YES Termination w.r.t. Q proof of Zantema_05_z22.ari

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

0 QTRS
↳1 DependencyPairsProof (⇔, 11 ms)
↳2 QDP
↳3 MRRProof (⇔, 0 ms)
↳4 QDP
↳5 QDPOrderProof (⇔, 21 ms)
↳6 QDP
↳7 SemLabProof (⇒, 154 ms)
↳8 QDP
↳9 DependencyGraphProof (⇔, 0 ms)
↳10 QDP
↳11 MRRProof (⇔, 0 ms)
↳12 QDP
↳13 UsableRulesReductionPairsProof (⇔, 4 ms)
↳14 QDP
↳15 MRRProof (⇔, 0 ms)
↳16 QDP
↳17 QDPOrderProof (⇔, 0 ms)
↳18 QDP
↳19 DependencyGraphProof (⇔, 0 ms)
↳20 AND
    ↳21 QDP
        ↳22 QDPOrderProof (⇔, 0 ms)
        ↳23 QDP
        ↳24 PisEmptyProof (⇔, 0 ms)
        ↳25 YES
    ↳26 QDP
        ↳27 UsableRulesReductionPairsProof (⇔, 0 ms)
        ↳28 QDP
        ↳29 PisEmptyProof (⇔, 0 ms)
        ↳30 YES

(0) Obligation:

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

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


Used ordering: Polynomial interpretation [POLO]:

POL(F(x1, x2)) = 2·x1 + 2·x2   
POL(a) = 1   
POL(b) = 0   
POL(c) = 1   
POL(f(x1, x2)) = x1 + x2   

(4) Obligation:

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

F(f(f(a, b), c), x) → F(b, f(a, f(c, f(b, x))))
F(f(f(a, b), c), x) → F(a, f(c, f(b, 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(f(a, b), c), x) → f(b, f(a, f(c, f(b, 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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


F(f(f(a, b), c), x) → F(b, f(a, f(c, f(b, x))))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2)  =  x1
f(x1, x2)  =  x1
a  =  a
b  =  b

Knuth-Bendix order [KBO] with precedence:
a > b

and weight map:

a=1
b=1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

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

(6) Obligation:

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

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

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

(7) SemLabProof (SOUND transformation)

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

(8) Obligation:

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

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

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

(9) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

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

The TRS R consists of the following rules:

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

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

(11) 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.1-0(x, f.1-0(y, z)) → F.0-0(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.0-0(x, f.1-0(y, z)) → F.0-1(x, y)

Strictly oriented rules of the TRS R:

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.1-1(y, z)) → f.0-1(f.1-1(x, y), z)

Used ordering: Polynomial interpretation [POLO]:

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

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

The TRS R consists of the following rules:

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

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

(13) 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.0-0(x, f.1-1(y, z)) → F.0-1(x, y)
F.0-0(x, f.1-1(y, z)) → F.0-1(f.0-1(x, y), z)
The following rules are removed from R:

f.0-0(x, f.1-1(y, z)) → f.0-1(f.0-1(x, y), z)
Used ordering: POLO with Polynomial interpretation [POLO]:

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

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

The TRS R 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(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-0(b., x))))
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-1(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-1(b., x))))
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)

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

(15) 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.0-0(x, f.0-1(y, z)) → F.0-0(x, y)
F.1-0(x, f.0-1(y, z)) → F.1-0(x, y)


Used ordering: Polynomial interpretation [POLO]:

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

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

The TRS R 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(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-0(b., x))))
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-1(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-1(b., x))))
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)

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

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


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

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

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

(18) 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.1-0(x, f.0-1(y, z)) → F.0-1(f.1-0(x, y), z)
F.1-0(x, f.0-0(y, z)) → F.1-0(x, y)
F.0-0(x, f.0-0(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.0-1(y, z)) → F.0-1(f.0-0(x, y), z)

The TRS R 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(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-0(b., x))))
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-1(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-1(b., x))))
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)

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

(19) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.

(20) Complex Obligation (AND)

(21) 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)

The TRS R 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(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-0(b., x))))
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-1(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-1(b., x))))
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)

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

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


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

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none

(23) Obligation:

Q DP problem:
P is empty.
The TRS R 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(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-0(b., x))))
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-1(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-1(b., x))))
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)

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

(24) PisEmptyProof (EQUIVALENT transformation)

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

(25) YES

(26) 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)

The TRS R 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(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-0(b., x))))
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-1(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-1(b., x))))
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)

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

(27) 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)
The following rules are removed from R:

f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
f.0-0(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-0(b., x))))
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-1(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-1(b., x))))
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)
Used ordering: POLO with Polynomial interpretation [POLO]:

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

(28) Obligation:

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

(29) PisEmptyProof (EQUIVALENT transformation)

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

(30) YES