Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

MINUS(s(x), s(y)) -> MINUS(x, y)
DOUBLE(s(x)) -> DOUBLE(x)
PLUS(s(x), y) -> PLUS(x, y)
PLUS(s(x), y) -> PLUS(x, s(y))
PLUS(s(x), y) -> PLUS(minus(x, y), double(y))
PLUS(s(x), y) -> MINUS(x, y)
PLUS(s(x), y) -> DOUBLE(y)
PLUS(s(plus(x, y)), z) -> PLUS(plus(x, y), z)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

Dependency Pair:

MINUS(s(x), s(y)) -> MINUS(x, y)

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))

Strategy:

innermost

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

MINUS(s(x), s(y)) -> MINUS(x, y)

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 5

                 ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

MINUS(s(s(s(x''''))), s(s(s(y'''')))) -> MINUS(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:

MINUS(x₁, x₂) -> MINUS(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 6

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

Dependency Pair:

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))

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

         ↳Nar

Dependency Pair:

DOUBLE(s(x)) -> DOUBLE(x)

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))

Strategy:

innermost

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

DOUBLE(s(x)) -> DOUBLE(x)

one new Dependency Pair is created:

DOUBLE(s(s(x''))) -> DOUBLE(s(x''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pair:

DOUBLE(s(s(x''))) -> DOUBLE(s(x''))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))

Strategy:

innermost

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

DOUBLE(s(s(x''))) -> DOUBLE(s(x''))

one new Dependency Pair is created:

DOUBLE(s(s(s(x'''')))) -> DOUBLE(s(s(x'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 8

                 ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳Nar

Dependency Pair:

DOUBLE(s(s(s(x'''')))) -> DOUBLE(s(s(x'''')))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

DOUBLE(s(s(s(x'''')))) -> DOUBLE(s(s(x'''')))

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:

DOUBLE(x₁) -> DOUBLE(x₁)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 9

                 ↳Dependency Graph

       →DP Problem 3

         ↳Nar

Dependency Pair:

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Narrowing Transformation

Dependency Pairs:

PLUS(s(x), y) -> PLUS(minus(x, y), double(y))
PLUS(s(x), y) -> PLUS(x, s(y))
PLUS(s(x), y) -> PLUS(x, y)

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))

Strategy:

innermost

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

PLUS(s(x), y) -> PLUS(minus(x, y), double(y))

four new Dependency Pairs are created:

PLUS(s(x''), 0) -> PLUS(x'', double(0))
PLUS(s(s(x'')), s(y'')) -> PLUS(minus(x'', y''), double(s(y'')))
PLUS(s(x), 0) -> PLUS(minus(x, 0), 0)
PLUS(s(x), s(x'')) -> PLUS(minus(x, s(x'')), s(s(double(x''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Rewriting Transformation

Dependency Pairs:

PLUS(s(x), s(x'')) -> PLUS(minus(x, s(x'')), s(s(double(x''))))
PLUS(s(x), 0) -> PLUS(minus(x, 0), 0)
PLUS(s(s(x'')), s(y'')) -> PLUS(minus(x'', y''), double(s(y'')))
PLUS(s(x''), 0) -> PLUS(x'', double(0))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(s(x), y) -> PLUS(x, s(y))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))

Strategy:

innermost

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

PLUS(s(x''), 0) -> PLUS(x'', double(0))

one new Dependency Pair is created:

PLUS(s(x''), 0) -> PLUS(x'', 0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Rw

...

               →DP Problem 11

                 ↳Rewriting Transformation

Dependency Pairs:

PLUS(s(x''), 0) -> PLUS(x'', 0)
PLUS(s(x), 0) -> PLUS(minus(x, 0), 0)
PLUS(s(s(x'')), s(y'')) -> PLUS(minus(x'', y''), double(s(y'')))
PLUS(s(x), y) -> PLUS(x, s(y))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(s(x), s(x'')) -> PLUS(minus(x, s(x'')), s(s(double(x''))))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))

Strategy:

innermost

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

PLUS(s(s(x'')), s(y'')) -> PLUS(minus(x'', y''), double(s(y'')))

one new Dependency Pair is created:

PLUS(s(s(x'')), s(y'')) -> PLUS(minus(x'', y''), s(s(double(y''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Rw

...

               →DP Problem 12

                 ↳Rewriting Transformation

Dependency Pairs:

PLUS(s(x), 0) -> PLUS(minus(x, 0), 0)
PLUS(s(s(x'')), s(y'')) -> PLUS(minus(x'', y''), s(s(double(y''))))
PLUS(s(x), s(x'')) -> PLUS(minus(x, s(x'')), s(s(double(x''))))
PLUS(s(x), y) -> PLUS(x, s(y))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(s(x''), 0) -> PLUS(x'', 0)

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))

Strategy:

innermost

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

PLUS(s(x), 0) -> PLUS(minus(x, 0), 0)

one new Dependency Pair is created:

PLUS(s(x), 0) -> PLUS(x, 0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Rw

...

               →DP Problem 13

                 ↳Narrowing Transformation

Dependency Pairs:

PLUS(s(x), 0) -> PLUS(x, 0)
PLUS(s(x''), 0) -> PLUS(x'', 0)
PLUS(s(x), s(x'')) -> PLUS(minus(x, s(x'')), s(s(double(x''))))
PLUS(s(x), y) -> PLUS(x, s(y))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(s(s(x'')), s(y'')) -> PLUS(minus(x'', y''), s(s(double(y''))))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))

Strategy:

innermost

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

PLUS(s(x), s(x'')) -> PLUS(minus(x, s(x'')), s(s(double(x''))))

three new Dependency Pairs are created:

PLUS(s(s(x'0)), s(x''')) -> PLUS(minus(x'0, x'''), s(s(double(x'''))))
PLUS(s(x), s(0)) -> PLUS(minus(x, s(0)), s(s(0)))
PLUS(s(x), s(s(x'''))) -> PLUS(minus(x, s(s(x'''))), s(s(s(s(double(x'''))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Rw

...

               →DP Problem 14

                 ↳Forward Instantiation Transformation

Dependency Pairs:

PLUS(s(x''), 0) -> PLUS(x'', 0)
PLUS(s(x), s(0)) -> PLUS(minus(x, s(0)), s(s(0)))
PLUS(s(x), s(s(x'''))) -> PLUS(minus(x, s(s(x'''))), s(s(s(s(double(x'''))))))
PLUS(s(s(x'0)), s(x''')) -> PLUS(minus(x'0, x'''), s(s(double(x'''))))
PLUS(s(s(x'')), s(y'')) -> PLUS(minus(x'', y''), s(s(double(y''))))
PLUS(s(x), y) -> PLUS(x, s(y))
PLUS(s(x), y) -> PLUS(x, y)
PLUS(s(x), 0) -> PLUS(x, 0)

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))

Strategy:

innermost

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

PLUS(s(x), y) -> PLUS(x, y)

seven new Dependency Pairs are created:

PLUS(s(s(x'')), y'') -> PLUS(s(x''), y'')
PLUS(s(s(x'''')), 0) -> PLUS(s(x''''), 0)
PLUS(s(s(s(x''''))), s(y'''')) -> PLUS(s(s(x'''')), s(y''''))
PLUS(s(s(x'')), 0) -> PLUS(s(x''), 0)
PLUS(s(s(s(x'0''))), s(x''''')) -> PLUS(s(s(x'0'')), s(x'''''))
PLUS(s(s(x'')), s(0)) -> PLUS(s(x''), s(0))
PLUS(s(s(x'')), s(s(x'''''))) -> PLUS(s(x''), s(s(x''''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Rw

...

               →DP Problem 15

                 ↳Forward Instantiation Transformation

Dependency Pairs:

PLUS(s(s(x'')), s(s(x'''''))) -> PLUS(s(x''), s(s(x''''')))
PLUS(s(s(x'')), s(0)) -> PLUS(s(x''), s(0))
PLUS(s(s(s(x'0''))), s(x''''')) -> PLUS(s(s(x'0'')), s(x'''''))
PLUS(s(s(s(x''''))), s(y'''')) -> PLUS(s(s(x'''')), s(y''''))
PLUS(s(x), s(0)) -> PLUS(minus(x, s(0)), s(s(0)))
PLUS(s(s(x'')), 0) -> PLUS(s(x''), 0)
PLUS(s(s(x'''')), 0) -> PLUS(s(x''''), 0)
PLUS(s(x), 0) -> PLUS(x, 0)
PLUS(s(s(x'')), y'') -> PLUS(s(x''), y'')
PLUS(s(x), s(s(x'''))) -> PLUS(minus(x, s(s(x'''))), s(s(s(s(double(x'''))))))
PLUS(s(s(x'0)), s(x''')) -> PLUS(minus(x'0, x'''), s(s(double(x'''))))
PLUS(s(s(x'')), s(y'')) -> PLUS(minus(x'', y''), s(s(double(y''))))
PLUS(s(x), y) -> PLUS(x, s(y))
PLUS(s(x''), 0) -> PLUS(x'', 0)

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))

