Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

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

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
logarithm(x) → ifa(lt(0, x), x)
ifa(true, x) → help(x, 1)
ifa(false, x) → logZeroError
help(x, y) → ifb(lt(y, x), x, y)
ifb(true, x, y) → help(half(x), s(y))
ifb(false, x, y) → y
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → 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:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
logarithm(x) → ifa(lt(0, x), x)
ifa(true, x) → help(x, 1)
ifa(false, x) → logZeroError
help(x, y) → ifb(lt(y, x), x, y)
ifb(true, x, y) → help(half(x), s(y))
ifb(false, x, y) → y
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(0))
half(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:

LT(s(x), s(y)) → LT(x, y)
LOGARITHM(x) → IFA(lt(0, x), x)
LOGARITHM(x) → LT(0, x)
IFA(true, x) → HELP(x, 1)
HELP(x, y) → IFB(lt(y, x), x, y)
HELP(x, y) → LT(y, x)
IFB(true, x, y) → HELP(half(x), s(y))
IFB(true, x, y) → HALF(x)
HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
logarithm(x) → ifa(lt(0, x), x)
ifa(true, x) → help(x, 1)
ifa(false, x) → logZeroError
help(x, y) → ifb(lt(y, x), x, y)
ifb(true, x, y) → help(half(x), s(y))
ifb(false, x, y) → y
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(0))
half(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 3 SCCs with 5 less nodes.

(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:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
logarithm(x) → ifa(lt(0, x), x)
ifa(true, x) → help(x, 1)
ifa(false, x) → logZeroError
help(x, y) → ifb(lt(y, x), x, y)
ifb(true, x, y) → help(half(x), s(y))
ifb(false, x, y) → y
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(0))
half(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:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
logarithm(x) → ifa(lt(0, x), x)
ifa(true, x) → help(x, 1)
ifa(false, x) → logZeroError
help(x, y) → ifb(lt(y, x), x, y)
ifb(true, x, y) → help(half(x), s(y))
ifb(false, x, y) → y
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(0))
half(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:

LT(s(x), s(y)) → LT(x, y)

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
logarithm(x) → ifa(lt(0, x), x)
ifa(true, x) → help(x, 1)
ifa(false, x) → logZeroError
help(x, y) → ifb(lt(y, x), x, y)
ifb(true, x, y) → help(half(x), s(y))
ifb(false, x, y) → y
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(0))
half(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.

LT(s(x), s(y)) → LT(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LT(x1, x2)  =  x1
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
logarithm(x) → ifa(lt(0, x), x)
ifa(true, x) → help(x, 1)
ifa(false, x) → logZeroError
help(x, y) → ifb(lt(y, x), x, y)
ifb(true, x, y) → help(half(x), s(y))
ifb(false, x, y) → y
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(0))
half(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

(17) Obligation:

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

HELP(x, y) → IFB(lt(y, x), x, y)
IFB(true, x, y) → HELP(half(x), s(y))

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
logarithm(x) → ifa(lt(0, x), x)
ifa(true, x) → help(x, 1)
ifa(false, x) → logZeroError
help(x, y) → ifb(lt(y, x), x, y)
ifb(true, x, y) → help(half(x), s(y))
ifb(false, x, y) → y
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(0))
half(s(s(x0)))

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