YES Termination w.r.t. Q proof of CiME_04_tree.ari

Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔, 0 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 QDPSizeChangeProof (⇔, 0 ms)
        ↳9 YES
    ↳10 QDP
        ↳11 UsableRulesProof (⇔, 0 ms)
        ↳12 QDP
        ↳13 QDPSizeChangeProof (⇔, 0 ms)
        ↳14 YES
    ↳15 QDP
        ↳16 UsableRulesProof (⇔, 0 ms)
        ↳17 QDP
        ↳18 QDPSizeChangeProof (⇔, 0 ms)
        ↳19 YES
    ↳20 QDP
        ↳21 UsableRulesProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 QDPSizeChangeProof (⇔, 0 ms)
        ↳24 YES
    ↳25 QDP
        ↳26 UsableRulesProof (⇔, 0 ms)
        ↳27 QDP
        ↳28 QDPSizeChangeProof (⇔, 0 ms)
        ↳29 YES
    ↳30 QDP
        ↳31 UsableRulesProof (⇔, 0 ms)
        ↳32 QDP
        ↳33 MRRProof (⇔, 0 ms)
        ↳34 QDP
        ↳35 PisEmptyProof (⇔, 0 ms)
        ↳36 YES
    ↳37 QDP
        ↳38 UsableRulesProof (⇔, 0 ms)
        ↳39 QDP
        ↳40 MRRProof (⇔, 12 ms)
        ↳41 QDP
        ↳42 QDPSizeChangeProof (⇔, 0 ms)
        ↳43 YES
    ↳44 QDP
        ↳45 UsableRulesProof (⇔, 0 ms)
        ↳46 QDP
        ↳47 QDPSizeChangeProof (⇔, 0 ms)
        ↳48 YES
    ↳49 QDP
        ↳50 UsableRulesProof (⇔, 0 ms)
        ↳51 QDP
        ↳52 QDPSizeChangeProof (⇔, 0 ms)
        ↳53 YES

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 9 SCCs with 24 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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.

(6) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(7) Obligation:

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.

(8) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • MAX(n(x, y, z)) → MAX(z)
    The graph contains the following edges 1 > 1

(9) YES

(10) Obligation:

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.

(11) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(12) Obligation:

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.

(13) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • MIN(n(x, y, z)) → MIN(y)
    The graph contains the following edges 1 > 1

(14) YES

(15) Obligation:

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.

(16) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(17) Obligation:

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.

(18) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • GE(#, 0(x)) → GE(#, x)
    The graph contains the following edges 1 >= 1, 2 > 2

(19) YES

(20) Obligation:

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

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

(21) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(22) Obligation:

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

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

(23) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • GE(0(x), 1(y)) → GE(y, x)
    The graph contains the following edges 2 > 1, 1 > 2

  • GE(0(x), 0(y)) → GE(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

  • GE(1(x), 0(y)) → GE(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

  • GE(1(x), 1(y)) → GE(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

(24) YES

(25) Obligation:

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

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

(26) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(27) Obligation:

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

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

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

(28) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • BS(n(x, y, z)) → BS(z)
    The graph contains the following edges 1 > 1

  • BS(n(x, y, z)) → BS(y)
    The graph contains the following edges 1 > 1

(29) YES

(30) Obligation:

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

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

The TRS R consists of the following rules:

0(#) → #
+(x, #) → x
+(#, x) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(x, +(y, z)) → +(+(x, y), z)
-(x, #) → x
-(#, x) → #
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(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.

(31) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(32) Obligation:

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

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

(33) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, 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), 1(y)) → -1(-(x, y), 1(#))
-1(0(x), 1(y)) → -1(x, y)
-1(0(x), 0(y)) → -1(x, y)
-1(1(x), 0(y)) → -1(x, y)
-1(1(x), 1(y)) → -1(x, y)

Strictly oriented rules of the TRS R:

-(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(#) → #

Used ordering: Knuth-Bendix order [KBO] with precedence:
-2 > -^12 > 01 > 11 > #

and weight map:

#=2
0_1=5
1_1=3
-_2=0
-^1_2=0

The variable weight is 1

(34) Obligation:

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

(35) PisEmptyProof (EQUIVALENT transformation)

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

(36) YES

(37) Obligation:

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

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

The TRS R consists of the following rules:

0(#) → #
+(x, #) → x
+(#, x) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(x, +(y, z)) → +(+(x, y), z)
-(x, #) → x
-(#, x) → #
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(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.

(38) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(39) Obligation:

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

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

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.

(40) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, 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), 1(y)) → +1(x, y)
+1(1(x), 0(y)) → +1(x, y)
+1(1(x), 1(y)) → +1(+(x, y), 1(#))
+1(1(x), 1(y)) → +1(x, y)


Used ordering: Polynomial interpretation [POLO]:

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

(41) Obligation:

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), z)
+1(x, +(y, z)) → +1(x, y)

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.

(42) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • +1(0(x), 0(y)) → +1(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

  • +1(x, +(y, z)) → +1(+(x, y), z)
    The graph contains the following edges 2 > 2

  • +1(x, +(y, z)) → +1(x, y)
    The graph contains the following edges 1 >= 1, 2 > 2

(43) YES

(44) Obligation:

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

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

(45) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(46) Obligation:

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

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

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

(47) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • SIZE(n(x, y, z)) → SIZE(y)
    The graph contains the following edges 1 > 1

  • SIZE(n(x, y, z)) → SIZE(x)
    The graph contains the following edges 1 > 1

(48) YES

(49) Obligation:

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.

(50) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(51) Obligation:

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.

(52) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • WB(n(x, y, z)) → WB(z)
    The graph contains the following edges 1 > 1

  • WB(n(x, y, z)) → WB(y)
    The graph contains the following edges 1 > 1

(53) YES