Left Termination proof of /tmp/tmpfBJXFm/flatten

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

↳ PROLOG

  ↳ PrologToPiTRSProof

flatten₂(atom₁(X), .₂(X, {}₀)).
flatten₂(cons₂(atom₁(X), U), .₂(X, Y)) :- flatten₂(U, Y).
flatten₂(cons₂(cons₂(U, V), W), X) :- flatten₂(cons₂(U, cons₂(V, W)), X).
count₂(atom₁(X), s₁(0₀)).
count₂(cons₂(atom₁(X), Y), s₁(Z)) :- count₂(Y, Z).
count₂(cons₂(cons₂(U, V), W), Z) :- flatten₂(cons₂(cons₂(U, V), W), X), count₂(X, Z).

With regard to the inferred argument filtering the predicates were used in the following modes:
count₂: (b,f)
flatten₂: (b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

count_2_in_ga₂(atom_1₁(X), s_1₁(0_0)) -> count_2_out_ga₂(atom_1₁(X), s_1₁(0_0))
count_2_in_ga₂(cons_2₂(atom_1₁(X), Y), s_1₁(Z)) -> if_count_2_in_1_ga₄(X, Y, Z, count_2_in_ga₂(Y, Z))
count_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), Z) -> if_count_2_in_2_ga₅(U, V, W, Z, flatten_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), X))
flatten_2_in_ga₂(atom_1₁(X), ._2₂(X, []_0)) -> flatten_2_out_ga₂(atom_1₁(X), ._2₂(X, []_0))
flatten_2_in_ga₂(cons_2₂(atom_1₁(X), U), ._2₂(X, Y)) -> if_flatten_2_in_1_ga₄(X, U, Y, flatten_2_in_ga₂(U, Y))
flatten_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), X) -> if_flatten_2_in_2_ga₅(U, V, W, X, flatten_2_in_ga₂(cons_2₂(U, cons_2₂(V, W)), X))
if_flatten_2_in_2_ga₅(U, V, W, X, flatten_2_out_ga₂(cons_2₂(U, cons_2₂(V, W)), X)) -> flatten_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), X)
if_flatten_2_in_1_ga₄(X, U, Y, flatten_2_out_ga₂(U, Y)) -> flatten_2_out_ga₂(cons_2₂(atom_1₁(X), U), ._2₂(X, Y))
if_count_2_in_2_ga₅(U, V, W, Z, flatten_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), X)) -> if_count_2_in_3_ga₆(U, V, W, Z, X, count_2_in_ga₂(X, Z))
if_count_2_in_3_ga₆(U, V, W, Z, X, count_2_out_ga₂(X, Z)) -> count_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), Z)
if_count_2_in_1_ga₄(X, Y, Z, count_2_out_ga₂(Y, Z)) -> count_2_out_ga₂(cons_2₂(atom_1₁(X), Y), s_1₁(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_2_in_ga₂(atom_1₁(X), s_1₁(0_0)) -> count_2_out_ga₂(atom_1₁(X), s_1₁(0_0))
count_2_in_ga₂(cons_2₂(atom_1₁(X), Y), s_1₁(Z)) -> if_count_2_in_1_ga₄(X, Y, Z, count_2_in_ga₂(Y, Z))
count_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), Z) -> if_count_2_in_2_ga₅(U, V, W, Z, flatten_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), X))
flatten_2_in_ga₂(atom_1₁(X), ._2₂(X, []_0)) -> flatten_2_out_ga₂(atom_1₁(X), ._2₂(X, []_0))
flatten_2_in_ga₂(cons_2₂(atom_1₁(X), U), ._2₂(X, Y)) -> if_flatten_2_in_1_ga₄(X, U, Y, flatten_2_in_ga₂(U, Y))
flatten_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), X) -> if_flatten_2_in_2_ga₅(U, V, W, X, flatten_2_in_ga₂(cons_2₂(U, cons_2₂(V, W)), X))
if_flatten_2_in_2_ga₅(U, V, W, X, flatten_2_out_ga₂(cons_2₂(U, cons_2₂(V, W)), X)) -> flatten_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), X)
if_flatten_2_in_1_ga₄(X, U, Y, flatten_2_out_ga₂(U, Y)) -> flatten_2_out_ga₂(cons_2₂(atom_1₁(X), U), ._2₂(X, Y))
if_count_2_in_2_ga₅(U, V, W, Z, flatten_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), X)) -> if_count_2_in_3_ga₆(U, V, W, Z, X, count_2_in_ga₂(X, Z))
if_count_2_in_3_ga₆(U, V, W, Z, X, count_2_out_ga₂(X, Z)) -> count_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), Z)
if_count_2_in_1_ga₄(X, Y, Z, count_2_out_ga₂(Y, Z)) -> count_2_out_ga₂(cons_2₂(atom_1₁(X), Y), s_1₁(Z))

