YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 8 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) QDP
(5) TransformationProof [EQUIVALENT, 0 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) QDP
(9) UsableRulesProof [EQUIVALENT, 0 ms]
(10) QDP
(11) QReductionProof [EQUIVALENT, 0 ms]
(12) QDP
(13) TransformationProof [EQUIVALENT, 0 ms]
(14) QDP
(15) TransformationProof [EQUIVALENT, 0 ms]
(16) QDP
(17) DependencyGraphProof [EQUIVALENT, 0 ms]
(18) TRUE


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

   f(a(x), y, s(z), u) -> f(a(b), y, z, g(x, y, s(z), u))
   g(x, y, z, u) -> h(x, y, z, u)
   h(b, y, z, u) -> f(y, y, z, u)
   a(b) -> c

The set Q consists of the following terms:

   f(a(x0), x1, s(x2), x3)
   g(x0, x1, x2, x3)
   h(b, x0, x1, x2)
   a(b)


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

(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(a(x), y, s(z), u) -> F(a(b), y, z, g(x, y, s(z), u))
   F(a(x), y, s(z), u) -> A(b)
   F(a(x), y, s(z), u) -> G(x, y, s(z), u)
   G(x, y, z, u) -> H(x, y, z, u)
   H(b, y, z, u) -> F(y, y, z, u)

The TRS R consists of the following rules:

   f(a(x), y, s(z), u) -> f(a(b), y, z, g(x, y, s(z), u))
   g(x, y, z, u) -> h(x, y, z, u)
   h(b, y, z, u) -> f(y, y, z, u)
   a(b) -> c

The set Q consists of the following terms:

   f(a(x0), x1, s(x2), x3)
   g(x0, x1, x2, x3)
   h(b, x0, x1, x2)
   a(b)

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

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

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

   F(a(x), y, s(z), u) -> G(x, y, s(z), u)
   G(x, y, z, u) -> H(x, y, z, u)
   H(b, y, z, u) -> F(y, y, z, u)
   F(a(x), y, s(z), u) -> F(a(b), y, z, g(x, y, s(z), u))

The TRS R consists of the following rules:

   f(a(x), y, s(z), u) -> f(a(b), y, z, g(x, y, s(z), u))
   g(x, y, z, u) -> h(x, y, z, u)
   h(b, y, z, u) -> f(y, y, z, u)
   a(b) -> c

The set Q consists of the following terms:

   f(a(x0), x1, s(x2), x3)
   g(x0, x1, x2, x3)
   h(b, x0, x1, x2)
   a(b)

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

(5) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule F(a(x), y, s(z), u) -> F(a(b), y, z, g(x, y, s(z), u)) at position [0] we obtained the following new rules [LPAR04]:

   (F(a(x), y, s(z), u) -> F(c, y, z, g(x, y, s(z), u)),F(a(x), y, s(z), u) -> F(c, y, z, g(x, y, s(z), u)))


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

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

   F(a(x), y, s(z), u) -> G(x, y, s(z), u)
   G(x, y, z, u) -> H(x, y, z, u)
   H(b, y, z, u) -> F(y, y, z, u)
   F(a(x), y, s(z), u) -> F(c, y, z, g(x, y, s(z), u))

The TRS R consists of the following rules:

   f(a(x), y, s(z), u) -> f(a(b), y, z, g(x, y, s(z), u))
   g(x, y, z, u) -> h(x, y, z, u)
   h(b, y, z, u) -> f(y, y, z, u)
   a(b) -> c

The set Q consists of the following terms:

   f(a(x0), x1, s(x2), x3)
   g(x0, x1, x2, x3)
   h(b, x0, x1, x2)
   a(b)

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

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

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

   G(x, y, z, u) -> H(x, y, z, u)
   H(b, y, z, u) -> F(y, y, z, u)
   F(a(x), y, s(z), u) -> G(x, y, s(z), u)

The TRS R consists of the following rules:

   f(a(x), y, s(z), u) -> f(a(b), y, z, g(x, y, s(z), u))
   g(x, y, z, u) -> h(x, y, z, u)
   h(b, y, z, u) -> f(y, y, z, u)
   a(b) -> c

The set Q consists of the following terms:

   f(a(x0), x1, s(x2), x3)
   g(x0, x1, x2, x3)
   h(b, x0, x1, x2)
   a(b)

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

(9) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

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

   G(x, y, z, u) -> H(x, y, z, u)
   H(b, y, z, u) -> F(y, y, z, u)
   F(a(x), y, s(z), u) -> G(x, y, s(z), u)

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

   f(a(x0), x1, s(x2), x3)
   g(x0, x1, x2, x3)
   h(b, x0, x1, x2)
   a(b)

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

(11) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   f(a(x0), x1, s(x2), x3)
   g(x0, x1, x2, x3)
   h(b, x0, x1, x2)


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

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

   G(x, y, z, u) -> H(x, y, z, u)
   H(b, y, z, u) -> F(y, y, z, u)
   F(a(x), y, s(z), u) -> G(x, y, s(z), u)

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

   a(b)

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

(13) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule F(a(x), y, s(z), u) -> G(x, y, s(z), u) we obtained the following new rules [LPAR04]:

   (F(a(x0), a(x0), s(x2), z2) -> G(x0, a(x0), s(x2), z2),F(a(x0), a(x0), s(x2), z2) -> G(x0, a(x0), s(x2), z2))


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

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

   G(x, y, z, u) -> H(x, y, z, u)
   H(b, y, z, u) -> F(y, y, z, u)
   F(a(x0), a(x0), s(x2), z2) -> G(x0, a(x0), s(x2), z2)

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

   a(b)

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

(15) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule G(x, y, z, u) -> H(x, y, z, u) we obtained the following new rules [LPAR04]:

   (G(z0, a(z0), s(z1), z2) -> H(z0, a(z0), s(z1), z2),G(z0, a(z0), s(z1), z2) -> H(z0, a(z0), s(z1), z2))


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

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

   H(b, y, z, u) -> F(y, y, z, u)
   F(a(x0), a(x0), s(x2), z2) -> G(x0, a(x0), s(x2), z2)
   G(z0, a(z0), s(z1), z2) -> H(z0, a(z0), s(z1), z2)

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

   a(b)

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

(17) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 3 less nodes.
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(18)
TRUE