Termination w.r.t. Q proof of /tmp/tmpAC

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:

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(false) → true
not(true) → false
and(x, true) → x
and(x, false) → false
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(#, 1(x)) → false
ge(#, 0(x)) → ge(#, x)
val(l(x)) → x
val(n(x, y, z)) → x
min(l(x)) → x
min(n(x, y, z)) → min(y)
max(l(x)) → x
max(n(x, y, z)) → max(z)
bs(l(x)) → true
bs(n(x, y, z)) → and(and(ge(x, max(y)), ge(min(z), x)), and(bs(y), bs(z)))
size(l(x)) → 1(#)
size(n(x, y, z)) → +(+(size(x), size(y)), 1(#))
wb(l(x)) → true
wb(n(x, y, z)) → and(if(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y)))), and(wb(y), wb(z)))

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(false) → true
not(true) → false
and(x, true) → x
and(x, false) → false
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(#, 1(x)) → false
ge(#, 0(x)) → ge(#, x)
val(l(x)) → x
val(n(x, y, z)) → x
min(l(x)) → x
min(n(x, y, z)) → min(y)
max(l(x)) → x
max(n(x, y, z)) → max(z)
bs(l(x)) → true
bs(n(x, y, z)) → and(and(ge(x, max(y)), ge(min(z), x)), and(bs(y), bs(z)))
size(l(x)) → 1(#)
size(n(x, y, z)) → +(+(size(x), size(y)), 1(#))
wb(l(x)) → true
wb(n(x, y, z)) → and(if(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y)))), and(wb(y), wb(z)))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

WB(n(x, y, z)) → AND(wb(y), wb(z))
GE(#, 0(x)) → GE(#, x)
MAX(n(x, y, z)) → MAX(z)
+¹(x, +(y, z)) → +¹(x, y)
BS(n(x, y, z)) → AND(ge(x, max(y)), ge(min(z), x))
BS(n(x, y, z)) → MIN(z)
WB(n(x, y, z)) → WB(y)
GE(0(x), 1(y)) → NOT(ge(y, x))
+¹(1(x), 1(y)) → 0¹(+(+(x, y), 1(#)))
SIZE(n(x, y, z)) → +¹(+(size(x), size(y)), 1(#))
BS(n(x, y, z)) → AND(and(ge(x, max(y)), ge(min(z), x)), and(bs(y), bs(z)))
WB(n(x, y, z)) → IF(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y))))
SIZE(n(x, y, z)) → SIZE(y)
MIN(n(x, y, z)) → MIN(y)
WB(n(x, y, z)) → GE(size(y), size(z))
WB(n(x, y, z)) → AND(if(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y)))), and(wb(y), wb(z)))
+¹(0(x), 0(y)) → +¹(x, y)
GE(0(x), 0(y)) → GE(x, y)
BS(n(x, y, z)) → GE(min(z), x)
WB(n(x, y, z)) → WB(z)
GE(1(x), 0(y)) → GE(x, y)
GE(0(x), 1(y)) → GE(y, x)
-¹(1(x), 0(y)) → -¹(x, y)
-¹(0(x), 1(y)) → -¹(x, y)
SIZE(n(x, y, z)) → +¹(size(x), size(y))
+¹(0(x), 0(y)) → 0¹(+(x, y))
BS(n(x, y, z)) → GE(x, max(y))
-¹(0(x), 0(y)) → 0¹(-(x, y))
BS(n(x, y, z)) → BS(y)
+¹(1(x), 1(y)) → +¹(x, y)
WB(n(x, y, z)) → SIZE(z)
-¹(0(x), 1(y)) → -¹(-(x, y), 1(#))
BS(n(x, y, z)) → MAX(y)
-¹(1(x), 1(y)) → 0¹(-(x, y))
+¹(x, +(y, z)) → +¹(+(x, y), z)
+¹(1(x), 1(y)) → +¹(+(x, y), 1(#))
GE(1(x), 1(y)) → GE(x, y)
WB(n(x, y, z)) → -¹(size(y), size(z))
WB(n(x, y, z)) → -¹(size(z), size(y))
-¹(1(x), 1(y)) → -¹(x, y)
BS(n(x, y, z)) → AND(bs(y), bs(z))
-¹(0(x), 0(y)) → -¹(x, y)
WB(n(x, y, z)) → SIZE(y)
WB(n(x, y, z)) → GE(1(#), -(size(y), size(z)))
WB(n(x, y, z)) → GE(1(#), -(size(z), size(y)))
+¹(0(x), 1(y)) → +¹(x, y)
+¹(1(x), 0(y)) → +¹(x, y)
SIZE(n(x, y, z)) → SIZE(x)
BS(n(x, y, z)) → BS(z)

