/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) QDP
(7) TransformationProof [EQUIVALENT, 0 ms]
(8) QDP
(9) TransformationProof [EQUIVALENT, 0 ms]
(10) QDP
(11) TransformationProof [EQUIVALENT, 0 ms]
(12) QDP
(13) TransformationProof [EQUIVALENT, 0 ms]
(14) QDP
(15) TransformationProof [EQUIVALENT, 0 ms]
(16) QDP
(17) TransformationProof [EQUIVALENT, 0 ms]
(18) QDP
(19) TransformationProof [EQUIVALENT, 0 ms]
(20) QDP
(21) SemLabProof [SOUND, 174 ms]
(22) QDP
(23) DependencyGraphProof [EQUIVALENT, 0 ms]
(24) QDP
(25) MRRProof [EQUIVALENT, 13 ms]
(26) QDP
(27) DependencyGraphProof [EQUIVALENT, 0 ms]
(28) QDP
(29) MRRProof [EQUIVALENT, 8 ms]
(30) QDP
(31) PisEmptyProof [EQUIVALENT, 0 ms]
(32) YES


----------------------------------------

(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(x, f(y, a)) -> f(f(a, f(h(x), y)), a)

Q is empty.

----------------------------------------

(1) Overlay + Local Confluence (EQUIVALENT)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(x, f(y, a)) -> f(f(a, f(h(x), y)), a)

The set Q consists of the following terms:

   f(x0, f(x1, a))


----------------------------------------

(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(x, f(y, a)) -> F(f(a, f(h(x), y)), a)
   F(x, f(y, a)) -> F(a, f(h(x), y))
   F(x, f(y, a)) -> F(h(x), y)

The TRS R consists of the following rules:

   f(x, f(y, a)) -> f(f(a, f(h(x), y)), a)

The set Q consists of the following terms:

   f(x0, f(x1, a))

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
----------------------------------------

(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(x, f(y, a)) -> F(h(x), y)
   F(x, f(y, a)) -> F(a, f(h(x), y))

The TRS R consists of the following rules:

   f(x, f(y, a)) -> f(f(a, f(h(x), y)), a)

The set Q consists of the following terms:

   f(x0, f(x1, a))

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

(7) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule F(x, f(y, a)) -> F(h(x), y) we obtained the following new rules [LPAR04]:

   (F(h(z0), f(x1, a)) -> F(h(h(z0)), x1),F(h(z0), f(x1, a)) -> F(h(h(z0)), x1))
   (F(a, f(x1, a)) -> F(h(a), x1),F(a, f(x1, a)) -> F(h(a), x1))


----------------------------------------

(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(x, f(y, a)) -> F(a, f(h(x), y))
   F(h(z0), f(x1, a)) -> F(h(h(z0)), x1)
   F(a, f(x1, a)) -> F(h(a), x1)

The TRS R consists of the following rules:

   f(x, f(y, a)) -> f(f(a, f(h(x), y)), a)

The set Q consists of the following terms:

   f(x0, f(x1, a))

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

(9) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule F(x, f(y, a)) -> F(a, f(h(x), y)) we obtained the following new rules [LPAR04]:

   (F(a, f(x1, a)) -> F(a, f(h(a), x1)),F(a, f(x1, a)) -> F(a, f(h(a), x1)))
   (F(h(h(z0)), f(x1, a)) -> F(a, f(h(h(h(z0))), x1)),F(h(h(z0)), f(x1, a)) -> F(a, f(h(h(h(z0))), x1)))
   (F(h(a), f(x1, a)) -> F(a, f(h(h(a)), x1)),F(h(a), f(x1, a)) -> F(a, f(h(h(a)), x1)))


----------------------------------------

(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(h(z0), f(x1, a)) -> F(h(h(z0)), x1)
   F(a, f(x1, a)) -> F(h(a), x1)
   F(a, f(x1, a)) -> F(a, f(h(a), x1))
   F(h(h(z0)), f(x1, a)) -> F(a, f(h(h(h(z0))), x1))
   F(h(a), f(x1, a)) -> F(a, f(h(h(a)), x1))

The TRS R consists of the following rules:

   f(x, f(y, a)) -> f(f(a, f(h(x), y)), a)

The set Q consists of the following terms:

   f(x0, f(x1, a))