Strategy:

innermost

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

PLUS(s(x), y) -> PLUS(x, s(y))

nine new Dependency Pairs are created:

PLUS(s(s(x'')), y'') -> PLUS(s(x''), s(y''))
PLUS(s(s(s(x''''))), y') -> PLUS(s(s(x'''')), s(y'))
PLUS(s(s(s(x'0''))), y') -> PLUS(s(s(x'0'')), s(y'))
PLUS(s(s(x'')), 0) -> PLUS(s(x''), s(0))
PLUS(s(s(x'')), s(x''''')) -> PLUS(s(x''), s(s(x''''')))
PLUS(s(s(s(s(x'''''')))), y') -> PLUS(s(s(s(x''''''))), s(y'))
PLUS(s(s(s(s(x'0'''')))), y') -> PLUS(s(s(s(x'0''''))), s(y'))
PLUS(s(s(s(x''''))), 0) -> PLUS(s(s(x'''')), s(0))
PLUS(s(s(s(x''''))), s(x''''''')) -> PLUS(s(s(x'''')), s(s(x''''''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Rw

...

               →DP Problem 16

                 ↳Forward Instantiation Transformation

Dependency Pairs:

PLUS(s(s(s(x''''))), 0) -> PLUS(s(s(x'''')), s(0))
PLUS(s(s(x'')), 0) -> PLUS(s(x''), s(0))
PLUS(s(s(s(x''''))), s(x''''''')) -> PLUS(s(s(x'''')), s(s(x''''''')))
PLUS(s(s(s(s(x'0'''')))), y') -> PLUS(s(s(s(x'0''''))), s(y'))
PLUS(s(s(s(s(x'''''')))), y') -> PLUS(s(s(s(x''''''))), s(y'))
PLUS(s(s(x'')), s(x''''')) -> PLUS(s(x''), s(s(x''''')))
PLUS(s(s(s(x'0''))), y') -> PLUS(s(s(x'0'')), s(y'))
PLUS(s(s(s(x''''))), y') -> PLUS(s(s(x'''')), s(y'))
PLUS(s(s(x'')), s(0)) -> PLUS(s(x''), s(0))
PLUS(s(s(s(x'0''))), s(x''''')) -> PLUS(s(s(x'0'')), s(x'''''))
PLUS(s(s(s(x''''))), s(y'''')) -> PLUS(s(s(x'''')), s(y''''))
PLUS(s(x), s(0)) -> PLUS(minus(x, s(0)), s(s(0)))
PLUS(s(s(x'')), y'') -> PLUS(s(x''), s(y''))
PLUS(s(s(x'')), 0) -> PLUS(s(x''), 0)
PLUS(s(s(x'''')), 0) -> PLUS(s(x''''), 0)
PLUS(s(x), 0) -> PLUS(x, 0)
PLUS(s(x''), 0) -> PLUS(x'', 0)
PLUS(s(s(x'')), y'') -> PLUS(s(x''), y'')
PLUS(s(x), s(s(x'''))) -> PLUS(minus(x, s(s(x'''))), s(s(s(s(double(x'''))))))
PLUS(s(s(x'0)), s(x''')) -> PLUS(minus(x'0, x'''), s(s(double(x'''))))
PLUS(s(s(x'')), s(y'')) -> PLUS(minus(x'', y''), s(s(double(y''))))
PLUS(s(s(x'')), s(s(x'''''))) -> PLUS(s(x''), s(s(x''''')))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))

Strategy:

innermost

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

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

20 new Dependency Pairs are created:

PLUS(s(s(x'''')), 0) -> PLUS(s(x''''), 0)
PLUS(s(s(s(x''''))), s(y'''')) -> PLUS(s(s(x'''')), s(y''''))
PLUS(s(s(x''')), 0) -> PLUS(s(x'''), 0)
PLUS(s(s(s(x'0''))), s(x''''')) -> PLUS(s(s(x'0'')), s(x'''''))
PLUS(s(s(x''')), s(0)) -> PLUS(s(x'''), s(0))
PLUS(s(s(x''')), s(s(x'''''))) -> PLUS(s(x'''), s(s(x''''')))
PLUS(s(s(s(x''''))), y'''') -> PLUS(s(s(x'''')), y'''')
PLUS(s(s(s(x''''''))), 0) -> PLUS(s(s(x'''''')), 0)
PLUS(s(s(s(s(x'''''')))), s(y'''''')) -> PLUS(s(s(s(x''''''))), s(y''''''))
PLUS(s(s(s(x''''))), 0) -> PLUS(s(s(x'''')), 0)
PLUS(s(s(s(s(x'0'''')))), s(x''''''')) -> PLUS(s(s(s(x'0''''))), s(x'''''''))
PLUS(s(s(s(x''''))), s(0)) -> PLUS(s(s(x'''')), s(0))
PLUS(s(s(s(x''''))), s(s(x'''''''))) -> PLUS(s(s(x'''')), s(s(x''''''')))
PLUS(s(s(s(s(x'''''')))), y'''') -> PLUS(s(s(s(x''''''))), y'''')
PLUS(s(s(s(s(x'0'''')))), y'''') -> PLUS(s(s(s(x'0''''))), y'''')
PLUS(s(s(s(x''''))), s(x''''''')) -> PLUS(s(s(x'''')), s(x'''''''))
PLUS(s(s(s(s(s(x''''''''))))), y'''') -> PLUS(s(s(s(s(x'''''''')))), y'''')
PLUS(s(s(s(s(s(x'0''''''))))), y'''') -> PLUS(s(s(s(s(x'0'''''')))), y'''')
PLUS(s(s(s(s(x'''''')))), 0) -> PLUS(s(s(s(x''''''))), 0)
PLUS(s(s(s(s(x'''''')))), s(x''''''''')) -> PLUS(s(s(s(x''''''))), s(x'''''''''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Rw

...

               →DP Problem 17

                 ↳Forward Instantiation Transformation

Dependency Pairs:

PLUS(s(s(s(s(x'''''')))), s(x''''''''')) -> PLUS(s(s(s(x''''''))), s(x'''''''''))
PLUS(s(s(s(s(x'''''')))), 0) -> PLUS(s(s(s(x''''''))), 0)
PLUS(s(s(s(s(s(x'0''''''))))), y'''') -> PLUS(s(s(s(s(x'0'''''')))), y'''')
PLUS(s(s(s(s(s(x''''''''))))), y'''') -> PLUS(s(s(s(s(x'''''''')))), y'''')
PLUS(s(s(s(x''''))), s(x''''''')) -> PLUS(s(s(x'''')), s(x'''''''))
PLUS(s(s(s(x''''))), s(s(x'''''''))) -> PLUS(s(s(x'''')), s(s(x''''''')))
PLUS(s(s(s(s(x'0'''')))), y'''') -> PLUS(s(s(s(x'0''''))), y'''')
PLUS(s(s(s(x''''))), 0) -> PLUS(s(s(x'''')), 0)
PLUS(s(s(s(x''''''))), 0) -> PLUS(s(s(x'''''')), 0)
PLUS(s(s(x''')), 0) -> PLUS(s(x'''), 0)
PLUS(s(s(x'''')), 0) -> PLUS(s(x''''), 0)
PLUS(s(s(s(s(x'''''')))), y'''') -> PLUS(s(s(s(x''''''))), y'''')
PLUS(s(s(s(x''''))), s(0)) -> PLUS(s(s(x'''')), s(0))
PLUS(s(s(s(s(x'0'''')))), s(x''''''')) -> PLUS(s(s(s(x'0''''))), s(x'''''''))
PLUS(s(s(x''')), s(s(x'''''))) -> PLUS(s(x'''), s(s(x''''')))
PLUS(s(s(s(s(x'''''')))), s(y'''''')) -> PLUS(s(s(s(x''''''))), s(y''''''))
PLUS(s(s(x'')), 0) -> PLUS(s(x''), s(0))
PLUS(s(s(x'')), 0) -> PLUS(s(x''), 0)
PLUS(s(s(x'''')), 0) -> PLUS(s(x''''), 0)
PLUS(s(x), 0) -> PLUS(x, 0)
PLUS(s(x''), 0) -> PLUS(x'', 0)
PLUS(s(s(s(x''''))), y'''') -> PLUS(s(s(x'''')), y'''')
PLUS(s(s(x''')), s(0)) -> PLUS(s(x'''), s(0))
PLUS(s(s(s(x'0''))), s(x''''')) -> PLUS(s(s(x'0'')), s(x'''''))
PLUS(s(s(s(x''''))), s(y'''')) -> PLUS(s(s(x'''')), s(y''''))
PLUS(s(s(s(x''''))), s(x''''''')) -> PLUS(s(s(x'''')), s(s(x''''''')))
PLUS(s(s(s(s(x'0'''')))), y') -> PLUS(s(s(s(x'0''''))), s(y'))
PLUS(s(s(s(s(x'''''')))), y') -> PLUS(s(s(s(x''''''))), s(y'))
PLUS(s(s(x'')), s(x''''')) -> PLUS(s(x''), s(s(x''''')))
PLUS(s(s(s(x'0''))), y') -> PLUS(s(s(x'0'')), s(y'))
PLUS(s(s(s(x''''))), y') -> PLUS(s(s(x'''')), s(y'))
PLUS(s(s(x'')), s(s(x'''''))) -> PLUS(s(x''), s(s(x''''')))
PLUS(s(s(x'')), y'') -> PLUS(s(x''), s(y''))
PLUS(s(s(x'')), s(0)) -> PLUS(s(x''), s(0))
PLUS(s(s(s(x'0''))), s(x''''')) -> PLUS(s(s(x'0'')), s(x'''''))
PLUS(s(x), s(0)) -> PLUS(minus(x, s(0)), s(s(0)))
PLUS(s(s(s(x''''))), s(y'''')) -> PLUS(s(s(x'''')), s(y''''))
PLUS(s(x), s(s(x'''))) -> PLUS(minus(x, s(s(x'''))), s(s(s(s(double(x'''))))))
PLUS(s(s(x'0)), s(x''')) -> PLUS(minus(x'0, x'''), s(s(double(x'''))))
PLUS(s(s(x'')), s(y'')) -> PLUS(minus(x'', y''), s(s(double(y''))))
PLUS(s(s(s(x''''))), 0) -> PLUS(s(s(x'''')), s(0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))

