Termination Proof using AProVE

Term Rewriting System R:
[x, y, l]
+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
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)
*'(s(x), s(y)) -> +'(*(x, y), +(x, y))
*'(s(x), s(y)) -> *'(x, y)
*'(s(x), s(y)) -> +'(x, y)
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

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

Dependency Pair:

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

Rules:

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 5

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

Dependency Pair:

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

Rules:

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 5

             ↳FwdInst

...

               →DP Problem 6

                 ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

Dependency Pair:

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

Rules:

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

There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.
Used Argument Filtering System:

+'(x₁, x₂) -> +'(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 5

             ↳FwdInst

...

               →DP Problem 7

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

Dependency Pair:

Rules:

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

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

Dependency Pair:

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

Rules:

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 8

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

Dependency Pair:

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

Rules:

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 8

             ↳FwdInst

...

               →DP Problem 9

                 ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

Dependency Pair:

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

Rules:

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

There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.
Used Argument Filtering System:

*'(x₁, x₂) -> *'(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 8

             ↳FwdInst

...

               →DP Problem 10

                 ↳Dependency Graph

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

Dependency Pair:

Rules:

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

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳FwdInst

Dependency Pair:

SUM(cons(x, l)) -> SUM(l)

Rules:

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

SUM(cons(x, l)) -> SUM(l)

one new Dependency Pair is created:

SUM(cons(x, cons(x'', l''))) -> SUM(cons(x'', l''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 11

             ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳FwdInst

Dependency Pair:

SUM(cons(x, cons(x'', l''))) -> SUM(cons(x'', l''))

Rules:

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

SUM(cons(x, cons(x'', l''))) -> SUM(cons(x'', l''))

one new Dependency Pair is created:

SUM(cons(x, cons(x'''', cons(x''''', l'''')))) -> SUM(cons(x'''', cons(x''''', l'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 11

             ↳FwdInst

...

               →DP Problem 12

                 ↳Argument Filtering and Ordering

       →DP Problem 4

         ↳FwdInst

Dependency Pair:

SUM(cons(x, cons(x'''', cons(x''''', l'''')))) -> SUM(cons(x'''', cons(x''''', l'''')))

Rules:

+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
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:

SUM(cons(x, cons(x'''', cons(x''''', l'''')))) -> SUM(cons(x'''', cons(x''''', l'''')))

There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.
Used Argument Filtering System:

SUM(x₁) -> SUM(x₁)
cons(x₁, x₂) -> cons(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 11

             ↳FwdInst

...

               →DP Problem 13

                 ↳Dependency Graph

       →DP Problem 4

         ↳FwdInst

Dependency Pair:

Rules:

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

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Forward Instantiation Transformation

Dependency Pair:

PROD(cons(x, l)) -> PROD(l)

Rules:

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

PROD(cons(x, l)) -> PROD(l)

one new Dependency Pair is created:

PROD(cons(x, cons(x'', l''))) -> PROD(cons(x'', l''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

           →DP Problem 14

             ↳Forward Instantiation Transformation

Dependency Pair:

PROD(cons(x, cons(x'', l''))) -> PROD(cons(x'', l''))

Rules:

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

PROD(cons(x, cons(x'', l''))) -> PROD(cons(x'', l''))

one new Dependency Pair is created:

PROD(cons(x, cons(x'''', cons(x''''', l'''')))) -> PROD(cons(x'''', cons(x''''', l'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

           →DP Problem 14

             ↳FwdInst

...

               →DP Problem 15

                 ↳Argument Filtering and Ordering

Dependency Pair:

PROD(cons(x, cons(x'''', cons(x''''', l'''')))) -> PROD(cons(x'''', cons(x''''', l'''')))

Rules:

+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
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:

PROD(cons(x, cons(x'''', cons(x''''', l'''')))) -> PROD(cons(x'''', cons(x''''', l'''')))

There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.
Used Argument Filtering System:

PROD(x₁) -> PROD(x₁)
cons(x₁, x₂) -> cons(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

           →DP Problem 14

             ↳FwdInst

...

               →DP Problem 16

                 ↳Dependency Graph

Dependency Pair:

Rules:

+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
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.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes