Left Termination proof of /tmp/tmpyckdzU/flatten

Left Termination of the query pattern count_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:

↳ Prolog

  ↳ PrologToPiTRSProof

Clauses:

flatten(atom(X), .(X, [])).
flatten(cons(atom(X), U), .(X, Y)) :- flatten(U, Y).
flatten(cons(cons(U, V), W), X) :- flatten(cons(U, cons(V, W)), X).
count(atom(X), s(0)).
count(cons(atom(X), Y), s(Z)) :- count(Y, Z).
count(cons(cons(U, V), W), Z) :- ','(flatten(cons(cons(U, V), W), X), count(X, Z)).

Queries:

count(g,a).

We use the technique of [30].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

count_in(cons(cons(U, V), W), Z) → U4(U, V, W, Z, flatten_in(cons(cons(U, V), W), X))
flatten_in(cons(cons(U, V), W), X) → U2(U, V, W, X, flatten_in(cons(U, cons(V, W)), X))
flatten_in(cons(atom(X), U), .(X, Y)) → U1(X, U, Y, flatten_in(U, Y))
flatten_in(atom(X), .(X, [])) → flatten_out(atom(X), .(X, []))
U1(X, U, Y, flatten_out(U, Y)) → flatten_out(cons(atom(X), U), .(X, Y))
U2(U, V, W, X, flatten_out(cons(U, cons(V, W)), X)) → flatten_out(cons(cons(U, V), W), X)
U4(U, V, W, Z, flatten_out(cons(cons(U, V), W), X)) → U5(U, V, W, Z, count_in(X, Z))
count_in(cons(atom(X), Y), s(Z)) → U3(X, Y, Z, count_in(Y, Z))
count_in(atom(X), s(0)) → count_out(atom(X), s(0))
U3(X, Y, Z, count_out(Y, Z)) → count_out(cons(atom(X), Y), s(Z))
U5(U, V, W, Z, count_out(X, Z)) → count_out(cons(cons(U, V), 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:

count_in(cons(cons(U, V), W), Z) → U4(U, V, W, Z, flatten_in(cons(cons(U, V), W), X))
flatten_in(cons(cons(U, V), W), X) → U2(U, V, W, X, flatten_in(cons(U, cons(V, W)), X))
flatten_in(cons(atom(X), U), .(X, Y)) → U1(X, U, Y, flatten_in(U, Y))
flatten_in(atom(X), .(X, [])) → flatten_out(atom(X), .(X, []))
U1(X, U, Y, flatten_out(U, Y)) → flatten_out(cons(atom(X), U), .(X, Y))
U2(U, V, W, X, flatten_out(cons(U, cons(V, W)), X)) → flatten_out(cons(cons(U, V), W), X)
U4(U, V, W, Z, flatten_out(cons(cons(U, V), W), X)) → U5(U, V, W, Z, count_in(X, Z))
count_in(cons(atom(X), Y), s(Z)) → U3(X, Y, Z, count_in(Y, Z))
count_in(atom(X), s(0)) → count_out(atom(X), s(0))
U3(X, Y, Z, count_out(Y, Z)) → count_out(cons(atom(X), Y), s(Z))
U5(U, V, W, Z, count_out(X, Z)) → count_out(cons(cons(U, V), W), Z)

The argument filtering Pi contains the following mapping:
count_in(x1, x2) = count_in(x1)
cons(x1, x2) = cons(x1, x2)
U4(x1, x2, x3, x4, x5) = U4(x5)
flatten_in(x1, x2) = flatten_in(x1)
U2(x1, x2, x3, x4, x5) = U2(x5)
atom(x1) = atom(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x1, x4)
[] = []
flatten_out(x1, x2) = flatten_out(x2)
U5(x1, x2, x3, x4, x5) = U5(x5)
s(x1) = s(x1)
U3(x1, x2, x3, x4) = U3(x4)
0 = 0
count_out(x1, x2) = count_out(x2)

Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

COUNT_IN(cons(cons(U, V), W), Z) → U4¹(U, V, W, Z, flatten_in(cons(cons(U, V), W), X))
COUNT_IN(cons(cons(U, V), W), Z) → FLATTEN_IN(cons(cons(U, V), W), X)
FLATTEN_IN(cons(cons(U, V), W), X) → U2¹(U, V, W, X, flatten_in(cons(U, cons(V, W)), X))
FLATTEN_IN(cons(cons(U, V), W), X) → FLATTEN_IN(cons(U, cons(V, W)), X)
FLATTEN_IN(cons(atom(X), U), .(X, Y)) → U1¹(X, U, Y, flatten_in(U, Y))
FLATTEN_IN(cons(atom(X), U), .(X, Y)) → FLATTEN_IN(U, Y)
U4¹(U, V, W, Z, flatten_out(cons(cons(U, V), W), X)) → U5¹(U, V, W, Z, count_in(X, Z))
U4¹(U, V, W, Z, flatten_out(cons(cons(U, V), W), X)) → COUNT_IN(X, Z)
COUNT_IN(cons(atom(X), Y), s(Z)) → U3¹(X, Y, Z, count_in(Y, Z))
COUNT_IN(cons(atom(X), Y), s(Z)) → COUNT_IN(Y, Z)

The TRS R consists of the following rules:

count_in(cons(cons(U, V), W), Z) → U4(U, V, W, Z, flatten_in(cons(cons(U, V), W), X))
flatten_in(cons(cons(U, V), W), X) → U2(U, V, W, X, flatten_in(cons(U, cons(V, W)), X))
flatten_in(cons(atom(X), U), .(X, Y)) → U1(X, U, Y, flatten_in(U, Y))
flatten_in(atom(X), .(X, [])) → flatten_out(atom(X), .(X, []))
U1(X, U, Y, flatten_out(U, Y)) → flatten_out(cons(atom(X), U), .(X, Y))
U2(U, V, W, X, flatten_out(cons(U, cons(V, W)), X)) → flatten_out(cons(cons(U, V), W), X)
U4(U, V, W, Z, flatten_out(cons(cons(U, V), W), X)) → U5(U, V, W, Z, count_in(X, Z))
count_in(cons(atom(X), Y), s(Z)) → U3(X, Y, Z, count_in(Y, Z))
count_in(atom(X), s(0)) → count_out(atom(X), s(0))
U3(X, Y, Z, count_out(Y, Z)) → count_out(cons(atom(X), Y), s(Z))
U5(U, V, W, Z, count_out(X, Z)) → count_out(cons(cons(U, V), W), Z)

The argument filtering Pi contains the following mapping:
count_in(x1, x2) = count_in(x1)
cons(x1, x2) = cons(x1, x2)
U4(x1, x2, x3, x4, x5) = U4(x5)
flatten_in(x1, x2) = flatten_in(x1)
U2(x1, x2, x3, x4, x5) = U2(x5)
atom(x1) = atom(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x1, x4)
[] = []
flatten_out(x1, x2) = flatten_out(x2)
U5(x1, x2, x3, x4, x5) = U5(x5)
s(x1) = s(x1)
U3(x1, x2, x3, x4) = U3(x4)
0 = 0
count_out(x1, x2) = count_out(x2)
COUNT_IN(x1, x2) = COUNT_IN(x1)
U3¹(x1, x2, x3, x4) = U3¹(x4)
U5¹(x1, x2, x3, x4, x5) = U5¹(x5)
U4¹(x1, x2, x3, x4, x5) = U4¹(x5)
FLATTEN_IN(x1, x2) = FLATTEN_IN(x1)
U2¹(x1, x2, x3, x4, x5) = U2¹(x5)
U1¹(x1, x2, x3, x4) = U1¹(x1, x4)

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:

COUNT_IN(cons(cons(U, V), W), Z) → U4¹(U, V, W, Z, flatten_in(cons(cons(U, V), W), X))
COUNT_IN(cons(cons(U, V), W), Z) → FLATTEN_IN(cons(cons(U, V), W), X)
FLATTEN_IN(cons(cons(U, V), W), X) → U2¹(U, V, W, X, flatten_in(cons(U, cons(V, W)), X))
FLATTEN_IN(cons(cons(U, V), W), X) → FLATTEN_IN(cons(U, cons(V, W)), X)
FLATTEN_IN(cons(atom(X), U), .(X, Y)) → U1¹(X, U, Y, flatten_in(U, Y))
FLATTEN_IN(cons(atom(X), U), .(X, Y)) → FLATTEN_IN(U, Y)
U4¹(U, V, W, Z, flatten_out(cons(cons(U, V), W), X)) → U5¹(U, V, W, Z, count_in(X, Z))
U4¹(U, V, W, Z, flatten_out(cons(cons(U, V), W), X)) → COUNT_IN(X, Z)
COUNT_IN(cons(atom(X), Y), s(Z)) → U3¹(X, Y, Z, count_in(Y, Z))
COUNT_IN(cons(atom(X), Y), s(Z)) → COUNT_IN(Y, Z)