Strategy:

innermost

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

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

21 new Dependency Pairs are created:

PLUS(s(s(s(x''''))), y'''') -> PLUS(s(s(x'''')), s(y''''))
PLUS(s(s(s(x'0''))), y''') -> PLUS(s(s(x'0'')), s(y'''))
PLUS(s(s(x''')), 0) -> PLUS(s(x'''), s(0))
PLUS(s(s(x''')), s(x''''')) -> PLUS(s(x'''), s(s(x''''')))
PLUS(s(s(s(s(x'''''')))), y''') -> PLUS(s(s(s(x''''''))), s(y'''))
PLUS(s(s(s(s(x'0'''')))), y''') -> PLUS(s(s(s(x'0''''))), s(y'''))
PLUS(s(s(s(x''''))), 0) -> PLUS(s(s(x'''')), s(0))
PLUS(s(s(s(x''''))), s(x''''''')) -> PLUS(s(s(x'''')), s(s(x''''''')))
PLUS(s(s(s(s(x'''''')))), y'''') -> PLUS(s(s(s(x''''''))), s(y''''))
PLUS(s(s(s(s(x'0'''')))), y'''') -> PLUS(s(s(s(x'0''''))), s(y''''))
PLUS(s(s(s(x''''))), y''') -> PLUS(s(s(x'''')), s(y'''))
PLUS(s(s(s(s(s(x''''''''))))), y'''') -> PLUS(s(s(s(s(x'''''''')))), s(y''''))
PLUS(s(s(s(s(s(x'0''''''))))), y'''') -> PLUS(s(s(s(s(x'0'''''')))), s(y''''))
PLUS(s(s(s(x'''''))), 0) -> PLUS(s(s(x''''')), s(0))
PLUS(s(s(s(x'''''))), s(x''''''')) -> PLUS(s(s(x''''')), s(s(x''''''')))
PLUS(s(s(s(s(s(x''''''''))))), y''') -> PLUS(s(s(s(s(x'''''''')))), s(y'''))
PLUS(s(s(s(s(s(x'0''''''))))), y''') -> PLUS(s(s(s(s(x'0'''''')))), s(y'''))
PLUS(s(s(s(s(x'''''')))), 0) -> PLUS(s(s(s(x''''''))), s(0))
PLUS(s(s(s(s(x'''''')))), s(x''''''''')) -> PLUS(s(s(s(x''''''))), s(s(x''''''''')))
PLUS(s(s(s(s(s(s(x'''''''''')))))), y''') -> PLUS(s(s(s(s(s(x''''''''''))))), s(y'''))
PLUS(s(s(s(s(s(s(x'0'''''''')))))), y''') -> PLUS(s(s(s(s(s(x'0''''''''))))), s(y'''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Rw

...

