Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

DOUBLE(s(x)) -> DOUBLE(x)
HALF(s(s(x))) -> HALF(x)
-'(s(x), s(y)) -> -'(x, y)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

Dependency Pair:

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

Rules:

double(0) -> 0
double(s(x)) -> s(s(double(x)))
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
half(double(x)) -> x
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
if(0, y, z) -> y
if(s(x), y, z) -> z

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

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

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

Dependency Pair:

Rules:

double(0) -> 0
double(s(x)) -> s(s(double(x)))
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
half(double(x)) -> x
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
if(0, y, z) -> y
if(s(x), y, z) -> z

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

Dependency Pair:

HALF(s(s(x))) -> HALF(x)

Rules:

double(0) -> 0
double(s(x)) -> s(s(double(x)))
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
half(double(x)) -> x
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
if(0, y, z) -> y
if(s(x), y, z) -> z

Strategy:

innermost

The following dependency pair can be strictly oriented:

HALF(s(s(x))) -> HALF(x)

HALF(x₁) -> HALF(x₁)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 3

         ↳AFS

Dependency Pair:

Rules:

double(0) -> 0
double(s(x)) -> s(s(double(x)))
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
half(double(x)) -> x
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
if(0, y, z) -> y
if(s(x), y, z) -> z

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

Dependency Pair:

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

Rules:

double(0) -> 0
double(s(x)) -> s(s(double(x)))
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
half(double(x)) -> x
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
if(0, y, z) -> y
if(s(x), y, z) -> z

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 6

             ↳Dependency Graph

Dependency Pair:

Rules:

double(0) -> 0
double(s(x)) -> s(s(double(x)))
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
half(double(x)) -> x
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
if(0, y, z) -> y
if(s(x), y, z) -> z

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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