Left Termination proof of /tmp/tmpTniAp0/flat.pl

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

↳ PROLOG

  ↳ PrologToPiTRSProof

flat₂(niltree₀, nil₀).
flat₂(tree₃(X, niltree₀, T), cons₂(X, Xs)) :- flat₂(T, Xs).
flat₂(tree₃(X, tree₃(Y, T1, T2), T3), Xs) :- flat₂(tree₃(Y, T1, tree₃(X, T2, T3)), Xs).

With regard to the inferred argument filtering the predicates were used in the following modes:
flat₂: (b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

flat_2_in_ga₂(niltree_0, nil_0) -> flat_2_out_ga₂(niltree_0, nil_0)
flat_2_in_ga₂(tree_3₃(X, niltree_0, T), cons_2₂(X, Xs)) -> if_flat_2_in_1_ga₄(X, T, Xs, flat_2_in_ga₂(T, Xs))
flat_2_in_ga₂(tree_3₃(X, tree_3₃(Y, T1, T2), T3), Xs) -> if_flat_2_in_2_ga₇(X, Y, T1, T2, T3, Xs, flat_2_in_ga₂(tree_3₃(Y, T1, tree_3₃(X, T2, T3)), Xs))
if_flat_2_in_2_ga₇(X, Y, T1, T2, T3, Xs, flat_2_out_ga₂(tree_3₃(Y, T1, tree_3₃(X, T2, T3)), Xs)) -> flat_2_out_ga₂(tree_3₃(X, tree_3₃(Y, T1, T2), T3), Xs)
if_flat_2_in_1_ga₄(X, T, Xs, flat_2_out_ga₂(T, Xs)) -> flat_2_out_ga₂(tree_3₃(X, niltree_0, T), cons_2₂(X, Xs))

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:

flat_2_in_ga₂(niltree_0, nil_0) -> flat_2_out_ga₂(niltree_0, nil_0)
flat_2_in_ga₂(tree_3₃(X, niltree_0, T), cons_2₂(X, Xs)) -> if_flat_2_in_1_ga₄(X, T, Xs, flat_2_in_ga₂(T, Xs))
flat_2_in_ga₂(tree_3₃(X, tree_3₃(Y, T1, T2), T3), Xs) -> if_flat_2_in_2_ga₇(X, Y, T1, T2, T3, Xs, flat_2_in_ga₂(tree_3₃(Y, T1, tree_3₃(X, T2, T3)), Xs))
if_flat_2_in_2_ga₇(X, Y, T1, T2, T3, Xs, flat_2_out_ga₂(tree_3₃(Y, T1, tree_3₃(X, T2, T3)), Xs)) -> flat_2_out_ga₂(tree_3₃(X, tree_3₃(Y, T1, T2), T3), Xs)
if_flat_2_in_1_ga₄(X, T, Xs, flat_2_out_ga₂(T, Xs)) -> flat_2_out_ga₂(tree_3₃(X, niltree_0, T), cons_2₂(X, Xs))

FLAT_2_IN_GA₂(tree_3₃(X, niltree_0, T), cons_2₂(X, Xs)) -> IF_FLAT_2_IN_1_GA₄(X, T, Xs, flat_2_in_ga₂(T, Xs))
FLAT_2_IN_GA₂(tree_3₃(X, niltree_0, T), cons_2₂(X, Xs)) -> FLAT_2_IN_GA₂(T, Xs)
FLAT_2_IN_GA₂(tree_3₃(X, tree_3₃(Y, T1, T2), T3), Xs) -> IF_FLAT_2_IN_2_GA₇(X, Y, T1, T2, T3, Xs, flat_2_in_ga₂(tree_3₃(Y, T1, tree_3₃(X, T2, T3)), Xs))
FLAT_2_IN_GA₂(tree_3₃(X, tree_3₃(Y, T1, T2), T3), Xs) -> FLAT_2_IN_GA₂(tree_3₃(Y, T1, tree_3₃(X, T2, T3)), Xs)

