(X, YS2, XS)), ZS)

The TRS R consists of the following rules:

flat_in_ag(niltree, nil) → flat_out_ag(niltree, nil)
flat_in_ag(tree(X, niltree, XS), cons(X, YS)) → U1_ag(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
right_in_aa(tree(X, XS1, XS2), XS2) → right_out_aa(tree(X, XS1, XS2), XS2)
U1_ag(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → U2_ag(X, XS, YS, flat_in_ag(ZS, YS))
flat_in_ag(tree(X, tree(Y, YS1, YS2), XS), ZS) → U3_ag(X, Y, YS1, YS2, XS, ZS, flat_in_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS))
U3_ag(X, Y, YS1, YS2, XS, ZS, flat_out_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS)) → flat_out_ag(tree(X, tree(Y, YS1, YS2), XS), ZS)
U2_ag(X, XS, YS, flat_out_ag(ZS, YS)) → flat_out_ag(tree(X, niltree, XS), cons(X, YS))

The argument filtering Pi contains the following mapping:
flat_in_ag(x1, x2) = flat_in_ag(x2)
nil = nil
flat_out_ag(x1, x2) = flat_out_ag(x2)
cons(x1, x2) = cons(x1, x2)
U1_ag(x1, x2, x3, x4) = U1_ag(x1, x3, x4)
right_in_aa(x1, x2) = right_in_aa
right_out_aa(x1, x2) = right_out_aa
U2_ag(x1, x2, x3, x4) = U2_ag(x1, x3, x4)
U3_ag(x1, x2, x3, x4, x5, x6, x7) = U3_ag(x6, x7)
tree(x1, x2, x3) = tree(x1, x2, x3)
U3_AG(x1, x2, x3, x4, x5, x6, x7) = U3_AG(x6, x7)
FLAT_IN_AG(x1, x2) = FLAT_IN_AG(x2)
RIGHT_IN_AA(x1, x2) = RIGHT_IN_AA
U2_AG(x1, x2, x3, x4) = U2_AG(x1, x3, x4)
U1_AG(x1, x2, x3, x4) = U1_AG(x1, x3, x4)

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:

FLAT_IN_AG(tree(X, niltree, XS), cons(X, YS)) → U1_AG(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
FLAT_IN_AG(tree(X, niltree, XS), cons(X, YS)) → RIGHT_IN_AA(tree(X, niltree, XS), ZS)
U1_AG(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → U2_AG(X, XS, YS, flat_in_ag(ZS, YS))
U1_AG(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → FLAT_IN_AG(ZS, YS)
FLAT_IN_AG(tree(X, tree(Y, YS1, YS2), XS), ZS) → U3_AG(X, Y, YS1, YS2, XS, ZS, flat_in_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS))
FLAT_IN_AG(tree(X, tree(Y, YS1, YS2), XS), ZS) → FLAT_IN_AG(tree(Y, YS1, tree(X, YS2, XS)), ZS)

The TRS R consists of the following rules:

flat_in_ag(niltree, nil) → flat_out_ag(niltree, nil)
flat_in_ag(tree(X, niltree, XS), cons(X, YS)) → U1_ag(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
right_in_aa(tree(X, XS1, XS2), XS2) → right_out_aa(tree(X, XS1, XS2), XS2)
U1_ag(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → U2_ag(X, XS, YS, flat_in_ag(ZS, YS))
flat_in_ag(tree(X, tree(Y, YS1, YS2), XS), ZS) → U3_ag(X, Y, YS1, YS2, XS, ZS, flat_in_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS))
U3_ag(X, Y, YS1, YS2, XS, ZS, flat_out_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS)) → flat_out_ag(tree(X, tree(Y, YS1, YS2), XS), ZS)
U2_ag(X, XS, YS, flat_out_ag(ZS, YS)) → flat_out_ag(tree(X, niltree, XS), cons(X, YS))

The argument filtering Pi contains the following mapping:
flat_in_ag(x1, x2) = flat_in_ag(x2)
nil = nil
flat_out_ag(x1, x2) = flat_out_ag(x2)
cons(x1, x2) = cons(x1, x2)
U1_ag(x1, x2, x3, x4) = U1_ag(x1, x3, x4)
right_in_aa(x1, x2) = right_in_aa
right_out_aa(x1, x2) = right_out_aa
U2_ag(x1, x2, x3, x4) = U2_ag(x1, x3, x4)
U3_ag(x1, x2, x3, x4, x5, x6, x7) = U3_ag(x6, x7)
tree(x1, x2, x3) = tree(x1, x2, x3)
U3_AG(x1, x2, x3, x4, x5, x6, x7) = U3_AG(x6, x7)
FLAT_IN_AG(x1, x2) = FLAT_IN_AG(x2)
RIGHT_IN_AA(x1, x2) = RIGHT_IN_AA
U2_AG(x1, x2, x3, x4) = U2_AG(x1, x3, x4)
U1_AG(x1, x2, x3, x4) = U1_AG(x1, x3, x4)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 3 less nodes.