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

(11) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule F(h(z0), f(x1, a)) -> F(h(h(z0)), x1) we obtained the following new rules [LPAR04]:

   (F(h(h(z0)), f(x1, a)) -> F(h(h(h(z0))), x1),F(h(h(z0)), f(x1, a)) -> F(h(h(h(z0))), x1))
   (F(h(a), f(x1, a)) -> F(h(h(a)), x1),F(h(a), f(x1, a)) -> F(h(h(a)), x1))


----------------------------------------

(12)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(a, f(x1, a)) -> F(h(a), x1)
   F(a, f(x1, a)) -> F(a, f(h(a), x1))
   F(h(h(z0)), f(x1, a)) -> F(a, f(h(h(h(z0))), x1))
   F(h(a), f(x1, a)) -> F(a, f(h(h(a)), x1))
   F(h(h(z0)), f(x1, a)) -> F(h(h(h(z0))), x1)
   F(h(a), f(x1, a)) -> F(h(h(a)), x1)

The TRS R consists of the following rules:

   f(x, f(y, a)) -> f(f(a, f(h(x), y)), a)

The set Q consists of the following terms:

   f(x0, f(x1, a))

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

(13) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule F(h(h(z0)), f(x1, a)) -> F(a, f(h(h(h(z0))), x1)) we obtained the following new rules [LPAR04]:

   (F(h(h(h(z0))), f(x1, a)) -> F(a, f(h(h(h(h(z0)))), x1)),F(h(h(h(z0))), f(x1, a)) -> F(a, f(h(h(h(h(z0)))), x1)))
   (F(h(h(a)), f(x1, a)) -> F(a, f(h(h(h(a))), x1)),F(h(h(a)), f(x1, a)) -> F(a, f(h(h(h(a))), x1)))


----------------------------------------

(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(a, f(x1, a)) -> F(h(a), x1)
   F(a, f(x1, a)) -> F(a, f(h(a), x1))
   F(h(a), f(x1, a)) -> F(a, f(h(h(a)), x1))
   F(h(h(z0)), f(x1, a)) -> F(h(h(h(z0))), x1)
   F(h(a), f(x1, a)) -> F(h(h(a)), x1)
   F(h(h(h(z0))), f(x1, a)) -> F(a, f(h(h(h(h(z0)))), x1))
   F(h(h(a)), f(x1, a)) -> F(a, f(h(h(h(a))), x1))

The TRS R consists of the following rules:

   f(x, f(y, a)) -> f(f(a, f(h(x), y)), a)

The set Q consists of the following terms:

   f(x0, f(x1, a))

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

(15) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule F(a, f(x1, a)) -> F(h(a), x1) we obtained the following new rules [LPAR04]:

   (F(a, f(f(y_0, a), a)) -> F(h(a), f(y_0, a)),F(a, f(f(y_0, a), a)) -> F(h(a), f(y_0, a)))


----------------------------------------

(16)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(a, f(x1, a)) -> F(a, f(h(a), x1))
   F(h(a), f(x1, a)) -> F(a, f(h(h(a)), x1))
   F(h(h(z0)), f(x1, a)) -> F(h(h(h(z0))), x1)
   F(h(a), f(x1, a)) -> F(h(h(a)), x1)
   F(h(h(h(z0))), f(x1, a)) -> F(a, f(h(h(h(h(z0)))), x1))
   F(h(h(a)), f(x1, a)) -> F(a, f(h(h(h(a))), x1))
   F(a, f(f(y_0, a), a)) -> F(h(a), f(y_0, a))

The TRS R consists of the following rules:

   f(x, f(y, a)) -> f(f(a, f(h(x), y)), a)

The set Q consists of the following terms:

   f(x0, f(x1, a))

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

(17) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule F(h(h(z0)), f(x1, a)) -> F(h(h(h(z0))), x1) we obtained the following new rules [LPAR04]:

   (F(h(h(x0)), f(f(y_1, a), a)) -> F(h(h(h(x0))), f(y_1, a)),F(h(h(x0)), f(f(y_1, a), a)) -> F(h(h(h(x0))), f(y_1, a)))


