Term Rewriting System R:
[x, y, z, l]
+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

+'(s(x), s(y)) -> +'(x, y)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(+(x, y), z) -> +'(y, z)
*'(s(x), s(y)) -> +'(*(x, y), +(x, y))
*'(s(x), s(y)) -> *'(x, y)
*'(s(x), s(y)) -> +'(x, y)
*'(*(x, y), z) -> *'(x, *(y, z))
*'(*(x, y), z) -> *'(y, z)
SUM(cons(x, l)) -> +'(x, sum(l))
SUM(cons(x, l)) -> SUM(l)
PROD(cons(x, l)) -> *'(x, prod(l))
PROD(cons(x, l)) -> PROD(l)

Furthermore, R contains four SCCs.


   R
DPs
       →DP Problem 1
Polynomial Ordering
       →DP Problem 2
Nar
       →DP Problem 3
Remaining
       →DP Problem 4
Remaining


Dependency Pairs:

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


Rules:


+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


Additionally, the following usable rules for innermost can be oriented:

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


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  1  
  POL(s(x1))=  1 + x1  
  POL(+(x1, x2))=  x1 + x2  
  POL(+'(x1, x2))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 5
Dependency Graph
       →DP Problem 2
Nar
       →DP Problem 3
Remaining
       →DP Problem 4
Remaining


Dependency Pairs:

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


Rules:


+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))


Strategy:

innermost




Using the Dependency Graph the DP problem was split into 1 DP problems.


   R
DPs
       →DP Problem 1
Polo
           →DP Problem 5
DGraph
             ...
               →DP Problem 6
Polynomial Ordering
       →DP Problem 2
Nar
       →DP Problem 3
Remaining
       →DP Problem 4
Remaining


Dependency Pair:

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


Rules:


+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(+(x1, x2))=  1 + x2  
  POL(+'(x1, x2))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 5
DGraph
             ...
               →DP Problem 7
Dependency Graph
       →DP Problem 2
Nar
       →DP Problem 3
Remaining
       →DP Problem 4
Remaining


Dependency Pair:


Rules:


+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Narrowing Transformation
       →DP Problem 3
Remaining
       →DP Problem 4
Remaining


Dependency Pairs:

*'(*(x, y), z) -> *'(y, z)
*'(*(x, y), z) -> *'(x, *(y, z))
*'(s(x), s(y)) -> *'(x, y)


Rules:


+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

*'(*(x, y), z) -> *'(x, *(y, z))
three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Nar
           →DP Problem 8
Forward Instantiation Transformation
       →DP Problem 3
Remaining
       →DP Problem 4
Remaining


Dependency Pairs:

*'(*(x, *(x'', y'')), z'') -> *'(x, *(x'', *(y'', z'')))
*'(s(x), s(y)) -> *'(x, y)
*'(*(x, y), z) -> *'(y, z)


Rules:


+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))


Strategy:

innermost




On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

*'(s(x), s(y)) -> *'(x, y)
three new Dependency Pairs are created:

*'(s(s(x'')), s(s(y''))) -> *'(s(x''), s(y''))
*'(s(*(x'', y'')), s(y0)) -> *'(*(x'', y''), y0)
*'(s(*(x'', *(x'''', y''''))), s(y')) -> *'(*(x'', *(x'''', y'''')), y')

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Nar
           →DP Problem 8
FwdInst
             ...
               →DP Problem 9
Forward Instantiation Transformation
       →DP Problem 3
Remaining
       →DP Problem 4
Remaining


Dependency Pairs:

*'(s(*(x'', *(x'''', y''''))), s(y')) -> *'(*(x'', *(x'''', y'''')), y')
*'(*(x, y), z) -> *'(y, z)
*'(s(*(x'', y'')), s(y0)) -> *'(*(x'', y''), y0)
*'(s(s(x'')), s(s(y''))) -> *'(s(x''), s(y''))
*'(*(x, *(x'', y'')), z'') -> *'(x, *(x'', *(y'', z'')))


Rules:


+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))


Strategy:

innermost




On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

*'(*(x, y), z) -> *'(y, z)
five new Dependency Pairs are created:

*'(*(x, *(x'', y'')), z'') -> *'(*(x'', y''), z'')
*'(*(x, *(x'', *(x'''', y''''))), z') -> *'(*(x'', *(x'''', y'''')), z')
*'(*(x, s(s(x''''))), s(s(y''''))) -> *'(s(s(x'''')), s(s(y'''')))
*'(*(x, s(*(x'''', y''''))), s(y0'')) -> *'(s(*(x'''', y'''')), s(y0''))
*'(*(x, s(*(x'''', *(x'''''', y'''''')))), s(y''')) -> *'(s(*(x'''', *(x'''''', y''''''))), s(y'''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Nar
       →DP Problem 3
Remaining Obligation(s)
       →DP Problem 4
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Nar
       →DP Problem 3
Remaining Obligation(s)
       →DP Problem 4
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Nar
       →DP Problem 3
Remaining Obligation(s)
       →DP Problem 4
Remaining Obligation(s)




The following remains to be proven:

Innermost Termination of R could not be shown.
Duration:
0:01 minutes