Left Termination proof of /tmp/tmpQBa1-i/flat.pl

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

↳ PROLOG

  ↳ PrologToPiTRSProof

right₂(tree₃(X, XS1, XS2), XS2).
flat₂(niltree₀, nil₀).
flat₂(tree₃(X, niltree₀, XS), cons₂(X, YS)) :- right₂(tree₃(X, niltree₀, XS), ZS), flat₂(ZS, YS).
flat₂(tree₃(X, tree₃(Y, YS1, YS2), XS), ZS) :- flat₂(tree₃(Y, YS1, tree₃(X, YS2, XS)), ZS).

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, XS), cons_2₂(X, YS)) -> if_flat_2_in_1_ga₄(X, XS, YS, right_2_in_ga₂(tree_3₃(X, niltree_0, XS), ZS))
right_2_in_ga₂(tree_3₃(X, XS1, XS2), XS2) -> right_2_out_ga₂(tree_3₃(X, XS1, XS2), XS2)
if_flat_2_in_1_ga₄(X, XS, YS, right_2_out_ga₂(tree_3₃(X, niltree_0, XS), ZS)) -> if_flat_2_in_2_ga₅(X, XS, YS, ZS, flat_2_in_ga₂(ZS, YS))
flat_2_in_ga₂(tree_3₃(X, tree_3₃(Y, YS1, YS2), XS), ZS) -> if_flat_2_in_3_ga₇(X, Y, YS1, YS2, XS, ZS, flat_2_in_ga₂(tree_3₃(Y, YS1, tree_3₃(X, YS2, XS)), ZS))
if_flat_2_in_3_ga₇(X, Y, YS1, YS2, XS, ZS, flat_2_out_ga₂(tree_3₃(Y, YS1, tree_3₃(X, YS2, XS)), ZS)) -> flat_2_out_ga₂(tree_3₃(X, tree_3₃(Y, YS1, YS2), XS), ZS)
if_flat_2_in_2_ga₅(X, XS, YS, ZS, flat_2_out_ga₂(ZS, YS)) -> flat_2_out_ga₂(tree_3₃(X, niltree_0, XS), cons_2₂(X, YS))

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, XS), cons_2₂(X, YS)) -> if_flat_2_in_1_ga₄(X, XS, YS, right_2_in_ga₂(tree_3₃(X, niltree_0, XS), ZS))
right_2_in_ga₂(tree_3₃(X, XS1, XS2), XS2) -> right_2_out_ga₂(tree_3₃(X, XS1, XS2), XS2)
if_flat_2_in_1_ga₄(X, XS, YS, right_2_out_ga₂(tree_3₃(X, niltree_0, XS), ZS)) -> if_flat_2_in_2_ga₅(X, XS, YS, ZS, flat_2_in_ga₂(ZS, YS))
flat_2_in_ga₂(tree_3₃(X, tree_3₃(Y, YS1, YS2), XS), ZS) -> if_flat_2_in_3_ga₇(X, Y, YS1, YS2, XS, ZS, flat_2_in_ga₂(tree_3₃(Y, YS1, tree_3₃(X, YS2, XS)), ZS))
if_flat_2_in_3_ga₇(X, Y, YS1, YS2, XS, ZS, flat_2_out_ga₂(tree_3₃(Y, YS1, tree_3₃(X, YS2, XS)), ZS)) -> flat_2_out_ga₂(tree_3₃(X, tree_3₃(Y, YS1, YS2), XS), ZS)
if_flat_2_in_2_ga₅(X, XS, YS, ZS, flat_2_out_ga₂(ZS, YS)) -> flat_2_out_ga₂(tree_3₃(X, niltree_0, XS), cons_2₂(X, YS))

