Termination Proof using AProVE

Term Rewriting System R:
[N, X, Y, Z]
terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
sqr(s(X)) -> s(add(sqr(X), dbl(X)))
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

TERMS(N) -> SQR(N)
TERMS(N) -> TERMS(s(N))
SQR(s(X)) -> ADD(sqr(X), dbl(X))
SQR(s(X)) -> SQR(X)
SQR(s(X)) -> DBL(X)
DBL(s(X)) -> DBL(X)
ADD(s(X), Y) -> ADD(X, Y)
FIRST(s(X), cons(Y, Z)) -> FIRST(X, Z)
HALF(s(s(X))) -> HALF(X)

Furthermore, R contains six SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Remaining

Dependency Pair:

ADD(s(X), Y) -> ADD(X, Y)

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
sqr(s(X)) -> s(add(sqr(X), dbl(X)))
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X

The following dependency pair can be strictly oriented:

ADD(s(X), Y) -> ADD(X, Y)

There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(s(x₁)) = 1 + x₁

POL(ADD(x₁, x₂)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 7

             ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Remaining

Dependency Pair:

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
sqr(s(X)) -> s(add(sqr(X), dbl(X)))
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polynomial Ordering

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Remaining

Dependency Pair:

DBL(s(X)) -> DBL(X)

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
sqr(s(X)) -> s(add(sqr(X), dbl(X)))
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X

The following dependency pair can be strictly oriented:

DBL(s(X)) -> DBL(X)

There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(s(x₁)) = 1 + x₁

POL(DBL(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 8

             ↳Dependency Graph

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Remaining

Dependency Pair:

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
sqr(s(X)) -> s(add(sqr(X), dbl(X)))
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polynomial Ordering

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Remaining

Dependency Pair:

FIRST(s(X), cons(Y, Z)) -> FIRST(X, Z)

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
sqr(s(X)) -> s(add(sqr(X), dbl(X)))
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X

The following dependency pair can be strictly oriented:

FIRST(s(X), cons(Y, Z)) -> FIRST(X, Z)

There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(cons(x₁, x₂)) = 0

POL(FIRST(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 9

             ↳Dependency Graph

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Remaining

Dependency Pair:

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
sqr(s(X)) -> s(add(sqr(X), dbl(X)))
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polynomial Ordering

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Remaining

Dependency Pair:

HALF(s(s(X))) -> HALF(X)

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
sqr(s(X)) -> s(add(sqr(X), dbl(X)))
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X

The following dependency pair can be strictly oriented:

HALF(s(s(X))) -> HALF(X)

There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(HALF(x₁)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

           →DP Problem 10

             ↳Dependency Graph

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Remaining

Dependency Pair:

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
sqr(s(X)) -> s(add(sqr(X), dbl(X)))
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polynomial Ordering

       →DP Problem 6

         ↳Remaining

Dependency Pair:

SQR(s(X)) -> SQR(X)

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
sqr(s(X)) -> s(add(sqr(X), dbl(X)))
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X

The following dependency pair can be strictly oriented:

SQR(s(X)) -> SQR(X)

There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(s(x₁)) = 1 + x₁

POL(SQR(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

           →DP Problem 11

             ↳Dependency Graph

       →DP Problem 6

         ↳Remaining

Dependency Pair:

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
sqr(s(X)) -> s(add(sqr(X), dbl(X)))
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pair:

TERMS(N) -> TERMS(s(N))

Rules:

terms(N) -> cons(recip(sqr(N)), terms(s(N)))
sqr(0) -> 0
sqr(s(X)) -> s(add(sqr(X), dbl(X)))
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X

Termination of R could not be shown.
Duration:
0:00 minutes

POL(cons(x₁, x₂))	= 0
POL(FIRST(x₁, x₂))	= x₁
POL(s(x₁))	= 1 + x₁