Term Rewriting System R:
[x, y, z, l, l1, l2]
+(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))
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) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))

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

Furthermore, R contains five SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation
       →DP Problem 2
         ↳Remaining
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining
       →DP Problem 5
         ↳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))
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) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))


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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 6
             ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳Remaining
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining
       →DP Problem 5
         ↳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))
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) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))


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
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 7
                 ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳Remaining
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining
       →DP Problem 5
         ↳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))
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) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))


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
         ↳Nar
       →DP Problem 2
         ↳Remaining Obligation(s)
       →DP Problem 3
         ↳Remaining Obligation(s)
       →DP Problem 4
         ↳Remaining Obligation(s)
       →DP Problem 5
         ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳Nar
       →DP Problem 2
         ↳Remaining Obligation(s)
       →DP Problem 3
         ↳Remaining Obligation(s)
       →DP Problem 4
         ↳Remaining Obligation(s)
       →DP Problem 5
         ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳Nar
       →DP Problem 2
         ↳Remaining Obligation(s)
       →DP Problem 3
         ↳Remaining Obligation(s)
       →DP Problem 4
         ↳Remaining Obligation(s)
       →DP Problem 5
         ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳Nar
       →DP Problem 2
         ↳Remaining Obligation(s)
       →DP Problem 3
         ↳Remaining Obligation(s)
       →DP Problem 4
         ↳Remaining Obligation(s)
       →DP Problem 5
         ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳Nar
       →DP Problem 2
         ↳Remaining Obligation(s)
       →DP Problem 3
         ↳Remaining Obligation(s)
       →DP Problem 4
         ↳Remaining Obligation(s)
       →DP Problem 5
         ↳Remaining Obligation(s)




The following remains to be proven:

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