Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

01(#) -> #
+2(#, x) -> x
+2(x, #) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> x
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

01(#) -> #
+2(#, x) -> x
+2(x, #) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> x
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)

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(11(x)) -> +12(log'1(x), 11(#))
-12(01(x), 11(y)) -> -12(-2(x, y), 11(#))
LOG'1(01(x)) -> LOG'1(x)
-12(11(x), 11(y)) -> 011(-2(x, y))
GE2(01(x), 11(y)) -> GE2(y, x)
GE2(11(x), 01(y)) -> GE2(x, y)
+12(11(x), 11(y)) -> 011(+2(+2(x, y), 11(#)))
+12(01(x), 01(y)) -> +12(x, y)
-12(01(x), 01(y)) -> 011(-2(x, y))
LOG1(x) -> LOG'1(x)
-12(01(x), 11(y)) -> -12(x, y)
-12(11(x), 01(y)) -> -12(x, y)
+12(+2(x, y), z) -> +12(y, z)
+12(01(x), 01(y)) -> 011(+2(x, y))
LOG'1(11(x)) -> LOG'1(x)
LOG1(x) -> -12(log'1(x), 11(#))
LOG'1(01(x)) -> GE2(x, 11(#))
GE2(11(x), 11(y)) -> GE2(x, y)
+12(11(x), 11(y)) -> +12(x, y)
-12(01(x), 01(y)) -> -12(x, y)
GE2(01(x), 11(y)) -> NOT1(ge2(y, x))
+12(+2(x, y), z) -> +12(x, +2(y, z))
LOG'1(01(x)) -> +12(log'1(x), 11(#))
LOG'1(01(x)) -> IF3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
GE2(#, 01(x)) -> GE2(#, x)
GE2(01(x), 01(y)) -> GE2(x, y)
+12(11(x), 01(y)) -> +12(x, y)
+12(01(x), 11(y)) -> +12(x, y)
-12(11(x), 11(y)) -> -12(x, y)
+12(11(x), 11(y)) -> +12(+2(x, y), 11(#))

The TRS R consists of the following rules:

01(#) -> #
+2(#, x) -> x
+2(x, #) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> x
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

LOG'1(11(x)) -> +12(log'1(x), 11(#))
-12(01(x), 11(y)) -> -12(-2(x, y), 11(#))
LOG'1(01(x)) -> LOG'1(x)
-12(11(x), 11(y)) -> 011(-2(x, y))
GE2(01(x), 11(y)) -> GE2(y, x)
GE2(11(x), 01(y)) -> GE2(x, y)
+12(11(x), 11(y)) -> 011(+2(+2(x, y), 11(#)))
+12(01(x), 01(y)) -> +12(x, y)
-12(01(x), 01(y)) -> 011(-2(x, y))
LOG1(x) -> LOG'1(x)
-12(01(x), 11(y)) -> -12(x, y)
-12(11(x), 01(y)) -> -12(x, y)
+12(+2(x, y), z) -> +12(y, z)
+12(01(x), 01(y)) -> 011(+2(x, y))
LOG'1(11(x)) -> LOG'1(x)
LOG1(x) -> -12(log'1(x), 11(#))
LOG'1(01(x)) -> GE2(x, 11(#))
GE2(11(x), 11(y)) -> GE2(x, y)
+12(11(x), 11(y)) -> +12(x, y)
-12(01(x), 01(y)) -> -12(x, y)
GE2(01(x), 11(y)) -> NOT1(ge2(y, x))
+12(+2(x, y), z) -> +12(x, +2(y, z))
LOG'1(01(x)) -> +12(log'1(x), 11(#))
LOG'1(01(x)) -> IF3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
GE2(#, 01(x)) -> GE2(#, x)
GE2(01(x), 01(y)) -> GE2(x, y)
+12(11(x), 01(y)) -> +12(x, y)
+12(01(x), 11(y)) -> +12(x, y)
-12(11(x), 11(y)) -> -12(x, y)
+12(11(x), 11(y)) -> +12(+2(x, y), 11(#))

The TRS R consists of the following rules:

01(#) -> #
+2(#, x) -> x
+2(x, #) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> x
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)

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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

GE2(#, 01(x)) -> GE2(#, x)

The TRS R consists of the following rules:

01(#) -> #
+2(#, x) -> x
+2(x, #) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> x
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)

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.