----------------------------------------

(18)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(a, f(x1, a)) -> F(a, f(h(a), x1))
   F(h(a), f(x1, a)) -> F(a, f(h(h(a)), x1))
   F(h(a), f(x1, a)) -> F(h(h(a)), x1)
   F(h(h(h(z0))), f(x1, a)) -> F(a, f(h(h(h(h(z0)))), x1))
   F(h(h(a)), f(x1, a)) -> F(a, f(h(h(h(a))), x1))
   F(a, f(f(y_0, a), a)) -> F(h(a), f(y_0, a))
   F(h(h(x0)), f(f(y_1, a), a)) -> F(h(h(h(x0))), f(y_1, a))

The TRS R consists of the following rules:

   f(x, f(y, a)) -> f(f(a, f(h(x), y)), a)

The set Q consists of the following terms:

   f(x0, f(x1, a))

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

(19) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule F(h(a), f(x1, a)) -> F(h(h(a)), x1) we obtained the following new rules [LPAR04]:

   (F(h(a), f(f(y_0, a), a)) -> F(h(h(a)), f(y_0, a)),F(h(a), f(f(y_0, a), a)) -> F(h(h(a)), f(y_0, a)))
   (F(h(a), f(f(f(y_1, a), a), a)) -> F(h(h(a)), f(f(y_1, a), a)),F(h(a), f(f(f(y_1, a), a), a)) -> F(h(h(a)), f(f(y_1, a), a)))


----------------------------------------

(20)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(a, f(x1, a)) -> F(a, f(h(a), x1))
   F(h(a), f(x1, a)) -> F(a, f(h(h(a)), x1))
   F(h(h(h(z0))), f(x1, a)) -> F(a, f(h(h(h(h(z0)))), x1))
   F(h(h(a)), f(x1, a)) -> F(a, f(h(h(h(a))), x1))
   F(a, f(f(y_0, a), a)) -> F(h(a), f(y_0, a))
   F(h(h(x0)), f(f(y_1, a), a)) -> F(h(h(h(x0))), f(y_1, a))
   F(h(a), f(f(y_0, a), a)) -> F(h(h(a)), f(y_0, a))
   F(h(a), f(f(f(y_1, a), a), a)) -> F(h(h(a)), f(f(y_1, a), a))

The TRS R consists of the following rules:

   f(x, f(y, a)) -> f(f(a, f(h(x), y)), a)

The set Q consists of the following terms:

   f(x0, f(x1, a))

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

(21) SemLabProof (SOUND)
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
f: 0
h: 0
F: 0
By semantic labelling [SEMLAB] we obtain the following labelled QDP problem.
----------------------------------------

(22)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F.1-0(a., f.0-1(x1, a.)) -> F.1-0(a., f.0-0(h.1(a.), x1))
   F.1-0(a., f.1-1(x1, a.)) -> F.1-0(a., f.0-1(h.1(a.), x1))
   F.1-0(a., f.0-1(f.0-1(y_0, a.), a.)) -> F.0-0(h.1(a.), f.0-1(y_0, a.))
   F.1-0(a., f.0-1(f.1-1(y_0, a.), a.)) -> F.0-0(h.1(a.), f.1-1(y_0, a.))
   F.0-0(h.1(a.), f.0-1(x1, a.)) -> F.1-0(a., f.0-0(h.0(h.1(a.)), x1))
   F.0-0(h.1(a.), f.1-1(x1, a.)) -> F.1-0(a., f.0-1(h.0(h.1(a.)), x1))
   F.0-0(h.0(h.0(h.0(z0))), f.0-1(x1, a.)) -> F.1-0(a., f.0-0(h.0(h.0(h.0(h.0(z0)))), x1))
   F.0-0(h.0(h.0(h.0(z0))), f.1-1(x1, a.)) -> F.1-0(a., f.0-1(h.0(h.0(h.0(h.0(z0)))), x1))
   F.0-0(h.0(h.0(h.1(z0))), f.0-1(x1, a.)) -> F.1-0(a., f.0-0(h.0(h.0(h.0(h.1(z0)))), x1))
   F.0-0(h.0(h.0(h.1(z0))), f.1-1(x1, a.)) -> F.1-0(a., f.0-1(h.0(h.0(h.0(h.1(z0)))), x1))
   F.0-0(h.0(h.1(a.)), f.0-1(x1, a.)) -> F.1-0(a., f.0-0(h.0(h.0(h.1(a.))), x1))
   F.0-0(h.0(h.1(a.)), f.1-1(x1, a.)) -> F.1-0(a., f.0-1(h.0(h.0(h.1(a.))), x1))
   F.0-0(h.1(a.), f.0-1(f.0-1(y_0, a.), a.)) -> F.0-0(h.0(h.1(a.)), f.0-1(y_0, a.))
   F.0-0(h.1(a.), f.0-1(f.1-1(y_0, a.), a.)) -> F.0-0(h.0(h.1(a.)), f.1-1(y_0, a.))
   F.0-0(h.1(a.), f.0-1(f.0-1(f.0-1(y_1, a.), a.), a.)) -> F.0-0(h.0(h.1(a.)), f.0-1(f.0-1(y_1, a.), a.))
   F.0-0(h.1(a.), f.0-1(f.0-1(f.1-1(y_1, a.), a.), a.)) -> F.0-0(h.0(h.1(a.)), f.0-1(f.1-1(y_1, a.), a.))
   F.0-0(h.0(h.0(x0)), f.0-1(f.0-1(y_1, a.), a.)) -> F.0-0(h.0(h.0(h.0(x0))), f.0-1(y_1, a.))
   F.0-0(h.0(h.0(x0)), f.0-1(f.1-1(y_1, a.), a.)) -> F.0-0(h.0(h.0(h.0(x0))), f.1-1(y_1, a.))
   F.0-0(h.0(h.1(x0)), f.0-1(f.0-1(y_1, a.), a.)) -> F.0-0(h.0(h.0(h.1(x0))), f.0-1(y_1, a.))
   F.0-0(h.0(h.1(x0)), f.0-1(f.1-1(y_1, a.), a.)) -> F.0-0(h.0(h.0(h.1(x0))), f.1-1(y_1, a.))

