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)

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) -> 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

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Remaining

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)

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Remaining

Dependency Pair:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 5

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Remaining

Dependency Pair:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(SEL(x₁, x₂)) = x₁

POL(cons(x₁, x₂)) = 0

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 6

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Remaining

Dependency Pair:

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)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Remaining

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)

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Remaining

Dependency Pair:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 8

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳Remaining

Dependency Pair:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(s(x₁)) = 1 + x₁

POL(ADD(x₁, x₂)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 9

                 ↳Dependency Graph

       →DP Problem 3

         ↳Remaining

Dependency Pair:

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)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:
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)

Strategy:

innermost

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

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

POL(s(x₁))	= 1 + x₁
POL(ADD(x₁, x₂))	= 1 + x₁