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
            ↳ UsableRulesProof
          ↳ 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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QDPSizeChangeProof
          ↳ 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)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
QDP
            ↳ UsableRulesProof
          ↳ 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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QDPSizeChangeProof
          ↳ 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)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
QDP
            ↳ UsableRulesProof
          ↳ 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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QDPSizeChangeProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

GE(#, 0(x)) → GE(#, x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ UsableRulesProof
          ↳ 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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QDPSizeChangeProof
          ↳ 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(0(x), 1(y)) → GE(y, x)
GE(1(x), 0(y)) → GE(x, y)
GE(1(x), 1(y)) → GE(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ UsableRulesProof
          ↳ 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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QDPSizeChangeProof
          ↳ 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)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ UsableRulesProof
          ↳ 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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ RuleRemovalProof
          ↳ 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(0(x), 1(y)) → -1(-(x, y), 1(#))
-1(0(x), 1(y)) → -1(x, y)
-1(1(x), 0(y)) → -1(x, y)

The TRS R consists of the following rules:

-(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))
0(#) → #

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

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

Strictly oriented rules of the TRS R:

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

Used ordering: POLO with Polynomial interpretation [25]:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ RuleRemovalProof
QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ UsableRulesProof
          ↳ 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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ RuleRemovalProof
          ↳ QDP
          ↳ QDP

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

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

+(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)
0(#) → #

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

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

Strictly oriented rules of the TRS R:

+(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))
0(#) → #

Used ordering: POLO with Polynomial interpretation [25]:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ RuleRemovalProof
QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

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

+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(x, +(y, z)) → +(+(x, y), 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
            ↳ UsableRulesProof
          ↳ 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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QDPSizeChangeProof
          ↳ 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)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ UsableRulesProof

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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QDPSizeChangeProof

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)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: