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

(0) Obligation:

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

f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

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, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

The set Q consists of the following terms:

f(x0, 0, 0)
f(0, x0, 0)
f(0, 0, x0)
f(s(0), x0, x1)
f(s(x0), s(x1), 0)
f(s(x0), 0, s(x1))
f(0, s(0), s(0))
f(s(x0), s(x1), s(x2))
f(0, s(s(x0)), s(0))
f(0, s(0), s(s(x0)))
f(0, s(s(x0)), s(s(x1)))

(3) DependencyPairsProof (EQUIVALENT transformation)

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(s(0), y, z) → F(0, s(y), s(z))
F(s(x), s(y), 0) → F(x, y, s(0))
F(s(x), 0, s(z)) → F(x, s(0), z)
F(s(x), s(y), s(z)) → F(x, y, f(s(x), s(y), z))
F(s(x), s(y), s(z)) → F(s(x), s(y), z)
F(0, s(s(y)), s(0)) → F(0, y, s(0))
F(0, s(0), s(s(z))) → F(0, s(0), z)
F(0, s(s(y)), s(s(z))) → F(0, y, f(0, s(s(y)), s(z)))
F(0, s(s(y)), s(s(z))) → F(0, s(s(y)), s(z))

The TRS R consists of the following rules:

f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

The set Q consists of the following terms:

f(x0, 0, 0)
f(0, x0, 0)
f(0, 0, x0)
f(s(0), x0, x1)
f(s(x0), s(x1), 0)
f(s(x0), 0, s(x1))
f(0, s(0), s(0))
f(s(x0), s(x1), s(x2))
f(0, s(s(x0)), s(0))
f(0, s(0), s(s(x0)))
f(0, s(s(x0)), s(s(x1)))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 1 less node.

(6) Complex Obligation (AND)

(7) Obligation:

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

F(0, s(0), s(s(z))) → F(0, s(0), z)

The TRS R consists of the following rules:

f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

The set Q consists of the following terms:

f(x0, 0, 0)
f(0, x0, 0)
f(0, 0, x0)
f(s(0), x0, x1)
f(s(x0), s(x1), 0)
f(s(x0), 0, s(x1))
f(0, s(0), s(0))
f(s(x0), s(x1), s(x2))
f(0, s(s(x0)), s(0))
f(0, s(0), s(s(x0)))
f(0, s(s(x0)), s(s(x1)))

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

F(0, s(0), s(s(z))) → F(0, s(0), z)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3)  =  F(x3)
0  =  0
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:

0 > [F₁, s₁]

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

The set Q consists of the following terms:

f(x0, 0, 0)
f(0, x0, 0)
f(0, 0, x0)
f(s(0), x0, x1)
f(s(x0), s(x1), 0)
f(s(x0), 0, s(x1))
f(0, s(0), s(0))
f(s(x0), s(x1), s(x2))
f(0, s(s(x0)), s(0))
f(0, s(0), s(s(x0)))
f(0, s(s(x0)), s(s(x1)))

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

F(0, s(s(y)), s(0)) → F(0, y, s(0))

The TRS R consists of the following rules:

f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

The set Q consists of the following terms:

f(x0, 0, 0)
f(0, x0, 0)
f(0, 0, x0)
f(s(0), x0, x1)
f(s(x0), s(x1), 0)
f(s(x0), 0, s(x1))
f(0, s(0), s(0))
f(s(x0), s(x1), s(x2))
f(0, s(s(x0)), s(0))
f(0, s(0), s(s(x0)))
f(0, s(s(x0)), s(s(x1)))

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, s(s(y)), s(0)) → F(0, y, s(0))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3)  =  F(x2, x3)
0  =  0
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:

s₁ > [F₂, 0]

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

The set Q consists of the following terms:

f(x0, 0, 0)
f(0, x0, 0)
f(0, 0, x0)
f(s(0), x0, x1)
f(s(x0), s(x1), 0)
f(s(x0), 0, s(x1))
f(0, s(0), s(0))
f(s(x0), s(x1), s(x2))
f(0, s(s(x0)), s(0))
f(0, s(0), s(s(x0)))
f(0, s(s(x0)), s(s(x1)))

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) Obligation:

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

F(0, s(s(y)), s(s(z))) → F(0, s(s(y)), s(z))
F(0, s(s(y)), s(s(z))) → F(0, y, f(0, s(s(y)), s(z)))