               →DP Problem 18

                 ↳Argument Filtering and Ordering

Dependency Pairs:

PLUS(s(s(s(s(x'''''')))), 0) -> PLUS(s(s(s(x''''''))), s(0))
PLUS(s(s(s(x'''''))), 0) -> PLUS(s(s(x''''')), s(0))
PLUS(s(s(s(x''''))), 0) -> PLUS(s(s(x'''')), s(0))
PLUS(s(s(x''')), 0) -> PLUS(s(x'''), s(0))
PLUS(s(s(s(s(s(s(x'0'''''''')))))), y''') -> PLUS(s(s(s(s(s(x'0''''''''))))), s(y'''))
PLUS(s(s(s(s(s(s(x'''''''''')))))), y''') -> PLUS(s(s(s(s(s(x''''''''''))))), s(y'''))
PLUS(s(s(s(s(x'''''')))), s(x''''''''')) -> PLUS(s(s(s(x''''''))), s(s(x''''''''')))
PLUS(s(s(s(s(s(x'0''''''))))), y''') -> PLUS(s(s(s(s(x'0'''''')))), s(y'''))
PLUS(s(s(s(s(s(x''''''''))))), y''') -> PLUS(s(s(s(s(x'''''''')))), s(y'''))
PLUS(s(s(s(x'''''))), s(x''''''')) -> PLUS(s(s(x''''')), s(s(x''''''')))
PLUS(s(s(s(s(s(x'0''''''))))), y'''') -> PLUS(s(s(s(s(x'0'''''')))), s(y''''))
PLUS(s(s(s(s(s(x''''''''))))), y'''') -> PLUS(s(s(s(s(x'''''''')))), s(y''''))
PLUS(s(s(s(x''''))), y''') -> PLUS(s(s(x'''')), s(y'''))
PLUS(s(s(s(s(x'0'''')))), y'''') -> PLUS(s(s(s(x'0''''))), s(y''''))
PLUS(s(s(s(s(x'''''')))), y'''') -> PLUS(s(s(s(x''''''))), s(y''''))
PLUS(s(s(s(x''''))), s(x''''''')) -> PLUS(s(s(x'''')), s(s(x''''''')))
PLUS(s(s(s(s(x'0'''')))), y''') -> PLUS(s(s(s(x'0''''))), s(y'''))
PLUS(s(s(s(s(x'''''')))), y''') -> PLUS(s(s(s(x''''''))), s(y'''))
PLUS(s(s(x''')), s(x''''')) -> PLUS(s(x'''), s(s(x''''')))
PLUS(s(s(s(x'0''))), y''') -> PLUS(s(s(x'0'')), s(y'''))
PLUS(s(s(s(x''''))), y'''') -> PLUS(s(s(x'''')), s(y''''))
PLUS(s(s(s(s(x'''''')))), 0) -> PLUS(s(s(s(x''''''))), 0)
PLUS(s(s(s(s(s(x'0''''''))))), y'''') -> PLUS(s(s(s(s(x'0'''''')))), y'''')
PLUS(s(s(s(x''''))), s(x''''''')) -> PLUS(s(s(x'''')), s(x'''''''))
PLUS(s(s(s(x''''))), s(s(x'''''''))) -> PLUS(s(s(x'''')), s(s(x''''''')))
PLUS(s(s(s(s(s(x''''''''))))), y'''') -> PLUS(s(s(s(s(x'''''''')))), y'''')
PLUS(s(s(s(x''''))), 0) -> PLUS(s(s(x'''')), 0)
PLUS(s(s(s(x''''''))), 0) -> PLUS(s(s(x'''''')), 0)
