Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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.


QTRS
  ↳ DependencyPairsProof

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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ EdgeDeletionProof

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

LOG'(1(x)) → +1(log'(x), 1(#))
-1(0(x), 1(y)) → -1(-(x, y), 1(#))
LOG'(0(x)) → LOG'(x)
-1(1(x), 1(y)) → 01(-(x, y))
GE(0(x), 1(y)) → GE(y, x)
GE(1(x), 0(y)) → GE(x, y)
+1(1(x), 1(y)) → 01(+(+(x, y), 1(#)))
+1(0(x), 0(y)) → +1(x, y)
-1(0(x), 0(y)) → 01(-(x, y))
LOG(x) → LOG'(x)
-1(0(x), 1(y)) → -1(x, y)
-1(1(x), 0(y)) → -1(x, y)
+1(+(x, y), z) → +1(y, z)
+1(0(x), 0(y)) → 01(+(x, y))
LOG'(1(x)) → LOG'(x)
LOG(x) → -1(log'(x), 1(#))
LOG'(0(x)) → GE(x, 1(#))
GE(1(x), 1(y)) → GE(x, y)
+1(1(x), 1(y)) → +1(x, y)
-1(0(x), 0(y)) → -1(x, y)
GE(0(x), 1(y)) → NOT(ge(y, x))
+1(+(x, y), z) → +1(x, +(y, z))
LOG'(0(x)) → +1(log'(x), 1(#))
LOG'(0(x)) → IF(ge(x, 1(#)), +(log'(x), 1(#)), #)
GE(#, 0(x)) → GE(#, x)
GE(0(x), 0(y)) → GE(x, y)
+1(1(x), 0(y)) → +1(x, y)
+1(0(x), 1(y)) → +1(x, y)
-1(1(x), 1(y)) → -1(x, y)
+1(1(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.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
QDP
          ↳ DependencyGraphProof

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

LOG'(1(x)) → +1(log'(x), 1(#))
-1(0(x), 1(y)) → -1(-(x, y), 1(#))
LOG'(0(x)) → LOG'(x)
GE(1(x), 0(y)) → GE(x, y)
GE(0(x), 1(y)) → GE(y, x)
-1(1(x), 1(y)) → 01(-(x, y))
+1(0(x), 0(y)) → +1(x, y)
+1(1(x), 1(y)) → 01(+(+(x, y), 1(#)))
LOG(x) → LOG'(x)
-1(0(x), 0(y)) → 01(-(x, y))
-1(1(x), 0(y)) → -1(x, y)
-1(0(x), 1(y)) → -1(x, y)
+1(+(x, y), z) → +1(y, z)
+1(0(x), 0(y)) → 01(+(x, y))
LOG'(1(x)) → LOG'(x)
LOG(x) → -1(log'(x), 1(#))
GE(1(x), 1(y)) → GE(x, y)
LOG'(0(x)) → GE(x, 1(#))
+1(1(x), 1(y)) → +1(x, y)
-1(0(x), 0(y)) → -1(x, y)
GE(0(x), 1(y)) → NOT(ge(y, x))
+1(+(x, y), z) → +1(x, +(y, z))
LOG'(0(x)) → IF(ge(x, 1(#)), +(log'(x), 1(#)), #)
LOG'(0(x)) → +1(log'(x), 1(#))
GE(#, 0(x)) → GE(#, x)
GE(0(x), 0(y)) → GE(x, y)
+1(0(x), 1(y)) → +1(x, y)
+1(1(x), 0(y)) → +1(x, y)
-1(1(x), 1(y)) → -1(x, y)
+1(1(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.
The approximation of the Dependency Graph [13,14,18] contains 5 SCCs with 11 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
QDP
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


GE(#, 0(x)) → GE(#, x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
GE(x1, x2)  =  x2
0(x1)  =  0(x1)

Lexicographic path order with status [19].
Precedence:
trivial

Status:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
QDP
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP

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

-1(1(x), 0(y)) → -1(x, y)
-1(0(x), 1(y)) → -1(x, y)
-1(0(x), 0(y)) → -1(x, y)
-1(0(x), 1(y)) → -1(-(x, y), 1(#))
-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.
We use the reduction pair processor [13].


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.

-1(1(x), 0(y)) → -1(x, y)
-1(0(x), 0(y)) → -1(x, y)
-1(0(x), 1(y)) → -1(-(x, y), 1(#))
Used ordering: Combined order from the following AFS and order.
-1(x1, x2)  =  x2
0(x1)  =  x1
1(x1)  =  1(x1)
#  =  #

Lexicographic path order with status [19].
Precedence:
11 > #

Status:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ DependencyGraphProof
              ↳ QDP
              ↳ QDP

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

-1(1(x), 0(y)) → -1(x, y)
-1(0(x), 0(y)) → -1(x, y)
-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.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ AND
QDP
                        ↳ QDP
              ↳ QDP
              ↳ QDP

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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ AND
                        ↳ QDP
QDP
                          ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP

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

-1(1(x), 0(y)) → -1(x, y)
-1(0(x), 0(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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


-1(1(x), 0(y)) → -1(x, y)
-1(0(x), 0(y)) → -1(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
-1(x1, x2)  =  x2
0(x1)  =  0(x1)

Lexicographic path order with status [19].
Precedence:
trivial

Status:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ AND
                        ↳ QDP
                        ↳ QDP
                          ↳ QDPOrderProof
QDP
                              ↳ PisEmptyProof
              ↳ QDP
              ↳ QDP

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
QDP
              ↳ QDP

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
QDP
                ↳ QDPOrderProof

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

LOG'(1(x)) → LOG'(x)
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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


LOG'(0(x)) → LOG'(x)
The remaining pairs can at least be oriented weakly.

LOG'(1(x)) → LOG'(x)
Used ordering: Combined order from the following AFS and order.
LOG'(x1)  =  x1
1(x1)  =  x1
0(x1)  =  0(x1)

Lexicographic path order with status [19].
Precedence:
trivial

Status:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ QDPOrderProof

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

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


LOG'(1(x)) → LOG'(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
LOG'(x1)  =  x1
1(x1)  =  1(x1)

Lexicographic path order with status [19].
Precedence:
trivial

Status:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ PisEmptyProof

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