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

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

FLAT_IN_AG(tree(X, niltree, XS), cons(X, YS)) → U1_AG(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
U1_AG(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → FLAT_IN_AG(ZS, YS)
FLAT_IN_AG(tree(X, tree(Y, YS1, YS2), XS), ZS) → FLAT_IN_AG(tree(Y, YS1, tree(X, YS2, XS)), ZS)

The TRS R consists of the following rules:

flat_in_ag(niltree, nil) → flat_out_ag(niltree, nil)
flat_in_ag(tree(X, niltree, XS), cons(X, YS)) → U1_ag(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
right_in_aa(tree(X, XS1, XS2), XS2) → right_out_aa(tree(X, XS1, XS2), XS2)
U1_ag(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → U2_ag(X, XS, YS, flat_in_ag(ZS, YS))
flat_in_ag(tree(X, tree(Y, YS1, YS2), XS), ZS) → U3_ag(X, Y, YS1, YS2, XS, ZS, flat_in_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS))
U3_ag(X, Y, YS1, YS2, XS, ZS, flat_out_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS)) → flat_out_ag(tree(X, tree(Y, YS1, YS2), XS), ZS)
U2_ag(X, XS, YS, flat_out_ag(ZS, YS)) → flat_out_ag(tree(X, niltree, XS), cons(X, YS))

The argument filtering Pi contains the following mapping:
flat_in_ag(x1, x2) = flat_in_ag(x2)
nil = nil
flat_out_ag(x1, x2) = flat_out_ag(x2)
cons(x1, x2) = cons(x1, x2)
U1_ag(x1, x2, x3, x4) = U1_ag(x1, x3, x4)
right_in_aa(x1, x2) = right_in_aa
right_out_aa(x1, x2) = right_out_aa
U2_ag(x1, x2, x3, x4) = U2_ag(x1, x3, x4)
U3_ag(x1, x2, x3, x4, x5, x6, x7) = U3_ag(x6, x7)
tree(x1, x2, x3) = tree(x1, x2, x3)
FLAT_IN_AG(x1, x2) = FLAT_IN_AG(x2)
U1_AG(x1, x2, x3, x4) = U1_AG(x1, x3, x4)

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:

FLAT_IN_AG(tree(X, niltree, XS), cons(X, YS)) → U1_AG(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
U1_AG(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → FLAT_IN_AG(ZS, YS)
FLAT_IN_AG(tree(X, tree(Y, YS1, YS2), XS), ZS) → FLAT_IN_AG(tree(Y, YS1, tree(X, YS2, XS)), ZS)

The TRS R consists of the following rules:

right_in_aa(tree(X, XS1, XS2), XS2) → right_out_aa(tree(X, XS1, XS2), XS2)

The argument filtering Pi contains the following mapping:
cons(x1, x2) = cons(x1, x2)
right_in_aa(x1, x2) = right_in_aa
right_out_aa(x1, x2) = right_out_aa
tree(x1, x2, x3) = tree(x1, x2, x3)
FLAT_IN_AG(x1, x2) = FLAT_IN_AG(x2)
U1_AG(x1, x2, x3, x4) = U1_AG(x1, x3, x4)

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

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

                    ↳ QDP

                      ↳ UsableRulesReductionPairsProof

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

FLAT_IN_AG(cons(X, YS)) → U1_AG(X, YS, right_in_aa)
U1_AG(X, YS, right_out_aa) → FLAT_IN_AG(YS)
FLAT_IN_AG(ZS) → FLAT_IN_AG(ZS)

The TRS R consists of the following rules:

right_in_aa → right_out_aa

The set Q consists of the following terms:

right_in_aa

FLAT_IN_AG(cons(X, YS)) → U1_AG(X, YS, right_in_aa)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(FLAT_IN_AG(x₁)) = x₁
POL(U1_AG(x₁, x₂, x₃)) = x₁ + 2·x₂ + 2·x₃
POL(cons(x₁, x₂)) = 1 + 2·x₁ + 2·x₂
POL(right_in_aa) = 0
POL(right_out_aa) = 0

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

                    ↳ QDP

                      ↳ UsableRulesReductionPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

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

U1_AG(X, YS, right_out_aa) → FLAT_IN_AG(YS)
FLAT_IN_AG(ZS) → FLAT_IN_AG(ZS)

The TRS R consists of the following rules:

right_in_aa → right_out_aa

The set Q consists of the following terms:

right_in_aa

We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

                    ↳ QDP

                      ↳ UsableRulesReductionPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ UsableRulesProof

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

FLAT_IN_AG(ZS) → FLAT_IN_AG(ZS)

The TRS R consists of the following rules:

right_in_aa → right_out_aa

The set Q consists of the following terms:

right_in_aa

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

                    ↳ QDP

                      ↳ UsableRulesReductionPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ UsableRulesProof

                                ↳ QDP

                                  ↳ QReductionProof

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

FLAT_IN_AG(ZS) → FLAT_IN_AG(ZS)

R is empty.
The set Q consists of the following terms:

right_in_aa

We have to consider all (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

right_in_aa

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

                    ↳ QDP

                      ↳ UsableRulesReductionPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ UsableRulesProof

                                ↳ QDP

                                  ↳ QReductionProof

                                    ↳ QDP

                                      ↳ NonTerminationProof

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

FLAT_IN_AG(ZS) → FLAT_IN_AG(ZS)

The TRS R consists of the following rules:none

s = FLAT_IN_AG(ZS) evaluates to t =FLAT_IN_AG(ZS)

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 FLAT_IN_AG(ZS) to FLAT_IN_AG(ZS).