The TRS R consists of the following rules:

count_in(cons(cons(U, V), W), Z) → U4(U, V, W, Z, flatten_in(cons(cons(U, V), W), X))
flatten_in(cons(cons(U, V), W), X) → U2(U, V, W, X, flatten_in(cons(U, cons(V, W)), X))
flatten_in(cons(atom(X), U), .(X, Y)) → U1(X, U, Y, flatten_in(U, Y))
flatten_in(atom(X), .(X, [])) → flatten_out(atom(X), .(X, []))
U1(X, U, Y, flatten_out(U, Y)) → flatten_out(cons(atom(X), U), .(X, Y))
U2(U, V, W, X, flatten_out(cons(U, cons(V, W)), X)) → flatten_out(cons(cons(U, V), W), X)
U4(U, V, W, Z, flatten_out(cons(cons(U, V), W), X)) → U5(U, V, W, Z, count_in(X, Z))
count_in(cons(atom(X), Y), s(Z)) → U3(X, Y, Z, count_in(Y, Z))
count_in(atom(X), s(0)) → count_out(atom(X), s(0))
U3(X, Y, Z, count_out(Y, Z)) → count_out(cons(atom(X), Y), s(Z))
U5(U, V, W, Z, count_out(X, Z)) → count_out(cons(cons(U, V), W), Z)

The argument filtering Pi contains the following mapping:
count_in(x1, x2) = count_in(x1)
cons(x1, x2) = cons(x1, x2)
U4(x1, x2, x3, x4, x5) = U4(x5)
flatten_in(x1, x2) = flatten_in(x1)
U2(x1, x2, x3, x4, x5) = U2(x5)
atom(x1) = atom(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x1, x4)
[] = []
flatten_out(x1, x2) = flatten_out(x2)
U5(x1, x2, x3, x4, x5) = U5(x5)
s(x1) = s(x1)
U3(x1, x2, x3, x4) = U3(x4)
0 = 0
count_out(x1, x2) = count_out(x2)
COUNT_IN(x1, x2) = COUNT_IN(x1)
U3¹(x1, x2, x3, x4) = U3¹(x4)
U5¹(x1, x2, x3, x4, x5) = U5¹(x5)
U4¹(x1, x2, x3, x4, x5) = U4¹(x5)
FLATTEN_IN(x1, x2) = FLATTEN_IN(x1)
U2¹(x1, x2, x3, x4, x5) = U2¹(x5)
U1¹(x1, x2, x3, x4) = U1¹(x1, x4)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 5 less nodes.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

FLATTEN_IN(cons(atom(X), U), .(X, Y)) → FLATTEN_IN(U, Y)
FLATTEN_IN(cons(cons(U, V), W), X) → FLATTEN_IN(cons(U, cons(V, W)), X)

The TRS R consists of the following rules:

count_in(cons(cons(U, V), W), Z) → U4(U, V, W, Z, flatten_in(cons(cons(U, V), W), X))
flatten_in(cons(cons(U, V), W), X) → U2(U, V, W, X, flatten_in(cons(U, cons(V, W)), X))
flatten_in(cons(atom(X), U), .(X, Y)) → U1(X, U, Y, flatten_in(U, Y))
flatten_in(atom(X), .(X, [])) → flatten_out(atom(X), .(X, []))
U1(X, U, Y, flatten_out(U, Y)) → flatten_out(cons(atom(X), U), .(X, Y))
U2(U, V, W, X, flatten_out(cons(U, cons(V, W)), X)) → flatten_out(cons(cons(U, V), W), X)
U4(U, V, W, Z, flatten_out(cons(cons(U, V), W), X)) → U5(U, V, W, Z, count_in(X, Z))
count_in(cons(atom(X), Y), s(Z)) → U3(X, Y, Z, count_in(Y, Z))
count_in(atom(X), s(0)) → count_out(atom(X), s(0))
U3(X, Y, Z, count_out(Y, Z)) → count_out(cons(atom(X), Y), s(Z))
U5(U, V, W, Z, count_out(X, Z)) → count_out(cons(cons(U, V), W), Z)