FLAT_2_IN_GA₂(tree_3₃(X, niltree_0, XS), cons_2₂(X, YS)) -> IF_FLAT_2_IN_1_GA₄(X, XS, YS, right_2_in_ga₂(tree_3₃(X, niltree_0, XS), ZS))
FLAT_2_IN_GA₂(tree_3₃(X, niltree_0, XS), cons_2₂(X, YS)) -> RIGHT_2_IN_GA₂(tree_3₃(X, niltree_0, XS), ZS)
IF_FLAT_2_IN_1_GA₄(X, XS, YS, right_2_out_ga₂(tree_3₃(X, niltree_0, XS), ZS)) -> IF_FLAT_2_IN_2_GA₅(X, XS, YS, ZS, flat_2_in_ga₂(ZS, YS))
IF_FLAT_2_IN_1_GA₄(X, XS, YS, right_2_out_ga₂(tree_3₃(X, niltree_0, XS), ZS)) -> FLAT_2_IN_GA₂(ZS, YS)
FLAT_2_IN_GA₂(tree_3₃(X, tree_3₃(Y, YS1, YS2), XS), ZS) -> IF_FLAT_2_IN_3_GA₇(X, Y, YS1, YS2, XS, ZS, flat_2_in_ga₂(tree_3₃(Y, YS1, tree_3₃(X, YS2, XS)), ZS))
FLAT_2_IN_GA₂(tree_3₃(X, tree_3₃(Y, YS1, YS2), XS), ZS) -> FLAT_2_IN_GA₂(tree_3₃(Y, YS1, tree_3₃(X, YS2, XS)), ZS)

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, XS), cons_2₂(X, YS)) -> if_flat_2_in_1_ga₄(X, XS, YS, right_2_in_ga₂(tree_3₃(X, niltree_0, XS), ZS))
right_2_in_ga₂(tree_3₃(X, XS1, XS2), XS2) -> right_2_out_ga₂(tree_3₃(X, XS1, XS2), XS2)
if_flat_2_in_1_ga₄(X, XS, YS, right_2_out_ga₂(tree_3₃(X, niltree_0, XS), ZS)) -> if_flat_2_in_2_ga₅(X, XS, YS, ZS, flat_2_in_ga₂(ZS, YS))
flat_2_in_ga₂(tree_3₃(X, tree_3₃(Y, YS1, YS2), XS), ZS) -> if_flat_2_in_3_ga₇(X, Y, YS1, YS2, XS, ZS, flat_2_in_ga₂(tree_3₃(Y, YS1, tree_3₃(X, YS2, XS)), ZS))
if_flat_2_in_3_ga₇(X, Y, YS1, YS2, XS, ZS, flat_2_out_ga₂(tree_3₃(Y, YS1, tree_3₃(X, YS2, XS)), ZS)) -> flat_2_out_ga₂(tree_3₃(X, tree_3₃(Y, YS1, YS2), XS), ZS)
if_flat_2_in_2_ga₅(X, XS, YS, ZS, flat_2_out_ga₂(ZS, YS)) -> flat_2_out_ga₂(tree_3₃(X, niltree_0, XS), cons_2₂(X, YS))

The argument filtering Pi contains the following mapping:
flat_2_in_ga₂(x1, x2) = flat_2_in_ga₁(x1)
tree_3₃(x1, x2, x3) = tree_3₃(x1, x2, x3)
niltree_0 = niltree_0
nil_0 = nil_0
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)
right_2_in_ga₂(x1, x2) = right_2_in_ga₁(x1)
right_2_out_ga₂(x1, x2) = right_2_out_ga₁(x2)
if_flat_2_in_2_ga₅(x1, x2, x3, x4, x5) = if_flat_2_in_2_ga₂(x1, x5)
if_flat_2_in_3_ga₇(x1, x2, x3, x4, x5, x6, x7) = if_flat_2_in_3_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_3_GA₇(x1, x2, x3, x4, x5, x6, x7) = IF_FLAT_2_IN_3_GA₁(x7)
RIGHT_2_IN_GA₂(x1, x2) = RIGHT_2_IN_GA₁(x1)
IF_FLAT_2_IN_2_GA₅(x1, x2, x3, x4, x5) = IF_FLAT_2_IN_2_GA₂(x1, x5)

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, XS), cons_2₂(X, YS)) -> IF_FLAT_2_IN_1_GA₄(X, XS, YS, right_2_in_ga₂(tree_3₃(X, niltree_0, XS), ZS))
FLAT_2_IN_GA₂(tree_3₃(X, niltree_0, XS), cons_2₂(X, YS)) -> RIGHT_2_IN_GA₂(tree_3₃(X, niltree_0, XS), ZS)
IF_FLAT_2_IN_1_GA₄(X, XS, YS, right_2_out_ga₂(tree_3₃(X, niltree_0, XS), ZS)) -> IF_FLAT_2_IN_2_GA₅(X, XS, YS, ZS, flat_2_in_ga₂(ZS, YS))
IF_FLAT_2_IN_1_GA₄(X, XS, YS, right_2_out_ga₂(tree_3₃(X, niltree_0, XS), ZS)) -> FLAT_2_IN_GA₂(ZS, YS)
FLAT_2_IN_GA₂(tree_3₃(X, tree_3₃(Y, YS1, YS2), XS), ZS) -> IF_FLAT_2_IN_3_GA₇(X, Y, YS1, YS2, XS, ZS, flat_2_in_ga₂(tree_3₃(Y, YS1, tree_3₃(X, YS2, XS)), ZS))
FLAT_2_IN_GA₂(tree_3₃(X, tree_3₃(Y, YS1, YS2), XS), ZS) -> FLAT_2_IN_GA₂(tree_3₃(Y, YS1, tree_3₃(X, YS2, XS)), ZS)

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, XS), cons_2₂(X, YS)) -> if_flat_2_in_1_ga₄(X, XS, YS, right_2_in_ga₂(tree_3₃(X, niltree_0, XS), ZS))
right_2_in_ga₂(tree_3₃(X, XS1, XS2), XS2) -> right_2_out_ga₂(tree_3₃(X, XS1, XS2), XS2)
if_flat_2_in_1_ga₄(X, XS, YS, right_2_out_ga₂(tree_3₃(X, niltree_0, XS), ZS)) -> if_flat_2_in_2_ga₅(X, XS, YS, ZS, flat_2_in_ga₂(ZS, YS))
flat_2_in_ga₂(tree_3₃(X, tree_3₃(Y, YS1, YS2), XS), ZS) -> if_flat_2_in_3_ga₇(X, Y, YS1, YS2, XS, ZS, flat_2_in_ga₂(tree_3₃(Y, YS1, tree_3₃(X, YS2, XS)), ZS))
if_flat_2_in_3_ga₇(X, Y, YS1, YS2, XS, ZS, flat_2_out_ga₂(tree_3₃(Y, YS1, tree_3₃(X, YS2, XS)), ZS)) -> flat_2_out_ga₂(tree_3₃(X, tree_3₃(Y, YS1, YS2), XS), ZS)
if_flat_2_in_2_ga₅(X, XS, YS, ZS, flat_2_out_ga₂(ZS, YS)) -> flat_2_out_ga₂(tree_3₃(X, niltree_0, XS), cons_2₂(X, YS))

↳ 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, XS), cons_2₂(X, YS)) -> IF_FLAT_2_IN_1_GA₄(X, XS, YS, right_2_in_ga₂(tree_3₃(X, niltree_0, XS), ZS))
IF_FLAT_2_IN_1_GA₄(X, XS, YS, right_2_out_ga₂(tree_3₃(X, niltree_0, XS), ZS)) -> FLAT_2_IN_GA₂(ZS, YS)
FLAT_2_IN_GA₂(tree_3₃(X, tree_3₃(Y, YS1, YS2), XS), ZS) -> FLAT_2_IN_GA₂(tree_3₃(Y, YS1, tree_3₃(X, YS2, XS)), ZS)