The TRS R consists of the following rules:

   f.0-0(x, f.0-1(y, a.)) -> f.0-1(f.1-0(a., f.0-0(h.0(x), y)), a.)
   f.0-0(x, f.1-1(y, a.)) -> f.0-1(f.1-0(a., f.0-1(h.0(x), y)), a.)
   f.1-0(x, f.0-1(y, a.)) -> f.0-1(f.1-0(a., f.0-0(h.1(x), y)), a.)
   f.1-0(x, f.1-1(y, a.)) -> f.0-1(f.1-0(a., f.0-1(h.1(x), y)), a.)

The set Q consists of the following terms:

   f.0-0(x0, f.0-1(x1, a.))
   f.0-0(x0, f.1-1(x1, a.))
   f.1-0(x0, f.0-1(x1, a.))
   f.1-0(x0, f.1-1(x1, a.))

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

(23) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.
----------------------------------------

(24)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F.1-0(a., f.0-1(x1, a.)) -> F.1-0(a., f.0-0(h.1(a.), x1))
   F.1-0(a., f.0-1(f.0-1(y_0, a.), a.)) -> F.0-0(h.1(a.), f.0-1(y_0, a.))
   F.0-0(h.1(a.), f.0-1(x1, a.)) -> F.1-0(a., f.0-0(h.0(h.1(a.)), x1))
   F.0-0(h.1(a.), f.0-1(f.0-1(y_0, a.), a.)) -> F.0-0(h.0(h.1(a.)), f.0-1(y_0, a.))
   F.0-0(h.0(h.1(a.)), f.0-1(x1, a.)) -> F.1-0(a., f.0-0(h.0(h.0(h.1(a.))), x1))
   F.0-0(h.0(h.1(x0)), f.0-1(f.0-1(y_1, a.), a.)) -> F.0-0(h.0(h.0(h.1(x0))), f.0-1(y_1, a.))
   F.0-0(h.0(h.0(h.1(z0))), f.0-1(x1, a.)) -> F.1-0(a., f.0-0(h.0(h.0(h.0(h.1(z0)))), x1))
   F.0-0(h.0(h.0(x0)), f.0-1(f.0-1(y_1, a.), a.)) -> F.0-0(h.0(h.0(h.0(x0))), f.0-1(y_1, a.))
   F.0-0(h.0(h.0(h.0(z0))), f.0-1(x1, a.)) -> F.1-0(a., f.0-0(h.0(h.0(h.0(h.0(z0)))), x1))
   F.0-0(h.0(h.0(x0)), f.0-1(f.1-1(y_1, a.), a.)) -> F.0-0(h.0(h.0(h.0(x0))), f.1-1(y_1, a.))
   F.0-0(h.0(h.0(h.0(z0))), f.1-1(x1, a.)) -> F.1-0(a., f.0-1(h.0(h.0(h.0(h.0(z0)))), x1))
   F.0-0(h.0(h.1(x0)), f.0-1(f.1-1(y_1, a.), a.)) -> F.0-0(h.0(h.0(h.1(x0))), f.1-1(y_1, a.))
   F.0-0(h.0(h.0(h.1(z0))), f.1-1(x1, a.)) -> F.1-0(a., f.0-1(h.0(h.0(h.0(h.1(z0)))), x1))
   F.0-0(h.1(a.), f.0-1(f.1-1(y_0, a.), a.)) -> F.0-0(h.0(h.1(a.)), f.1-1(y_0, a.))
   F.0-0(h.0(h.1(a.)), f.1-1(x1, a.)) -> F.1-0(a., f.0-1(h.0(h.0(h.1(a.))), x1))
   F.0-0(h.1(a.), f.0-1(f.0-1(f.0-1(y_1, a.), a.), a.)) -> F.0-0(h.0(h.1(a.)), f.0-1(f.0-1(y_1, a.), a.))
   F.0-0(h.1(a.), f.0-1(f.0-1(f.1-1(y_1, a.), a.), a.)) -> F.0-0(h.0(h.1(a.)), f.0-1(f.1-1(y_1, a.), a.))

