Termination Proof using AProVE

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.

     ↳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.

     ↳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.

     ↳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.

     ↳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.

     ↳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.

     ↳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:

     ↳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:

     ↳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:

     ↳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:

Dependency Pairs:
*'(*(x, s(*(x'''', *(x'''''', y'''''')))), s(y''')) -> *'(s(*(x'''', *(x'''''', y''''''))), s(y'''))
*'(*(x, s(*(x'''', y''''))), s(y0'')) -> *'(s(*(x'''', y'''')), s(y0''))
*'(*(x, s(s(x''''))), s(s(y''''))) -> *'(s(s(x'''')), s(s(y'''')))
*'(*(x, *(x'', *(x'''', y''''))), z') -> *'(*(x'', *(x'''', y'''')), z')
*'(*(x, *(x'', y'')), z'') -> *'(*(x'', 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'')))
*'(s(*(x'', *(x'''', y''''))), s(y')) -> *'(*(x'', *(x'''', y'''')), 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
Dependency Pair:
SUM(cons(x, l)) -> SUM(l)

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
Dependency Pair:
PROD(cons(x, l)) -> PROD(l)

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

     ↳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:

Dependency Pairs:
*'(*(x, s(*(x'''', *(x'''''', y'''''')))), s(y''')) -> *'(s(*(x'''', *(x'''''', y''''''))), s(y'''))
*'(*(x, s(*(x'''', y''''))), s(y0'')) -> *'(s(*(x'''', y'''')), s(y0''))
*'(*(x, s(s(x''''))), s(s(y''''))) -> *'(s(s(x'''')), s(s(y'''')))
*'(*(x, *(x'', *(x'''', y''''))), z') -> *'(*(x'', *(x'''', y'''')), z')
*'(*(x, *(x'', y'')), z'') -> *'(*(x'', 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'')))
*'(s(*(x'', *(x'''', y''''))), s(y')) -> *'(*(x'', *(x'''', y'''')), 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
Dependency Pair:
SUM(cons(x, l)) -> SUM(l)

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
Dependency Pair:
PROD(cons(x, l)) -> PROD(l)

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

     ↳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:

Dependency Pairs:
*'(*(x, s(*(x'''', *(x'''''', y'''''')))), s(y''')) -> *'(s(*(x'''', *(x'''''', y''''''))), s(y'''))
*'(*(x, s(*(x'''', y''''))), s(y0'')) -> *'(s(*(x'''', y'''')), s(y0''))
*'(*(x, s(s(x''''))), s(s(y''''))) -> *'(s(s(x'''')), s(s(y'''')))
*'(*(x, *(x'', *(x'''', y''''))), z') -> *'(*(x'', *(x'''', y'''')), z')
*'(*(x, *(x'', y'')), z'') -> *'(*(x'', 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'')))
*'(s(*(x'', *(x'''', y''''))), s(y')) -> *'(*(x'', *(x'''', y'''')), 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
Dependency Pair:
SUM(cons(x, l)) -> SUM(l)

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
Dependency Pair:
PROD(cons(x, l)) -> PROD(l)

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

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

POL(0)	= 1
POL(s(x₁))	= 1 + x₁
POL(+(x₁, x₂))	= x₁ + x₂
POL(+'(x₁, x₂))	= x₁

POL(+(x₁, x₂))	= 1 + x₂
POL(+'(x₁, x₂))	= x₁