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, XS), cons_2₂(X, YS)) -> if_flat_2_in_1_ga₄(X, XS, YS, right_2_in_ga₂(tree_3₃(X, niltree_0, XS), ZS))
right_2_in_ga₂(tree_3₃(X, XS1, XS2), XS2) -> right_2_out_ga₂(tree_3₃(X, XS1, XS2), XS2)
if_flat_2_in_1_ga₄(X, XS, YS, right_2_out_ga₂(tree_3₃(X, niltree_0, XS), ZS)) -> if_flat_2_in_2_ga₅(X, XS, YS, ZS, flat_2_in_ga₂(ZS, YS))
flat_2_in_ga₂(tree_3₃(X, tree_3₃(Y, YS1, YS2), XS), ZS) -> if_flat_2_in_3_ga₇(X, Y, YS1, YS2, XS, ZS, flat_2_in_ga₂(tree_3₃(Y, YS1, tree_3₃(X, YS2, XS)), ZS))
if_flat_2_in_3_ga₇(X, Y, YS1, YS2, XS, ZS, flat_2_out_ga₂(tree_3₃(Y, YS1, tree_3₃(X, YS2, XS)), ZS)) -> flat_2_out_ga₂(tree_3₃(X, tree_3₃(Y, YS1, YS2), XS), ZS)
if_flat_2_in_2_ga₅(X, XS, YS, ZS, flat_2_out_ga₂(ZS, YS)) -> flat_2_out_ga₂(tree_3₃(X, niltree_0, XS), cons_2₂(X, YS))

The argument filtering Pi contains the following mapping:
flat_2_in_ga₂(x1, x2) = flat_2_in_ga₁(x1)
tree_3₃(x1, x2, x3) = tree_3₃(x1, x2, x3)
niltree_0 = niltree_0
nil_0 = nil_0
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)
right_2_in_ga₂(x1, x2) = right_2_in_ga₁(x1)
right_2_out_ga₂(x1, x2) = right_2_out_ga₁(x2)
if_flat_2_in_2_ga₅(x1, x2, x3, x4, x5) = if_flat_2_in_2_ga₂(x1, x5)
if_flat_2_in_3_ga₇(x1, x2, x3, x4, x5, x6, x7) = if_flat_2_in_3_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)

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, XS), cons_2₂(X, YS)) -> IF_FLAT_2_IN_1_GA₄(X, XS, YS, right_2_in_ga₂(tree_3₃(X, niltree_0, XS), ZS))
IF_FLAT_2_IN_1_GA₄(X, XS, YS, right_2_out_ga₂(tree_3₃(X, niltree_0, XS), ZS)) -> FLAT_2_IN_GA₂(ZS, YS)
FLAT_2_IN_GA₂(tree_3₃(X, tree_3₃(Y, YS1, YS2), XS), ZS) -> FLAT_2_IN_GA₂(tree_3₃(Y, YS1, tree_3₃(X, YS2, XS)), ZS)

The TRS R consists of the following rules:

right_2_in_ga₂(tree_3₃(X, XS1, XS2), XS2) -> right_2_out_ga₂(tree_3₃(X, XS1, XS2), XS2)

The argument filtering Pi contains the following mapping:
tree_3₃(x1, x2, x3) = tree_3₃(x1, x2, x3)
niltree_0 = niltree_0
cons_2₂(x1, x2) = cons_2₂(x1, x2)
right_2_in_ga₂(x1, x2) = right_2_in_ga₁(x1)
right_2_out_ga₂(x1, x2) = right_2_out_ga₁(x2)
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)

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

                      ↳ QDPPoloProof

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

FLAT_2_IN_GA₁(tree_3₃(X, niltree_0, XS)) -> IF_FLAT_2_IN_1_GA₂(X, right_2_in_ga₁(tree_3₃(X, niltree_0, XS)))
IF_FLAT_2_IN_1_GA₂(X, right_2_out_ga₁(ZS)) -> FLAT_2_IN_GA₁(ZS)
FLAT_2_IN_GA₁(tree_3₃(X, tree_3₃(Y, YS1, YS2), XS)) -> FLAT_2_IN_GA₁(tree_3₃(Y, YS1, tree_3₃(X, YS2, XS)))

The TRS R consists of the following rules:

right_2_in_ga₁(tree_3₃(X, XS1, XS2)) -> right_2_out_ga₁(XS2)

The set Q consists of the following terms:

right_2_in_ga₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_FLAT_2_IN_1_GA₂, FLAT_2_IN_GA₁}.
By using a polynomial ordering, the following set of Dependency Pairs of this DP problem can be strictly oriented.