COUNT_2_IN_GA₂(cons_2₂(atom_1₁(X), Y), s_1₁(Z)) -> IF_COUNT_2_IN_1_GA₄(X, Y, Z, count_2_in_ga₂(Y, Z))
COUNT_2_IN_GA₂(cons_2₂(atom_1₁(X), Y), s_1₁(Z)) -> COUNT_2_IN_GA₂(Y, Z)
COUNT_2_IN_GA₂(cons_2₂(cons_2₂(U, V), W), Z) -> IF_COUNT_2_IN_2_GA₅(U, V, W, Z, flatten_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), X))
COUNT_2_IN_GA₂(cons_2₂(cons_2₂(U, V), W), Z) -> FLATTEN_2_IN_GA₂(cons_2₂(cons_2₂(U, V), W), X)
FLATTEN_2_IN_GA₂(cons_2₂(atom_1₁(X), U), ._2₂(X, Y)) -> IF_FLATTEN_2_IN_1_GA₄(X, U, Y, flatten_2_in_ga₂(U, Y))
FLATTEN_2_IN_GA₂(cons_2₂(atom_1₁(X), U), ._2₂(X, Y)) -> FLATTEN_2_IN_GA₂(U, Y)
FLATTEN_2_IN_GA₂(cons_2₂(cons_2₂(U, V), W), X) -> IF_FLATTEN_2_IN_2_GA₅(U, V, W, X, flatten_2_in_ga₂(cons_2₂(U, cons_2₂(V, W)), X))
FLATTEN_2_IN_GA₂(cons_2₂(cons_2₂(U, V), W), X) -> FLATTEN_2_IN_GA₂(cons_2₂(U, cons_2₂(V, W)), X)
IF_COUNT_2_IN_2_GA₅(U, V, W, Z, flatten_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), X)) -> IF_COUNT_2_IN_3_GA₆(U, V, W, Z, X, count_2_in_ga₂(X, Z))
IF_COUNT_2_IN_2_GA₅(U, V, W, Z, flatten_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), X)) -> COUNT_2_IN_GA₂(X, Z)

The TRS R consists of the following rules:

count_2_in_ga₂(atom_1₁(X), s_1₁(0_0)) -> count_2_out_ga₂(atom_1₁(X), s_1₁(0_0))
count_2_in_ga₂(cons_2₂(atom_1₁(X), Y), s_1₁(Z)) -> if_count_2_in_1_ga₄(X, Y, Z, count_2_in_ga₂(Y, Z))
count_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), Z) -> if_count_2_in_2_ga₅(U, V, W, Z, flatten_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), X))
flatten_2_in_ga₂(atom_1₁(X), ._2₂(X, []_0)) -> flatten_2_out_ga₂(atom_1₁(X), ._2₂(X, []_0))
flatten_2_in_ga₂(cons_2₂(atom_1₁(X), U), ._2₂(X, Y)) -> if_flatten_2_in_1_ga₄(X, U, Y, flatten_2_in_ga₂(U, Y))
flatten_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), X) -> if_flatten_2_in_2_ga₅(U, V, W, X, flatten_2_in_ga₂(cons_2₂(U, cons_2₂(V, W)), X))
if_flatten_2_in_2_ga₅(U, V, W, X, flatten_2_out_ga₂(cons_2₂(U, cons_2₂(V, W)), X)) -> flatten_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), X)
if_flatten_2_in_1_ga₄(X, U, Y, flatten_2_out_ga₂(U, Y)) -> flatten_2_out_ga₂(cons_2₂(atom_1₁(X), U), ._2₂(X, Y))
if_count_2_in_2_ga₅(U, V, W, Z, flatten_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), X)) -> if_count_2_in_3_ga₆(U, V, W, Z, X, count_2_in_ga₂(X, Z))
if_count_2_in_3_ga₆(U, V, W, Z, X, count_2_out_ga₂(X, Z)) -> count_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), Z)
if_count_2_in_1_ga₄(X, Y, Z, count_2_out_ga₂(Y, Z)) -> count_2_out_ga₂(cons_2₂(atom_1₁(X), Y), s_1₁(Z))

The argument filtering Pi contains the following mapping:
count_2_in_ga₂(x1, x2) = count_2_in_ga₁(x1)
atom_1₁(x1) = atom_1₁(x1)
._2₂(x1, x2) = ._2₂(x1, x2)
[]_0 = []_0
cons_2₂(x1, x2) = cons_2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
0_0 = 0_0
count_2_out_ga₂(x1, x2) = count_2_out_ga₁(x2)
if_count_2_in_1_ga₄(x1, x2, x3, x4) = if_count_2_in_1_ga₁(x4)
if_count_2_in_2_ga₅(x1, x2, x3, x4, x5) = if_count_2_in_2_ga₁(x5)
flatten_2_in_ga₂(x1, x2) = flatten_2_in_ga₁(x1)
flatten_2_out_ga₂(x1, x2) = flatten_2_out_ga₁(x2)
if_flatten_2_in_1_ga₄(x1, x2, x3, x4) = if_flatten_2_in_1_ga₂(x1, x4)
if_flatten_2_in_2_ga₅(x1, x2, x3, x4, x5) = if_flatten_2_in_2_ga₁(x5)
if_count_2_in_3_ga₆(x1, x2, x3, x4, x5, x6) = if_count_2_in_3_ga₁(x6)
FLATTEN_2_IN_GA₂(x1, x2) = FLATTEN_2_IN_GA₁(x1)
IF_FLATTEN_2_IN_2_GA₅(x1, x2, x3, x4, x5) = IF_FLATTEN_2_IN_2_GA₁(x5)
IF_COUNT_2_IN_1_GA₄(x1, x2, x3, x4) = IF_COUNT_2_IN_1_GA₁(x4)
IF_FLATTEN_2_IN_1_GA₄(x1, x2, x3, x4) = IF_FLATTEN_2_IN_1_GA₂(x1, x4)
COUNT_2_IN_GA₂(x1, x2) = COUNT_2_IN_GA₁(x1)
IF_COUNT_2_IN_2_GA₅(x1, x2, x3, x4, x5) = IF_COUNT_2_IN_2_GA₁(x5)
IF_COUNT_2_IN_3_GA₆(x1, x2, x3, x4, x5, x6) = IF_COUNT_2_IN_3_GA₁(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:

COUNT_2_IN_GA₂(cons_2₂(atom_1₁(X), Y), s_1₁(Z)) -> IF_COUNT_2_IN_1_GA₄(X, Y, Z, count_2_in_ga₂(Y, Z))
COUNT_2_IN_GA₂(cons_2₂(atom_1₁(X), Y), s_1₁(Z)) -> COUNT_2_IN_GA₂(Y, Z)
COUNT_2_IN_GA₂(cons_2₂(cons_2₂(U, V), W), Z) -> IF_COUNT_2_IN_2_GA₅(U, V, W, Z, flatten_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), X))
COUNT_2_IN_GA₂(cons_2₂(cons_2₂(U, V), W), Z) -> FLATTEN_2_IN_GA₂(cons_2₂(cons_2₂(U, V), W), X)
FLATTEN_2_IN_GA₂(cons_2₂(atom_1₁(X), U), ._2₂(X, Y)) -> IF_FLATTEN_2_IN_1_GA₄(X, U, Y, flatten_2_in_ga₂(U, Y))
FLATTEN_2_IN_GA₂(cons_2₂(atom_1₁(X), U), ._2₂(X, Y)) -> FLATTEN_2_IN_GA₂(U, Y)
FLATTEN_2_IN_GA₂(cons_2₂(cons_2₂(U, V), W), X) -> IF_FLATTEN_2_IN_2_GA₅(U, V, W, X, flatten_2_in_ga₂(cons_2₂(U, cons_2₂(V, W)), X))
FLATTEN_2_IN_GA₂(cons_2₂(cons_2₂(U, V), W), X) -> FLATTEN_2_IN_GA₂(cons_2₂(U, cons_2₂(V, W)), X)
IF_COUNT_2_IN_2_GA₅(U, V, W, Z, flatten_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), X)) -> IF_COUNT_2_IN_3_GA₆(U, V, W, Z, X, count_2_in_ga₂(X, Z))
IF_COUNT_2_IN_2_GA₅(U, V, W, Z, flatten_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), X)) -> COUNT_2_IN_GA₂(X, Z)

The TRS R consists of the following rules:

count_2_in_ga₂(atom_1₁(X), s_1₁(0_0)) -> count_2_out_ga₂(atom_1₁(X), s_1₁(0_0))
count_2_in_ga₂(cons_2₂(atom_1₁(X), Y), s_1₁(Z)) -> if_count_2_in_1_ga₄(X, Y, Z, count_2_in_ga₂(Y, Z))
count_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), Z) -> if_count_2_in_2_ga₅(U, V, W, Z, flatten_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), X))
flatten_2_in_ga₂(atom_1₁(X), ._2₂(X, []_0)) -> flatten_2_out_ga₂(atom_1₁(X), ._2₂(X, []_0))
flatten_2_in_ga₂(cons_2₂(atom_1₁(X), U), ._2₂(X, Y)) -> if_flatten_2_in_1_ga₄(X, U, Y, flatten_2_in_ga₂(U, Y))
flatten_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), X) -> if_flatten_2_in_2_ga₅(U, V, W, X, flatten_2_in_ga₂(cons_2₂(U, cons_2₂(V, W)), X))
if_flatten_2_in_2_ga₅(U, V, W, X, flatten_2_out_ga₂(cons_2₂(U, cons_2₂(V, W)), X)) -> flatten_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), X)
if_flatten_2_in_1_ga₄(X, U, Y, flatten_2_out_ga₂(U, Y)) -> flatten_2_out_ga₂(cons_2₂(atom_1₁(X), U), ._2₂(X, Y))
if_count_2_in_2_ga₅(U, V, W, Z, flatten_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), X)) -> if_count_2_in_3_ga₆(U, V, W, Z, X, count_2_in_ga₂(X, Z))
if_count_2_in_3_ga₆(U, V, W, Z, X, count_2_out_ga₂(X, Z)) -> count_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), Z)
if_count_2_in_1_ga₄(X, Y, Z, count_2_out_ga₂(Y, Z)) -> count_2_out_ga₂(cons_2₂(atom_1₁(X), Y), s_1₁(Z))

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

FLATTEN_2_IN_GA₂(cons_2₂(atom_1₁(X), U), ._2₂(X, Y)) -> FLATTEN_2_IN_GA₂(U, Y)
FLATTEN_2_IN_GA₂(cons_2₂(cons_2₂(U, V), W), X) -> FLATTEN_2_IN_GA₂(cons_2₂(U, cons_2₂(V, W)), X)

The TRS R consists of the following rules:

count_2_in_ga₂(atom_1₁(X), s_1₁(0_0)) -> count_2_out_ga₂(atom_1₁(X), s_1₁(0_0))
count_2_in_ga₂(cons_2₂(atom_1₁(X), Y), s_1₁(Z)) -> if_count_2_in_1_ga₄(X, Y, Z, count_2_in_ga₂(Y, Z))
count_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), Z) -> if_count_2_in_2_ga₅(U, V, W, Z, flatten_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), X))
flatten_2_in_ga₂(atom_1₁(X), ._2₂(X, []_0)) -> flatten_2_out_ga₂(atom_1₁(X), ._2₂(X, []_0))
flatten_2_in_ga₂(cons_2₂(atom_1₁(X), U), ._2₂(X, Y)) -> if_flatten_2_in_1_ga₄(X, U, Y, flatten_2_in_ga₂(U, Y))
flatten_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), X) -> if_flatten_2_in_2_ga₅(U, V, W, X, flatten_2_in_ga₂(cons_2₂(U, cons_2₂(V, W)), X))
if_flatten_2_in_2_ga₅(U, V, W, X, flatten_2_out_ga₂(cons_2₂(U, cons_2₂(V, W)), X)) -> flatten_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), X)
if_flatten_2_in_1_ga₄(X, U, Y, flatten_2_out_ga₂(U, Y)) -> flatten_2_out_ga₂(cons_2₂(atom_1₁(X), U), ._2₂(X, Y))
if_count_2_in_2_ga₅(U, V, W, Z, flatten_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), X)) -> if_count_2_in_3_ga₆(U, V, W, Z, X, count_2_in_ga₂(X, Z))
if_count_2_in_3_ga₆(U, V, W, Z, X, count_2_out_ga₂(X, Z)) -> count_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), Z)
if_count_2_in_1_ga₄(X, Y, Z, count_2_out_ga₂(Y, Z)) -> count_2_out_ga₂(cons_2₂(atom_1₁(X), Y), s_1₁(Z))

The argument filtering Pi contains the following mapping:
count_2_in_ga₂(x1, x2) = count_2_in_ga₁(x1)
atom_1₁(x1) = atom_1₁(x1)
._2₂(x1, x2) = ._2₂(x1, x2)
[]_0 = []_0
cons_2₂(x1, x2) = cons_2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
0_0 = 0_0
count_2_out_ga₂(x1, x2) = count_2_out_ga₁(x2)
if_count_2_in_1_ga₄(x1, x2, x3, x4) = if_count_2_in_1_ga₁(x4)
if_count_2_in_2_ga₅(x1, x2, x3, x4, x5) = if_count_2_in_2_ga₁(x5)
flatten_2_in_ga₂(x1, x2) = flatten_2_in_ga₁(x1)
flatten_2_out_ga₂(x1, x2) = flatten_2_out_ga₁(x2)
if_flatten_2_in_1_ga₄(x1, x2, x3, x4) = if_flatten_2_in_1_ga₂(x1, x4)
if_flatten_2_in_2_ga₅(x1, x2, x3, x4, x5) = if_flatten_2_in_2_ga₁(x5)
if_count_2_in_3_ga₆(x1, x2, x3, x4, x5, x6) = if_count_2_in_3_ga₁(x6)
FLATTEN_2_IN_GA₂(x1, x2) = FLATTEN_2_IN_GA₁(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

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

FLATTEN_2_IN_GA₂(cons_2₂(atom_1₁(X), U), ._2₂(X, Y)) -> FLATTEN_2_IN_GA₂(U, Y)
FLATTEN_2_IN_GA₂(cons_2₂(cons_2₂(U, V), W), X) -> FLATTEN_2_IN_GA₂(cons_2₂(U, cons_2₂(V, W)), X)

R is empty.
The argument filtering Pi contains the following mapping:
atom_1₁(x1) = atom_1₁(x1)
._2₂(x1, x2) = ._2₂(x1, x2)
cons_2₂(x1, x2) = cons_2₂(x1, x2)
FLATTEN_2_IN_GA₂(x1, x2) = FLATTEN_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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPPoloProof

              ↳ PiDP

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

FLATTEN_2_IN_GA₁(cons_2₂(atom_1₁(X), U)) -> FLATTEN_2_IN_GA₁(U)
FLATTEN_2_IN_GA₁(cons_2₂(cons_2₂(U, V), W)) -> FLATTEN_2_IN_GA₁(cons_2₂(U, cons_2₂(V, W)))

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {FLATTEN_2_IN_GA₁}.
By using a polynomial ordering, the following set of Dependency Pairs of this DP problem can be strictly oriented.

FLATTEN_2_IN_GA₁(cons_2₂(atom_1₁(X), U)) -> FLATTEN_2_IN_GA₁(U)

The remaining Dependency Pairs were at least non-strictly be oriented.

FLATTEN_2_IN_GA₁(cons_2₂(cons_2₂(U, V), W)) -> FLATTEN_2_IN_GA₁(cons_2₂(U, cons_2₂(V, W)))

With the implicit AFS there is no usable rule.

Used ordering: POLO with Polynomial interpretation:

POL(FLATTEN_2_IN_GA₁(x₁)) = x₁
POL(atom_1₁(x₁)) = 1 + x₁
POL(cons_2₂(x₁, x₂)) = x₁ + x₂

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPPoloProof

                          ↳ QDP

                            ↳ QDPSizeChangeProof

              ↳ PiDP

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

FLATTEN_2_IN_GA₁(cons_2₂(cons_2₂(U, V), W)) -> FLATTEN_2_IN_GA₁(cons_2₂(U, cons_2₂(V, W)))

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {FLATTEN_2_IN_GA₁}.
We used the following order and afs together with the size-change analysis to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
cons_2₂(x1, x2) = cons_2₁(x1)

From the DPs we obtained the following set of size-change graphs:

FLATTEN_2_IN_GA₁(cons_2₂(cons_2₂(U, V), W)) -> FLATTEN_2_IN_GA₁(cons_2₂(U, cons_2₂(V, W))) (allowed arguments on rhs = {1})
The graph contains the following edges 1 > 1

We oriented the following set of usable rules. none

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

IF_COUNT_2_IN_2_GA₅(U, V, W, Z, flatten_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), X)) -> COUNT_2_IN_GA₂(X, Z)
COUNT_2_IN_GA₂(cons_2₂(atom_1₁(X), Y), s_1₁(Z)) -> COUNT_2_IN_GA₂(Y, Z)
COUNT_2_IN_GA₂(cons_2₂(cons_2₂(U, V), W), Z) -> IF_COUNT_2_IN_2_GA₅(U, V, W, Z, flatten_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), X))

The TRS R consists of the following rules:

count_2_in_ga₂(atom_1₁(X), s_1₁(0_0)) -> count_2_out_ga₂(atom_1₁(X), s_1₁(0_0))
count_2_in_ga₂(cons_2₂(atom_1₁(X), Y), s_1₁(Z)) -> if_count_2_in_1_ga₄(X, Y, Z, count_2_in_ga₂(Y, Z))
count_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), Z) -> if_count_2_in_2_ga₅(U, V, W, Z, flatten_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), X))
flatten_2_in_ga₂(atom_1₁(X), ._2₂(X, []_0)) -> flatten_2_out_ga₂(atom_1₁(X), ._2₂(X, []_0))
flatten_2_in_ga₂(cons_2₂(atom_1₁(X), U), ._2₂(X, Y)) -> if_flatten_2_in_1_ga₄(X, U, Y, flatten_2_in_ga₂(U, Y))
flatten_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), X) -> if_flatten_2_in_2_ga₅(U, V, W, X, flatten_2_in_ga₂(cons_2₂(U, cons_2₂(V, W)), X))
if_flatten_2_in_2_ga₅(U, V, W, X, flatten_2_out_ga₂(cons_2₂(U, cons_2₂(V, W)), X)) -> flatten_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), X)
if_flatten_2_in_1_ga₄(X, U, Y, flatten_2_out_ga₂(U, Y)) -> flatten_2_out_ga₂(cons_2₂(atom_1₁(X), U), ._2₂(X, Y))
if_count_2_in_2_ga₅(U, V, W, Z, flatten_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), X)) -> if_count_2_in_3_ga₆(U, V, W, Z, X, count_2_in_ga₂(X, Z))
if_count_2_in_3_ga₆(U, V, W, Z, X, count_2_out_ga₂(X, Z)) -> count_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), Z)
if_count_2_in_1_ga₄(X, Y, Z, count_2_out_ga₂(Y, Z)) -> count_2_out_ga₂(cons_2₂(atom_1₁(X), Y), s_1₁(Z))

