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))

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

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

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))

The following dependency pair can be strictly oriented:

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

The following rules 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)))
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))

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:

MINUS(x₁, x₂) -> MINUS(x₁, x₂)
s(x₁) -> s(x₁)
minus(x₁, x₂) -> x₁
double(x₁) -> double(x₁)
plus(x₁, x₂) -> plus(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

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))

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

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))

The following dependency pair can be strictly oriented:

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

The following rules 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)))
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))

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:

DOUBLE(x₁) -> DOUBLE(x₁)
s(x₁) -> s(x₁)
minus(x₁, x₂) -> x₁
double(x₁) -> double(x₁)
plus(x₁, x₂) -> plus(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 3

         ↳AFS

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))

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

Dependency Pairs:

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

The following dependency pairs can be strictly oriented:

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

The following rules 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)))
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))

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

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

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 6

             ↳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))

Using the Dependency Graph resulted in no new DP problems.

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