Termination Proof using AProVE

Term Rewriting System R:
[x, y]
sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

SQR(s(x)) -> +'(sqr(x), s(double(x)))
SQR(s(x)) -> SQR(x)
SQR(s(x)) -> DOUBLE(x)
SQR(s(x)) -> +'(sqr(x), double(x))
DOUBLE(s(x)) -> DOUBLE(x)
+'(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:

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

Rules:

sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

The following dependency pair can be strictly oriented:

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

The following rules can be oriented:

sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

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

{double, sqr} > + > s

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

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

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

Dependency Pair:

Rules:

sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

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:

sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

The following dependency pair can be strictly oriented:

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

The following rules can be oriented:

sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

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

{double, sqr} > + > s

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

DOUBLE(x₁) -> DOUBLE(x₁)
s(x₁) -> s(x₁)
sqr(x₁) -> sqr(x₁)
+(x₁, x₂) -> +(x₁, x₂)
double(x₁) -> double(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 3

         ↳AFS

Dependency Pair:

Rules:

sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

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:

SQR(s(x)) -> SQR(x)

Rules:

sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

The following dependency pair can be strictly oriented:

SQR(s(x)) -> SQR(x)

The following rules can be oriented:

sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

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

{double, sqr} > + > s

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

SQR(x₁) -> SQR(x₁)
s(x₁) -> s(x₁)
sqr(x₁) -> sqr(x₁)
+(x₁, x₂) -> +(x₁, x₂)
double(x₁) -> double(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 6

             ↳Dependency Graph

Dependency Pair:

Rules:

sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Using the Dependency Graph resulted in no new DP problems.

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