The TRS R consists of the following rules:

0(#) → #
+(x, #) → x
+(#, x) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(x, +(y, z)) → +(+(x, y), z)
-(x, #) → x
-(#, x) → #
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(false) → true
not(true) → false
and(x, true) → x
and(x, false) → false
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(#, 1(x)) → false
ge(#, 0(x)) → ge(#, x)
val(l(x)) → x
val(n(x, y, z)) → x
min(l(x)) → x
min(n(x, y, z)) → min(y)
max(l(x)) → x
max(n(x, y, z)) → max(z)
bs(l(x)) → true
bs(n(x, y, z)) → and(and(ge(x, max(y)), ge(min(z), x)), and(bs(y), bs(z)))
size(l(x)) → 1(#)
size(n(x, y, z)) → +(+(size(x), size(y)), 1(#))
wb(l(x)) → true
wb(n(x, y, z)) → and(if(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y)))), and(wb(y), wb(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:

WB(n(x, y, z)) → AND(wb(y), wb(z))
GE(#, 0(x)) → GE(#, x)
MAX(n(x, y, z)) → MAX(z)
+¹(x, +(y, z)) → +¹(x, y)
BS(n(x, y, z)) → AND(ge(x, max(y)), ge(min(z), x))
BS(n(x, y, z)) → MIN(z)
WB(n(x, y, z)) → WB(y)
GE(0(x), 1(y)) → NOT(ge(y, x))
+¹(1(x), 1(y)) → 0¹(+(+(x, y), 1(#)))
SIZE(n(x, y, z)) → +¹(+(size(x), size(y)), 1(#))
BS(n(x, y, z)) → AND(and(ge(x, max(y)), ge(min(z), x)), and(bs(y), bs(z)))
WB(n(x, y, z)) → IF(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y))))
SIZE(n(x, y, z)) → SIZE(y)
MIN(n(x, y, z)) → MIN(y)
WB(n(x, y, z)) → GE(size(y), size(z))
WB(n(x, y, z)) → AND(if(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y)))), and(wb(y), wb(z)))
+¹(0(x), 0(y)) → +¹(x, y)
GE(0(x), 0(y)) → GE(x, y)
BS(n(x, y, z)) → GE(min(z), x)
WB(n(x, y, z)) → WB(z)
GE(1(x), 0(y)) → GE(x, y)
GE(0(x), 1(y)) → GE(y, x)
-¹(1(x), 0(y)) → -¹(x, y)
-¹(0(x), 1(y)) → -¹(x, y)
SIZE(n(x, y, z)) → +¹(size(x), size(y))
+¹(0(x), 0(y)) → 0¹(+(x, y))
BS(n(x, y, z)) → GE(x, max(y))
-¹(0(x), 0(y)) → 0¹(-(x, y))
BS(n(x, y, z)) → BS(y)
+¹(1(x), 1(y)) → +¹(x, y)
WB(n(x, y, z)) → SIZE(z)
-¹(0(x), 1(y)) → -¹(-(x, y), 1(#))
BS(n(x, y, z)) → MAX(y)
-¹(1(x), 1(y)) → 0¹(-(x, y))
+¹(x, +(y, z)) → +¹(+(x, y), z)
+¹(1(x), 1(y)) → +¹(+(x, y), 1(#))
GE(1(x), 1(y)) → GE(x, y)
WB(n(x, y, z)) → -¹(size(y), size(z))
WB(n(x, y, z)) → -¹(size(z), size(y))
-¹(1(x), 1(y)) → -¹(x, y)
BS(n(x, y, z)) → AND(bs(y), bs(z))
-¹(0(x), 0(y)) → -¹(x, y)
WB(n(x, y, z)) → SIZE(y)
WB(n(x, y, z)) → GE(1(#), -(size(y), size(z)))
WB(n(x, y, z)) → GE(1(#), -(size(z), size(y)))
+¹(0(x), 1(y)) → +¹(x, y)
+¹(1(x), 0(y)) → +¹(x, y)
SIZE(n(x, y, z)) → SIZE(x)
BS(n(x, y, z)) → BS(z)

