Termination Proof using AProVE

Term Rewriting System R:
[x, y]
fib(0) -> 0
fib(s(0)) -> s(0)
fib(s(s(0))) -> s(0)
fib(s(s(x))) -> sp(g(x))
g(0) -> pair(s(0), 0)
g(s(0)) -> pair(s(0), s(0))
g(s(x)) -> np(g(x))
sp(pair(x, y)) -> +(x, y)
np(pair(x, y)) -> pair(+(x, y), x)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

FIB(s(s(x))) -> SP(g(x))
FIB(s(s(x))) -> G(x)
G(s(x)) -> NP(g(x))
G(s(x)) -> G(x)
SP(pair(x, y)) -> +'(x, y)
NP(pair(x, y)) -> +'(x, y)
+'(x, s(y)) -> +'(x, y)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

+'(x, s(y)) -> +'(x, y)

Rules:

fib(0) -> 0
fib(s(0)) -> s(0)
fib(s(s(0))) -> s(0)
fib(s(s(x))) -> sp(g(x))
g(0) -> pair(s(0), 0)
g(s(0)) -> pair(s(0), s(0))
g(s(x)) -> np(g(x))
sp(pair(x, y)) -> +(x, y)
np(pair(x, y)) -> pair(+(x, y), x)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

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

+'(x, s(y)) -> +'(x, y)

one new Dependency Pair is created:

+'(x'', s(s(y''))) -> +'(x'', s(y''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

+'(x'', s(s(y''))) -> +'(x'', s(y''))

Rules:

fib(0) -> 0
fib(s(0)) -> s(0)
fib(s(s(0))) -> s(0)
fib(s(s(x))) -> sp(g(x))
g(0) -> pair(s(0), 0)
g(s(0)) -> pair(s(0), s(0))
g(s(x)) -> np(g(x))
sp(pair(x, y)) -> +(x, y)
np(pair(x, y)) -> pair(+(x, y), x)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

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

+'(x'', s(s(y''))) -> +'(x'', s(y''))

one new Dependency Pair is created:

+'(x'''', s(s(s(y'''')))) -> +'(x'''', s(s(y'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 4

                 ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

+'(x'''', s(s(s(y'''')))) -> +'(x'''', s(s(y'''')))

Rules:

fib(0) -> 0
fib(s(0)) -> s(0)
fib(s(s(0))) -> s(0)
fib(s(s(x))) -> sp(g(x))
g(0) -> pair(s(0), 0)
g(s(0)) -> pair(s(0), s(0))
g(s(x)) -> np(g(x))
sp(pair(x, y)) -> +(x, y)
np(pair(x, y)) -> pair(+(x, y), x)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

+'(x'''', s(s(s(y'''')))) -> +'(x'''', s(s(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:

+'(x₁, x₂) -> +'(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 5

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

Rules:

fib(0) -> 0
fib(s(0)) -> s(0)
fib(s(s(0))) -> s(0)
fib(s(s(x))) -> sp(g(x))
g(0) -> pair(s(0), 0)
g(s(0)) -> pair(s(0), s(0))
g(s(x)) -> np(g(x))
sp(pair(x, y)) -> +(x, y)
np(pair(x, y)) -> pair(+(x, y), x)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

Dependency Pair:

G(s(x)) -> G(x)

Rules:

fib(0) -> 0
fib(s(0)) -> s(0)
fib(s(s(0))) -> s(0)
fib(s(s(x))) -> sp(g(x))
g(0) -> pair(s(0), 0)
g(s(0)) -> pair(s(0), s(0))
g(s(x)) -> np(g(x))
sp(pair(x, y)) -> +(x, y)
np(pair(x, y)) -> pair(+(x, y), x)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

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

G(s(x)) -> G(x)

one new Dependency Pair is created:

G(s(s(x''))) -> G(s(x''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 6

             ↳Forward Instantiation Transformation

Dependency Pair:

G(s(s(x''))) -> G(s(x''))

Rules:

fib(0) -> 0
fib(s(0)) -> s(0)
fib(s(s(0))) -> s(0)
fib(s(s(x))) -> sp(g(x))
g(0) -> pair(s(0), 0)
g(s(0)) -> pair(s(0), s(0))
g(s(x)) -> np(g(x))
sp(pair(x, y)) -> +(x, y)
np(pair(x, y)) -> pair(+(x, y), x)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

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

G(s(s(x''))) -> G(s(x''))

one new Dependency Pair is created:

G(s(s(s(x'''')))) -> G(s(s(x'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 7

                 ↳Argument Filtering and Ordering

Dependency Pair:

G(s(s(s(x'''')))) -> G(s(s(x'''')))

Rules:

fib(0) -> 0
fib(s(0)) -> s(0)
fib(s(s(0))) -> s(0)
fib(s(s(x))) -> sp(g(x))
g(0) -> pair(s(0), 0)
g(s(0)) -> pair(s(0), s(0))
g(s(x)) -> np(g(x))
sp(pair(x, y)) -> +(x, y)
np(pair(x, y)) -> pair(+(x, y), x)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

G(s(s(s(x'''')))) -> G(s(s(x'''')))

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:

G(x₁) -> G(x₁)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 8

                 ↳Dependency Graph

Dependency Pair:

Rules:

fib(0) -> 0
fib(s(0)) -> s(0)
fib(s(s(0))) -> s(0)
fib(s(s(x))) -> sp(g(x))
g(0) -> pair(s(0), 0)
g(s(0)) -> pair(s(0), s(0))
g(s(x)) -> np(g(x))
sp(pair(x, y)) -> +(x, y)
np(pair(x, y)) -> pair(+(x, y), x)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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