Term Rewriting System R:
[x, y, z, l, l1, l2, l3]
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(true) -> false
not(false) -> true
if(true, x, y) -> x
if(false, x, y) -> y
eq(#, #) -> true
eq(#, 1(y)) -> false
eq(1(x), #) -> false
eq(#, 0(y)) -> eq(#, y)
eq(0(x), #) -> eq(x, #)
eq(1(x), 1(y)) -> eq(x, y)
eq(0(x), 1(y)) -> false
eq(1(x), 0(y)) -> false
eq(0(x), 0(y)) -> eq(x, 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(#, 0(x)) -> ge(#, x)
ge(#, 1(x)) -> false
log(x) -> -(log'(x), 1(#))
log'(#) -> #
log'(1(x)) -> +(log'(x), 1(#))
log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))
mem(x, nil) -> false
mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
inter(x, nil) -> nil
inter(nil, x) -> nil
inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
ifinter(false, x, l1, l2) -> inter(l1, l2)

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

+'(0(x), 0(y)) -> 0'(+(x, y))
+'(0(x), 0(y)) -> +'(x, y)
+'(0(x), 1(y)) -> +'(x, y)
+'(1(x), 0(y)) -> +'(x, y)
+'(1(x), 1(y)) -> 0'(+(+(x, y), 1(#)))
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(1(x), 1(y)) -> +'(x, y)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(+(x, y), z) -> +'(y, z)
-'(0(x), 0(y)) -> 0'(-(x, y))
-'(0(x), 0(y)) -> -'(x, y)
-'(0(x), 1(y)) -> -'(-(x, y), 1(#))
-'(0(x), 1(y)) -> -'(x, y)
-'(1(x), 0(y)) -> -'(x, y)
-'(1(x), 1(y)) -> 0'(-(x, y))
-'(1(x), 1(y)) -> -'(x, y)
EQ(#, 0(y)) -> EQ(#, y)
EQ(0(x), #) -> EQ(x, #)
EQ(1(x), 1(y)) -> EQ(x, y)
EQ(0(x), 0(y)) -> EQ(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)
LOG(x) -> -'(log'(x), 1(#))
LOG(x) -> LOG'(x)
LOG'(1(x)) -> +'(log'(x), 1(#))
LOG'(1(x)) -> LOG'(x)
LOG'(0(x)) -> IF(ge(x, 1(#)), +(log'(x), 1(#)), #)
LOG'(0(x)) -> GE(x, 1(#))
LOG'(0(x)) -> +'(log'(x), 1(#))
LOG'(0(x)) -> LOG'(x)
*'(0(x), y) -> 0'(*(x, y))
*'(0(x), y) -> *'(x, y)
*'(1(x), y) -> +'(0(*(x, y)), y)
*'(1(x), y) -> 0'(*(x, y))
*'(1(x), y) -> *'(x, y)
*'(*(x, y), z) -> *'(x, *(y, z))
*'(*(x, y), z) -> *'(y, z)
*'(x, +(y, z)) -> +'(*(x, y), *(x, z))
*'(x, +(y, z)) -> *'(x, y)
*'(x, +(y, z)) -> *'(x, z)
APP(cons(x, l1), l2) -> APP(l1, l2)
SUM(nil) -> 0'(#)
SUM(cons(x, l)) -> +'(x, sum(l))
SUM(cons(x, l)) -> SUM(l)
SUM(app(l1, l2)) -> +'(sum(l1), sum(l2))
SUM(app(l1, l2)) -> SUM(l1)
SUM(app(l1, l2)) -> SUM(l2)
PROD(cons(x, l)) -> *'(x, prod(l))
PROD(cons(x, l)) -> PROD(l)
PROD(app(l1, l2)) -> *'(prod(l1), prod(l2))
PROD(app(l1, l2)) -> PROD(l1)
PROD(app(l1, l2)) -> PROD(l2)
MEM(x, cons(y, l)) -> IF(eq(x, y), true, mem(x, l))
MEM(x, cons(y, l)) -> EQ(x, y)
MEM(x, cons(y, l)) -> MEM(x, l)
INTER(app(l1, l2), l3) -> APP(inter(l1, l3), inter(l2, l3))
INTER(app(l1, l2), l3) -> INTER(l1, l3)
INTER(app(l1, l2), l3) -> INTER(l2, l3)
INTER(l1, app(l2, l3)) -> APP(inter(l1, l2), inter(l1, l3))
INTER(l1, app(l2, l3)) -> INTER(l1, l2)
INTER(l1, app(l2, l3)) -> INTER(l1, l3)
INTER(cons(x, l1), l2) -> IFINTER(mem(x, l2), x, l1, l2)
INTER(cons(x, l1), l2) -> MEM(x, l2)
INTER(l1, cons(x, l2)) -> IFINTER(mem(x, l1), x, l2, l1)
INTER(l1, cons(x, l2)) -> MEM(x, l1)
IFINTER(true, x, l1, l2) -> INTER(l1, l2)
IFINTER(false, x, l1, l2) -> INTER(l1, l2)

Furthermore, R contains 14 SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Usable Rules (Innermost)
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules


Dependency Pairs:

+'(+(x, y), z) -> +'(y, z)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(1(x), 1(y)) -> +'(x, y)
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(1(x), 0(y)) -> +'(x, y)
+'(0(x), 1(y)) -> +'(x, y)
+'(0(x), 0(y)) -> +'(x, y)


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(true) -> false
not(false) -> true
if(true, x, y) -> x
if(false, x, y) -> y
eq(#, #) -> true
eq(#, 1(y)) -> false
eq(1(x), #) -> false
eq(#, 0(y)) -> eq(#, y)
eq(0(x), #) -> eq(x, #)
eq(1(x), 1(y)) -> eq(x, y)
eq(0(x), 1(y)) -> false
eq(1(x), 0(y)) -> false
eq(0(x), 0(y)) -> eq(x, 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(#, 0(x)) -> ge(#, x)
ge(#, 1(x)) -> false
log(x) -> -(log'(x), 1(#))
log'(#) -> #
log'(1(x)) -> +(log'(x), 1(#))
log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))
mem(x, nil) -> false
mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
inter(x, nil) -> nil
inter(nil, x) -> nil
inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
ifinter(false, x, l1, l2) -> inter(l1, l2)


Strategy:

innermost




As we are in the innermost case, we can delete all 53 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
           →DP Problem 15
             ↳Modular Removal of Rules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules


Dependency Pairs:

+'(+(x, y), z) -> +'(y, z)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(1(x), 1(y)) -> +'(x, y)
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(1(x), 0(y)) -> +'(x, y)
+'(0(x), 1(y)) -> +'(x, y)
+'(0(x), 0(y)) -> +'(x, y)


Rules:


+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(#, x) -> x
+(+(x, y), z) -> +(x, +(y, z))
+(1(x), 0(y)) -> 1(+(x, y))
0(#) -> #


Strategy:

innermost




We have the following set of usable rules:

+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(#, x) -> x
+(+(x, y), z) -> +(x, +(y, z))
+(1(x), 0(y)) -> 1(+(x, y))
0(#) -> #
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(#)=  0  
  POL(0(x1))=  x1  
  POL(1(x1))=  1 + x1  
  POL(+(x1, x2))=  x1 + x2  
  POL(+'(x1, x2))=  1 + x1 + x2  

We have the following set D of usable symbols: {#, 0, 1, +, +'}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

+'(1(x), 1(y)) -> +'(x, y)
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(1(x), 0(y)) -> +'(x, y)
+'(0(x), 1(y)) -> +'(x, y)

The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))


The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
           →DP Problem 15
             ↳MRR
             ...
               →DP Problem 16
                 ↳Modular Removal of Rules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules


Dependency Pairs:

+'(+(x, y), z) -> +'(y, z)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(0(x), 0(y)) -> +'(x, y)


Rules:


+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(#, x) -> x
+(+(x, y), z) -> +(x, +(y, z))
+(1(x), 0(y)) -> 1(+(x, y))
0(#) -> #


Strategy:

innermost




We have the following set of usable rules:

+(x, #) -> x
0(#) -> #
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(#, x) -> x
+(+(x, y), z) -> +(x, +(y, z))
+(1(x), 0(y)) -> 1(+(x, y))
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(#)=  0  
  POL(0(x1))=  1 + x1  
  POL(1(x1))=  x1  
  POL(+(x1, x2))=  x1 + x2  
  POL(+'(x1, x2))=  x1 + x2  

We have the following set D of usable symbols: {#, 0, 1, +, +'}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

+'(0(x), 0(y)) -> +'(x, y)

The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

0(#) -> #
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))


The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
           →DP Problem 15
             ↳MRR
             ...
               →DP Problem 17
                 ↳Modular Removal of Rules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules


Dependency Pairs:

+'(+(x, y), z) -> +'(y, z)
+'(+(x, y), z) -> +'(x, +(y, z))


Rules:


+(x, #) -> x
+(#, x) -> x
+(+(x, y), z) -> +(x, +(y, z))


Strategy:

innermost




We have the following set of usable rules:

+(x, #) -> x
+(#, x) -> x
+(+(x, y), z) -> +(x, +(y, z))
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(#)=  0  
  POL(+(x1, x2))=  x1 + x2  
  POL(+'(x1, x2))=  1 + x1 + x2  

We have the following set D of usable symbols: {+, +'}
No Dependency Pairs can be deleted.
The following rules can be deleted as they contain symbols in their lhs which do not occur in D:

+(x, #) -> x
+(#, x) -> x


The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
           →DP Problem 15
             ↳MRR
             ...
               →DP Problem 18
                 ↳Modular Removal of Rules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules


Dependency Pairs:

+'(+(x, y), z) -> +'(y, z)
+'(+(x, y), z) -> +'(x, +(y, z))


Rule:


+(+(x, y), z) -> +(x, +(y, z))


Strategy:

innermost




We have the following set of usable rules:

+(+(x, y), z) -> +(x, +(y, z))
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(+(x1, x2))=  1 + x1 + x2  
  POL(+'(x1, x2))=  1 + x1 + x2  

We have the following set D of usable symbols: {+, +'}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

+'(+(x, y), z) -> +'(y, z)

No Rules can be deleted.

