Termination w.r.t. Q proof of /tmp/tmpbvDU9F/z19.trs

Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 MRRProof (⇔)

↳4 QDP

↳5 SemLabProof (⇐)

↳6 QDP

↳7 DependencyGraphProof (⇔)

↳8 QDP

↳9 QDPOrderProof (⇔)

↳10 QDP

↳11 QDPOrderProof (⇔)

↳12 QDP

↳13 QDPOrderProof (⇔)

↳14 QDP

↳15 DependencyGraphProof (⇔)

↳16 AND

  ↳17 QDP

    ↳18 QDPOrderProof (⇔)

    ↳19 QDP

    ↳20 PisEmptyProof (⇔)

    ↳21 TRUE

  ↳22 QDP

    ↳23 UsableRulesReductionPairsProof (⇔)

    ↳24 QDP

    ↳25 PisEmptyProof (⇔)

    ↳26 TRUE

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

f(f(a, a), x) → f(a, f(b, f(a, x)))
f(x, f(y, z)) → f(f(x, y), z)

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

F(f(a, a), x) → F(a, f(b, f(a, x)))
F(f(a, a), x) → F(b, f(a, x))
F(f(a, a), x) → F(a, x)
F(x, f(y, z)) → F(f(x, y), z)
F(x, f(y, z)) → F(x, y)

The TRS R consists of the following rules:

f(f(a, a), x) → f(a, f(b, f(a, x)))
f(x, f(y, z)) → f(f(x, y), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

F(f(a, a), x) → F(b, f(a, x))
F(f(a, a), x) → F(a, x)

Used ordering: Polynomial interpretation [POLO]:

POL(F(x₁, x₂)) = x₁ + x₂
POL(a) = 1
POL(b) = 0
POL(f(x₁, x₂)) = x₁ + x₂

(4) Obligation:

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

F(f(a, a), x) → F(a, f(b, f(a, x)))
F(x, f(y, z)) → F(f(x, y), z)
F(x, f(y, z)) → F(x, y)

The TRS R consists of the following rules:

f(f(a, a), x) → f(a, f(b, f(a, x)))
f(x, f(y, z)) → f(f(x, y), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(5) SemLabProof (SOUND transformation)

We found the following model for the rules of the TRS R. Interpretation over the domain with elements from 0 to 1.a: 1
f: 0
b: 0
F: 0
By semantic labelling [SEMLAB] we obtain the following labelled TRS.

(6) Obligation:

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

F.0-0(f.1-1(a., a.), x) → F.1-0(a., f.0-0(b., f.1-0(a., x)))
F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)
F.0-0(x, f.0-1(y, z)) → F.0-1(f.0-0(x, y), z)
F.0-0(x, f.1-0(y, z)) → F.0-0(f.0-1(x, y), z)
F.0-0(x, f.1-1(y, z)) → F.0-1(f.0-1(x, y), z)
F.1-0(x, f.0-0(y, z)) → F.0-0(f.1-0(x, y), z)
F.1-0(x, f.0-1(y, z)) → F.0-1(f.1-0(x, y), z)
F.1-0(x, f.1-0(y, z)) → F.0-0(f.1-1(x, y), z)
F.1-0(x, f.1-1(y, z)) → F.0-1(f.1-1(x, y), z)
F.0-1(f.1-1(a., a.), x) → F.1-0(a., f.0-0(b., f.1-1(a., x)))
F.0-0(x, f.0-0(y, z)) → F.0-0(x, y)
F.0-0(x, f.0-1(y, z)) → F.0-0(x, y)
F.0-0(x, f.1-0(y, z)) → F.0-1(x, y)
F.0-0(x, f.1-1(y, z)) → F.0-1(x, y)
F.1-0(x, f.0-0(y, z)) → F.1-0(x, y)
F.1-0(x, f.0-1(y, z)) → F.1-0(x, y)
F.1-0(x, f.1-0(y, z)) → F.1-1(x, y)
F.1-0(x, f.1-1(y, z)) → F.1-1(x, y)

The TRS R consists of the following rules:

f.1-0(x, f.0-0(y, z)) → f.0-0(f.1-0(x, y), z)
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-1(f.1-1(a., a.), x) → f.1-0(a., f.0-0(b., f.1-1(a., x)))
f.1-0(x, f.1-1(y, z)) → f.0-1(f.1-1(x, y), z)
f.0-0(x, f.1-0(y, z)) → f.0-0(f.0-1(x, y), z)
f.1-0(x, f.1-0(y, z)) → f.0-0(f.1-1(x, y), z)
f.1-0(x, f.0-1(y, z)) → f.0-1(f.1-0(x, y), z)
f.0-0(x, f.1-1(y, z)) → f.0-1(f.0-1(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
f.0-0(f.1-1(a., a.), x) → f.1-0(a., f.0-0(b., f.1-0(a., x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 5 less nodes.

(8) Obligation:

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

F.1-0(x, f.0-0(y, z)) → F.0-0(f.1-0(x, y), z)
F.0-0(x, f.0-0(y, z)) → F.0-0(x, y)
F.0-0(f.1-1(a., a.), x) → F.1-0(a., f.0-0(b., f.1-0(a., x)))
F.1-0(x, f.1-0(y, z)) → F.0-0(f.1-1(x, y), z)
F.0-0(x, f.0-1(y, z)) → F.0-0(x, y)
F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)
F.0-0(x, f.1-0(y, z)) → F.0-1(x, y)
F.0-1(f.1-1(a., a.), x) → F.1-0(a., f.0-0(b., f.1-1(a., x)))
F.1-0(x, f.0-0(y, z)) → F.1-0(x, y)
F.1-0(x, f.1-1(y, z)) → F.0-1(f.1-1(x, y), z)
F.1-0(x, f.0-1(y, z)) → F.1-0(x, y)
F.0-0(x, f.1-1(y, z)) → F.0-1(x, y)
F.0-0(x, f.1-0(y, z)) → F.0-0(f.0-1(x, y), z)

The TRS R consists of the following rules:

f.1-0(x, f.0-0(y, z)) → f.0-0(f.1-0(x, y), z)
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-1(f.1-1(a., a.), x) → f.1-0(a., f.0-0(b., f.1-1(a., x)))
f.1-0(x, f.1-1(y, z)) → f.0-1(f.1-1(x, y), z)
f.0-0(x, f.1-0(y, z)) → f.0-0(f.0-1(x, y), z)
f.1-0(x, f.1-0(y, z)) → f.0-0(f.1-1(x, y), z)
f.1-0(x, f.0-1(y, z)) → f.0-1(f.1-0(x, y), z)
f.0-0(x, f.1-1(y, z)) → f.0-1(f.0-1(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
f.0-0(f.1-1(a., a.), x) → f.1-0(a., f.0-0(b., f.1-0(a., x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

F.1-0(x, f.1-0(y, z)) → F.0-0(f.1-1(x, y), z)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(F.0-0(x₁, x₂)) = 0
POL(F.0-1(x₁, x₂)) = 0
POL(F.1-0(x₁, x₂)) = x₂
POL(a.) = 1
POL(b.) = 0
POL(f.0-0(x₁, x₂)) = x₁
POL(f.0-1(x₁, x₂)) = x₁
POL(f.1-0(x₁, x₂)) = 1 + x₁ + x₂
POL(f.1-1(x₁, x₂)) = x₁ + x₂

The following usable rules [FROCOS05] were oriented:

f.1-0(x, f.0-0(y, z)) → f.0-0(f.1-0(x, y), z)
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-1(f.1-1(a., a.), x) → f.1-0(a., f.0-0(b., f.1-1(a., x)))
f.1-0(x, f.1-1(y, z)) → f.0-1(f.1-1(x, y), z)
f.0-0(x, f.1-0(y, z)) → f.0-0(f.0-1(x, y), z)
f.1-0(x, f.1-0(y, z)) → f.0-0(f.1-1(x, y), z)
f.1-0(x, f.0-1(y, z)) → f.0-1(f.1-0(x, y), z)
f.0-0(x, f.1-1(y, z)) → f.0-1(f.0-1(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
f.0-0(f.1-1(a., a.), x) → f.1-0(a., f.0-0(b., f.1-0(a., x)))

(10) Obligation:

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

F.1-0(x, f.0-0(y, z)) → F.0-0(f.1-0(x, y), z)
F.0-0(x, f.0-0(y, z)) → F.0-0(x, y)
F.0-0(f.1-1(a., a.), x) → F.1-0(a., f.0-0(b., f.1-0(a., x)))
F.0-0(x, f.0-1(y, z)) → F.0-0(x, y)
F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)
F.0-0(x, f.1-0(y, z)) → F.0-1(x, y)
F.0-1(f.1-1(a., a.), x) → F.1-0(a., f.0-0(b., f.1-1(a., x)))
F.1-0(x, f.0-0(y, z)) → F.1-0(x, y)
F.1-0(x, f.1-1(y, z)) → F.0-1(f.1-1(x, y), z)
F.1-0(x, f.0-1(y, z)) → F.1-0(x, y)
F.0-0(x, f.1-1(y, z)) → F.0-1(x, y)
F.0-0(x, f.1-0(y, z)) → F.0-0(f.0-1(x, y), z)

The TRS R consists of the following rules:

f.1-0(x, f.0-0(y, z)) → f.0-0(f.1-0(x, y), z)
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-1(f.1-1(a., a.), x) → f.1-0(a., f.0-0(b., f.1-1(a., x)))
f.1-0(x, f.1-1(y, z)) → f.0-1(f.1-1(x, y), z)
f.0-0(x, f.1-0(y, z)) → f.0-0(f.0-1(x, y), z)
f.1-0(x, f.1-0(y, z)) → f.0-0(f.1-1(x, y), z)
f.1-0(x, f.0-1(y, z)) → f.0-1(f.1-0(x, y), z)
f.0-0(x, f.1-1(y, z)) → f.0-1(f.0-1(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
f.0-0(f.1-1(a., a.), x) → f.1-0(a., f.0-0(b., f.1-0(a., x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

F.0-0(f.1-1(a., a.), x) → F.1-0(a., f.0-0(b., f.1-0(a., x)))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(F.0-0(x₁, x₂)) = x₁
POL(F.0-1(x₁, x₂)) = 0
POL(F.1-0(x₁, x₂)) = 0
POL(a.) = 1
POL(b.) = 0
POL(f.0-0(x₁, x₂)) = 0
POL(f.0-1(x₁, x₂)) = 0
POL(f.1-0(x₁, x₂)) = 0
POL(f.1-1(x₁, x₂)) = 1 + x₁

The following usable rules [FROCOS05] were oriented:

f.1-0(x, f.0-0(y, z)) → f.0-0(f.1-0(x, y), z)
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-1(f.1-1(a., a.), x) → f.1-0(a., f.0-0(b., f.1-1(a., x)))
f.1-0(x, f.1-1(y, z)) → f.0-1(f.1-1(x, y), z)
f.0-0(x, f.1-0(y, z)) → f.0-0(f.0-1(x, y), z)
f.1-0(x, f.1-0(y, z)) → f.0-0(f.1-1(x, y), z)
f.1-0(x, f.0-1(y, z)) → f.0-1(f.1-0(x, y), z)
f.0-0(x, f.1-1(y, z)) → f.0-1(f.0-1(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
f.0-0(f.1-1(a., a.), x) → f.1-0(a., f.0-0(b., f.1-0(a., x)))

(12) Obligation:

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

F.1-0(x, f.0-0(y, z)) → F.0-0(f.1-0(x, y), z)
F.0-0(x, f.0-0(y, z)) → F.0-0(x, y)
F.0-0(x, f.0-1(y, z)) → F.0-0(x, y)
F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)
F.0-0(x, f.1-0(y, z)) → F.0-1(x, y)
F.0-1(f.1-1(a., a.), x) → F.1-0(a., f.0-0(b., f.1-1(a., x)))
F.1-0(x, f.0-0(y, z)) → F.1-0(x, y)
F.1-0(x, f.1-1(y, z)) → F.0-1(f.1-1(x, y), z)
F.1-0(x, f.0-1(y, z)) → F.1-0(x, y)
F.0-0(x, f.1-1(y, z)) → F.0-1(x, y)
F.0-0(x, f.1-0(y, z)) → F.0-0(f.0-1(x, y), z)

The TRS R consists of the following rules:

f.1-0(x, f.0-0(y, z)) → f.0-0(f.1-0(x, y), z)
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-1(f.1-1(a., a.), x) → f.1-0(a., f.0-0(b., f.1-1(a., x)))
f.1-0(x, f.1-1(y, z)) → f.0-1(f.1-1(x, y), z)
f.0-0(x, f.1-0(y, z)) → f.0-0(f.0-1(x, y), z)
f.1-0(x, f.1-0(y, z)) → f.0-0(f.1-1(x, y), z)
f.1-0(x, f.0-1(y, z)) → f.0-1(f.1-0(x, y), z)
f.0-0(x, f.1-1(y, z)) → f.0-1(f.0-1(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
f.0-0(f.1-1(a., a.), x) → f.1-0(a., f.0-0(b., f.1-0(a., x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