The TRS R consists of the following rules:

0(#) → #
+(x, #) → x
+(#, x) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(x, +(y, z)) → +(+(x, y), z)
-(x, #) → x
-(#, x) → #
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(false) → true
not(true) → false
and(x, true) → x
and(x, false) → false
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(#, 1(x)) → false
ge(#, 0(x)) → ge(#, x)
val(l(x)) → x
val(n(x, y, z)) → x
min(l(x)) → x
min(n(x, y, z)) → min(y)
max(l(x)) → x
max(n(x, y, z)) → max(z)
bs(l(x)) → true
bs(n(x, y, z)) → and(and(ge(x, max(y)), ge(min(z), x)), and(bs(y), bs(z)))
size(l(x)) → 1(#)
size(n(x, y, z)) → +(+(size(x), size(y)), 1(#))
wb(l(x)) → true
wb(n(x, y, z)) → and(if(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y)))), and(wb(y), wb(z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] 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:

MAX(n(x, y, z)) → MAX(z)

The TRS R consists of the following rules:

0(#) → #
+(x, #) → x
+(#, x) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(x, +(y, z)) → +(+(x, y), z)
-(x, #) → x
-(#, x) → #
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(false) → true
not(true) → false
and(x, true) → x
and(x, false) → false
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(#, 1(x)) → false
ge(#, 0(x)) → ge(#, x)
val(l(x)) → x
val(n(x, y, z)) → x
min(l(x)) → x
min(n(x, y, z)) → min(y)
max(l(x)) → x
max(n(x, y, z)) → max(z)
bs(l(x)) → true
bs(n(x, y, z)) → and(and(ge(x, max(y)), ge(min(z), x)), and(bs(y), bs(z)))
size(l(x)) → 1(#)
size(n(x, y, z)) → +(+(size(x), size(y)), 1(#))
wb(l(x)) → true
wb(n(x, y, z)) → and(if(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y)))), and(wb(y), wb(z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

MAX(n(x, y, z)) → MAX(z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(n(x₁, x₂, x₃)) = 1 + (4)x_3
POL(MAX(x₁)) = (4)x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] 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:

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(false) → true
not(true) → false
and(x, true) → x
and(x, false) → false
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(#, 1(x)) → false
ge(#, 0(x)) → ge(#, x)
val(l(x)) → x
val(n(x, y, z)) → x
min(l(x)) → x
min(n(x, y, z)) → min(y)
max(l(x)) → x
max(n(x, y, z)) → max(z)
bs(l(x)) → true
bs(n(x, y, z)) → and(and(ge(x, max(y)), ge(min(z), x)), and(bs(y), bs(z)))
size(l(x)) → 1(#)
size(n(x, y, z)) → +(+(size(x), size(y)), 1(#))
wb(l(x)) → true
wb(n(x, y, z)) → and(if(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y)))), and(wb(y), wb(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:

MIN(n(x, y, z)) → MIN(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(false) → true
not(true) → false
and(x, true) → x
and(x, false) → false
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(#, 1(x)) → false
ge(#, 0(x)) → ge(#, x)
val(l(x)) → x
val(n(x, y, z)) → x
min(l(x)) → x
min(n(x, y, z)) → min(y)
max(l(x)) → x
max(n(x, y, z)) → max(z)
bs(l(x)) → true
bs(n(x, y, z)) → and(and(ge(x, max(y)), ge(min(z), x)), and(bs(y), bs(z)))
size(l(x)) → 1(#)
size(n(x, y, z)) → +(+(size(x), size(y)), 1(#))
wb(l(x)) → true
wb(n(x, y, z)) → and(if(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y)))), and(wb(y), wb(z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

MIN(n(x, y, z)) → MIN(y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(n(x₁, x₂, x₃)) = 1 + (4)x_2
POL(MIN(x₁)) = (4)x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] 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:

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(false) → true
not(true) → false
and(x, true) → x
and(x, false) → false
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(#, 1(x)) → false
ge(#, 0(x)) → ge(#, x)
val(l(x)) → x
val(n(x, y, z)) → x
min(l(x)) → x
min(n(x, y, z)) → min(y)
max(l(x)) → x
max(n(x, y, z)) → max(z)
bs(l(x)) → true
bs(n(x, y, z)) → and(and(ge(x, max(y)), ge(min(z), x)), and(bs(y), bs(z)))
size(l(x)) → 1(#)
size(n(x, y, z)) → +(+(size(x), size(y)), 1(#))
wb(l(x)) → true
wb(n(x, y, z)) → and(if(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y)))), and(wb(y), wb(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:

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(false) → true
not(true) → false
and(x, true) → x
and(x, false) → false
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(#, 1(x)) → false
ge(#, 0(x)) → ge(#, x)
val(l(x)) → x
val(n(x, y, z)) → x
min(l(x)) → x
min(n(x, y, z)) → min(y)
max(l(x)) → x
max(n(x, y, z)) → max(z)
bs(l(x)) → true
bs(n(x, y, z)) → and(and(ge(x, max(y)), ge(min(z), x)), and(bs(y), bs(z)))
size(l(x)) → 1(#)
size(n(x, y, z)) → +(+(size(x), size(y)), 1(#))
wb(l(x)) → true
wb(n(x, y, z)) → and(if(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y)))), and(wb(y), wb(z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

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: Polynomial interpretation [25,35]:

POL(GE(x₁, x₂)) = (4)x_2
POL(0(x₁)) = 1 + (4)x_1
POL(#) = 0

The value of delta used in the strict ordering is 4.
The following usable rules [17] 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:

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(false) → true
not(true) → false
and(x, true) → x
and(x, false) → false
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(#, 1(x)) → false
ge(#, 0(x)) → ge(#, x)
val(l(x)) → x
val(n(x, y, z)) → x
min(l(x)) → x
min(n(x, y, z)) → min(y)
max(l(x)) → x
max(n(x, y, z)) → max(z)
bs(l(x)) → true
bs(n(x, y, z)) → and(and(ge(x, max(y)), ge(min(z), x)), and(bs(y), bs(z)))
size(l(x)) → 1(#)
size(n(x, y, z)) → +(+(size(x), size(y)), 1(#))
wb(l(x)) → true
wb(n(x, y, z)) → and(if(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y)))), and(wb(y), wb(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:

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

The TRS R consists of the following rules:

0(#) → #
+(x, #) → x
+(#, x) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(x, +(y, z)) → +(+(x, y), z)
-(x, #) → x
-(#, x) → #
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(false) → true
not(true) → false
and(x, true) → x
and(x, false) → false
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(#, 1(x)) → false
ge(#, 0(x)) → ge(#, x)
val(l(x)) → x
val(n(x, y, z)) → x
min(l(x)) → x
min(n(x, y, z)) → min(y)
max(l(x)) → x
max(n(x, y, z)) → max(z)
bs(l(x)) → true
bs(n(x, y, z)) → and(and(ge(x, max(y)), ge(min(z), x)), and(bs(y), bs(z)))
size(l(x)) → 1(#)
size(n(x, y, z)) → +(+(size(x), size(y)), 1(#))
wb(l(x)) → true
wb(n(x, y, z)) → and(if(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y)))), and(wb(y), wb(z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(1(x₁)) = 1 + (4)x_1
POL(GE(x₁, x₂)) = (4)x_1 + (3)x_2
POL(0(x₁)) = 1 + x_1

The value of delta used in the strict ordering is 7.
The following usable rules [17] 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:

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(false) → true
not(true) → false
and(x, true) → x
and(x, false) → false
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(#, 1(x)) → false
ge(#, 0(x)) → ge(#, x)
val(l(x)) → x
val(n(x, y, z)) → x
min(l(x)) → x
min(n(x, y, z)) → min(y)
max(l(x)) → x
max(n(x, y, z)) → max(z)
bs(l(x)) → true
bs(n(x, y, z)) → and(and(ge(x, max(y)), ge(min(z), x)), and(bs(y), bs(z)))
size(l(x)) → 1(#)
size(n(x, y, z)) → +(+(size(x), size(y)), 1(#))
wb(l(x)) → true
wb(n(x, y, z)) → and(if(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y)))), and(wb(y), wb(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:

BS(n(x, y, z)) → BS(y)
BS(n(x, y, z)) → BS(z)

The TRS R consists of the following rules:

0(#) → #
+(x, #) → x
+(#, x) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(x, +(y, z)) → +(+(x, y), z)
-(x, #) → x
-(#, x) → #
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(false) → true
not(true) → false
and(x, true) → x
and(x, false) → false
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(#, 1(x)) → false
ge(#, 0(x)) → ge(#, x)
val(l(x)) → x
val(n(x, y, z)) → x
min(l(x)) → x
min(n(x, y, z)) → min(y)
max(l(x)) → x
max(n(x, y, z)) → max(z)
bs(l(x)) → true
bs(n(x, y, z)) → and(and(ge(x, max(y)), ge(min(z), x)), and(bs(y), bs(z)))
size(l(x)) → 1(#)
size(n(x, y, z)) → +(+(size(x), size(y)), 1(#))
wb(l(x)) → true
wb(n(x, y, z)) → and(if(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y)))), and(wb(y), wb(z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

BS(n(x, y, z)) → BS(y)
BS(n(x, y, z)) → BS(z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(n(x₁, x₂, x₃)) = 1 + (4)x_2 + (4)x_3
POL(BS(x₁)) = (4)x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] 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:

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(false) → true
not(true) → false
and(x, true) → x
and(x, false) → false
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(#, 1(x)) → false
ge(#, 0(x)) → ge(#, x)
val(l(x)) → x
val(n(x, y, z)) → x
min(l(x)) → x
min(n(x, y, z)) → min(y)
max(l(x)) → x
max(n(x, y, z)) → max(z)
bs(l(x)) → true
bs(n(x, y, z)) → and(and(ge(x, max(y)), ge(min(z), x)), and(bs(y), bs(z)))
size(l(x)) → 1(#)
size(n(x, y, z)) → +(+(size(x), size(y)), 1(#))
wb(l(x)) → true
wb(n(x, y, z)) → and(if(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y)))), and(wb(y), wb(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:

-¹(1(x), 1(y)) → -¹(x, y)
-¹(0(x), 0(y)) → -¹(x, y)
-¹(1(x), 0(y)) → -¹(x, y)
-¹(0(x), 1(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(false) → true
not(true) → false
and(x, true) → x
and(x, false) → false
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(#, 1(x)) → false
ge(#, 0(x)) → ge(#, x)
val(l(x)) → x
val(n(x, y, z)) → x
min(l(x)) → x
min(n(x, y, z)) → min(y)
max(l(x)) → x
max(n(x, y, z)) → max(z)
bs(l(x)) → true
bs(n(x, y, z)) → and(and(ge(x, max(y)), ge(min(z), x)), and(bs(y), bs(z)))
size(l(x)) → 1(#)
size(n(x, y, z)) → +(+(size(x), size(y)), 1(#))
wb(l(x)) → true
wb(n(x, y, z)) → and(if(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y)))), and(wb(y), wb(z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

-¹(0(x), 1(y)) → -¹(-(x, y), 1(#))

Used ordering: Polynomial interpretation [25,35]:

POL(1(x₁)) = 1/4 + (4)x_1
POL(-(x₁, x₂)) = 0
POL(-¹(x₁, x₂)) = (1/4)x_2
POL(0(x₁)) = 13/4 + (4)x_1
POL(#) = 0

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] 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:

-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(false) → true
not(true) → false
and(x, true) → x
and(x, false) → false
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(#, 1(x)) → false
ge(#, 0(x)) → ge(#, x)
val(l(x)) → x
val(n(x, y, z)) → x
min(l(x)) → x
min(n(x, y, z)) → min(y)
max(l(x)) → x
max(n(x, y, z)) → max(z)
bs(l(x)) → true
bs(n(x, y, z)) → and(and(ge(x, max(y)), ge(min(z), x)), and(bs(y), bs(z)))
size(l(x)) → 1(#)
size(n(x, y, z)) → +(+(size(x), size(y)), 1(#))
wb(l(x)) → true
wb(n(x, y, z)) → and(if(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y)))), and(wb(y), wb(z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

-¹(0(x), 1(y)) → -¹(-(x, y), 1(#))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(1(x₁)) = 1/2 + (4)x_1
POL(-(x₁, x₂)) = x_1
POL(-¹(x₁, x₂)) = (4)x_1 + (1/4)x_2
POL(0(x₁)) = 1/2 + (4)x_1
POL(#) = 0

The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented:

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

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(false) → true
not(true) → false
and(x, true) → x
and(x, false) → false
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(#, 1(x)) → false
ge(#, 0(x)) → ge(#, x)
val(l(x)) → x
val(n(x, y, z)) → x
min(l(x)) → x
min(n(x, y, z)) → min(y)
max(l(x)) → x
max(n(x, y, z)) → max(z)
bs(l(x)) → true
bs(n(x, y, z)) → and(and(ge(x, max(y)), ge(min(z), x)), and(bs(y), bs(z)))
size(l(x)) → 1(#)
size(n(x, y, z)) → +(+(size(x), size(y)), 1(#))
wb(l(x)) → true
wb(n(x, y, z)) → and(if(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y)))), and(wb(y), wb(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:

+¹(x, +(y, z)) → +¹(x, y)
+¹(0(x), 0(y)) → +¹(x, y)
+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(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(false) → true
not(true) → false
and(x, true) → x
and(x, false) → false
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(#, 1(x)) → false
ge(#, 0(x)) → ge(#, x)
val(l(x)) → x
val(n(x, y, z)) → x
min(l(x)) → x
min(n(x, y, z)) → min(y)
max(l(x)) → x
max(n(x, y, z)) → max(z)
bs(l(x)) → true
bs(n(x, y, z)) → and(and(ge(x, max(y)), ge(min(z), x)), and(bs(y), bs(z)))
size(l(x)) → 1(#)
size(n(x, y, z)) → +(+(size(x), size(y)), 1(#))
wb(l(x)) → true
wb(n(x, y, z)) → and(if(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y)))), and(wb(y), wb(z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

+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(#))

Used ordering: Polynomial interpretation [25,35]:

POL(1(x₁)) = (4)x_1
POL(0(x₁)) = 4 + (5/2)x_1
POL(+¹(x₁, x₂)) = (2)x_2
POL(+(x₁, x₂)) = 2 + x_1 + (3/2)x_2
POL(#) = 0

The value of delta used in the strict ordering is 4.
The following usable rules [17] 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:

+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(false) → true
not(true) → false
and(x, true) → x
and(x, false) → false
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(#, 1(x)) → false
ge(#, 0(x)) → ge(#, x)
val(l(x)) → x
val(n(x, y, z)) → x
min(l(x)) → x
min(n(x, y, z)) → min(y)
max(l(x)) → x
max(n(x, y, z)) → max(z)
bs(l(x)) → true
bs(n(x, y, z)) → and(and(ge(x, max(y)), ge(min(z), x)), and(bs(y), bs(z)))
size(l(x)) → 1(#)
size(n(x, y, z)) → +(+(size(x), size(y)), 1(#))
wb(l(x)) → true
wb(n(x, y, z)) → and(if(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y)))), and(wb(y), wb(z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

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), 1(y)) → +¹(+(x, y), 1(#))

Used ordering: Polynomial interpretation [25,35]:

POL(1(x₁)) = 2 + (4)x_1
POL(0(x₁)) = 7/4 + (15/4)x_1
POL(+¹(x₁, x₂)) = (1/4)x_2
POL(+(x₁, x₂)) = 0
POL(#) = 0

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

+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(false) → true
not(true) → false
and(x, true) → x
and(x, false) → false
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(#, 1(x)) → false
ge(#, 0(x)) → ge(#, x)
val(l(x)) → x
val(n(x, y, z)) → x
min(l(x)) → x
min(n(x, y, z)) → min(y)
max(l(x)) → x
max(n(x, y, z)) → max(z)
bs(l(x)) → true
bs(n(x, y, z)) → and(and(ge(x, max(y)), ge(min(z), x)), and(bs(y), bs(z)))
size(l(x)) → 1(#)
size(n(x, y, z)) → +(+(size(x), size(y)), 1(#))
wb(l(x)) → true
wb(n(x, y, z)) → and(if(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y)))), and(wb(y), wb(z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

+¹(1(x), 1(y)) → +¹(+(x, y), 1(#))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(1(x₁)) = 9/4 + x_1
POL(0(x₁)) = x_1
POL(+¹(x₁, x₂)) = (1/2)x_1 + (1/2)x_2
POL(+(x₁, x₂)) = 1/2 + x_1 + x_2
POL(#) = 3/4