The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
           →DP Problem 15
             ↳MRR
             ...
               →DP Problem 19
                 ↳Dependency Graph
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules


Dependency Pair:

+'(+(x, y), z) -> +'(x, +(y, z))


Rule:


+(+(x, y), z) -> +(x, +(y, z))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳Usable Rules (Innermost)
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules


Dependency Pairs:

-'(1(x), 1(y)) -> -'(x, y)
-'(1(x), 0(y)) -> -'(x, y)
-'(0(x), 1(y)) -> -'(x, y)
-'(0(x), 1(y)) -> -'(-(x, y), 1(#))
-'(0(x), 0(y)) -> -'(x, y)


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(true) -> false
not(false) -> true
if(true, x, y) -> x
if(false, x, y) -> y
eq(#, #) -> true
eq(#, 1(y)) -> false
eq(1(x), #) -> false
eq(#, 0(y)) -> eq(#, y)
eq(0(x), #) -> eq(x, #)
eq(1(x), 1(y)) -> eq(x, y)
eq(0(x), 1(y)) -> false
eq(1(x), 0(y)) -> false
eq(0(x), 0(y)) -> eq(x, 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(#, 0(x)) -> ge(#, x)
ge(#, 1(x)) -> false
log(x) -> -(log'(x), 1(#))
log'(#) -> #
log'(1(x)) -> +(log'(x), 1(#))
log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))
mem(x, nil) -> false
mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
inter(x, nil) -> nil
inter(nil, x) -> nil
inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
ifinter(false, x, l1, l2) -> inter(l1, l2)


Strategy:

innermost




As we are in the innermost case, we can delete all 54 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
           →DP Problem 20
             ↳Modular Removal of Rules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules


Dependency Pairs:

-'(1(x), 1(y)) -> -'(x, y)
-'(1(x), 0(y)) -> -'(x, y)
-'(0(x), 1(y)) -> -'(x, y)
-'(0(x), 1(y)) -> -'(-(x, y), 1(#))
-'(0(x), 0(y)) -> -'(x, y)


Rules:


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


Strategy:

innermost




We have the following set of usable rules:

-(x, #) -> x
-(1(x), 0(y)) -> 1(-(x, y))
-(#, x) -> #
-(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
-(0(x), 0(y)) -> 0(-(x, y))
-(1(x), 1(y)) -> 0(-(x, y))
0(#) -> #
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(#)=  0  
  POL(-'(x1, x2))=  1 + x1 + x2  
  POL(0(x1))=  1 + x1  
  POL(1(x1))=  1 + x1  
  POL(-(x1, x2))=  x1 + x2  

We have the following set D of usable symbols: {#, -', 0, 1, -}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

-'(1(x), 1(y)) -> -'(x, y)
-'(1(x), 0(y)) -> -'(x, y)
-'(0(x), 1(y)) -> -'(x, y)
-'(0(x), 1(y)) -> -'(-(x, y), 1(#))
-'(0(x), 0(y)) -> -'(x, y)

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved. 



   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳Usable Rules (Innermost)
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules


Dependency Pair:

EQ(#, 0(y)) -> EQ(#, y)


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(true) -> false
not(false) -> true
if(true, x, y) -> x
if(false, x, y) -> y
eq(#, #) -> true
eq(#, 1(y)) -> false
eq(1(x), #) -> false
eq(#, 0(y)) -> eq(#, y)
eq(0(x), #) -> eq(x, #)
eq(1(x), 1(y)) -> eq(x, y)
eq(0(x), 1(y)) -> false
eq(1(x), 0(y)) -> false
eq(0(x), 0(y)) -> eq(x, 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(#, 0(x)) -> ge(#, x)
ge(#, 1(x)) -> false
log(x) -> -(log'(x), 1(#))
log'(#) -> #
log'(1(x)) -> +(log'(x), 1(#))
log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))
mem(x, nil) -> false
mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
inter(x, nil) -> nil
inter(nil, x) -> nil
inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
ifinter(false, x, l1, l2) -> inter(l1, l2)


Strategy:

innermost




As we are in the innermost case, we can delete all 61 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
           →DP Problem 21
             ↳Size-Change Principle
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules


Dependency Pair:

EQ(#, 0(y)) -> EQ(#, y)


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. EQ(#, 0(y)) -> EQ(#, y)
and get the following Size-Change Graph(s):
{1} , {1}
1=1
2>2

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1=1
2>2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
0(x1) -> 0(x1)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳Usable Rules (Innermost)
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules


Dependency Pair:

EQ(0(x), #) -> EQ(x, #)


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(true) -> false
not(false) -> true
if(true, x, y) -> x
if(false, x, y) -> y
eq(#, #) -> true
eq(#, 1(y)) -> false
eq(1(x), #) -> false
eq(#, 0(y)) -> eq(#, y)
eq(0(x), #) -> eq(x, #)
eq(1(x), 1(y)) -> eq(x, y)
eq(0(x), 1(y)) -> false
eq(1(x), 0(y)) -> false
eq(0(x), 0(y)) -> eq(x, 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(#, 0(x)) -> ge(#, x)
ge(#, 1(x)) -> false
log(x) -> -(log'(x), 1(#))
log'(#) -> #
log'(1(x)) -> +(log'(x), 1(#))
log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))
mem(x, nil) -> false
mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
inter(x, nil) -> nil
inter(nil, x) -> nil
inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
ifinter(false, x, l1, l2) -> inter(l1, l2)


