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:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
-(#, x) → #
-(x, #) → x
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(0(x), 0(y)) → ge(x, y)
ge(0(x), 1(y)) → not(ge(y, x))
ge(1(x), 0(y)) → ge(x, y)
ge(1(x), 1(y)) → ge(x, y)
ge(x, #) → true
ge(#, 0(x)) → ge(#, x)
ge(#, 1(x)) → false
log(x) → -(log'(x), 1(#))
log'(#) → #
log'(1(x)) → +(log'(x), 1(#))
log'(0(x)) → if(ge(x, 1(#)), +(log'(x), 1(#)), #)

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

+¹(0(x), 0(y)) → 0¹(+(x, y))
+¹(0(x), 0(y)) → +¹(x, y)
+¹(0(x), 1(y)) → +¹(x, y)
+¹(1(x), 0(y)) → +¹(x, y)
+¹(1(x), 1(y)) → 0¹(+(+(x, y), 1(#)))
+¹(1(x), 1(y)) → +¹(+(x, y), 1(#))
+¹(1(x), 1(y)) → +¹(x, y)
+¹(+(x, y), z) → +¹(x, +(y, z))
+¹(+(x, y), z) → +¹(y, z)
-¹(0(x), 0(y)) → 0¹(-(x, y))
-¹(0(x), 0(y)) → -¹(x, y)
-¹(0(x), 1(y)) → -¹(-(x, y), 1(#))
-¹(0(x), 1(y)) → -¹(x, y)
-¹(1(x), 0(y)) → -¹(x, y)
-¹(1(x), 1(y)) → 0¹(-(x, y))
-¹(1(x), 1(y)) → -¹(x, y)
GE(0(x), 0(y)) → GE(x, y)
GE(0(x), 1(y)) → NOT(ge(y, x))
GE(0(x), 1(y)) → GE(y, x)
GE(1(x), 0(y)) → GE(x, y)
GE(1(x), 1(y)) → GE(x, y)
GE(#, 0(x)) → GE(#, x)
LOG(x) → -¹(log'(x), 1(#))
LOG(x) → LOG'(x)
LOG'(1(x)) → +¹(log'(x), 1(#))
LOG'(1(x)) → LOG'(x)
LOG'(0(x)) → IF(ge(x, 1(#)), +(log'(x), 1(#)), #)
LOG'(0(x)) → GE(x, 1(#))
LOG'(0(x)) → +¹(log'(x), 1(#))
LOG'(0(x)) → LOG'(x)

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
-(#, x) → #
-(x, #) → x
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(0(x), 0(y)) → ge(x, y)
ge(0(x), 1(y)) → not(ge(y, x))
ge(1(x), 0(y)) → ge(x, y)
ge(1(x), 1(y)) → ge(x, y)
ge(x, #) → true
ge(#, 0(x)) → ge(#, x)
ge(#, 1(x)) → false
log(x) → -(log'(x), 1(#))
log'(#) → #
log'(1(x)) → +(log'(x), 1(#))
log'(0(x)) → if(ge(x, 1(#)), +(log'(x), 1(#)), #)

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 11 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

GE(#, 0(x)) → GE(#, x)

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
-(#, x) → #
-(x, #) → x
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(0(x), 0(y)) → ge(x, y)
ge(0(x), 1(y)) → not(ge(y, x))
ge(1(x), 0(y)) → ge(x, y)
ge(1(x), 1(y)) → ge(x, y)
ge(x, #) → true
ge(#, 0(x)) → ge(#, x)
ge(#, 1(x)) → false
log(x) → -(log'(x), 1(#))
log'(#) → #
log'(1(x)) → +(log'(x), 1(#))
log'(0(x)) → if(ge(x, 1(#)), +(log'(x), 1(#)), #)

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

GE(#, 0(x)) → GE(#, x)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
-(#, x) → #
-(x, #) → x
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(0(x), 0(y)) → ge(x, y)
ge(0(x), 1(y)) → not(ge(y, x))
ge(1(x), 0(y)) → ge(x, y)
ge(1(x), 1(y)) → ge(x, y)
ge(x, #) → true
ge(#, 0(x)) → ge(#, x)
ge(#, 1(x)) → false
log(x) → -(log'(x), 1(#))
log'(#) → #
log'(1(x)) → +(log'(x), 1(#))
log'(0(x)) → if(ge(x, 1(#)), +(log'(x), 1(#)), #)