GE2(#, 01(x)) -> GE2(#, x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(#) = 0   
POL(01(x1)) = 1 + 2·x1   
POL(GE2(x1, x2)) = 2·x2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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:

01(#) -> #
+2(#, x) -> x
+2(x, #) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> x
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)

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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

GE2(11(x), 11(y)) -> GE2(x, y)
GE2(01(x), 01(y)) -> GE2(x, y)
GE2(11(x), 01(y)) -> GE2(x, y)
GE2(01(x), 11(y)) -> GE2(y, x)

The TRS R consists of the following rules:

01(#) -> #
+2(#, x) -> x
+2(x, #) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> x
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)

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.


GE2(01(x), 01(y)) -> GE2(x, y)
GE2(11(x), 01(y)) -> GE2(x, y)
GE2(01(x), 11(y)) -> GE2(y, x)
The remaining pairs can at least be oriented weakly.

GE2(11(x), 11(y)) -> GE2(x, y)
Used ordering: Polynomial interpretation [21]:

POL(01(x1)) = 2 + 2·x1   
POL(11(x1)) = 2·x1   
POL(GE2(x1, x2)) = 2·x1 + 2·x2   

The following usable rules [14] were oriented: none



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

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

GE2(11(x), 11(y)) -> GE2(x, y)

The TRS R consists of the following rules:

01(#) -> #
+2(#, x) -> x
+2(x, #) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> x
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)

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.


GE2(11(x), 11(y)) -> GE2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(11(x1)) = 2 + 2·x1   
POL(GE2(x1, x2)) = 2·x2   

The following usable rules [14] were oriented: none



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

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

01(#) -> #
+2(#, x) -> x
+2(x, #) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> x
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)

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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

-12(11(x), 01(y)) -> -12(x, y)
-12(01(x), 11(y)) -> -12(x, y)
-12(01(x), 01(y)) -> -12(x, y)
-12(01(x), 11(y)) -> -12(-2(x, y), 11(#))
-12(11(x), 11(y)) -> -12(x, y)

The TRS R consists of the following rules:

01(#) -> #
+2(#, x) -> x
+2(x, #) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> x
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)

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.


-12(01(x), 11(y)) -> -12(x, y)
-12(11(x), 11(y)) -> -12(x, y)
The remaining pairs can at least be oriented weakly.

-12(11(x), 01(y)) -> -12(x, y)
-12(01(x), 01(y)) -> -12(x, y)
-12(01(x), 11(y)) -> -12(-2(x, y), 11(#))
Used ordering: Polynomial interpretation [21]:

POL(#) = 0   
POL(-2(x1, x2)) = 0   
POL(-12(x1, x2)) = 2·x2   
POL(01(x1)) = 2·x1   
POL(11(x1)) = 2 + 2·x1   

The following usable rules [14] were oriented: none



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

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

-12(11(x), 01(y)) -> -12(x, y)
-12(01(x), 01(y)) -> -12(x, y)
-12(01(x), 11(y)) -> -12(-2(x, y), 11(#))

The TRS R consists of the following rules:

01(#) -> #
+2(#, x) -> x
+2(x, #) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> x
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)

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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ DependencyGraphProof
                  ↳ AND
                    ↳ QDP
                      ↳ QDPOrderProof
                    ↳ QDP
          ↳ QDP
          ↳ QDP

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

-12(01(x), 11(y)) -> -12(-2(x, y), 11(#))

The TRS R consists of the following rules:

01(#) -> #
+2(#, x) -> x
+2(x, #) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> x
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)

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.


-12(01(x), 11(y)) -> -12(-2(x, y), 11(#))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(#) = 0   
POL(-2(x1, x2)) = x1   
POL(-12(x1, x2)) = 2·x1   
POL(01(x1)) = 2 + 2·x1   
POL(11(x1)) = 2 + 2·x1   

The following usable rules [14] were oriented:

01(#) -> #
-2(#, x) -> #
-2(11(x), 11(y)) -> 01(-2(x, y))
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(x, #) -> x



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

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

01(#) -> #
+2(#, x) -> x
+2(x, #) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> x
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)

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
      ↳ 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:

-12(11(x), 01(y)) -> -12(x, y)
-12(01(x), 01(y)) -> -12(x, y)

The TRS R consists of the following rules:

01(#) -> #
+2(#, x) -> x
+2(x, #) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> x
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)

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.


