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 QDPOrderProof (⇔)
        ↳19 QDP
        ↳20 PisEmptyProof (⇔)
        ↳21 TRUE
    ↳22 QDP
        ↳23 QDPOrderProof (⇔)
        ↳24 QDP
        ↳25 DependencyGraphProof (⇔)
        ↳26 QDP
        ↳27 QDPOrderProof (⇔)
        ↳28 QDP
        ↳29 PisEmptyProof (⇔)
        ↳30 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(x0)

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

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(F(x1, x2, x3)) = x3   
POL(s(x1)) = 1 + x1   

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 [0,3,2,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(0) = 1   
POL(F(x1, x2, x3)) = 1 + x1 + x3   
POL(s(x1)) = 1 + x1   

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, 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(x2)

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

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(F(x1, x2, x3)) = 1 + x2   
POL(f(x1, x2, x3)) = 1 + x2 + x3   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(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))

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

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

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(F(x1, x2, x3)) = 1 + x2 + x3   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(19) 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.

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE

(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), 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.

(23) 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 [2,0,0,1] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(F(x1, x2, x3)) = 1   
POL(f(x1, x2, x3)) = 1   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(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), 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) DependencyGraphProof (EQUIVALENT transformation)

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

(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)))

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

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:
Polynomial interpretation [POLO]:

POL(F(x1, x2, x3)) = 0   
POL(s(x1)) = 1 + x1   

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)))

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) TRUE