The TRS R consists of the following rules:

   f.0-0(x, f.0-1(y, a.)) -> f.0-1(f.1-0(a., f.0-0(h.0(x), y)), a.)
   f.0-0(x, f.1-1(y, a.)) -> f.0-1(f.1-0(a., f.0-1(h.0(x), y)), a.)
   f.1-0(x, f.0-1(y, a.)) -> f.0-1(f.1-0(a., f.0-0(h.1(x), y)), a.)
   f.1-0(x, f.1-1(y, a.)) -> f.0-1(f.1-0(a., f.0-1(h.1(x), y)), a.)

The set Q consists of the following terms:

   f.0-0(x0, f.0-1(x1, a.))
   f.0-0(x0, f.1-1(x1, a.))
   f.1-0(x0, f.0-1(x1, a.))
   f.1-0(x0, f.1-1(x1, a.))

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

(25) MRRProof (EQUIVALENT)
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(h.0(h.0(h.0(z0))), f.1-1(x1, a.)) -> F.1-0(a., f.0-1(h.0(h.0(h.0(h.0(z0)))), x1))
   F.0-0(h.0(h.0(h.1(z0))), f.1-1(x1, a.)) -> F.1-0(a., f.0-1(h.0(h.0(h.0(h.1(z0)))), x1))
   F.0-0(h.0(h.1(a.)), f.1-1(x1, a.)) -> F.1-0(a., f.0-1(h.0(h.0(h.1(a.))), x1))

Strictly oriented rules of the TRS R:

   f.0-0(x, f.1-1(y, a.)) -> f.0-1(f.1-0(a., f.0-1(h.0(x), y)), a.)
   f.1-0(x, f.1-1(y, a.)) -> f.0-1(f.1-0(a., f.0-1(h.1(x), y)), a.)

Used ordering: Polynomial interpretation [POLO]:

   POL(F.0-0(x_1, x_2)) = 1 + x_1 + x_2
   POL(F.1-0(x_1, x_2)) = 1 + x_1 + x_2
   POL(a.) = 0
   POL(f.0-0(x_1, x_2)) = x_1 + x_2
   POL(f.0-1(x_1, x_2)) = x_1 + x_2
   POL(f.1-0(x_1, x_2)) = x_1 + x_2
   POL(f.1-1(x_1, x_2)) = 1 + x_1 + x_2
   POL(h.0(x_1)) = x_1
   POL(h.1(x_1)) = x_1


