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 QDPSizeChangeProof (⇔)
        ↳7 TRUE
    ↳8 QDP
        ↳9 QDPSizeChangeProof (⇔)
        ↳10 TRUE
    ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔)
        ↳13 TRUE
    ↳14 QDP
        ↳15 QDPSizeChangeProof (⇔)
        ↳16 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) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
0  =  0
s(x1)  =  s(x1)

From the DPs we obtained the following set of size-change graphs:

  • F(0, s(0), s(s(z))) → F(0, s(0), z) (allowed arguments on rhs = {1, 2, 3})
    The graph contains the following edges 1 >= 1, 2 > 1, 2 >= 2, 3 > 3

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(7) TRUE

(8) 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.

(9) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
0  =  0
s(x1)  =  s(x1)

From the DPs we obtained the following set of size-change graphs:

  • F(0, s(s(y)), s(0)) → F(0, y, s(0)) (allowed arguments on rhs = {1, 2, 3})
    The graph contains the following edges 1 >= 1, 3 > 1, 2 > 2, 3 >= 3

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(10) TRUE

(11) 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.

(12) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
0  =  0
s(x1)  =  s(x1)

From the DPs we obtained the following set of size-change graphs:

  • F(0, s(s(y)), s(s(z))) → F(0, s(s(y)), s(z)) (allowed arguments on rhs = {1, 2, 3})
    The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3

  • F(0, s(s(y)), s(s(z))) → F(0, y, f(0, s(s(y)), s(z))) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 >= 1, 2 > 2

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(13) TRUE

(14) 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.

(15) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
0  =  0
s(x1)  =  s(x1)

From the DPs we obtained the following set of size-change graphs:

  • F(s(x), s(y), s(z)) → F(s(x), s(y), z) (allowed arguments on rhs = {1, 2, 3})
    The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3

  • F(s(x), 0, s(z)) → F(x, s(0), z) (allowed arguments on rhs = {1, 2, 3})
    The graph contains the following edges 1 > 1, 3 > 3

  • F(s(x), s(y), s(z)) → F(x, y, f(s(x), s(y), z)) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 > 1, 2 > 2

  • F(s(x), s(y), 0) → F(x, y, s(0)) (allowed arguments on rhs = {1, 2, 3})
    The graph contains the following edges 1 > 1, 2 > 2

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(16) TRUE