The TRS R consists of the following rules:

flat_2_in_ga₂(niltree_0, nil_0) -> flat_2_out_ga₂(niltree_0, nil_0)
flat_2_in_ga₂(tree_3₃(X, niltree_0, T), cons_2₂(X, Xs)) -> if_flat_2_in_1_ga₄(X, T, Xs, flat_2_in_ga₂(T, Xs))
flat_2_in_ga₂(tree_3₃(X, tree_3₃(Y, T1, T2), T3), Xs) -> if_flat_2_in_2_ga₇(X, Y, T1, T2, T3, Xs, flat_2_in_ga₂(tree_3₃(Y, T1, tree_3₃(X, T2, T3)), Xs))
if_flat_2_in_2_ga₇(X, Y, T1, T2, T3, Xs, flat_2_out_ga₂(tree_3₃(Y, T1, tree_3₃(X, T2, T3)), Xs)) -> flat_2_out_ga₂(tree_3₃(X, tree_3₃(Y, T1, T2), T3), Xs)
if_flat_2_in_1_ga₄(X, T, Xs, flat_2_out_ga₂(T, Xs)) -> flat_2_out_ga₂(tree_3₃(X, niltree_0, T), cons_2₂(X, Xs))

The argument filtering Pi contains the following mapping:
flat_2_in_ga₂(x1, x2) = flat_2_in_ga₁(x1)
niltree_0 = niltree_0
nil_0 = nil_0
tree_3₃(x1, x2, x3) = tree_3₃(x1, x2, x3)
cons_2₂(x1, x2) = cons_2₂(x1, x2)
flat_2_out_ga₂(x1, x2) = flat_2_out_ga₁(x2)
if_flat_2_in_1_ga₄(x1, x2, x3, x4) = if_flat_2_in_1_ga₂(x1, x4)
if_flat_2_in_2_ga₇(x1, x2, x3, x4, x5, x6, x7) = if_flat_2_in_2_ga₁(x7)
FLAT_2_IN_GA₂(x1, x2) = FLAT_2_IN_GA₁(x1)
IF_FLAT_2_IN_1_GA₄(x1, x2, x3, x4) = IF_FLAT_2_IN_1_GA₂(x1, x4)
IF_FLAT_2_IN_2_GA₇(x1, x2, x3, x4, x5, x6, x7) = IF_FLAT_2_IN_2_GA₁(x7)

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:

FLAT_2_IN_GA₂(tree_3₃(X, niltree_0, T), cons_2₂(X, Xs)) -> IF_FLAT_2_IN_1_GA₄(X, T, Xs, flat_2_in_ga₂(T, Xs))
FLAT_2_IN_GA₂(tree_3₃(X, niltree_0, T), cons_2₂(X, Xs)) -> FLAT_2_IN_GA₂(T, Xs)
FLAT_2_IN_GA₂(tree_3₃(X, tree_3₃(Y, T1, T2), T3), Xs) -> IF_FLAT_2_IN_2_GA₇(X, Y, T1, T2, T3, Xs, flat_2_in_ga₂(tree_3₃(Y, T1, tree_3₃(X, T2, T3)), Xs))
FLAT_2_IN_GA₂(tree_3₃(X, tree_3₃(Y, T1, T2), T3), Xs) -> FLAT_2_IN_GA₂(tree_3₃(Y, T1, tree_3₃(X, T2, T3)), Xs)

The TRS R consists of the following rules:

flat_2_in_ga₂(niltree_0, nil_0) -> flat_2_out_ga₂(niltree_0, nil_0)
flat_2_in_ga₂(tree_3₃(X, niltree_0, T), cons_2₂(X, Xs)) -> if_flat_2_in_1_ga₄(X, T, Xs, flat_2_in_ga₂(T, Xs))
flat_2_in_ga₂(tree_3₃(X, tree_3₃(Y, T1, T2), T3), Xs) -> if_flat_2_in_2_ga₇(X, Y, T1, T2, T3, Xs, flat_2_in_ga₂(tree_3₃(Y, T1, tree_3₃(X, T2, T3)), Xs))
if_flat_2_in_2_ga₇(X, Y, T1, T2, T3, Xs, flat_2_out_ga₂(tree_3₃(Y, T1, tree_3₃(X, T2, T3)), Xs)) -> flat_2_out_ga₂(tree_3₃(X, tree_3₃(Y, T1, T2), T3), Xs)
if_flat_2_in_1_ga₄(X, T, Xs, flat_2_out_ga₂(T, Xs)) -> flat_2_out_ga₂(tree_3₃(X, niltree_0, T), cons_2₂(X, Xs))

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

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

FLAT_2_IN_GA₂(tree_3₃(X, niltree_0, T), cons_2₂(X, Xs)) -> FLAT_2_IN_GA₂(T, Xs)
FLAT_2_IN_GA₂(tree_3₃(X, tree_3₃(Y, T1, T2), T3), Xs) -> FLAT_2_IN_GA₂(tree_3₃(Y, T1, tree_3₃(X, T2, T3)), Xs)