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(26)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F.1-0(a., f.0-1(x1, a.)) -> F.1-0(a., f.0-0(h.1(a.), x1))
   F.1-0(a., f.0-1(f.0-1(y_0, a.), a.)) -> F.0-0(h.1(a.), f.0-1(y_0, a.))
   F.0-0(h.1(a.), f.0-1(x1, a.)) -> F.1-0(a., f.0-0(h.0(h.1(a.)), x1))
   F.0-0(h.1(a.), f.0-1(f.0-1(y_0, a.), a.)) -> F.0-0(h.0(h.1(a.)), f.0-1(y_0, a.))
   F.0-0(h.0(h.1(a.)), f.0-1(x1, a.)) -> F.1-0(a., f.0-0(h.0(h.0(h.1(a.))), x1))
   F.0-0(h.0(h.1(x0)), f.0-1(f.0-1(y_1, a.), a.)) -> F.0-0(h.0(h.0(h.1(x0))), f.0-1(y_1, a.))
   F.0-0(h.0(h.0(h.1(z0))), f.0-1(x1, a.)) -> F.1-0(a., f.0-0(h.0(h.0(h.0(h.1(z0)))), x1))
   F.0-0(h.0(h.0(x0)), f.0-1(f.0-1(y_1, a.), a.)) -> F.0-0(h.0(h.0(h.0(x0))), f.0-1(y_1, a.))
   F.0-0(h.0(h.0(h.0(z0))), f.0-1(x1, a.)) -> F.1-0(a., f.0-0(h.0(h.0(h.0(h.0(z0)))), x1))
   F.0-0(h.0(h.0(x0)), f.0-1(f.1-1(y_1, a.), a.)) -> F.0-0(h.0(h.0(h.0(x0))), f.1-1(y_1, a.))
   F.0-0(h.0(h.1(x0)), f.0-1(f.1-1(y_1, a.), a.)) -> F.0-0(h.0(h.0(h.1(x0))), f.1-1(y_1, a.))
   F.0-0(h.1(a.), f.0-1(f.1-1(y_0, a.), a.)) -> F.0-0(h.0(h.1(a.)), f.1-1(y_0, a.))
   F.0-0(h.1(a.), f.0-1(f.0-1(f.0-1(y_1, a.), a.), a.)) -> F.0-0(h.0(h.1(a.)), f.0-1(f.0-1(y_1, a.), a.))
   F.0-0(h.1(a.), f.0-1(f.0-1(f.1-1(y_1, a.), a.), a.)) -> F.0-0(h.0(h.1(a.)), f.0-1(f.1-1(y_1, a.), a.))

The TRS R consists of the following rules:

   f.0-0(x, f.0-1(y, a.)) -> f.0-1(f.1-0(a., f.0-0(h.0(x), y)), a.)
   f.1-0(x, f.0-1(y, a.)) -> f.0-1(f.1-0(a., f.0-0(h.1(x), y)), a.)

The set Q consists of the following terms:

   f.0-0(x0, f.0-1(x1, a.))
   f.0-0(x0, f.1-1(x1, a.))
   f.1-0(x0, f.0-1(x1, a.))
   f.1-0(x0, f.1-1(x1, a.))

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

(27) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.
----------------------------------------

(28)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F.1-0(a., f.0-1(x1, a.)) -> F.1-0(a., f.0-0(h.1(a.), x1))
   F.1-0(a., f.0-1(f.0-1(y_0, a.), a.)) -> F.0-0(h.1(a.), f.0-1(y_0, a.))
   F.0-0(h.1(a.), f.0-1(x1, a.)) -> F.1-0(a., f.0-0(h.0(h.1(a.)), x1))
   F.0-0(h.1(a.), f.0-1(f.0-1(y_0, a.), a.)) -> F.0-0(h.0(h.1(a.)), f.0-1(y_0, a.))
   F.0-0(h.0(h.1(a.)), f.0-1(x1, a.)) -> F.1-0(a., f.0-0(h.0(h.0(h.1(a.))), x1))
   F.0-0(h.0(h.1(x0)), f.0-1(f.0-1(y_1, a.), a.)) -> F.0-0(h.0(h.0(h.1(x0))), f.0-1(y_1, a.))
   F.0-0(h.0(h.0(h.1(z0))), f.0-1(x1, a.)) -> F.1-0(a., f.0-0(h.0(h.0(h.0(h.1(z0)))), x1))
   F.0-0(h.0(h.0(x0)), f.0-1(f.0-1(y_1, a.), a.)) -> F.0-0(h.0(h.0(h.0(x0))), f.0-1(y_1, a.))
   F.0-0(h.0(h.0(h.0(z0))), f.0-1(x1, a.)) -> F.1-0(a., f.0-0(h.0(h.0(h.0(h.0(z0)))), x1))
   F.0-0(h.1(a.), f.0-1(f.0-1(f.0-1(y_1, a.), a.), a.)) -> F.0-0(h.0(h.1(a.)), f.0-1(f.0-1(y_1, a.), a.))
   F.0-0(h.1(a.), f.0-1(f.0-1(f.1-1(y_1, a.), a.), a.)) -> F.0-0(h.0(h.1(a.)), f.0-1(f.1-1(y_1, a.), a.))