IF_FLAT_2_IN_1_GA₂(X, right_2_out_ga₁(ZS)) -> FLAT_2_IN_GA₁(ZS)

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

FLAT_2_IN_GA₁(tree_3₃(X, niltree_0, XS)) -> IF_FLAT_2_IN_1_GA₂(X, right_2_in_ga₁(tree_3₃(X, niltree_0, XS)))
FLAT_2_IN_GA₁(tree_3₃(X, tree_3₃(Y, YS1, YS2), XS)) -> FLAT_2_IN_GA₁(tree_3₃(Y, YS1, tree_3₃(X, YS2, XS)))

With the implicit AFS we had to orient the following set of usable rules non-strictly.

right_2_in_ga₁(tree_3₃(X, XS1, XS2)) -> right_2_out_ga₁(XS2)

Used ordering: POLO with Polynomial interpretation:

POL(niltree_0) = 0
POL(FLAT_2_IN_GA₁(x₁)) = x₁
POL(right_2_out_ga₁(x₁)) = 1 + x₁
POL(IF_FLAT_2_IN_1_GA₂(x₁, x₂)) = x₂
POL(right_2_in_ga₁(x₁)) = x₁
POL(tree_3₃(x₁, x₂, x₃)) = 1 + x₂ + x₃

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

                    ↳ QDP

                      ↳ QDPPoloProof

                        ↳ QDP

                          ↳ DependencyGraphProof

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

FLAT_2_IN_GA₁(tree_3₃(X, niltree_0, XS)) -> IF_FLAT_2_IN_1_GA₂(X, right_2_in_ga₁(tree_3₃(X, niltree_0, XS)))
FLAT_2_IN_GA₁(tree_3₃(X, tree_3₃(Y, YS1, YS2), XS)) -> FLAT_2_IN_GA₁(tree_3₃(Y, YS1, tree_3₃(X, YS2, XS)))

The TRS R consists of the following rules:

right_2_in_ga₁(tree_3₃(X, XS1, XS2)) -> right_2_out_ga₁(XS2)

The set Q consists of the following terms:

right_2_in_ga₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_FLAT_2_IN_1_GA₂, FLAT_2_IN_GA₁}.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

                    ↳ QDP

                      ↳ QDPPoloProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ UsableRulesProof

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

FLAT_2_IN_GA₁(tree_3₃(X, tree_3₃(Y, YS1, YS2), XS)) -> FLAT_2_IN_GA₁(tree_3₃(Y, YS1, tree_3₃(X, YS2, XS)))

The TRS R consists of the following rules:

right_2_in_ga₁(tree_3₃(X, XS1, XS2)) -> right_2_out_ga₁(XS2)

The set Q consists of the following terms:

right_2_in_ga₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {FLAT_2_IN_GA₁}.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, we can delete all non-usable rules from R.

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

                    ↳ QDP

                      ↳ QDPPoloProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ UsableRulesProof

                                ↳ QDP

                                  ↳ QDPSizeChangeProof

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

FLAT_2_IN_GA₁(tree_3₃(X, tree_3₃(Y, YS1, YS2), XS)) -> FLAT_2_IN_GA₁(tree_3₃(Y, YS1, tree_3₃(X, YS2, XS)))

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

right_2_in_ga₁(x0)

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, YS1, YS2), XS)) -> FLAT_2_IN_GA₁(tree_3₃(Y, YS1, tree_3₃(X, YS2, XS))) (allowed arguments on rhs = {1})
The graph contains the following edges 1 > 1

We oriented the following set of usable rules. none