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(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, 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(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
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(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))

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(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, 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(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
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(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))

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:

BS1(n3(x, y, z)) -> GE2(x, max1(y))
-12(01(x), 11(y)) -> -12(-2(x, y), 11(#))
GE2(11(x), 01(y)) -> GE2(x, y)
GE2(01(x), 11(y)) -> GE2(y, x)
SIZE1(n3(x, y, z)) -> +12(+2(size1(x), size1(y)), 11(#))
+12(x, +2(y, z)) -> +12(+2(x, y), z)
+12(11(x), 11(y)) -> 011(+2(+2(x, y), 11(#)))
-12(11(x), 01(y)) -> -12(x, y)
-12(01(x), 11(y)) -> -12(x, y)
SIZE1(n3(x, y, z)) -> +12(size1(x), size1(y))
BS1(n3(x, y, z)) -> BS1(z)
WB1(n3(x, y, z)) -> GE2(11(#), -2(size1(y), size1(z)))
WB1(n3(x, y, z)) -> GE2(11(#), -2(size1(z), size1(y)))
WB1(n3(x, y, z)) -> AND2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))
BS1(n3(x, y, z)) -> AND2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
GE2(01(x), 11(y)) -> NOT1(ge2(y, x))
SIZE1(n3(x, y, z)) -> SIZE1(y)
+12(x, +2(y, z)) -> +12(x, y)
SIZE1(n3(x, y, z)) -> SIZE1(x)
WB1(n3(x, y, z)) -> SIZE1(y)
WB1(n3(x, y, z)) -> SIZE1(z)
BS1(n3(x, y, z)) -> BS1(y)
MIN1(n3(x, y, z)) -> MIN1(y)
+12(01(x), 11(y)) -> +12(x, y)
+12(11(x), 01(y)) -> +12(x, y)
+12(11(x), 11(y)) -> +12(+2(x, y), 11(#))
WB1(n3(x, y, z)) -> AND2(wb1(y), wb1(z))
-12(11(x), 11(y)) -> 011(-2(x, y))
BS1(n3(x, y, z)) -> MAX1(y)
WB1(n3(x, y, z)) -> GE2(size1(y), size1(z))
+12(01(x), 01(y)) -> +12(x, y)
-12(01(x), 01(y)) -> 011(-2(x, y))
MAX1(n3(x, y, z)) -> MAX1(z)
WB1(n3(x, y, z)) -> -12(size1(y), size1(z))
WB1(n3(x, y, z)) -> -12(size1(z), size1(y))
BS1(n3(x, y, z)) -> AND2(ge2(x, max1(y)), ge2(min1(z), x))
+12(01(x), 01(y)) -> 011(+2(x, y))
WB1(n3(x, y, z)) -> WB1(z)
GE2(11(x), 11(y)) -> GE2(x, y)
+12(11(x), 11(y)) -> +12(x, y)
-12(01(x), 01(y)) -> -12(x, y)
WB1(n3(x, y, z)) -> WB1(y)
GE2(#, 01(x)) -> GE2(#, x)
BS1(n3(x, y, z)) -> MIN1(z)
BS1(n3(x, y, z)) -> GE2(min1(z), x)
BS1(n3(x, y, z)) -> AND2(bs1(y), bs1(z))
WB1(n3(x, y, z)) -> IF3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y))))
GE2(01(x), 01(y)) -> GE2(x, y)
-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(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, 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(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
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(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))

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:

BS1(n3(x, y, z)) -> GE2(x, max1(y))
-12(01(x), 11(y)) -> -12(-2(x, y), 11(#))
GE2(11(x), 01(y)) -> GE2(x, y)
GE2(01(x), 11(y)) -> GE2(y, x)
SIZE1(n3(x, y, z)) -> +12(+2(size1(x), size1(y)), 11(#))
+12(x, +2(y, z)) -> +12(+2(x, y), z)
+12(11(x), 11(y)) -> 011(+2(+2(x, y), 11(#)))
-12(11(x), 01(y)) -> -12(x, y)
-12(01(x), 11(y)) -> -12(x, y)
SIZE1(n3(x, y, z)) -> +12(size1(x), size1(y))
BS1(n3(x, y, z)) -> BS1(z)
WB1(n3(x, y, z)) -> GE2(11(#), -2(size1(y), size1(z)))
WB1(n3(x, y, z)) -> GE2(11(#), -2(size1(z), size1(y)))
WB1(n3(x, y, z)) -> AND2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))
BS1(n3(x, y, z)) -> AND2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
GE2(01(x), 11(y)) -> NOT1(ge2(y, x))
SIZE1(n3(x, y, z)) -> SIZE1(y)
+12(x, +2(y, z)) -> +12(x, y)
SIZE1(n3(x, y, z)) -> SIZE1(x)
WB1(n3(x, y, z)) -> SIZE1(y)
WB1(n3(x, y, z)) -> SIZE1(z)
BS1(n3(x, y, z)) -> BS1(y)
MIN1(n3(x, y, z)) -> MIN1(y)
+12(01(x), 11(y)) -> +12(x, y)
+12(11(x), 01(y)) -> +12(x, y)
+12(11(x), 11(y)) -> +12(+2(x, y), 11(#))
WB1(n3(x, y, z)) -> AND2(wb1(y), wb1(z))
-12(11(x), 11(y)) -> 011(-2(x, y))
BS1(n3(x, y, z)) -> MAX1(y)
WB1(n3(x, y, z)) -> GE2(size1(y), size1(z))
+12(01(x), 01(y)) -> +12(x, y)
-12(01(x), 01(y)) -> 011(-2(x, y))
MAX1(n3(x, y, z)) -> MAX1(z)
WB1(n3(x, y, z)) -> -12(size1(y), size1(z))
WB1(n3(x, y, z)) -> -12(size1(z), size1(y))
BS1(n3(x, y, z)) -> AND2(ge2(x, max1(y)), ge2(min1(z), x))
+12(01(x), 01(y)) -> 011(+2(x, y))
WB1(n3(x, y, z)) -> WB1(z)
GE2(11(x), 11(y)) -> GE2(x, y)
+12(11(x), 11(y)) -> +12(x, y)
-12(01(x), 01(y)) -> -12(x, y)
WB1(n3(x, y, z)) -> WB1(y)
GE2(#, 01(x)) -> GE2(#, x)
BS1(n3(x, y, z)) -> MIN1(z)
BS1(n3(x, y, z)) -> GE2(min1(z), x)
BS1(n3(x, y, z)) -> AND2(bs1(y), bs1(z))
WB1(n3(x, y, z)) -> IF3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y))))
GE2(01(x), 01(y)) -> GE2(x, y)
-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(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, 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(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
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(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 9 SCCs with 24 less nodes.

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

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

MAX1(n3(x, y, z)) -> MAX1(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(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, 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(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
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(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))

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.


MAX1(n3(x, y, z)) -> MAX1(z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MAX1(x1)) = x1   
POL(n3(x1, x2, x3)) = 1 + x3   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ 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(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, 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(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
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(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))

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
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

MIN1(n3(x, y, z)) -> MIN1(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(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, 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(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
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(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))

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.


MIN1(n3(x, y, z)) -> MIN1(y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MIN1(x1)) = x1   
POL(n3(x1, x2, x3)) = 1 + x2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ 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(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, 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(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
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(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))

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
          ↳ 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(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, 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(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
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(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))

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 + x1   
POL(GE2(x1, x2)) = x2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ 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(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, 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(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
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(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))

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
          ↳ QDP
          ↳ 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(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, 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(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
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(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))

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)
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.
none
Used ordering: Polynomial interpretation [21]:

POL(01(x1)) = 1 + x1   
POL(11(x1)) = 1 + x1   
POL(GE2(x1, x2)) = x1 + x2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ 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(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, 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(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
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(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))

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
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

BS1(n3(x, y, z)) -> BS1(y)
BS1(n3(x, y, z)) -> BS1(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(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, 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(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
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(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))

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.


BS1(n3(x, y, z)) -> BS1(y)
BS1(n3(x, y, z)) -> BS1(z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(BS1(x1)) = x1   
POL(n3(x1, x2, x3)) = 1 + x2 + x3   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ 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(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, 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(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
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(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))

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
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ 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(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, 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(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
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(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))

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

-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)) = x2   
POL(01(x1)) = 1 + x1   
POL(11(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ 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(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, 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(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
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(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))

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 + x2   
POL(-12(x1, x2)) = x1 + x2   
POL(01(x1)) = 1 + x1   
POL(11(x1)) = 1 + x1   

The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ 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(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, 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(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
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(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))

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
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ 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(x, +2(y, z)) -> +12(x, y)
+12(11(x), 01(y)) -> +12(x, y)
+12(01(x), 11(y)) -> +12(x, y)
+12(x, +2(y, z)) -> +12(+2(x, 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(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, 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(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
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(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))

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

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

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

The following usable rules [14] were oriented: none



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

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

+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(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, 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(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
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(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))

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(+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 + x2   
POL(+12(x1, x2)) = x1 + x2   
POL(01(x1)) = x1   
POL(11(x1)) = 1 + x1   

The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ 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(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, 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(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
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(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))

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
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

SIZE1(n3(x, y, z)) -> SIZE1(y)
SIZE1(n3(x, y, z)) -> SIZE1(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(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, 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(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
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(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))

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.


SIZE1(n3(x, y, z)) -> SIZE1(y)
SIZE1(n3(x, y, z)) -> SIZE1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(SIZE1(x1)) = x1   
POL(n3(x1, x2, x3)) = 1 + x1 + x2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ 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(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, 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(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
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(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))

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
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

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

WB1(n3(x, y, z)) -> WB1(y)
WB1(n3(x, y, z)) -> WB1(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(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, 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(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
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(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))

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.


WB1(n3(x, y, z)) -> WB1(y)
WB1(n3(x, y, z)) -> WB1(z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(WB1(x1)) = x1   
POL(n3(x1, x2, x3)) = 1 + x2 + x3   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ 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(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, 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(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
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(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))

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.