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:

log(x, s(s(y))) → cond(le(x, s(s(y))), x, y)
cond(true, x, y) → s(0)
cond(false, x, y) → double(log(x, square(s(s(y)))))
le(0, v) → true
le(s(u), 0) → false
le(s(u), s(v)) → le(u, v)
double(0) → 0
double(s(x)) → s(s(double(x)))
square(0) → 0
square(s(x)) → s(plus(square(x), double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

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:

log(x, s(s(y))) → cond(le(x, s(s(y))), x, y)
cond(true, x, y) → s(0)
cond(false, x, y) → double(log(x, square(s(s(y)))))
le(0, v) → true
le(s(u), 0) → false
le(s(u), s(v)) → le(u, v)
double(0) → 0
double(s(x)) → s(s(double(x)))
square(0) → 0
square(s(x)) → s(plus(square(x), double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

log(x0, s(s(x1)))
cond(true, x0, x1)
cond(false, x0, x1)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
double(0)
double(s(x0))
square(0)
square(s(x0))
plus(x0, 0)
plus(x0, s(x1))

(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:

LOG(x, s(s(y))) → COND(le(x, s(s(y))), x, y)
LOG(x, s(s(y))) → LE(x, s(s(y)))
COND(false, x, y) → DOUBLE(log(x, square(s(s(y)))))
COND(false, x, y) → LOG(x, square(s(s(y))))
COND(false, x, y) → SQUARE(s(s(y)))
LE(s(u), s(v)) → LE(u, v)
DOUBLE(s(x)) → DOUBLE(x)
SQUARE(s(x)) → PLUS(square(x), double(x))
SQUARE(s(x)) → SQUARE(x)
SQUARE(s(x)) → DOUBLE(x)
PLUS(n, s(m)) → PLUS(n, m)

The TRS R consists of the following rules:

log(x, s(s(y))) → cond(le(x, s(s(y))), x, y)
cond(true, x, y) → s(0)
cond(false, x, y) → double(log(x, square(s(s(y)))))
le(0, v) → true
le(s(u), 0) → false
le(s(u), s(v)) → le(u, v)
double(0) → 0
double(s(x)) → s(s(double(x)))
square(0) → 0
square(s(x)) → s(plus(square(x), double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

log(x0, s(s(x1)))
cond(true, x0, x1)
cond(false, x0, x1)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
double(0)
double(s(x0))
square(0)
square(s(x0))
plus(x0, 0)
plus(x0, s(x1))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 5 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

PLUS(n, s(m)) → PLUS(n, m)

The TRS R consists of the following rules:

log(x, s(s(y))) → cond(le(x, s(s(y))), x, y)
cond(true, x, y) → s(0)
cond(false, x, y) → double(log(x, square(s(s(y)))))
le(0, v) → true
le(s(u), 0) → false
le(s(u), s(v)) → le(u, v)
double(0) → 0
double(s(x)) → s(s(double(x)))
square(0) → 0
square(s(x)) → s(plus(square(x), double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

log(x0, s(s(x1)))
cond(true, x0, x1)
cond(false, x0, x1)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
double(0)
double(s(x0))
square(0)
square(s(x0))
plus(x0, 0)
plus(x0, s(x1))

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.

PLUS(n, s(m)) → PLUS(n, m)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

s₁ > PLUS₂

Status:

PLUS₂: [1,2]
s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

log(x, s(s(y))) → cond(le(x, s(s(y))), x, y)
cond(true, x, y) → s(0)
cond(false, x, y) → double(log(x, square(s(s(y)))))
le(0, v) → true
le(s(u), 0) → false
le(s(u), s(v)) → le(u, v)
double(0) → 0
double(s(x)) → s(s(double(x)))
square(0) → 0
square(s(x)) → s(plus(square(x), double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

log(x0, s(s(x1)))
cond(true, x0, x1)
cond(false, x0, x1)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
double(0)
double(s(x0))
square(0)
square(s(x0))
plus(x0, 0)
plus(x0, s(x1))

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:

DOUBLE(s(x)) → DOUBLE(x)

The TRS R consists of the following rules:

log(x, s(s(y))) → cond(le(x, s(s(y))), x, y)
cond(true, x, y) → s(0)
cond(false, x, y) → double(log(x, square(s(s(y)))))
le(0, v) → true
le(s(u), 0) → false
le(s(u), s(v)) → le(u, v)
double(0) → 0
double(s(x)) → s(s(double(x)))
square(0) → 0
square(s(x)) → s(plus(square(x), double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

log(x0, s(s(x1)))
cond(true, x0, x1)
cond(false, x0, x1)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
double(0)
double(s(x0))
square(0)
square(s(x0))
plus(x0, 0)
plus(x0, s(x1))

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.