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ 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(false) → true
not(true) → false
and(x, true) → x
and(x, false) → false
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(#, 1(x)) → false
ge(#, 0(x)) → ge(#, x)
val(l(x)) → x
val(n(x, y, z)) → x
min(l(x)) → x
min(n(x, y, z)) → min(y)
max(l(x)) → x
max(n(x, y, z)) → max(z)
bs(l(x)) → true
bs(n(x, y, z)) → and(and(ge(x, max(y)), ge(min(z), x)), and(bs(y), bs(z)))
size(l(x)) → 1(#)
size(n(x, y, z)) → +(+(size(x), size(y)), 1(#))
wb(l(x)) → true
wb(n(x, y, z)) → and(if(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y)))), and(wb(y), wb(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:

SIZE(n(x, y, z)) → SIZE(x)
SIZE(n(x, y, z)) → SIZE(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(false) → true
not(true) → false
and(x, true) → x
and(x, false) → false
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(#, 1(x)) → false
ge(#, 0(x)) → ge(#, x)
val(l(x)) → x
val(n(x, y, z)) → x
min(l(x)) → x
min(n(x, y, z)) → min(y)
max(l(x)) → x
max(n(x, y, z)) → max(z)
bs(l(x)) → true
bs(n(x, y, z)) → and(and(ge(x, max(y)), ge(min(z), x)), and(bs(y), bs(z)))
size(l(x)) → 1(#)
size(n(x, y, z)) → +(+(size(x), size(y)), 1(#))
wb(l(x)) → true
wb(n(x, y, z)) → and(if(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y)))), and(wb(y), wb(z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

SIZE(n(x, y, z)) → SIZE(x)
SIZE(n(x, y, z)) → SIZE(y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(n(x₁, x₂, x₃)) = 1 + (4)x_1 + (4)x_2
POL(SIZE(x₁)) = (4)x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] 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:

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(false) → true
not(true) → false
and(x, true) → x
and(x, false) → false
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(#, 1(x)) → false
ge(#, 0(x)) → ge(#, x)
val(l(x)) → x
val(n(x, y, z)) → x
min(l(x)) → x
min(n(x, y, z)) → min(y)
max(l(x)) → x
max(n(x, y, z)) → max(z)
bs(l(x)) → true
bs(n(x, y, z)) → and(and(ge(x, max(y)), ge(min(z), x)), and(bs(y), bs(z)))
size(l(x)) → 1(#)
size(n(x, y, z)) → +(+(size(x), size(y)), 1(#))
wb(l(x)) → true
wb(n(x, y, z)) → and(if(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y)))), and(wb(y), wb(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:

WB(n(x, y, z)) → WB(z)
WB(n(x, y, z)) → WB(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(false) → true
not(true) → false
and(x, true) → x
and(x, false) → false
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(#, 1(x)) → false
ge(#, 0(x)) → ge(#, x)
val(l(x)) → x
val(n(x, y, z)) → x
min(l(x)) → x
min(n(x, y, z)) → min(y)
max(l(x)) → x
max(n(x, y, z)) → max(z)
bs(l(x)) → true
bs(n(x, y, z)) → and(and(ge(x, max(y)), ge(min(z), x)), and(bs(y), bs(z)))
size(l(x)) → 1(#)
size(n(x, y, z)) → +(+(size(x), size(y)), 1(#))
wb(l(x)) → true
wb(n(x, y, z)) → and(if(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y)))), and(wb(y), wb(z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

WB(n(x, y, z)) → WB(z)
WB(n(x, y, z)) → WB(y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(WB(x₁)) = (4)x_1
POL(n(x₁, x₂, x₃)) = 1 + (4)x_2 + (4)x_3

The value of delta used in the strict ordering is 4.
The following usable rules [17] 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:

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(false) → true
not(true) → false
and(x, true) → x
and(x, false) → false
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(#, 1(x)) → false
ge(#, 0(x)) → ge(#, x)
val(l(x)) → x
val(n(x, y, z)) → x
min(l(x)) → x
min(n(x, y, z)) → min(y)
max(l(x)) → x
max(n(x, y, z)) → max(z)
bs(l(x)) → true
bs(n(x, y, z)) → and(and(ge(x, max(y)), ge(min(z), x)), and(bs(y), bs(z)))
size(l(x)) → 1(#)
size(n(x, y, z)) → +(+(size(x), size(y)), 1(#))
wb(l(x)) → true
wb(n(x, y, z)) → and(if(ge(size(y), size(z)), ge(1(#), -(size(y), size(z))), ge(1(#), -(size(z), size(y)))), and(wb(y), wb(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.