Termination Proof using AProVE

Term Rewriting System R:
[N, X, Y, Z, X1, X2]
terms(N) -> cons(recip(sqr(N)), n_{_terms}(s(N)))
terms(X) -> n_{_terms}(X)
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, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_terms}(X)) -> terms(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

TERMS(N) -> SQR(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)) -> ACTIVATE(Z)
ACTIVATE(n_{_terms}(X)) -> TERMS(X)
ACTIVATE(n_{_first}(X1, X2)) -> FIRST(X1, X2)

Furthermore, R contains four SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), n_{_terms}(s(N)))
terms(X) -> n_{_terms}(X)
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, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_terms}(X)) -> terms(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the implicit AFS 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 5

             ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

Dependency Pair:

Rules:

terms(N) -> cons(recip(sqr(N)), n_{_terms}(s(N)))
terms(X) -> n_{_terms}(X)
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, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_terms}(X)) -> terms(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X

Strategy:

innermost

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

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), n_{_terms}(s(N)))
terms(X) -> n_{_terms}(X)
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, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_terms}(X)) -> terms(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the implicit AFS 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 6

             ↳Dependency Graph

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

Dependency Pair:

Rules:

terms(N) -> cons(recip(sqr(N)), n_{_terms}(s(N)))
terms(X) -> n_{_terms}(X)
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, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_terms}(X)) -> terms(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X

Strategy:

innermost

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

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), n_{_terms}(s(N)))
terms(X) -> n_{_terms}(X)
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, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_terms}(X)) -> terms(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the implicit AFS 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 7

             ↳Dependency Graph

       →DP Problem 4

         ↳Polo

Dependency Pair:

Rules:

terms(N) -> cons(recip(sqr(N)), n_{_terms}(s(N)))
terms(X) -> n_{_terms}(X)
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, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_terms}(X)) -> terms(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X

Strategy:

innermost

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

Dependency Pairs:

ACTIVATE(n_{_first}(X1, X2)) -> FIRST(X1, X2)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)

Rules:

terms(N) -> cons(recip(sqr(N)), n_{_terms}(s(N)))
terms(X) -> n_{_terms}(X)
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, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_terms}(X)) -> terms(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X

Strategy:

innermost

The following dependency pair can be strictly oriented:

ACTIVATE(n_{_first}(X1, X2)) -> FIRST(X1, X2)

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(cons(x₁, x₂)) = x₁ + x₂

POL(FIRST(x₁, x₂)) = x₂

POL(s(x₁)) = 0

POL(ACTIVATE(x₁)) = x₁

POL(n__first(x₁, x₂)) = 1 + x₂

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 8

             ↳Dependency Graph

Dependency Pair:

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

Rules:

terms(N) -> cons(recip(sqr(N)), n_{_terms}(s(N)))
terms(X) -> n_{_terms}(X)
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, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
activate(n_{_terms}(X)) -> terms(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(X) -> X

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

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