(0) Obligation:

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

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
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))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), 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:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
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))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(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:

HALF(s(s(x))) → HALF(x)
LE(s(x), s(y)) → LE(x, y)
INC(s(x)) → INC(x)
LOG(x) → LOG2(x, 0)
LOG2(x, y) → IF(le(x, s(0)), x, inc(y))
LOG2(x, y) → LE(x, s(0))
LOG2(x, y) → INC(y)
IF(false, x, y) → LOG2(half(x), y)
IF(false, x, y) → HALF(x)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
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))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(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 4 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) 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:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
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))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(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.


INC(s(x)) → INC(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
INC(x1)  =  x1
s(x1)  =  s(x1)
half(x1)  =  half(x1)
0  =  0
le(x1, x2)  =  le(x1, x2)
true  =  true
false  =  false
inc(x1)  =  x1
log(x1)  =  log
log2(x1, x2)  =  x2
if(x1, x2, x3)  =  x3

Lexicographic path order with status [LPO].
Quasi-Precedence:
s1 > half1 > 0
s1 > le2 > true > 0
s1 > false > 0
log > 0

Status:
half1: [1]
log: []
le2: [1,2]
true: []
false: []
s1: [1]
0: []


The following usable rules [FROCOS05] were oriented:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
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))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

(9) Obligation:

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

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
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))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(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:

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
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))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(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.


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(x1, x2)
s(x1)  =  s(x1)
half(x1)  =  half(x1)
0  =  0
le(x1, x2)  =  le
true  =  true
false  =  false
inc(x1)  =  x1
log(x1)  =  log
log2(x1, x2)  =  x2
if(x1, x2, x3)  =  x3

Lexicographic path order with status [LPO].
Quasi-Precedence:
LE2 > [true, false]
s1 > half1 > 0 > [true, false]
s1 > le > [true, false]
log > 0 > [true, false]

Status:
half1: [1]
log: []
true: []
false: []
s1: [1]
0: []
le: []
LE2: [2,1]


The following usable rules [FROCOS05] were oriented:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
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))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

(14) Obligation:

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

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
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))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(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:

HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
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))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(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.


HALF(s(s(x))) → HALF(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
HALF(x1)  =  x1
s(x1)  =  s(x1)
half(x1)  =  x1
0  =  0
le(x1, x2)  =  le(x1, x2)
true  =  true
false  =  false
inc(x1)  =  x1
log(x1)  =  log(x1)
log2(x1, x2)  =  log2(x1, x2)
if(x1, x2, x3)  =  if(x2, x3)

Lexicographic path order with status [LPO].
Quasi-Precedence:
[s1, false] > [le2, log22, if2] > [0, true]
log1 > [le2, log22, if2] > [0, true]

Status:
log22: [2,1]
if2: [2,1]
le2: [2,1]
log1: [1]
true: []
false: []
s1: [1]
0: []


The following usable rules [FROCOS05] were oriented:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
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))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

(19) Obligation:

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

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
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))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(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:

IF(false, x, y) → LOG2(half(x), y)
LOG2(x, y) → IF(le(x, s(0)), x, inc(y))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
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))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(false, x0, x1)

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