The TRS R consists of the following rules:

flat_2_in_ga₂(niltree_0, nil_0) -> flat_2_out_ga₂(niltree_0, nil_0)
flat_2_in_ga₂(tree_3₃(X, niltree_0, T), cons_2₂(X, Xs)) -> if_flat_2_in_1_ga₄(X, T, Xs, flat_2_in_ga₂(T, Xs))
flat_2_in_ga₂(tree_3₃(X, tree_3₃(Y, T1, T2), T3), Xs) -> if_flat_2_in_2_ga₇(X, Y, T1, T2, T3, Xs, flat_2_in_ga₂(tree_3₃(Y, T1, tree_3₃(X, T2, T3)), Xs))
if_flat_2_in_2_ga₇(X, Y, T1, T2, T3, Xs, flat_2_out_ga₂(tree_3₃(Y, T1, tree_3₃(X, T2, T3)), Xs)) -> flat_2_out_ga₂(tree_3₃(X, tree_3₃(Y, T1, T2), T3), Xs)
if_flat_2_in_1_ga₄(X, T, Xs, flat_2_out_ga₂(T, Xs)) -> flat_2_out_ga₂(tree_3₃(X, niltree_0, T), cons_2₂(X, Xs))

The argument filtering Pi contains the following mapping:
flat_2_in_ga₂(x1, x2) = flat_2_in_ga₁(x1)
niltree_0 = niltree_0
nil_0 = nil_0
tree_3₃(x1, x2, x3) = tree_3₃(x1, x2, x3)
cons_2₂(x1, x2) = cons_2₂(x1, x2)
flat_2_out_ga₂(x1, x2) = flat_2_out_ga₁(x2)
if_flat_2_in_1_ga₄(x1, x2, x3, x4) = if_flat_2_in_1_ga₂(x1, x4)
if_flat_2_in_2_ga₇(x1, x2, x3, x4, x5, x6, x7) = if_flat_2_in_2_ga₁(x7)
FLAT_2_IN_GA₂(x1, x2) = FLAT_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

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

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

FLAT_2_IN_GA₂(tree_3₃(X, niltree_0, T), cons_2₂(X, Xs)) -> FLAT_2_IN_GA₂(T, Xs)
FLAT_2_IN_GA₂(tree_3₃(X, tree_3₃(Y, T1, T2), T3), Xs) -> FLAT_2_IN_GA₂(tree_3₃(Y, T1, tree_3₃(X, T2, T3)), Xs)

R is empty.
The argument filtering Pi contains the following mapping:
niltree_0 = niltree_0
tree_3₃(x1, x2, x3) = tree_3₃(x1, x2, x3)
cons_2₂(x1, x2) = cons_2₂(x1, x2)
FLAT_2_IN_GA₂(x1, x2) = FLAT_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

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

                    ↳ QDP

                      ↳ RuleRemovalProof

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

FLAT_2_IN_GA₁(tree_3₃(X, niltree_0, T)) -> FLAT_2_IN_GA₁(T)
FLAT_2_IN_GA₁(tree_3₃(X, tree_3₃(Y, T1, T2), T3)) -> FLAT_2_IN_GA₁(tree_3₃(Y, T1, tree_3₃(X, T2, T3)))

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {FLAT_2_IN_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:

FLAT_2_IN_GA₁(tree_3₃(X, niltree_0, T)) -> FLAT_2_IN_GA₁(T)

Used ordering: POLO with Polynomial interpretation:

POL(niltree_0) = 1
POL(FLAT_2_IN_GA₁(x₁)) = 1 + x₁
POL(tree_3₃(x₁, x₂, x₃)) = x₁ + x₂ + x₃

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ QDPSizeChangeProof

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

FLAT_2_IN_GA₁(tree_3₃(X, tree_3₃(Y, T1, T2), T3)) -> FLAT_2_IN_GA₁(tree_3₃(Y, T1, tree_3₃(X, T2, T3)))

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {FLAT_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:
tree_3₃(x1, x2, x3) = tree_3₁(x2)

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

FLAT_2_IN_GA₁(tree_3₃(X, tree_3₃(Y, T1, T2), T3)) -> FLAT_2_IN_GA₁(tree_3₃(Y, T1, tree_3₃(X, T2, T3))) (allowed arguments on rhs = {1})
The graph contains the following edges 1 > 1

We oriented the following set of usable rules. none