Strategy:

innermost




As we are in the innermost case, we can delete all 61 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
           →DP Problem 22
             ↳Size-Change Principle
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules


Dependency Pair:

EQ(0(x), #) -> EQ(x, #)


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. EQ(0(x), #) -> EQ(x, #)
and get the following Size-Change Graph(s):
{1} , {1}
1>1
2=2

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1>1
2=2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
0(x1) -> 0(x1)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳Usable Rules (Innermost)
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules


Dependency Pair:

GE(#, 0(x)) -> GE(#, x)


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(true) -> false
not(false) -> true
if(true, x, y) -> x
if(false, x, y) -> y
eq(#, #) -> true
eq(#, 1(y)) -> false
eq(1(x), #) -> false
eq(#, 0(y)) -> eq(#, y)
eq(0(x), #) -> eq(x, #)
eq(1(x), 1(y)) -> eq(x, y)
eq(0(x), 1(y)) -> false
eq(1(x), 0(y)) -> false
eq(0(x), 0(y)) -> eq(x, 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(#, 0(x)) -> ge(#, x)
ge(#, 1(x)) -> false
log(x) -> -(log'(x), 1(#))
log'(#) -> #
log'(1(x)) -> +(log'(x), 1(#))
log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))
mem(x, nil) -> false
mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
inter(x, nil) -> nil
inter(nil, x) -> nil
inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
ifinter(false, x, l1, l2) -> inter(l1, l2)


Strategy:

innermost




As we are in the innermost case, we can delete all 61 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
           →DP Problem 23
             ↳Size-Change Principle
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules


Dependency Pair:

GE(#, 0(x)) -> GE(#, x)


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. GE(#, 0(x)) -> GE(#, x)
and get the following Size-Change Graph(s):
{1} , {1}
1=1
2>2

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1=1
2>2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
0(x1) -> 0(x1)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳Usable Rules (Innermost)
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules


Dependency Pair:

APP(cons(x, l1), l2) -> APP(l1, l2)


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(true) -> false
not(false) -> true
if(true, x, y) -> x
if(false, x, y) -> y
eq(#, #) -> true
eq(#, 1(y)) -> false
eq(1(x), #) -> false
eq(#, 0(y)) -> eq(#, y)
eq(0(x), #) -> eq(x, #)
eq(1(x), 1(y)) -> eq(x, y)
eq(0(x), 1(y)) -> false
eq(1(x), 0(y)) -> false
eq(0(x), 0(y)) -> eq(x, 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(#, 0(x)) -> ge(#, x)
ge(#, 1(x)) -> false
log(x) -> -(log'(x), 1(#))
log'(#) -> #
log'(1(x)) -> +(log'(x), 1(#))
log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))
mem(x, nil) -> false
mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
inter(x, nil) -> nil
inter(nil, x) -> nil
inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
ifinter(false, x, l1, l2) -> inter(l1, l2)


Strategy:

innermost




As we are in the innermost case, we can delete all 61 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
           →DP Problem 24
             ↳Size-Change Principle
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules


Dependency Pair:

APP(cons(x, l1), l2) -> APP(l1, l2)


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. APP(cons(x, l1), l2) -> APP(l1, l2)
and get the following Size-Change Graph(s):
{1} , {1}
1>1
2=2

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1>1
2=2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
cons(x1, x2) -> cons(x1, x2)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳Usable Rules (Innermost)
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules


Dependency Pairs:

*'(x, +(y, z)) -> *'(x, z)
*'(*(x, y), z) -> *'(y, z)
*'(x, +(y, z)) -> *'(x, y)
*'(*(x, y), z) -> *'(x, *(y, z))
*'(1(x), y) -> *'(x, y)
*'(0(x), y) -> *'(x, y)


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(true) -> false
not(false) -> true
if(true, x, y) -> x
if(false, x, y) -> y
eq(#, #) -> true
eq(#, 1(y)) -> false
eq(1(x), #) -> false
eq(#, 0(y)) -> eq(#, y)
eq(0(x), #) -> eq(x, #)
eq(1(x), 1(y)) -> eq(x, y)
eq(0(x), 1(y)) -> false
eq(1(x), 0(y)) -> false
eq(0(x), 0(y)) -> eq(x, 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(#, 0(x)) -> ge(#, x)
ge(#, 1(x)) -> false
log(x) -> -(log'(x), 1(#))
log'(#) -> #
log'(1(x)) -> +(log'(x), 1(#))
log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))
mem(x, nil) -> false
mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
inter(x, nil) -> nil
inter(nil, x) -> nil
inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
ifinter(false, x, l1, l2) -> inter(l1, l2)


Strategy:

innermost




As we are in the innermost case, we can delete all 48 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
           →DP Problem 25
             ↳Size-Change Principle
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules


Dependency Pairs:

*'(x, +(y, z)) -> *'(x, z)
*'(*(x, y), z) -> *'(y, z)
*'(x, +(y, z)) -> *'(x, y)
*'(*(x, y), z) -> *'(x, *(y, z))
*'(1(x), y) -> *'(x, y)
*'(0(x), y) -> *'(x, y)


Rules:


*(1(x), y) -> +(0(*(x, y)), y)
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(*(x, y), z) -> *(x, *(y, z))
*(0(x), y) -> 0(*(x, y))
*(#, x) -> #
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(#, x) -> x
+(+(x, y), z) -> +(x, +(y, z))
0(#) -> #


Strategy:

innermost




We number the DPs as follows:
  1. *'(x, +(y, z)) -> *'(x, z)
  2. *'(*(x, y), z) -> *'(y, z)
  3. *'(x, +(y, z)) -> *'(x, y)
  4. *'(*(x, y), z) -> *'(x, *(y, z))
  5. *'(1(x), y) -> *'(x, y)
  6. *'(0(x), y) -> *'(x, y)
and get the following Size-Change Graph(s):
{1} , {1}
1=1
2>2
{2, 5, 6} , {2, 5, 6}
1>1
2=2
{3} , {3}
1=1
2>2
{4} , {4}
1>1

which lead(s) to this/these maximal multigraph(s):
{2, 5, 6} , {2, 5, 6}
1>1
2=2
{1} , {1}
1=1
2>2
{3} , {2, 5, 6}
1>1
2>2
{3} , {4}
1>1
{2, 5, 6} , {1}
1>1
2>2
{1} , {4}
1>1
{4} , {1}
1>1
{1} , {2, 5, 6}
1>1
2>2
{2, 5, 6} , {3}
1>1
2>2
{4} , {3}
1>1
{4} , {2, 5, 6}
1>1
{2, 5, 6} , {1}
1>1
{1} , {1}
1>1
2>2
{3} , {1}
1>1
2>2
{2, 5, 6} , {3}
1>1
{1} , {1}
1>1
{3} , {1}
1>1
{2, 5, 6} , {2, 5, 6}
1>1
2>2
{2, 5, 6} , {2, 5, 6}
1>1
{3} , {2, 5, 6}
1>1
{1} , {2, 5, 6}
1>1

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
0(x1) -> 0(x1)
1(x1) -> 1(x1)
+(x1, x2) -> +(x1, x2)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳Usable Rules (Innermost)
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules


Dependency Pairs:

SUM(app(l1, l2)) -> SUM(l2)
SUM(app(l1, l2)) -> SUM(l1)
SUM(cons(x, l)) -> SUM(l)


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(true) -> false
not(false) -> true
if(true, x, y) -> x
if(false, x, y) -> y
eq(#, #) -> true
eq(#, 1(y)) -> false
eq(1(x), #) -> false
eq(#, 0(y)) -> eq(#, y)
eq(0(x), #) -> eq(x, #)
eq(1(x), 1(y)) -> eq(x, y)
eq(0(x), 1(y)) -> false
eq(1(x), 0(y)) -> false
eq(0(x), 0(y)) -> eq(x, 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(#, 0(x)) -> ge(#, x)
ge(#, 1(x)) -> false
log(x) -> -(log'(x), 1(#))
log'(#) -> #
log'(1(x)) -> +(log'(x), 1(#))
log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))
mem(x, nil) -> false
mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
inter(x, nil) -> nil
inter(nil, x) -> nil
inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
ifinter(false, x, l1, l2) -> inter(l1, l2)


Strategy:

innermost




As we are in the innermost case, we can delete all 61 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
           →DP Problem 26
             ↳Size-Change Principle
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules


Dependency Pairs:

SUM(app(l1, l2)) -> SUM(l2)
SUM(app(l1, l2)) -> SUM(l1)
SUM(cons(x, l)) -> SUM(l)


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. SUM(app(l1, l2)) -> SUM(l2)
  2. SUM(app(l1, l2)) -> SUM(l1)
  3. SUM(cons(x, l)) -> SUM(l)
and get the following Size-Change Graph(s):
{1} , {1}
1>1
{2} , {2}
1>1
{3} , {3}
1>1

which lead(s) to this/these maximal multigraph(s):
{3} , {3}
1>1
{1} , {1}
1>1
{2} , {2}
1>1
{1} , {2}
1>1
{3} , {1}
1>1
{1} , {3}
1>1
{2} , {1}
1>1
{2} , {3}
1>1

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
cons(x1, x2) -> cons(x1, x2)
app(x1, x2) -> app(x1, x2)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳Usable Rules (Innermost)
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules


Dependency Pairs:

EQ(0(x), 0(y)) -> EQ(x, y)
EQ(1(x), 1(y)) -> EQ(x, y)


