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

(0) Obligation:

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

half(0) → 0
half(s(s(x))) → s(half(x))
log(s(0)) → 0
log(s(s(x))) → s(log(s(half(x))))

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:

HALF(s(s(x))) → HALF(x)
LOG(s(s(x))) → LOG(s(half(x)))
LOG(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

half(0) → 0
half(s(s(x))) → s(half(x))
log(s(0)) → 0
log(s(s(x))) → s(log(s(half(x))))

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 2 SCCs with 1 less node.

(4) Complex Obligation (AND)

(5) Obligation:

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

HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

half(0) → 0
half(s(s(x))) → s(half(x))
log(s(0)) → 0
log(s(s(x))) → s(log(s(half(x))))

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:
s(x1)  =  s(x1)

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

  • HALF(s(s(x))) → HALF(x) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

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:

LOG(s(s(x))) → LOG(s(half(x)))

The TRS R consists of the following rules:

half(0) → 0
half(s(s(x))) → s(half(x))
log(s(0)) → 0
log(s(s(x))) → s(log(s(half(x))))

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

(9) QDPSizeChangeProof (EQUIVALENT transformation)

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

Order:Polynomial interpretation [POLO]:


POL(0) = 0   
POL(half(x1)) = x1   
POL(s(x1)) = 1 + x1   

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

  • LOG(s(s(x))) → LOG(s(half(x))) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

We oriented the following set of usable rules [AAECC05,FROCOS05].


half(s(s(x))) → s(half(x))
half(0) → 0

(10) TRUE