-12(11(x), 01(y)) -> -12(x, y)
-12(01(x), 01(y)) -> -12(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(-12(x1, x2)) = 2·x2   
POL(01(x1)) = 2 + 2·x1   
POL(11(x1)) = 0   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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:

01(#) -> #
+2(#, x) -> x
+2(x, #) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> x
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)

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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

+12(01(x), 01(y)) -> +12(x, y)
+12(11(x), 11(y)) -> +12(x, y)
+12(+2(x, y), z) -> +12(y, z)
+12(11(x), 01(y)) -> +12(x, y)
+12(01(x), 11(y)) -> +12(x, y)
+12(+2(x, y), z) -> +12(x, +2(y, z))
+12(11(x), 11(y)) -> +12(+2(x, y), 11(#))

The TRS R consists of the following rules:

01(#) -> #
+2(#, x) -> x
+2(x, #) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> x
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)

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.


+12(11(x), 11(y)) -> +12(x, y)
+12(11(x), 01(y)) -> +12(x, y)
+12(01(x), 11(y)) -> +12(x, y)
+12(11(x), 11(y)) -> +12(+2(x, y), 11(#))
The remaining pairs can at least be oriented weakly.

+12(01(x), 01(y)) -> +12(x, y)
+12(+2(x, y), z) -> +12(y, z)
+12(+2(x, y), z) -> +12(x, +2(y, z))
Used ordering: Polynomial interpretation [21]:

POL(#) = 0   
POL(+2(x1, x2)) = x1 + x2   
POL(+12(x1, x2)) = 2·x1 + 2·x2   
POL(01(x1)) = 2·x1   
POL(11(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented:

01(#) -> #
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(#, x) -> x
+2(+2(x, y), z) -> +2(x, +2(y, z))
+2(x, #) -> x
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))



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

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

+12(01(x), 01(y)) -> +12(x, y)
+12(+2(x, y), z) -> +12(y, z)
+12(+2(x, y), z) -> +12(x, +2(y, z))

The TRS R consists of the following rules:

01(#) -> #
+2(#, x) -> x
+2(x, #) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> x
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)

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.


+12(+2(x, y), z) -> +12(y, z)
+12(+2(x, y), z) -> +12(x, +2(y, z))
The remaining pairs can at least be oriented weakly.

+12(01(x), 01(y)) -> +12(x, y)
Used ordering: Polynomial interpretation [21]:

POL(#) = 0   
POL(+2(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(+12(x1, x2)) = 2·x1   
POL(01(x1)) = 2·x1   
POL(11(x1)) = 0   

The following usable rules [14] were oriented: none



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

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

+12(01(x), 01(y)) -> +12(x, y)

The TRS R consists of the following rules:

01(#) -> #
+2(#, x) -> x
+2(x, #) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> x
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)

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.


+12(01(x), 01(y)) -> +12(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(+12(x1, x2)) = 2·x2   
POL(01(x1)) = 2 + 2·x1   

The following usable rules [14] were oriented: none



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

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

01(#) -> #
+2(#, x) -> x
+2(x, #) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> x
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)

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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

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

LOG'1(11(x)) -> LOG'1(x)
LOG'1(01(x)) -> LOG'1(x)

The TRS R consists of the following rules:

01(#) -> #
+2(#, x) -> x
+2(x, #) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> x
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)

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(01(x)) -> LOG'1(x)
The remaining pairs can at least be oriented weakly.

LOG'1(11(x)) -> LOG'1(x)
Used ordering: Polynomial interpretation [21]:

POL(01(x1)) = 2 + 2·x1   
POL(11(x1)) = 2·x1   
POL(LOG'1(x1)) = 2·x1   

The following usable rules [14] were oriented: none



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

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

LOG'1(11(x)) -> LOG'1(x)

The TRS R consists of the following rules:

01(#) -> #
+2(#, x) -> x
+2(x, #) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> x
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)

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(11(x)) -> LOG'1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(11(x1)) = 1 + 2·x1   
POL(LOG'1(x1)) = 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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:

01(#) -> #
+2(#, x) -> x
+2(x, #) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> x
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)

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.