PLUS(s(s(x''')), 0) -> PLUS(s(x'''), 0)
PLUS(s(s(x'''')), 0) -> PLUS(s(x''''), 0)
PLUS(s(s(s(s(x'0'''')))), y'''') -> PLUS(s(s(s(x'0''''))), y'''')
PLUS(s(s(s(x''''))), 0) -> PLUS(s(s(x'''')), s(0))
PLUS(s(s(s(s(x'''''')))), y'''') -> PLUS(s(s(s(x''''''))), y'''')
PLUS(s(s(s(x''''))), s(0)) -> PLUS(s(s(x'''')), s(0))
PLUS(s(s(s(s(x'0'''')))), s(x''''''')) -> PLUS(s(s(s(x'0''''))), s(x'''''''))
PLUS(s(s(x''')), s(s(x'''''))) -> PLUS(s(x'''), s(s(x''''')))
PLUS(s(s(s(s(x'''''')))), s(y'''''')) -> PLUS(s(s(s(x''''''))), s(y''''''))
PLUS(s(s(x'')), 0) -> PLUS(s(x''), s(0))
PLUS(s(s(x'')), 0) -> PLUS(s(x''), 0)
PLUS(s(s(x'''')), 0) -> PLUS(s(x''''), 0)
PLUS(s(x), 0) -> PLUS(x, 0)
PLUS(s(x''), 0) -> PLUS(x'', 0)
PLUS(s(s(s(x''''))), y'''') -> PLUS(s(s(x'''')), y'''')
PLUS(s(s(x''')), s(0)) -> PLUS(s(x'''), s(0))
PLUS(s(s(s(x'0''))), s(x''''')) -> PLUS(s(s(x'0'')), s(x'''''))
PLUS(s(s(s(x''''))), s(y'''')) -> PLUS(s(s(x'''')), s(y''''))
PLUS(s(s(s(x''''))), s(x''''''')) -> PLUS(s(s(x'''')), s(s(x''''''')))
PLUS(s(s(s(s(x'0'''')))), y') -> PLUS(s(s(s(x'0''''))), s(y'))
PLUS(s(s(s(s(x'''''')))), y') -> PLUS(s(s(s(x''''''))), s(y'))
PLUS(s(s(x'')), s(x''''')) -> PLUS(s(x''), s(s(x''''')))
PLUS(s(s(s(x'0''))), y') -> PLUS(s(s(x'0'')), s(y'))
PLUS(s(s(x'')), s(s(x'''''))) -> PLUS(s(x''), s(s(x''''')))
PLUS(s(s(s(x''''))), y') -> PLUS(s(s(x'''')), s(y'))
PLUS(s(s(x'')), s(0)) -> PLUS(s(x''), s(0))
PLUS(s(s(s(x'0''))), s(x''''')) -> PLUS(s(s(x'0'')), s(x'''''))
PLUS(s(x), s(0)) -> PLUS(minus(x, s(0)), s(s(0)))
PLUS(s(s(s(x''''))), s(y'''')) -> PLUS(s(s(x'''')), s(y''''))
PLUS(s(x), s(s(x'''))) -> PLUS(minus(x, s(s(x'''))), s(s(s(s(double(x'''))))))
PLUS(s(s(x'0)), s(x''')) -> PLUS(minus(x'0, x'''), s(s(double(x'''))))
PLUS(s(s(x'')), s(y'')) -> PLUS(minus(x'', y''), s(s(double(y''))))
PLUS(s(s(s(s(x'''''')))), s(x''''''''')) -> PLUS(s(s(s(x''''''))), s(x'''''''''))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

PLUS(s(s(s(s(x'''''')))), 0) -> PLUS(s(s(s(x''''''))), s(0))
PLUS(s(s(s(x'''''))), 0) -> PLUS(s(s(x''''')), s(0))
PLUS(s(s(s(x''''))), 0) -> PLUS(s(s(x'''')), s(0))
PLUS(s(s(x''')), 0) -> PLUS(s(x'''), s(0))
PLUS(s(s(s(s(s(s(x'0'''''''')))))), y''') -> PLUS(s(s(s(s(s(x'0''''''''))))), s(y'''))
PLUS(s(s(s(s(s(s(x'''''''''')))))), y''') -> PLUS(s(s(s(s(s(x''''''''''))))), s(y'''))
PLUS(s(s(s(s(x'''''')))), s(x''''''''')) -> PLUS(s(s(s(x''''''))), s(s(x''''''''')))
PLUS(s(s(s(s(s(x'0''''''))))), y''') -> PLUS(s(s(s(s(x'0'''''')))), s(y'''))
PLUS(s(s(s(s(s(x''''''''))))), y''') -> PLUS(s(s(s(s(x'''''''')))), s(y'''))
PLUS(s(s(s(x'''''))), s(x''''''')) -> PLUS(s(s(x''''')), s(s(x''''''')))
PLUS(s(s(s(s(s(x'0''''''))))), y'''') -> PLUS(s(s(s(s(x'0'''''')))), s(y''''))
PLUS(s(s(s(s(s(x''''''''))))), y'''') -> PLUS(s(s(s(s(x'''''''')))), s(y''''))
PLUS(s(s(s(x''''))), y''') -> PLUS(s(s(x'''')), s(y'''))
PLUS(s(s(s(s(x'0'''')))), y'''') -> PLUS(s(s(s(x'0''''))), s(y''''))
PLUS(s(s(s(s(x'''''')))), y'''') -> PLUS(s(s(s(x''''''))), s(y''''))
PLUS(s(s(s(x''''))), s(x''''''')) -> PLUS(s(s(x'''')), s(s(x''''''')))
PLUS(s(s(s(s(x'0'''')))), y''') -> PLUS(s(s(s(x'0''''))), s(y'''))
PLUS(s(s(s(s(x'''''')))), y''') -> PLUS(s(s(s(x''''''))), s(y'''))
PLUS(s(s(x''')), s(x''''')) -> PLUS(s(x'''), s(s(x''''')))
PLUS(s(s(s(x'0''))), y''') -> PLUS(s(s(x'0'')), s(y'''))
PLUS(s(s(s(x''''))), y'''') -> PLUS(s(s(x'''')), s(y''''))
PLUS(s(s(s(s(x'''''')))), 0) -> PLUS(s(s(s(x''''''))), 0)
PLUS(s(s(s(s(s(x'0''''''))))), y'''') -> PLUS(s(s(s(s(x'0'''''')))), y'''')
PLUS(s(s(s(x''''))), s(x''''''')) -> PLUS(s(s(x'''')), s(x'''''''))
PLUS(s(s(s(x''''))), s(s(x'''''''))) -> PLUS(s(s(x'''')), s(s(x''''''')))
PLUS(s(s(s(s(s(x''''''''))))), y'''') -> PLUS(s(s(s(s(x'''''''')))), y'''')
PLUS(s(s(s(x''''))), 0) -> PLUS(s(s(x'''')), 0)
PLUS(s(s(s(x''''''))), 0) -> PLUS(s(s(x'''''')), 0)
PLUS(s(s(x''')), 0) -> PLUS(s(x'''), 0)
PLUS(s(s(x'''')), 0) -> PLUS(s(x''''), 0)
PLUS(s(s(s(s(x'0'''')))), y'''') -> PLUS(s(s(s(x'0''''))), y'''')
PLUS(s(s(s(s(x'''''')))), y'''') -> PLUS(s(s(s(x''''''))), y'''')
PLUS(s(s(s(x''''))), s(0)) -> PLUS(s(s(x'''')), s(0))
PLUS(s(s(s(s(x'0'''')))), s(x''''''')) -> PLUS(s(s(s(x'0''''))), s(x'''''''))
PLUS(s(s(x''')), s(s(x'''''))) -> PLUS(s(x'''), s(s(x''''')))
PLUS(s(s(s(s(x'''''')))), s(y'''''')) -> PLUS(s(s(s(x''''''))), s(y''''''))
PLUS(s(s(x'')), 0) -> PLUS(s(x''), s(0))
PLUS(s(s(x'')), 0) -> PLUS(s(x''), 0)
PLUS(s(x), 0) -> PLUS(x, 0)
PLUS(s(x''), 0) -> PLUS(x'', 0)
PLUS(s(s(s(x''''))), y'''') -> PLUS(s(s(x'''')), y'''')
PLUS(s(s(x''')), s(0)) -> PLUS(s(x'''), s(0))
PLUS(s(s(s(x'0''))), s(x''''')) -> PLUS(s(s(x'0'')), s(x'''''))
PLUS(s(s(s(x''''))), s(y'''')) -> PLUS(s(s(x'''')), s(y''''))
PLUS(s(s(s(s(x'0'''')))), y') -> PLUS(s(s(s(x'0''''))), s(y'))
PLUS(s(s(s(s(x'''''')))), y') -> PLUS(s(s(s(x''''''))), s(y'))
PLUS(s(s(x'')), s(x''''')) -> PLUS(s(x''), s(s(x''''')))
PLUS(s(s(s(x'0''))), y') -> PLUS(s(s(x'0'')), s(y'))
PLUS(s(s(x'')), s(s(x'''''))) -> PLUS(s(x''), s(s(x''''')))
PLUS(s(s(s(x''''))), y') -> PLUS(s(s(x'''')), s(y'))
PLUS(s(s(x'')), s(0)) -> PLUS(s(x''), s(0))
PLUS(s(x), s(0)) -> PLUS(minus(x, s(0)), s(s(0)))
PLUS(s(x), s(s(x'''))) -> PLUS(minus(x, s(s(x'''))), s(s(s(s(double(x'''))))))
PLUS(s(s(x'0)), s(x''')) -> PLUS(minus(x'0, x'''), s(s(double(x'''))))
PLUS(s(s(x'')), s(y'')) -> PLUS(minus(x'', y''), s(s(double(y''))))
PLUS(s(s(s(s(x'''''')))), s(x''''''''')) -> PLUS(s(s(s(x''''''))), s(x'''''''''))

The following usable rules for innermost can be oriented:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

PLUS > double > s

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

PLUS(x₁, x₂) -> PLUS(x₁, x₂)
s(x₁) -> s(x₁)
minus(x₁, x₂) -> x₁
double(x₁) -> double(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Rw

...

               →DP Problem 19

                 ↳Dependency Graph

Dependency Pair:

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
double(0) -> 0
double(s(x)) -> s(s(double(x)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> plus(x, s(y))
plus(s(x), y) -> s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) -> s(plus(plus(x, y), z))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:48 minutes