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:

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) 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:

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

The set Q consists of the following terms:

half(0)
half(s(s(x0)))
log(s(0))
log(s(s(x0)))

(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:

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))))

The set Q consists of the following terms:

half(0)
half(s(s(x0)))
log(s(0))
log(s(s(x0)))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.

(6) Complex Obligation (AND)

(7) 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))))

The set Q consists of the following terms:

half(0)
half(s(s(x0)))
log(s(0))
log(s(s(x0)))

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.

HALF(s(s(x))) → HALF(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Recursive Path Order [RPO].
Precedence:

s₁ > HALF₁

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

Q DP problem:
P is empty.
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))))

The set Q consists of the following terms:

half(0)
half(s(s(x0)))
log(s(0))
log(s(s(x0)))

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:

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))))

The set Q consists of the following terms:

half(0)
half(s(s(x0)))
log(s(0))
log(s(s(x0)))

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.

LOG(s(s(x))) → LOG(s(half(x)))

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

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented:

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

(14) Obligation:

Q DP problem:
P is empty.
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))))

The set Q consists of the following terms:

half(0)
half(s(s(x0)))
log(s(0))
log(s(s(x0)))

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