The argument filtering Pi contains the following mapping:
count_in(x1, x2) = count_in(x1)
cons(x1, x2) = cons(x1, x2)
U4(x1, x2, x3, x4, x5) = U4(x5)
flatten_in(x1, x2) = flatten_in(x1)
U2(x1, x2, x3, x4, x5) = U2(x5)
atom(x1) = atom(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x1, x4)
[] = []
flatten_out(x1, x2) = flatten_out(x2)
U5(x1, x2, x3, x4, x5) = U5(x5)
s(x1) = s(x1)
U3(x1, x2, x3, x4) = U3(x4)
0 = 0
count_out(x1, x2) = count_out(x2)
FLATTEN_IN(x1, x2) = FLATTEN_IN(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

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

FLATTEN_IN(cons(atom(X), U), .(X, Y)) → FLATTEN_IN(U, Y)
FLATTEN_IN(cons(cons(U, V), W), X) → FLATTEN_IN(cons(U, cons(V, W)), X)

R is empty.
The argument filtering Pi contains the following mapping:
cons(x1, x2) = cons(x1, x2)
atom(x1) = atom(x1)
.(x1, x2) = .(x1, x2)
FLATTEN_IN(x1, x2) = FLATTEN_IN(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

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

              ↳ PiDP

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

FLATTEN_IN(cons(cons(U, V), W)) → FLATTEN_IN(cons(U, cons(V, W)))
FLATTEN_IN(cons(atom(X), U)) → FLATTEN_IN(U)

R is empty.
Q is empty.
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:

FLATTEN_IN(cons(cons(U, V), W)) → FLATTEN_IN(cons(U, cons(V, W)))
FLATTEN_IN(cons(atom(X), U)) → FLATTEN_IN(U)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(FLATTEN_IN(x₁)) = 2·x₁
POL(atom(x₁)) = x₁
POL(cons(x₁, x₂)) = 2 + 2·x₁ + x₂

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                          ↳ QDP

                            ↳ PisEmptyProof

              ↳ PiDP

Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

COUNT_IN(cons(atom(X), Y), s(Z)) → COUNT_IN(Y, Z)
U4¹(U, V, W, Z, flatten_out(cons(cons(U, V), W), X)) → COUNT_IN(X, Z)
COUNT_IN(cons(cons(U, V), W), Z) → U4¹(U, V, W, Z, flatten_in(cons(cons(U, V), W), X))

The TRS R consists of the following rules:

count_in(cons(cons(U, V), W), Z) → U4(U, V, W, Z, flatten_in(cons(cons(U, V), W), X))
flatten_in(cons(cons(U, V), W), X) → U2(U, V, W, X, flatten_in(cons(U, cons(V, W)), X))
flatten_in(cons(atom(X), U), .(X, Y)) → U1(X, U, Y, flatten_in(U, Y))
flatten_in(atom(X), .(X, [])) → flatten_out(atom(X), .(X, []))
U1(X, U, Y, flatten_out(U, Y)) → flatten_out(cons(atom(X), U), .(X, Y))
U2(U, V, W, X, flatten_out(cons(U, cons(V, W)), X)) → flatten_out(cons(cons(U, V), W), X)
U4(U, V, W, Z, flatten_out(cons(cons(U, V), W), X)) → U5(U, V, W, Z, count_in(X, Z))
count_in(cons(atom(X), Y), s(Z)) → U3(X, Y, Z, count_in(Y, Z))
count_in(atom(X), s(0)) → count_out(atom(X), s(0))
U3(X, Y, Z, count_out(Y, Z)) → count_out(cons(atom(X), Y), s(Z))
U5(U, V, W, Z, count_out(X, Z)) → count_out(cons(cons(U, V), W), Z)