The TRS R consists of the following rules:

f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

The set Q consists of the following terms:

f(x0, 0, 0)
f(0, x0, 0)
f(0, 0, x0)
f(s(0), x0, x1)
f(s(x0), s(x1), 0)
f(s(x0), 0, s(x1))
f(0, s(0), s(0))
f(s(x0), s(x1), s(x2))
f(0, s(s(x0)), s(0))
f(0, s(0), s(s(x0)))
f(0, s(s(x0)), s(s(x1)))

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, s(s(y)), s(s(z))) → F(0, y, f(0, s(s(y)), s(z)))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3)  =  F(x2)
0  =  0
s(x1)  =  s(x1)
f(x1, x2, x3)  =  x1

Recursive Path Order [RPO].
Precedence:

0 > s₁ > F₁

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

F(0, s(s(y)), s(s(z))) → F(0, s(s(y)), s(z))

The TRS R consists of the following rules:

f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

The set Q consists of the following terms:

f(x0, 0, 0)
f(0, x0, 0)
f(0, 0, x0)
f(s(0), x0, x1)
f(s(x0), s(x1), 0)
f(s(x0), 0, s(x1))
f(0, s(0), s(0))
f(s(x0), s(x1), s(x2))
f(0, s(s(x0)), s(0))
f(0, s(0), s(s(x0)))
f(0, s(s(x0)), s(s(x1)))

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

(20) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

F(0, s(s(y)), s(s(z))) → F(0, s(s(y)), s(z))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3)  =  F(x1, x3)
0  =  0
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:

[0, s₁] > F₂

The following usable rules [FROCOS05] were oriented: none

(21) Obligation:

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

f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

The set Q consists of the following terms:

f(x0, 0, 0)
f(0, x0, 0)
f(0, 0, x0)
f(s(0), x0, x1)
f(s(x0), s(x1), 0)
f(s(x0), 0, s(x1))
f(0, s(0), s(0))
f(s(x0), s(x1), s(x2))
f(0, s(s(x0)), s(0))
f(0, s(0), s(s(x0)))
f(0, s(s(x0)), s(s(x1)))

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

(22) PisEmptyProof (EQUIVALENT transformation)

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

(23) TRUE

(24) Obligation:

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

F(s(x), 0, s(z)) → F(x, s(0), z)
F(s(x), s(y), 0) → F(x, y, s(0))
F(s(x), s(y), s(z)) → F(x, y, f(s(x), s(y), z))
F(s(x), s(y), s(z)) → F(s(x), s(y), z)

The TRS R consists of the following rules:

f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

The set Q consists of the following terms:

f(x0, 0, 0)
f(0, x0, 0)
f(0, 0, x0)
f(s(0), x0, x1)
f(s(x0), s(x1), 0)
f(s(x0), 0, s(x1))
f(0, s(0), s(0))
f(s(x0), s(x1), s(x2))
f(0, s(s(x0)), s(0))
f(0, s(0), s(s(x0)))
f(0, s(s(x0)), s(s(x1)))

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

(25) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

F(s(x), 0, s(z)) → F(x, s(0), z)
F(s(x), s(y), 0) → F(x, y, s(0))
F(s(x), s(y), s(z)) → F(x, y, f(s(x), s(y), z))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3)  =  F(x1)
s(x1)  =  s(x1)
0  =  0
f(x1, x2, x3)  =  f(x1, x3)

Recursive Path Order [RPO].
Precedence:

[F₁, s₁, f₂] > 0

The following usable rules [FROCOS05] were oriented: none

(26) Obligation:

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

F(s(x), s(y), s(z)) → F(s(x), s(y), z)

The TRS R consists of the following rules:

f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

The set Q consists of the following terms:

f(x0, 0, 0)
f(0, x0, 0)
f(0, 0, x0)
f(s(0), x0, x1)
f(s(x0), s(x1), 0)
f(s(x0), 0, s(x1))
f(0, s(0), s(0))
f(s(x0), s(x1), s(x2))
f(0, s(s(x0)), s(0))
f(0, s(0), s(s(x0)))
f(0, s(s(x0)), s(s(x1)))

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

(27) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

F(s(x), s(y), s(z)) → F(s(x), s(y), z)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3)  =  F(x3)
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:

s₁ > F₁

The following usable rules [FROCOS05] were oriented: none

(28) Obligation:

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

f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

The set Q consists of the following terms:

f(x0, 0, 0)
f(0, x0, 0)
f(0, 0, x0)
f(s(0), x0, x1)
f(s(x0), s(x1), 0)
f(s(x0), 0, s(x1))
f(0, s(0), s(0))
f(s(x0), s(x1), s(x2))
f(0, s(s(x0)), s(0))
f(0, s(0), s(s(x0)))
f(0, s(s(x0)), s(s(x1)))

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) TRUE