The argument filtering Pi contains the following mapping:
count_2_in_ga₂(x1, x2) = count_2_in_ga₁(x1)
atom_1₁(x1) = atom_1₁(x1)
._2₂(x1, x2) = ._2₂(x1, x2)
[]_0 = []_0
cons_2₂(x1, x2) = cons_2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
0_0 = 0_0
count_2_out_ga₂(x1, x2) = count_2_out_ga₁(x2)
if_count_2_in_1_ga₄(x1, x2, x3, x4) = if_count_2_in_1_ga₁(x4)
if_count_2_in_2_ga₅(x1, x2, x3, x4, x5) = if_count_2_in_2_ga₁(x5)
flatten_2_in_ga₂(x1, x2) = flatten_2_in_ga₁(x1)
flatten_2_out_ga₂(x1, x2) = flatten_2_out_ga₁(x2)
if_flatten_2_in_1_ga₄(x1, x2, x3, x4) = if_flatten_2_in_1_ga₂(x1, x4)
if_flatten_2_in_2_ga₅(x1, x2, x3, x4, x5) = if_flatten_2_in_2_ga₁(x5)
if_count_2_in_3_ga₆(x1, x2, x3, x4, x5, x6) = if_count_2_in_3_ga₁(x6)
COUNT_2_IN_GA₂(x1, x2) = COUNT_2_IN_GA₁(x1)
IF_COUNT_2_IN_2_GA₅(x1, x2, x3, x4, x5) = IF_COUNT_2_IN_2_GA₁(x5)

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

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

IF_COUNT_2_IN_2_GA₅(U, V, W, Z, flatten_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), X)) -> COUNT_2_IN_GA₂(X, Z)
COUNT_2_IN_GA₂(cons_2₂(atom_1₁(X), Y), s_1₁(Z)) -> COUNT_2_IN_GA₂(Y, Z)
COUNT_2_IN_GA₂(cons_2₂(cons_2₂(U, V), W), Z) -> IF_COUNT_2_IN_2_GA₅(U, V, W, Z, flatten_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), X))

The TRS R consists of the following rules:

flatten_2_in_ga₂(cons_2₂(cons_2₂(U, V), W), X) -> if_flatten_2_in_2_ga₅(U, V, W, X, flatten_2_in_ga₂(cons_2₂(U, cons_2₂(V, W)), X))
if_flatten_2_in_2_ga₅(U, V, W, X, flatten_2_out_ga₂(cons_2₂(U, cons_2₂(V, W)), X)) -> flatten_2_out_ga₂(cons_2₂(cons_2₂(U, V), W), X)
flatten_2_in_ga₂(cons_2₂(atom_1₁(X), U), ._2₂(X, Y)) -> if_flatten_2_in_1_ga₄(X, U, Y, flatten_2_in_ga₂(U, Y))
if_flatten_2_in_1_ga₄(X, U, Y, flatten_2_out_ga₂(U, Y)) -> flatten_2_out_ga₂(cons_2₂(atom_1₁(X), U), ._2₂(X, Y))
flatten_2_in_ga₂(atom_1₁(X), ._2₂(X, []_0)) -> flatten_2_out_ga₂(atom_1₁(X), ._2₂(X, []_0))

The argument filtering Pi contains the following mapping:
atom_1₁(x1) = atom_1₁(x1)
._2₂(x1, x2) = ._2₂(x1, x2)
[]_0 = []_0
cons_2₂(x1, x2) = cons_2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
flatten_2_in_ga₂(x1, x2) = flatten_2_in_ga₁(x1)
flatten_2_out_ga₂(x1, x2) = flatten_2_out_ga₁(x2)
if_flatten_2_in_1_ga₄(x1, x2, x3, x4) = if_flatten_2_in_1_ga₂(x1, x4)
if_flatten_2_in_2_ga₅(x1, x2, x3, x4, x5) = if_flatten_2_in_2_ga₁(x5)
COUNT_2_IN_GA₂(x1, x2) = COUNT_2_IN_GA₁(x1)
IF_COUNT_2_IN_2_GA₅(x1, x2, x3, x4, x5) = IF_COUNT_2_IN_2_GA₁(x5)

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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

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

IF_COUNT_2_IN_2_GA₁(flatten_2_out_ga₁(X)) -> COUNT_2_IN_GA₁(X)
COUNT_2_IN_GA₁(cons_2₂(atom_1₁(X), Y)) -> COUNT_2_IN_GA₁(Y)
COUNT_2_IN_GA₁(cons_2₂(cons_2₂(U, V), W)) -> IF_COUNT_2_IN_2_GA₁(flatten_2_in_ga₁(cons_2₂(cons_2₂(U, V), W)))