F.0-1(f.1-1(a., a.), x) → F.1-0(a., f.0-0(b., f.1-1(a., x)))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(F.0-0(x₁, x₂)) = x₁
POL(F.0-1(x₁, x₂)) = x₁
POL(F.1-0(x₁, x₂)) = x₁ + x₂
POL(a.) = 1
POL(b.) = 0
POL(f.0-0(x₁, x₂)) = x₁
POL(f.0-1(x₁, x₂)) = x₁
POL(f.1-0(x₁, x₂)) = x₁ + x₂
POL(f.1-1(x₁, x₂)) = x₁ + x₂

The following usable rules [FROCOS05] were oriented:

f.1-0(x, f.0-0(y, z)) → f.0-0(f.1-0(x, y), z)
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-1(f.1-1(a., a.), x) → f.1-0(a., f.0-0(b., f.1-1(a., x)))
f.1-0(x, f.1-1(y, z)) → f.0-1(f.1-1(x, y), z)
f.0-0(x, f.1-0(y, z)) → f.0-0(f.0-1(x, y), z)
f.1-0(x, f.1-0(y, z)) → f.0-0(f.1-1(x, y), z)
f.1-0(x, f.0-1(y, z)) → f.0-1(f.1-0(x, y), z)
f.0-0(x, f.1-1(y, z)) → f.0-1(f.0-1(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
f.0-0(f.1-1(a., a.), x) → f.1-0(a., f.0-0(b., f.1-0(a., x)))

(14) Obligation:

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

F.1-0(x, f.0-0(y, z)) → F.0-0(f.1-0(x, y), z)
F.0-0(x, f.0-0(y, z)) → F.0-0(x, y)
F.0-0(x, f.0-1(y, z)) → F.0-0(x, y)
F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)
F.0-0(x, f.1-0(y, z)) → F.0-1(x, y)
F.1-0(x, f.0-0(y, z)) → F.1-0(x, y)
F.1-0(x, f.1-1(y, z)) → F.0-1(f.1-1(x, y), z)
F.1-0(x, f.0-1(y, z)) → F.1-0(x, y)
F.0-0(x, f.1-1(y, z)) → F.0-1(x, y)
F.0-0(x, f.1-0(y, z)) → F.0-0(f.0-1(x, y), z)

The TRS R consists of the following rules:

f.1-0(x, f.0-0(y, z)) → f.0-0(f.1-0(x, y), z)
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-1(f.1-1(a., a.), x) → f.1-0(a., f.0-0(b., f.1-1(a., x)))
f.1-0(x, f.1-1(y, z)) → f.0-1(f.1-1(x, y), z)
f.0-0(x, f.1-0(y, z)) → f.0-0(f.0-1(x, y), z)
f.1-0(x, f.1-0(y, z)) → f.0-0(f.1-1(x, y), z)
f.1-0(x, f.0-1(y, z)) → f.0-1(f.1-0(x, y), z)
f.0-0(x, f.1-1(y, z)) → f.0-1(f.0-1(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
f.0-0(f.1-1(a., a.), x) → f.1-0(a., f.0-0(b., f.1-0(a., x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(15) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 4 less nodes.

(16) Complex Obligation (AND)

(17) Obligation:

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

F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)
F.0-0(x, f.0-0(y, z)) → F.0-0(x, y)
F.0-0(x, f.1-0(y, z)) → F.0-0(f.0-1(x, y), z)
F.0-0(x, f.0-1(y, z)) → F.0-0(x, y)

The TRS R consists of the following rules:

f.1-0(x, f.0-0(y, z)) → f.0-0(f.1-0(x, y), z)
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-1(f.1-1(a., a.), x) → f.1-0(a., f.0-0(b., f.1-1(a., x)))
f.1-0(x, f.1-1(y, z)) → f.0-1(f.1-1(x, y), z)
f.0-0(x, f.1-0(y, z)) → f.0-0(f.0-1(x, y), z)
f.1-0(x, f.1-0(y, z)) → f.0-0(f.1-1(x, y), z)
f.1-0(x, f.0-1(y, z)) → f.0-1(f.1-0(x, y), z)
f.0-0(x, f.1-1(y, z)) → f.0-1(f.0-1(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
f.0-0(f.1-1(a., a.), x) → f.1-0(a., f.0-0(b., f.1-0(a., x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)
F.0-0(x, f.0-0(y, z)) → F.0-0(x, y)
F.0-0(x, f.1-0(y, z)) → F.0-0(f.0-1(x, y), z)
F.0-0(x, f.0-1(y, z)) → F.0-0(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(F.0-0(x₁, x₂)) = x₂
POL(a.) = 0
POL(b.) = 0
POL(f.0-0(x₁, x₂)) = 1 + x₁ + x₂
POL(f.0-1(x₁, x₂)) = 1 + x₁
POL(f.1-0(x₁, x₂)) = 1 + x₁ + x₂
POL(f.1-1(x₁, x₂)) = 0

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f.1-0(x, f.0-0(y, z)) → f.0-0(f.1-0(x, y), z)
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-1(f.1-1(a., a.), x) → f.1-0(a., f.0-0(b., f.1-1(a., x)))
f.1-0(x, f.1-1(y, z)) → f.0-1(f.1-1(x, y), z)
f.0-0(x, f.1-0(y, z)) → f.0-0(f.0-1(x, y), z)
f.1-0(x, f.1-0(y, z)) → f.0-0(f.1-1(x, y), z)
f.1-0(x, f.0-1(y, z)) → f.0-1(f.1-0(x, y), z)
f.0-0(x, f.1-1(y, z)) → f.0-1(f.0-1(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
f.0-0(f.1-1(a., a.), x) → f.1-0(a., f.0-0(b., f.1-0(a., x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(20) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(21) TRUE

(22) Obligation:

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

F.1-0(x, f.0-0(y, z)) → F.1-0(x, y)
F.1-0(x, f.0-1(y, z)) → F.1-0(x, y)

The TRS R consists of the following rules:

f.1-0(x, f.0-0(y, z)) → f.0-0(f.1-0(x, y), z)
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-1(f.1-1(a., a.), x) → f.1-0(a., f.0-0(b., f.1-1(a., x)))
f.1-0(x, f.1-1(y, z)) → f.0-1(f.1-1(x, y), z)
f.0-0(x, f.1-0(y, z)) → f.0-0(f.0-1(x, y), z)
f.1-0(x, f.1-0(y, z)) → f.0-0(f.1-1(x, y), z)
f.1-0(x, f.0-1(y, z)) → f.0-1(f.1-0(x, y), z)
f.0-0(x, f.1-1(y, z)) → f.0-1(f.0-1(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
f.0-0(f.1-1(a., a.), x) → f.1-0(a., f.0-0(b., f.1-0(a., x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(23) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] 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:

F.1-0(x, f.0-0(y, z)) → F.1-0(x, y)
F.1-0(x, f.0-1(y, z)) → F.1-0(x, y)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(F.1-0(x₁, x₂)) = x₁ + x₂
POL(f.0-0(x₁, x₂)) = x₁ + x₂
POL(f.0-1(x₁, x₂)) = x₁ + x₂

(24) Obligation:

Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(25) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) MRRProof (EQUIVALENT transformation)

(4) Obligation:

(5) SemLabProof (SOUND transformation)

(6) Obligation:

(7) DependencyGraphProof (EQUIVALENT transformation)

(8) Obligation:

(9) QDPOrderProof (EQUIVALENT transformation)

(10) Obligation:

(11) QDPOrderProof (EQUIVALENT transformation)

(12) Obligation:

(13) QDPOrderProof (EQUIVALENT transformation)

(14) Obligation:

(15) DependencyGraphProof (EQUIVALENT transformation)

(16) Complex Obligation (AND)

(17) Obligation:

(18) QDPOrderProof (EQUIVALENT transformation)

(19) Obligation:

(20) PisEmptyProof (EQUIVALENT transformation)

(21) TRUE

(22) Obligation:

(23) UsableRulesReductionPairsProof (EQUIVALENT transformation)

(24) Obligation:

(25) PisEmptyProof (EQUIVALENT transformation)

(26) TRUE