DOUBLE(s(x)) → DOUBLE(x)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

log(x, s(s(y))) → cond(le(x, s(s(y))), x, y)
cond(true, x, y) → s(0)
cond(false, x, y) → double(log(x, square(s(s(y)))))
le(0, v) → true
le(s(u), 0) → false
le(s(u), s(v)) → le(u, v)
double(0) → 0
double(s(x)) → s(s(double(x)))
square(0) → 0
square(s(x)) → s(plus(square(x), double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

log(x0, s(s(x1)))
cond(true, x0, x1)
cond(false, x0, x1)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
double(0)
double(s(x0))
square(0)
square(s(x0))
plus(x0, 0)
plus(x0, s(x1))

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:

SQUARE(s(x)) → SQUARE(x)

The TRS R consists of the following rules:

log(x, s(s(y))) → cond(le(x, s(s(y))), x, y)
cond(true, x, y) → s(0)
cond(false, x, y) → double(log(x, square(s(s(y)))))
le(0, v) → true
le(s(u), 0) → false
le(s(u), s(v)) → le(u, v)
double(0) → 0
double(s(x)) → s(s(double(x)))
square(0) → 0
square(s(x)) → s(plus(square(x), double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

log(x0, s(s(x1)))
cond(true, x0, x1)
cond(false, x0, x1)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
double(0)
double(s(x0))
square(0)
square(s(x0))
plus(x0, 0)
plus(x0, s(x1))

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.

SQUARE(s(x)) → SQUARE(x)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

log(x, s(s(y))) → cond(le(x, s(s(y))), x, y)
cond(true, x, y) → s(0)
cond(false, x, y) → double(log(x, square(s(s(y)))))
le(0, v) → true
le(s(u), 0) → false
le(s(u), s(v)) → le(u, v)
double(0) → 0
double(s(x)) → s(s(double(x)))
square(0) → 0
square(s(x)) → s(plus(square(x), double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

log(x0, s(s(x1)))
cond(true, x0, x1)
cond(false, x0, x1)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
double(0)
double(s(x0))
square(0)
square(s(x0))
plus(x0, 0)
plus(x0, s(x1))

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:

LE(s(u), s(v)) → LE(u, v)

The TRS R consists of the following rules:

log(x, s(s(y))) → cond(le(x, s(s(y))), x, y)
cond(true, x, y) → s(0)
cond(false, x, y) → double(log(x, square(s(s(y)))))
le(0, v) → true
le(s(u), 0) → false
le(s(u), s(v)) → le(u, v)
double(0) → 0
double(s(x)) → s(s(double(x)))
square(0) → 0
square(s(x)) → s(plus(square(x), double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

log(x0, s(s(x1)))
cond(true, x0, x1)
cond(false, x0, x1)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
double(0)
double(s(x0))
square(0)
square(s(x0))
plus(x0, 0)
plus(x0, s(x1))

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

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

LE(s(u), s(v)) → LE(u, v)

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

Lexicographic path order with status [LPO].
Precedence:

s₁ > LE₁

Status:

s₁: [1]
LE₁: [1]

The following usable rules [FROCOS05] were oriented: none

(24) Obligation:

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

log(x, s(s(y))) → cond(le(x, s(s(y))), x, y)
cond(true, x, y) → s(0)
cond(false, x, y) → double(log(x, square(s(s(y)))))
le(0, v) → true
le(s(u), 0) → false
le(s(u), s(v)) → le(u, v)
double(0) → 0
double(s(x)) → s(s(double(x)))
square(0) → 0
square(s(x)) → s(plus(square(x), double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

log(x0, s(s(x1)))
cond(true, x0, x1)
cond(false, x0, x1)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
double(0)
double(s(x0))
square(0)
square(s(x0))
plus(x0, 0)
plus(x0, s(x1))

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

(25) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(26) TRUE

(27) Obligation:

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

COND(false, x, y) → LOG(x, square(s(s(y))))
LOG(x, s(s(y))) → COND(le(x, s(s(y))), x, y)

The TRS R consists of the following rules:

log(x, s(s(y))) → cond(le(x, s(s(y))), x, y)
cond(true, x, y) → s(0)
cond(false, x, y) → double(log(x, square(s(s(y)))))
le(0, v) → true
le(s(u), 0) → false
le(s(u), s(v)) → le(u, v)
double(0) → 0
double(s(x)) → s(s(double(x)))
square(0) → 0
square(s(x)) → s(plus(square(x), double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

log(x0, s(s(x1)))
cond(true, x0, x1)
cond(false, x0, x1)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
double(0)
double(s(x0))
square(0)
square(s(x0))
plus(x0, 0)
plus(x0, s(x1))

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