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

GE(0(x), 1(y)) → GE(y, x)
GE(0(x), 0(y)) → GE(x, y)
GE(1(x), 0(y)) → GE(x, y)
GE(1(x), 1(y)) → GE(x, y)

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
-(#, x) → #
-(x, #) → x
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(0(x), 0(y)) → ge(x, y)
ge(0(x), 1(y)) → not(ge(y, x))
ge(1(x), 0(y)) → ge(x, y)
ge(1(x), 1(y)) → ge(x, y)
ge(x, #) → true
ge(#, 0(x)) → ge(#, x)
ge(#, 1(x)) → false
log(x) → -(log'(x), 1(#))
log'(#) → #
log'(1(x)) → +(log'(x), 1(#))
log'(0(x)) → if(ge(x, 1(#)), +(log'(x), 1(#)), #)

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

(11) Obligation:

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

-¹(0(x), 1(y)) → -¹(-(x, y), 1(#))
-¹(0(x), 1(y)) → -¹(x, y)
-¹(0(x), 0(y)) → -¹(x, y)
-¹(1(x), 0(y)) → -¹(x, y)
-¹(1(x), 1(y)) → -¹(x, y)

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
-(#, x) → #
-(x, #) → x
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(0(x), 0(y)) → ge(x, y)
ge(0(x), 1(y)) → not(ge(y, x))
ge(1(x), 0(y)) → ge(x, y)
ge(1(x), 1(y)) → ge(x, y)
ge(x, #) → true
ge(#, 0(x)) → ge(#, x)
ge(#, 1(x)) → false
log(x) → -(log'(x), 1(#))
log'(#) → #
log'(1(x)) → +(log'(x), 1(#))
log'(0(x)) → if(ge(x, 1(#)), +(log'(x), 1(#)), #)

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

(12) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

-¹(0(x), 0(y)) → -¹(x, y)
-¹(1(x), 0(y)) → -¹(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
-¹(x1, x2)  =  x2
1(x1)  =  x1
#  =  #
0(x1)  =  0(x1)

Lexicographic Path Order [LPO].
Precedence:

0₁ > #

The following usable rules [FROCOS05] were oriented: none

(13) Obligation:

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

-¹(0(x), 1(y)) → -¹(-(x, y), 1(#))
-¹(0(x), 1(y)) → -¹(x, y)
-¹(1(x), 1(y)) → -¹(x, y)

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
-(#, x) → #
-(x, #) → x
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(0(x), 0(y)) → ge(x, y)
ge(0(x), 1(y)) → not(ge(y, x))
ge(1(x), 0(y)) → ge(x, y)
ge(1(x), 1(y)) → ge(x, y)
ge(x, #) → true
ge(#, 0(x)) → ge(#, x)
ge(#, 1(x)) → false
log(x) → -(log'(x), 1(#))
log'(#) → #
log'(1(x)) → +(log'(x), 1(#))
log'(0(x)) → if(ge(x, 1(#)), +(log'(x), 1(#)), #)

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

(14) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

-¹(0(x), 1(y)) → -¹(x, y)
-¹(1(x), 1(y)) → -¹(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
-¹(x1, x2)  =  x2
1(x1)  =  1(x1)
#  =  #

Lexicographic Path Order [LPO].
Precedence:

1₁ > #

The following usable rules [FROCOS05] were oriented: none

(15) Obligation:

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

-¹(0(x), 1(y)) → -¹(-(x, y), 1(#))

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
-(#, x) → #
-(x, #) → x
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(0(x), 0(y)) → ge(x, y)
ge(0(x), 1(y)) → not(ge(y, x))
ge(1(x), 0(y)) → ge(x, y)
ge(1(x), 1(y)) → ge(x, y)
ge(x, #) → true
ge(#, 0(x)) → ge(#, x)
ge(#, 1(x)) → false
log(x) → -(log'(x), 1(#))
log'(#) → #
log'(1(x)) → +(log'(x), 1(#))
log'(0(x)) → if(ge(x, 1(#)), +(log'(x), 1(#)), #)

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

-¹(0(x), 1(y)) → -¹(-(x, y), 1(#))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
-¹(x1, x2)  =  x1
0(x1)  =  0(x1)
-(x1, x2)  =  x1
#  =  #
1(x1)  =  1(x1)

Lexicographic Path Order [LPO].
Precedence:

[0₁, 1₁]

The following usable rules [FROCOS05] were oriented:

0(#) → #
-(1(x), 0(y)) → 1(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 1(y)) → 0(-(x, y))
-(#, x) → #
-(0(x), 0(y)) → 0(-(x, y))
-(x, #) → x

(17) Obligation:

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

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
-(#, x) → #
-(x, #) → x
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(0(x), 0(y)) → ge(x, y)
ge(0(x), 1(y)) → not(ge(y, x))
ge(1(x), 0(y)) → ge(x, y)
ge(1(x), 1(y)) → ge(x, y)
ge(x, #) → true
ge(#, 0(x)) → ge(#, x)
ge(#, 1(x)) → false
log(x) → -(log'(x), 1(#))
log'(#) → #
log'(1(x)) → +(log'(x), 1(#))
log'(0(x)) → if(ge(x, 1(#)), +(log'(x), 1(#)), #)

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

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE

(20) Obligation:

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

+¹(0(x), 1(y)) → +¹(x, y)
+¹(0(x), 0(y)) → +¹(x, y)
+¹(1(x), 0(y)) → +¹(x, y)
+¹(1(x), 1(y)) → +¹(+(x, y), 1(#))
+¹(1(x), 1(y)) → +¹(x, y)
+¹(+(x, y), z) → +¹(x, +(y, z))
+¹(+(x, y), z) → +¹(y, z)

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
-(#, x) → #
-(x, #) → x
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(0(x), 0(y)) → ge(x, y)
ge(0(x), 1(y)) → not(ge(y, x))
ge(1(x), 0(y)) → ge(x, y)
ge(1(x), 1(y)) → ge(x, y)
ge(x, #) → true
ge(#, 0(x)) → ge(#, x)
ge(#, 1(x)) → false
log(x) → -(log'(x), 1(#))
log'(#) → #
log'(1(x)) → +(log'(x), 1(#))
log'(0(x)) → if(ge(x, 1(#)), +(log'(x), 1(#)), #)

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

(21) Obligation:

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

LOG'(0(x)) → LOG'(x)
LOG'(1(x)) → LOG'(x)

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
-(#, x) → #
-(x, #) → x
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(0(x), 0(y)) → ge(x, y)
ge(0(x), 1(y)) → not(ge(y, x))
ge(1(x), 0(y)) → ge(x, y)
ge(1(x), 1(y)) → ge(x, y)
ge(x, #) → true
ge(#, 0(x)) → ge(#, x)
ge(#, 1(x)) → false
log(x) → -(log'(x), 1(#))
log'(#) → #
log'(1(x)) → +(log'(x), 1(#))
log'(0(x)) → if(ge(x, 1(#)), +(log'(x), 1(#)), #)

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

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

LOG'(1(x)) → LOG'(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LOG'(x1)  =  x1
0(x1)  =  x1
1(x1)  =  1(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(23) Obligation:

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

LOG'(0(x)) → LOG'(x)

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
-(#, x) → #
-(x, #) → x
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(0(x), 0(y)) → ge(x, y)
ge(0(x), 1(y)) → not(ge(y, x))
ge(1(x), 0(y)) → ge(x, y)
ge(1(x), 1(y)) → ge(x, y)
ge(x, #) → true
ge(#, 0(x)) → ge(#, x)
ge(#, 1(x)) → false
log(x) → -(log'(x), 1(#))
log'(#) → #
log'(1(x)) → +(log'(x), 1(#))
log'(0(x)) → if(ge(x, 1(#)), +(log'(x), 1(#)), #)

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

(24) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

LOG'(0(x)) → LOG'(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LOG'(x1)  =  x1
0(x1)  =  0(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(25) Obligation:

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

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
-(#, x) → #
-(x, #) → x
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(0(x), 0(y)) → ge(x, y)
ge(0(x), 1(y)) → not(ge(y, x))
ge(1(x), 0(y)) → ge(x, y)
ge(1(x), 1(y)) → ge(x, y)
ge(x, #) → true
ge(#, 0(x)) → ge(#, x)
ge(#, 1(x)) → false
log(x) → -(log'(x), 1(#))
log'(#) → #
log'(1(x)) → +(log'(x), 1(#))
log'(0(x)) → if(ge(x, 1(#)), +(log'(x), 1(#)), #)

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

(26) PisEmptyProof (EQUIVALENT transformation)

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

(27) TRUE