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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, 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:

LE(s(x), s(y)) → LE(x, y)
INC(s(x)) → INC(x)
MINUS(s(x), s(y)) → MINUS(x, y)
QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
QUOT(s(x), s(y)) → MINUS(x, y)
LOG(x) → LOG2(x, 0)
LOG2(x, y) → IF(le(x, 0), le(x, s(0)), x, inc(y))
LOG2(x, y) → LE(x, 0)
LOG2(x, y) → LE(x, s(0))
LOG2(x, y) → INC(y)
IF(false, b, x, y) → IF2(b, x, y)
IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)
IF2(false, x, y) → QUOT(x, s(s(0)))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, 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 6 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, 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.

MINUS(s(x), s(y)) → MINUS(x, y)

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

Recursive path order with status [RPO].
Quasi-Precedence:

[MINUS₁, s₁]

Status:

s₁: [1]
MINUS₁: multiset

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, 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:

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, 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.

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
QUOT(x1, x2)  =  QUOT(x1, x2)
s(x1)  =  s(x1)
minus(x1, x2)  =  x1
0  =  0

Recursive path order with status [RPO].
Quasi-Precedence:

s₁ > QUOT₂

Status:

QUOT₂: [2,1]
s₁: multiset
0: multiset

The following usable rules [FROCOS05] were oriented:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
minus(0, y) → 0

(14) Obligation:

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, 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:

INC(s(x)) → INC(x)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, 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.

INC(s(x)) → INC(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Quasi-Precedence:

s₁ > INC₁

Status:

INC₁: multiset
s₁: multiset

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, 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(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, 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(x), s(y)) → LE(x, y)

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

Recursive path order with status [RPO].
Quasi-Precedence:

[LE₁, s₁]

Status:

s₁: [1]
LE₁: multiset

The following usable rules [FROCOS05] were oriented: none

(24) Obligation:

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, 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:

LOG2(x, y) → IF(le(x, 0), le(x, s(0)), x, inc(y))
IF(false, b, x, y) → IF2(b, x, y)
IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

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