Termination Proof using AProVE

Term Rewriting System R:
[x, y, l]
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) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))

Innermost Termination of R to be shown.

     ↳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)
*'(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)
SUM(nil) -> 0'(#)
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

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

Dependency Pairs:

+'(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) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

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

+ > #
+' > #
0 > #
1 > #

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

+'(x₁, x₂) -> x₂
0(x₁) -> 0(x₁)
1(x₁) -> 1(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 5

             ↳Narrowing Transformation

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

Dependency Pair:

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

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) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> 1(#)
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

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

six new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 5

             ↳Nar

...

               →DP Problem 6

                 ↳Instantiation Transformation

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

Dependency Pairs:

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

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) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))

Strategy:

innermost

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

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

one new Dependency Pair is created:

+'(1(#), 1(#)) -> +'(#, 1(#))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 5

             ↳Nar

...

               →DP Problem 7

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

Dependency Pair:

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

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) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> 1(#)
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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 5

             ↳Nar

...

               →DP Problem 8

                 ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

Dependency Pair:

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

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) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the AFS 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)
1(x₁) -> 1(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 5

             ↳Nar

...

               →DP Problem 9

                 ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

Dependency Pair:

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) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

Dependency Pairs:

*'(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) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the AFS 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)
0(x₁) -> 0(x₁)
1(x₁) -> 1(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 10

             ↳Dependency Graph

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

Dependency Pair:

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) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Argument Filtering and Ordering

       →DP Problem 4

         ↳AFS

Dependency Pair:

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) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the AFS 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

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 11

             ↳Dependency Graph

       →DP Problem 4

         ↳AFS

Dependency Pair:

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) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Argument Filtering and Ordering

Dependency Pair:

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) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the AFS 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

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

           →DP Problem 12

             ↳Dependency Graph

Dependency Pair:

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) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> 1(#)
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:
0:01 minutes