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(true) -> false
not(false) -> true
if(true, x, y) -> x
if(false, x, y) -> y
eq(#, #) -> true
eq(#, 1(y)) -> false
eq(1(x), #) -> false
eq(#, 0(y)) -> eq(#, y)
eq(0(x), #) -> eq(x, #)
eq(1(x), 1(y)) -> eq(x, y)
eq(0(x), 1(y)) -> false
eq(1(x), 0(y)) -> false
eq(0(x), 0(y)) -> eq(x, 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(#, 0(x)) -> ge(#, x)
ge(#, 1(x)) -> false
log(x) -> -(log'(x), 1(#))
log'(#) -> #
log'(1(x)) -> +(log'(x), 1(#))
log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))
mem(x, nil) -> false
mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
inter(x, nil) -> nil
inter(nil, x) -> nil
inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
ifinter(false, x, l1, l2) -> inter(l1, l2)


Strategy:

innermost




As we are in the innermost case, we can delete all 61 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
           →DP Problem 27
             ↳Size-Change Principle
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules


Dependency Pairs:

EQ(0(x), 0(y)) -> EQ(x, y)
EQ(1(x), 1(y)) -> EQ(x, y)


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. EQ(0(x), 0(y)) -> EQ(x, y)
  2. EQ(1(x), 1(y)) -> EQ(x, y)
and get the following Size-Change Graph(s):
{1, 2} , {1, 2}
1>1
2>2

which lead(s) to this/these maximal multigraph(s):
{1, 2} , {1, 2}
1>1
2>2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
0(x1) -> 0(x1)
1(x1) -> 1(x1)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳Usable Rules (Innermost)
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules


Dependency Pairs:

GE(1(x), 1(y)) -> GE(x, y)
GE(1(x), 0(y)) -> GE(x, y)
GE(0(x), 1(y)) -> GE(y, x)
GE(0(x), 0(y)) -> GE(x, y)


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(true) -> false
not(false) -> true
if(true, x, y) -> x
if(false, x, y) -> y
eq(#, #) -> true
eq(#, 1(y)) -> false
eq(1(x), #) -> false
eq(#, 0(y)) -> eq(#, y)
eq(0(x), #) -> eq(x, #)
eq(1(x), 1(y)) -> eq(x, y)
eq(0(x), 1(y)) -> false
eq(1(x), 0(y)) -> false
eq(0(x), 0(y)) -> eq(x, 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(#, 0(x)) -> ge(#, x)
ge(#, 1(x)) -> false
log(x) -> -(log'(x), 1(#))
log'(#) -> #
log'(1(x)) -> +(log'(x), 1(#))
log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))
mem(x, nil) -> false
mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
inter(x, nil) -> nil
inter(nil, x) -> nil
inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
ifinter(false, x, l1, l2) -> inter(l1, l2)


Strategy:

innermost




As we are in the innermost case, we can delete all 61 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
           →DP Problem 28
             ↳Size-Change Principle
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules


Dependency Pairs:

GE(1(x), 1(y)) -> GE(x, y)
GE(1(x), 0(y)) -> GE(x, y)
GE(0(x), 1(y)) -> GE(y, x)
GE(0(x), 0(y)) -> GE(x, y)


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. GE(1(x), 1(y)) -> GE(x, y)
  2. GE(1(x), 0(y)) -> GE(x, y)
  3. GE(0(x), 1(y)) -> GE(y, x)
  4. GE(0(x), 0(y)) -> GE(x, y)
and get the following Size-Change Graph(s):
{1, 2, 3, 4} , {1, 2, 3, 4}
1>1
2>2
{1, 2, 3, 4} , {1, 2, 3, 4}
1>2
2>1

which lead(s) to this/these maximal multigraph(s):
{1, 2, 3, 4} , {1, 2, 3, 4}
1>1
2>2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
0(x1) -> 0(x1)
1(x1) -> 1(x1)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳Usable Rules (Innermost)
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules


Dependency Pairs:

PROD(app(l1, l2)) -> PROD(l2)
PROD(app(l1, l2)) -> PROD(l1)
PROD(cons(x, l)) -> PROD(l)


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(true) -> false
not(false) -> true
if(true, x, y) -> x
if(false, x, y) -> y
eq(#, #) -> true
eq(#, 1(y)) -> false
eq(1(x), #) -> false
eq(#, 0(y)) -> eq(#, y)
eq(0(x), #) -> eq(x, #)
eq(1(x), 1(y)) -> eq(x, y)
eq(0(x), 1(y)) -> false
eq(1(x), 0(y)) -> false
eq(0(x), 0(y)) -> eq(x, 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(#, 0(x)) -> ge(#, x)
ge(#, 1(x)) -> false
log(x) -> -(log'(x), 1(#))
log'(#) -> #
log'(1(x)) -> +(log'(x), 1(#))
log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))
mem(x, nil) -> false
mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
inter(x, nil) -> nil
inter(nil, x) -> nil
inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
ifinter(false, x, l1, l2) -> inter(l1, l2)


Strategy:

innermost




As we are in the innermost case, we can delete all 61 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
           →DP Problem 29
             ↳Size-Change Principle
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules


Dependency Pairs:

PROD(app(l1, l2)) -> PROD(l2)
PROD(app(l1, l2)) -> PROD(l1)
PROD(cons(x, l)) -> PROD(l)


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. PROD(app(l1, l2)) -> PROD(l2)
  2. PROD(app(l1, l2)) -> PROD(l1)
  3. PROD(cons(x, l)) -> PROD(l)
and get the following Size-Change Graph(s):
{1} , {1}
1>1
{2} , {2}
1>1
{3} , {3}
1>1

which lead(s) to this/these maximal multigraph(s):
{2} , {2}
1>1
{3} , {3}
1>1
{1} , {1}
1>1
{2} , {1}
1>1
{1} , {3}
1>1
{3} , {1}
1>1
{1} , {2}
1>1
{2} , {3}
1>1

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
cons(x1, x2) -> cons(x1, x2)
app(x1, x2) -> app(x1, x2)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳Usable Rules (Innermost)
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules


Dependency Pair:

MEM(x, cons(y, l)) -> MEM(x, l)


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(true) -> false
not(false) -> true
if(true, x, y) -> x
if(false, x, y) -> y
eq(#, #) -> true
eq(#, 1(y)) -> false
eq(1(x), #) -> false
eq(#, 0(y)) -> eq(#, y)
eq(0(x), #) -> eq(x, #)
eq(1(x), 1(y)) -> eq(x, y)
eq(0(x), 1(y)) -> false
eq(1(x), 0(y)) -> false
eq(0(x), 0(y)) -> eq(x, 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(#, 0(x)) -> ge(#, x)
ge(#, 1(x)) -> false
log(x) -> -(log'(x), 1(#))
log'(#) -> #
log'(1(x)) -> +(log'(x), 1(#))
log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))
mem(x, nil) -> false
mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
inter(x, nil) -> nil
inter(nil, x) -> nil
inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
ifinter(false, x, l1, l2) -> inter(l1, l2)


Strategy:

innermost




As we are in the innermost case, we can delete all 61 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
           →DP Problem 30
             ↳Size-Change Principle
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules


Dependency Pair:

MEM(x, cons(y, l)) -> MEM(x, l)


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. MEM(x, cons(y, l)) -> MEM(x, l)
and get the following Size-Change Graph(s):
{1} , {1}
1=1
2>2

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1=1
2>2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
cons(x1, x2) -> cons(x1, x2)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳Usable Rules (Innermost)
       →DP Problem 14
         ↳UsableRules


Dependency Pairs:

LOG'(0(x)) -> LOG'(x)
LOG'(1(x)) -> LOG'(x)


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(true) -> false
not(false) -> true
if(true, x, y) -> x
if(false, x, y) -> y
eq(#, #) -> true
eq(#, 1(y)) -> false
eq(1(x), #) -> false
eq(#, 0(y)) -> eq(#, y)
eq(0(x), #) -> eq(x, #)
eq(1(x), 1(y)) -> eq(x, y)
eq(0(x), 1(y)) -> false
eq(1(x), 0(y)) -> false
eq(0(x), 0(y)) -> eq(x, 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(#, 0(x)) -> ge(#, x)
ge(#, 1(x)) -> false
log(x) -> -(log'(x), 1(#))
log'(#) -> #
log'(1(x)) -> +(log'(x), 1(#))
log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))
mem(x, nil) -> false
mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
inter(x, nil) -> nil
inter(nil, x) -> nil
inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
ifinter(false, x, l1, l2) -> inter(l1, l2)


Strategy:

innermost




As we are in the innermost case, we can delete all 61 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
           →DP Problem 31
             ↳Size-Change Principle
       →DP Problem 14
         ↳UsableRules


Dependency Pairs:

LOG'(0(x)) -> LOG'(x)
LOG'(1(x)) -> LOG'(x)


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. LOG'(0(x)) -> LOG'(x)
  2. LOG'(1(x)) -> LOG'(x)
and get the following Size-Change Graph(s):
{1, 2} , {1, 2}
1>1

which lead(s) to this/these maximal multigraph(s):
{1, 2} , {1, 2}
1>1

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
0(x1) -> 0(x1)
1(x1) -> 1(x1)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳Usable Rules (Innermost)


Dependency Pairs:

IFINTER(false, x, l1, l2) -> INTER(l1, l2)
INTER(l1, cons(x, l2)) -> IFINTER(mem(x, l1), x, l2, l1)
IFINTER(true, x, l1, l2) -> INTER(l1, l2)
INTER(cons(x, l1), l2) -> IFINTER(mem(x, l2), x, l1, l2)
INTER(l1, app(l2, l3)) -> INTER(l1, l3)
INTER(l1, app(l2, l3)) -> INTER(l1, l2)
INTER(app(l1, l2), l3) -> INTER(l2, l3)
INTER(app(l1, l2), l3) -> INTER(l1, l3)


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(true) -> false
not(false) -> true
if(true, x, y) -> x
if(false, x, y) -> y
eq(#, #) -> true
eq(#, 1(y)) -> false
eq(1(x), #) -> false
eq(#, 0(y)) -> eq(#, y)
eq(0(x), #) -> eq(x, #)
eq(1(x), 1(y)) -> eq(x, y)
eq(0(x), 1(y)) -> false
eq(1(x), 0(y)) -> false
eq(0(x), 0(y)) -> eq(x, 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(#, 0(x)) -> ge(#, x)
ge(#, 1(x)) -> false
log(x) -> -(log'(x), 1(#))
log'(#) -> #
log'(1(x)) -> +(log'(x), 1(#))
log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))
mem(x, nil) -> false
mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
inter(x, nil) -> nil
inter(nil, x) -> nil
inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
ifinter(false, x, l1, l2) -> inter(l1, l2)


