Termination Proof using AProVE

Term Rewriting System R:
[N, X, Y, XS]
fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, fib1(Y, add(X, Y)))
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, XS)

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) -> FIB1(Y, add(X, Y))
FIB1(X, Y) -> ADD(X, Y)
ADD(s(X), Y) -> ADD(X, Y)
SEL(s(N), cons(X, XS)) -> SEL(N, XS)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
SEL(s(N), cons(X, XS)) -> SEL(N, XS)

Rules:

fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, fib1(Y, add(X, Y)))
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, XS)
Dependency Pair:
ADD(s(X), Y) -> ADD(X, Y)

Rules:

fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, fib1(Y, add(X, Y)))
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, XS)
Dependency Pair:
FIB1(X, Y) -> FIB1(Y, add(X, Y))

Rules:

fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, fib1(Y, add(X, Y)))
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, XS)

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
SEL(s(N), cons(X, XS)) -> SEL(N, XS)

Rules:

fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, fib1(Y, add(X, Y)))
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, XS)
Dependency Pair:
ADD(s(X), Y) -> ADD(X, Y)

Rules:

fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, fib1(Y, add(X, Y)))
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, XS)
Dependency Pair:
FIB1(X, Y) -> FIB1(Y, add(X, Y))

Rules:

fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, fib1(Y, add(X, Y)))
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, XS)

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
SEL(s(N), cons(X, XS)) -> SEL(N, XS)

Rules:

fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, fib1(Y, add(X, Y)))
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, XS)
Dependency Pair:
ADD(s(X), Y) -> ADD(X, Y)

Rules:

fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, fib1(Y, add(X, Y)))
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, XS)
Dependency Pair:
FIB1(X, Y) -> FIB1(Y, add(X, Y))

Rules:

fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, fib1(Y, add(X, Y)))
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, XS)

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