The argument filtering Pi contains the following mapping:
count_in(x1, x2) = count_in(x1)
cons(x1, x2) = cons(x1, x2)
U4(x1, x2, x3, x4, x5) = U4(x5)
flatten_in(x1, x2) = flatten_in(x1)
U2(x1, x2, x3, x4, x5) = U2(x5)
atom(x1) = atom(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x1, x4)
[] = []
flatten_out(x1, x2) = flatten_out(x2)
U5(x1, x2, x3, x4, x5) = U5(x5)
s(x1) = s(x1)
U3(x1, x2, x3, x4) = U3(x4)
0 = 0
count_out(x1, x2) = count_out(x2)
COUNT_IN(x1, x2) = COUNT_IN(x1)
U4¹(x1, x2, x3, x4, x5) = U4¹(x5)

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

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

COUNT_IN(cons(atom(X), Y), s(Z)) → COUNT_IN(Y, Z)
U4¹(U, V, W, Z, flatten_out(cons(cons(U, V), W), X)) → COUNT_IN(X, Z)
COUNT_IN(cons(cons(U, V), W), Z) → U4¹(U, V, W, Z, flatten_in(cons(cons(U, V), W), X))

The TRS R consists of the following rules:

flatten_in(cons(cons(U, V), W), X) → U2(U, V, W, X, flatten_in(cons(U, cons(V, W)), X))
U2(U, V, W, X, flatten_out(cons(U, cons(V, W)), X)) → flatten_out(cons(cons(U, V), W), X)
flatten_in(cons(atom(X), U), .(X, Y)) → U1(X, U, Y, flatten_in(U, Y))
U1(X, U, Y, flatten_out(U, Y)) → flatten_out(cons(atom(X), U), .(X, Y))
flatten_in(atom(X), .(X, [])) → flatten_out(atom(X), .(X, []))

The argument filtering Pi contains the following mapping:
cons(x1, x2) = cons(x1, x2)
flatten_in(x1, x2) = flatten_in(x1)
U2(x1, x2, x3, x4, x5) = U2(x5)
atom(x1) = atom(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x1, x4)
[] = []
flatten_out(x1, x2) = flatten_out(x2)
s(x1) = s(x1)
COUNT_IN(x1, x2) = COUNT_IN(x1)
U4¹(x1, x2, x3, x4, x5) = U4¹(x5)

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

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

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

COUNT_IN(cons(cons(U, V), W)) → U4¹(flatten_in(cons(cons(U, V), W)))
COUNT_IN(cons(atom(X), Y)) → COUNT_IN(Y)
U4¹(flatten_out(X)) → COUNT_IN(X)

The TRS R consists of the following rules:

flatten_in(cons(cons(U, V), W)) → U2(flatten_in(cons(U, cons(V, W))))
U2(flatten_out(X)) → flatten_out(X)
flatten_in(cons(atom(X), U)) → U1(X, flatten_in(U))
U1(X, flatten_out(Y)) → flatten_out(.(X, Y))
flatten_in(atom(X)) → flatten_out(.(X, []))

The set Q consists of the following terms:

flatten_in(x0)
U2(x0)
U1(x0, x1)

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:

COUNT_IN(cons(cons(U, V), W)) → U4¹(flatten_in(cons(cons(U, V), W)))
COUNT_IN(cons(atom(X), Y)) → COUNT_IN(Y)

The following rules are removed from R:

flatten_in(cons(cons(U, V), W)) → U2(flatten_in(cons(U, cons(V, W))))
flatten_in(cons(atom(X), U)) → U1(X, flatten_in(U))
U2(flatten_out(X)) → flatten_out(X)
flatten_in(atom(X)) → flatten_out(.(X, []))
U1(X, flatten_out(Y)) → flatten_out(.(X, Y))

Used ordering: POLO with Polynomial interpretation [25]:

POL(.(x₁, x₂)) = x₁ + x₂
POL(COUNT_IN(x₁)) = 2 + x₁
POL(U1(x₁, x₂)) = 1 + 2·x₁ + x₂
POL(U2(x₁)) = 1 + x₁
POL(U4¹(x₁)) = 1 + x₁
POL([]) = 0
POL(atom(x₁)) = 1 + 2·x₁
POL(cons(x₁, x₂)) = 2 + 2·x₁ + x₂
POL(flatten_in(x₁)) = x₁
POL(flatten_out(x₁)) = 1 + 2·x₁

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                          ↳ QDP

                            ↳ DependencyGraphProof

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

U4¹(flatten_out(X)) → COUNT_IN(X)

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

flatten_in(x0)
U2(x0)
U1(x0, x1)

We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.