Termination Proof using AProVE

Term Rewriting System R:
[N, X, Y, X1, X2, XS]
fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, n_{_fib1}(Y, add(X, Y)))
fib1(X1, X2) -> n_{_fib1}(X1, X2)
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(n_{_fib1}(X1, X2)) -> fib1(X1, X2)
activate(X) -> X

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

FIB(N) -> SEL(N, fib1(s(0), s(0)))
FIB(N) -> FIB1(s(0), s(0))
FIB1(X, Y) -> ADD(X, Y)
ADD(s(X), Y) -> ADD(X, Y)
SEL(s(N), cons(X, XS)) -> SEL(N, activate(XS))
SEL(s(N), cons(X, XS)) -> ACTIVATE(XS)
ACTIVATE(n_{_fib1}(X1, X2)) -> FIB1(X1, X2)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Nar

Dependency Pair:

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

Rules:

fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, n_{_fib1}(Y, add(X, Y)))
fib1(X1, X2) -> n_{_fib1}(X1, X2)
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(n_{_fib1}(X1, X2)) -> fib1(X1, X2)
activate(X) -> X

The following dependency pair can be strictly oriented:

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

There are no usable rules 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 3

             ↳Dependency Graph

       →DP Problem 2

         ↳Nar

Dependency Pair:

Rules:

fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, n_{_fib1}(Y, add(X, Y)))
fib1(X1, X2) -> n_{_fib1}(X1, X2)
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(n_{_fib1}(X1, X2)) -> fib1(X1, X2)
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Narrowing Transformation

Dependency Pair:

SEL(s(N), cons(X, XS)) -> SEL(N, activate(XS))

Rules:

fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, n_{_fib1}(Y, add(X, Y)))
fib1(X1, X2) -> n_{_fib1}(X1, X2)
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(n_{_fib1}(X1, X2)) -> fib1(X1, X2)
activate(X) -> X

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

SEL(s(N), cons(X, XS)) -> SEL(N, activate(XS))

two new Dependency Pairs are created:

SEL(s(N), cons(X, n_{_fib1}(X1', X2'))) -> SEL(N, fib1(X1', X2'))
SEL(s(N), cons(X, XS')) -> SEL(N, XS')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Narrowing Transformation

Dependency Pairs:

SEL(s(N), cons(X, XS')) -> SEL(N, XS')
SEL(s(N), cons(X, n_{_fib1}(X1', X2'))) -> SEL(N, fib1(X1', X2'))

Rules:

fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, n_{_fib1}(Y, add(X, Y)))
fib1(X1, X2) -> n_{_fib1}(X1, X2)
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(n_{_fib1}(X1, X2)) -> fib1(X1, X2)
activate(X) -> X

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

SEL(s(N), cons(X, n_{_fib1}(X1', X2'))) -> SEL(N, fib1(X1', X2'))

two new Dependency Pairs are created:

SEL(s(N), cons(X, n_{_fib1}(X1'', X2''))) -> SEL(N, cons(X1'', n_{_fib1}(X2'', add(X1'', X2''))))
SEL(s(N), cons(X, n_{_fib1}(X1'', X2''))) -> SEL(N, n_{_fib1}(X1'', X2''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 5

                 ↳Polynomial Ordering

Dependency Pairs:

SEL(s(N), cons(X, n_{_fib1}(X1'', X2''))) -> SEL(N, cons(X1'', n_{_fib1}(X2'', add(X1'', X2''))))
SEL(s(N), cons(X, XS')) -> SEL(N, XS')

Rules:

fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, n_{_fib1}(Y, add(X, Y)))
fib1(X1, X2) -> n_{_fib1}(X1, X2)
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(n_{_fib1}(X1, X2)) -> fib1(X1, X2)
activate(X) -> X

The following dependency pairs can be strictly oriented:

SEL(s(N), cons(X, n_{_fib1}(X1'', X2''))) -> SEL(N, cons(X1'', n_{_fib1}(X2'', add(X1'', X2''))))
SEL(s(N), cons(X, XS')) -> SEL(N, XS')

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(SEL(x₁, x₂)) = x₁

POL(cons(x₁, x₂)) = 0

POL(n__fib1(x₁, x₂)) = 0

POL(s(x₁)) = 1 + x₁

POL(add(x₁, x₂)) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 6

                 ↳Dependency Graph

Dependency Pair:

Rules:

fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, n_{_fib1}(Y, add(X, Y)))
fib1(X1, X2) -> n_{_fib1}(X1, X2)
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(n_{_fib1}(X1, X2)) -> fib1(X1, X2)
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

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

POL(0)	= 0
POL(SEL(x₁, x₂))	= x₁
POL(cons(x₁, x₂))	= 0
POL(n__fib1(x₁, x₂))	= 0
POL(s(x₁))	= 1 + x₁
POL(add(x₁, x₂))	= 0