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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
    ↳5 QDP
        ↳6 QDPOrderProof (⇔)
        ↳7 QDP
        ↳8 PisEmptyProof (⇔)
        ↳9 TRUE
    ↳10 QDP
        ↳11 QDPOrderProof (⇔)
        ↳12 QDP
        ↳13 PisEmptyProof (⇔)
        ↳14 TRUE
    ↳15 QDP
        ↳16 QDPOrderProof (⇔)
        ↳17 QDP
        ↳18 PisEmptyProof (⇔)
        ↳19 TRUE
    ↳20 QDP
        ↳21 QDPOrderProof (⇔)
        ↳22 QDP
        ↳23 DependencyGraphProof (⇔)
        ↳24 QDP
        ↳25 QDPOrderProof (⇔)
        ↳26 QDP
        ↳27 PisEmptyProof (⇔)
        ↳28 TRUE

(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) 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(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)))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

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

(4) Complex Obligation (AND)

(5) 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)))

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

(6) 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: SCNP Order with the following components:
Level mapping:
Top level AFS:
F(x0, x1, x2, x3)  =  F(x2, x3)

Tags:
F has argument tags [2,1,0,0] and root tag 0

Comparison: MS
Underlying order for the size change arcs and the rules of R:
Recursive path order with status [RPO].
Quasi-Precedence:
F3 > s1 > 0

Status:
F3: multiset
0: multiset
s1: multiset


The following usable rules [FROCOS05] were oriented: none

(7) 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)))

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) 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)))

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, s(s(y)), s(0)) → F(0, y, s(0))
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
F(x0, x1, x2, x3)  =  F(x2)

Tags:
F has argument tags [0,1,2,2] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Recursive path order with status [RPO].
Quasi-Precedence:
0 > [F3, s1]

Status:
F3: [3,1,2]
0: multiset
s1: multiset


The following usable rules [FROCOS05] were oriented: none

(12) 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)))

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) 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)))

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

(16) 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))
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: SCNP Order with the following components:
Level mapping:
Top level AFS:
F(x0, x1, x2, x3)  =  F(x0, x1, x2, x3)

Tags:
F has argument tags [0,1,1,1] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Recursive path order with status [RPO].
Quasi-Precedence:
0 > [F3, f3] > s1

Status:
F3: [1,2,3]
0: multiset
s1: multiset
f3: [1,2,3]


The following usable rules [FROCOS05] were oriented:

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

(17) 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)))

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

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE

(20) 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)))

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

(21) 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), 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: SCNP Order with the following components:
Level mapping:
Top level AFS:
F(x0, x1, x2, x3)  =  F(x1, x2)

Tags:
F has argument tags [0,3,3,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
F(x1, x2, x3)  =  F
s(x1)  =  s(x1)
0  =  0
f(x1, x2, x3)  =  f(x1, x3)

Recursive path order with status [RPO].
Quasi-Precedence:
[F, s1] > [0, f2]

Status:
F: multiset
s1: [1]
0: multiset
f2: multiset


The following usable rules [FROCOS05] were oriented: none

(22) 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), 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)))

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

(23) DependencyGraphProof (EQUIVALENT transformation)

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

(24) 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)))

Q is empty.
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), s(y), s(z)) → F(s(x), s(y), z)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
F(x0, x1, x2, x3)  =  F(x3)

Tags:
F has argument tags [1,2,2,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
F(x1, x2, x3)  =  x2
s(x1)  =  s(x1)

Recursive path order with status [RPO].
Quasi-Precedence:
trivial

Status:
s1: multiset


The following usable rules [FROCOS05] were oriented: none

(26) 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)))

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

(27) PisEmptyProof (EQUIVALENT transformation)

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

(28) TRUE