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:

tower(x) → f(a, x, s(0))
f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(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:

tower(x) → f(a, x, s(0))
f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

tower(x0)
f(a, 0, x0)
f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
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:

TOWER(x) → F(a, x, s(0))
F(a, s(x), y) → F(b, y, s(x))
F(b, y, x) → F(a, half(x), exp(y))
F(b, y, x) → HALF(x)
F(b, y, x) → EXP(y)
EXP(s(x)) → DOUBLE(exp(x))
EXP(s(x)) → EXP(x)
DOUBLE(s(x)) → DOUBLE(x)
HALF(0) → DOUBLE(0)
HALF(s(0)) → HALF(0)
HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

tower(x) → f(a, x, s(0))
f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

tower(x0)
f(a, 0, x0)
f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
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 4 SCCs with 6 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

DOUBLE(s(x)) → DOUBLE(x)

The TRS R consists of the following rules:

tower(x) → f(a, x, s(0))
f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

tower(x0)
f(a, 0, x0)
f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
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.

DOUBLE(s(x)) → DOUBLE(x)

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

s₁ > DOUBLE₁

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

tower(x) → f(a, x, s(0))
f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

tower(x0)
f(a, 0, x0)
f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
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:

HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

tower(x) → f(a, x, s(0))
f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

tower(x0)
f(a, 0, x0)
f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
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.

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

(14) Obligation:

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

tower(x) → f(a, x, s(0))
f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

tower(x0)
f(a, 0, x0)
f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
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:

EXP(s(x)) → EXP(x)

The TRS R consists of the following rules:

tower(x) → f(a, x, s(0))
f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

tower(x0)
f(a, 0, x0)
f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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.

EXP(s(x)) → EXP(x)

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

s₁ > EXP₁

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

tower(x) → f(a, x, s(0))
f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

tower(x0)
f(a, 0, x0)
f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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(b, y, x) → F(a, half(x), exp(y))
F(a, s(x), y) → F(b, y, s(x))

The TRS R consists of the following rules:

tower(x) → f(a, x, s(0))
f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

tower(x0)
f(a, 0, x0)
f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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