The TRS R consists of the following rules:

   f.0-0(x, f.0-1(y, a.)) -> f.0-1(f.1-0(a., f.0-0(h.0(x), y)), a.)
   f.1-0(x, f.0-1(y, a.)) -> f.0-1(f.1-0(a., f.0-0(h.1(x), y)), a.)

The set Q consists of the following terms:

   f.0-0(x0, f.0-1(x1, a.))
   f.0-0(x0, f.1-1(x1, a.))
   f.1-0(x0, f.0-1(x1, a.))
   f.1-0(x0, f.1-1(x1, a.))

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

(29) MRRProof (EQUIVALENT)
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(a., f.0-1(x1, a.)) -> F.1-0(a., f.0-0(h.1(a.), x1))
   F.1-0(a., f.0-1(f.0-1(y_0, a.), a.)) -> F.0-0(h.1(a.), f.0-1(y_0, a.))
   F.0-0(h.1(a.), f.0-1(x1, a.)) -> F.1-0(a., f.0-0(h.0(h.1(a.)), x1))
   F.0-0(h.1(a.), f.0-1(f.0-1(y_0, a.), a.)) -> F.0-0(h.0(h.1(a.)), f.0-1(y_0, a.))
   F.0-0(h.0(h.1(a.)), f.0-1(x1, a.)) -> F.1-0(a., f.0-0(h.0(h.0(h.1(a.))), x1))
   F.0-0(h.0(h.1(x0)), f.0-1(f.0-1(y_1, a.), a.)) -> F.0-0(h.0(h.0(h.1(x0))), f.0-1(y_1, a.))
   F.0-0(h.0(h.0(h.1(z0))), f.0-1(x1, a.)) -> F.1-0(a., f.0-0(h.0(h.0(h.0(h.1(z0)))), x1))
   F.0-0(h.0(h.0(x0)), f.0-1(f.0-1(y_1, a.), a.)) -> F.0-0(h.0(h.0(h.0(x0))), f.0-1(y_1, a.))
   F.0-0(h.0(h.0(h.0(z0))), f.0-1(x1, a.)) -> F.1-0(a., f.0-0(h.0(h.0(h.0(h.0(z0)))), x1))
   F.0-0(h.1(a.), f.0-1(f.0-1(f.0-1(y_1, a.), a.), a.)) -> F.0-0(h.0(h.1(a.)), f.0-1(f.0-1(y_1, a.), a.))
   F.0-0(h.1(a.), f.0-1(f.0-1(f.1-1(y_1, a.), a.), a.)) -> F.0-0(h.0(h.1(a.)), f.0-1(f.1-1(y_1, a.), a.))


Used ordering: Polynomial interpretation [POLO]:

   POL(F.0-0(x_1, x_2)) = 1 + x_1 + x_2
   POL(F.1-0(x_1, x_2)) = 1 + x_1 + x_2
   POL(a.) = 0
   POL(f.0-0(x_1, x_2)) = x_1 + x_2
   POL(f.0-1(x_1, x_2)) = 1 + x_1 + x_2
   POL(f.1-0(x_1, x_2)) = x_1 + x_2
   POL(f.1-1(x_1, x_2)) = x_1 + x_2
   POL(h.0(x_1)) = x_1
   POL(h.1(x_1)) = x_1


----------------------------------------

(30)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   f.0-0(x, f.0-1(y, a.)) -> f.0-1(f.1-0(a., f.0-0(h.0(x), y)), a.)
   f.1-0(x, f.0-1(y, a.)) -> f.0-1(f.1-0(a., f.0-0(h.1(x), y)), a.)

The set Q consists of the following terms:

   f.0-0(x0, f.0-1(x1, a.))
   f.0-0(x0, f.1-1(x1, a.))
   f.1-0(x0, f.0-1(x1, a.))
   f.1-0(x0, f.1-1(x1, a.))

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

(31) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(32)
YES