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

Innermost 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

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

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

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

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 3

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Argument Filtering and Ordering

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

The following usable rules for innermost can be oriented:

activate(n_{_fib1}(X1, X2)) -> fib1(X1, X2)
activate(X) -> X
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))

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

SEL > activate > fib1 > add > s
SEL > activate > fib1 > cons
SEL > activate > fib1 > n_{_fib1}

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

SEL(x₁, x₂) -> SEL(x₁, x₂)
s(x₁) -> s(x₁)
cons(x₁, x₂) -> cons(x₁, x₂)
activate(x₁) -> activate(x₁)
n_{_fib1}(x₁, x₂) -> n_{_fib1}(x₁, x₂)
fib1(x₁, x₂) -> fib1(x₁, x₂)
add(x₁, x₂) -> add(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 4

             ↳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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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