The TRS R consists of the following rules:

flatten_2_in_ga₁(cons_2₂(cons_2₂(U, V), W)) -> if_flatten_2_in_2_ga₁(flatten_2_in_ga₁(cons_2₂(U, cons_2₂(V, W))))
if_flatten_2_in_2_ga₁(flatten_2_out_ga₁(X)) -> flatten_2_out_ga₁(X)
flatten_2_in_ga₁(cons_2₂(atom_1₁(X), U)) -> if_flatten_2_in_1_ga₂(X, flatten_2_in_ga₁(U))
if_flatten_2_in_1_ga₂(X, flatten_2_out_ga₁(Y)) -> flatten_2_out_ga₁(._2₂(X, Y))
flatten_2_in_ga₁(atom_1₁(X)) -> flatten_2_out_ga₁(._2₂(X, []_0))

The set Q consists of the following terms:

flatten_2_in_ga₁(x0)
if_flatten_2_in_2_ga₁(x0)
if_flatten_2_in_1_ga₂(x0, x1)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {COUNT_2_IN_GA₁, IF_COUNT_2_IN_2_GA₁}.
By using a polynomial ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

COUNT_2_IN_GA₁(cons_2₂(atom_1₁(X), Y)) -> COUNT_2_IN_GA₁(Y)
COUNT_2_IN_GA₁(cons_2₂(cons_2₂(U, V), W)) -> IF_COUNT_2_IN_2_GA₁(flatten_2_in_ga₁(cons_2₂(cons_2₂(U, V), W)))

Strictly oriented rules of the TRS R:

flatten_2_in_ga₁(cons_2₂(cons_2₂(U, V), W)) -> if_flatten_2_in_2_ga₁(flatten_2_in_ga₁(cons_2₂(U, cons_2₂(V, W))))
if_flatten_2_in_2_ga₁(flatten_2_out_ga₁(X)) -> flatten_2_out_ga₁(X)
flatten_2_in_ga₁(cons_2₂(atom_1₁(X), U)) -> if_flatten_2_in_1_ga₂(X, flatten_2_in_ga₁(U))
flatten_2_in_ga₁(atom_1₁(X)) -> flatten_2_out_ga₁(._2₂(X, []_0))

Used ordering: POLO with Polynomial interpretation:

POL(COUNT_2_IN_GA₁(x₁)) = 2·x₁
POL(flatten_2_in_ga₁(x₁)) = 1 + x₁
POL(flatten_2_out_ga₁(x₁)) = 2·x₁
POL(._2₂(x₁, x₂)) = x₁ + x₂
POL(if_flatten_2_in_2_ga₁(x₁)) = 1 + x₁
POL(atom_1₁(x₁)) = 2·x₁
POL([]_0) = 0
POL(IF_COUNT_2_IN_2_GA₁(x₁)) = x₁
POL(cons_2₂(x₁, x₂)) = 2 + 2·x₁ + x₂
POL(if_flatten_2_in_1_ga₂(x₁, x₂)) = 2·x₁ + x₂

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ DependencyGraphProof

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

IF_COUNT_2_IN_2_GA₁(flatten_2_out_ga₁(X)) -> COUNT_2_IN_GA₁(X)

The TRS R consists of the following rules:

if_flatten_2_in_1_ga₂(X, flatten_2_out_ga₁(Y)) -> flatten_2_out_ga₁(._2₂(X, Y))

The set Q consists of the following terms:

flatten_2_in_ga₁(x0)
if_flatten_2_in_2_ga₁(x0)
if_flatten_2_in_1_ga₂(x0, x1)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {COUNT_2_IN_GA₁, IF_COUNT_2_IN_2_GA₁}.
The approximation of the Dependency Graph contains 1 SCC.

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

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

COUNT_2_IN_GA₁(cons_2₂(atom_1₁(X), Y)) -> COUNT_2_IN_GA₁(Y)

The TRS R consists of the following rules:

if_flatten_2_in_1_ga₂(X, flatten_2_out_ga₁(Y)) -> flatten_2_out_ga₁(._2₂(X, Y))

The set Q consists of the following terms:

flatten_2_in_ga₁(x0)
if_flatten_2_in_2_ga₁(x0)
if_flatten_2_in_1_ga₂(x0, x1)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {COUNT_2_IN_GA₁}.
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:

COUNT_2_IN_GA₁(cons_2₂(atom_1₁(X), Y)) -> COUNT_2_IN_GA₁(Y)
The graph contains the following edges 1 > 1