Strategy:

innermost




As we are in the innermost case, we can delete all 48 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
       →DP Problem 5
         ↳UsableRules
       →DP Problem 6
         ↳UsableRules
       →DP Problem 7
         ↳UsableRules
       →DP Problem 8
         ↳UsableRules
       →DP Problem 9
         ↳UsableRules
       →DP Problem 10
         ↳UsableRules
       →DP Problem 11
         ↳UsableRules
       →DP Problem 12
         ↳UsableRules
       →DP Problem 13
         ↳UsableRules
       →DP Problem 14
         ↳UsableRules
           →DP Problem 32
             ↳Size-Change Principle


Dependency Pairs:

IFINTER(false, x, l1, l2) -> INTER(l1, l2)
INTER(l1, cons(x, l2)) -> IFINTER(mem(x, l1), x, l2, l1)
IFINTER(true, x, l1, l2) -> INTER(l1, l2)
INTER(cons(x, l1), l2) -> IFINTER(mem(x, l2), x, l1, l2)
INTER(l1, app(l2, l3)) -> INTER(l1, l3)
INTER(l1, app(l2, l3)) -> INTER(l1, l2)
INTER(app(l1, l2), l3) -> INTER(l2, l3)
INTER(app(l1, l2), l3) -> INTER(l1, l3)


Rules:


eq(0(x), 1(y)) -> false
eq(#, 0(y)) -> eq(#, y)
eq(#, #) -> true
eq(1(x), 1(y)) -> eq(x, y)
eq(0(x), 0(y)) -> eq(x, y)
eq(#, 1(y)) -> false
eq(1(x), #) -> false
eq(1(x), 0(y)) -> false
eq(0(x), #) -> eq(x, #)
mem(x, nil) -> false
mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l))
if(false, x, y) -> y
if(true, x, y) -> x


Strategy:

innermost




We number the DPs as follows:
  1. IFINTER(false, x, l1, l2) -> INTER(l1, l2)
  2. INTER(l1, cons(x, l2)) -> IFINTER(mem(x, l1), x, l2, l1)
  3. IFINTER(true, x, l1, l2) -> INTER(l1, l2)
  4. INTER(cons(x, l1), l2) -> IFINTER(mem(x, l2), x, l1, l2)
  5. INTER(l1, app(l2, l3)) -> INTER(l1, l3)
  6. INTER(l1, app(l2, l3)) -> INTER(l1, l2)
  7. INTER(app(l1, l2), l3) -> INTER(l2, l3)
  8. INTER(app(l1, l2), l3) -> INTER(l1, l3)
and get the following Size-Change Graph(s):
{1, 3} , {1, 3}
3=1
4=2
{2} , {2}
1=4
2>2
2>3
{4} , {4}
1>2
1>3
2=4
{5, 7} , {5, 7}
1=1
2>2
{6} , {6}
1=1
2>2
{5, 7} , {5, 7}
1>1
2=2
{8} , {8}
1>1
2=2

which lead(s) to this/these maximal multigraph(s):
{8} , {8}
1>1
2=2
{5, 7} , {5, 7}
1>1
2=2
{5, 7} , {5, 7}
1=1
2>2
{6} , {6}
1=1
2>2
{5, 7} , {5, 7}
1>1
2>2
{8} , {5, 7}
1>1
2=2
{1, 3} , {4}
3>2
3>3
4=4
{5, 7} , {8}
1>1
2=2
{8} , {6}
1>1
2>2
{5, 7} , {6}
1>1
2>2
{6} , {8}
1>1
2>2
{5, 7} , {8}
1>1
2>2
{6} , {5, 7}
1>1
2>2
{5, 7} , {6}
1=1
2>2
{8} , {5, 7}
1>1
2>2
{4} , {1, 3}
1>1
2=2
{6} , {5, 7}
1=1
2>2
{5, 7} , {1, 3}
1>1
2=2
{5, 7} , {1, 3}
1>1
2>2
{8} , {8}
1>1
2>2
{6} , {6}
1>1
2>2
{4} , {6}
1>1
2>2
{4} , {5, 7}
1>1
2=2
{4} , {5, 7}
1>1
2>2
{1, 3} , {4}
3>2
3>3
4>4
{6} , {1, 3}
1>1
2>2
{8} , {1, 3}
1>1
2>2
{2} , {1, 3}
1>1
2>2
{1, 3} , {2}
3>2
3>3
4>4
{8} , {1, 3}
1>1
2=2
{2} , {5, 7}
1>1
2>2
{4} , {1, 3}
1>1
2>2
{2} , {8}
1>1
2>2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
cons(x1, x2) -> cons(x1, x2)
app(x1, x2) -> app(x1, x2)

We obtain no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:13 minutes