Termination Proof using AProVE

Term Rewriting System R:
[x]
half(0) -> 0
half(s(s(x))) -> s(half(x))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(half(x))))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

HALF(s(s(x))) -> HALF(x)
LOG(s(s(x))) -> LOG(s(half(x)))
LOG(s(s(x))) -> HALF(x)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Nar

Dependency Pair:

HALF(s(s(x))) -> HALF(x)

Rules:

half(0) -> 0
half(s(s(x))) -> s(half(x))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(half(x))))

Strategy:

innermost

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

HALF(s(s(x))) -> HALF(x)

one new Dependency Pair is created:

HALF(s(s(s(s(x''))))) -> HALF(s(s(x'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Nar

Dependency Pair:

HALF(s(s(s(s(x''))))) -> HALF(s(s(x'')))

Rules:

half(0) -> 0
half(s(s(x))) -> s(half(x))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(half(x))))

Strategy:

innermost

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

HALF(s(s(s(s(x''))))) -> HALF(s(s(x'')))

one new Dependency Pair is created:

HALF(s(s(s(s(s(s(x''''))))))) -> HALF(s(s(s(s(x'''')))))

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

         ↳Nar

Dependency Pair:

HALF(s(s(s(s(s(s(x''''))))))) -> HALF(s(s(s(s(x'''')))))

Rules:

half(0) -> 0
half(s(s(x))) -> s(half(x))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(half(x))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

HALF(s(s(s(s(s(s(x''''))))))) -> HALF(s(s(s(s(x'''')))))

There are no usable rules for innermost w.r.t. to the AFS 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:

HALF(x₁) -> HALF(x₁)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 5

                 ↳Dependency Graph

       →DP Problem 2

         ↳Nar

Dependency Pair:

Rules:

half(0) -> 0
half(s(s(x))) -> s(half(x))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(half(x))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Narrowing Transformation

Dependency Pair:

LOG(s(s(x))) -> LOG(s(half(x)))

Rules:

half(0) -> 0
half(s(s(x))) -> s(half(x))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(half(x))))

Strategy:

innermost

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

LOG(s(s(x))) -> LOG(s(half(x)))

two new Dependency Pairs are created:

LOG(s(s(0))) -> LOG(s(0))
LOG(s(s(s(s(x''))))) -> LOG(s(s(half(x''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Narrowing Transformation

Dependency Pair:

LOG(s(s(s(s(x''))))) -> LOG(s(s(half(x''))))

Rules:

half(0) -> 0
half(s(s(x))) -> s(half(x))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(half(x))))

Strategy:

innermost

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

LOG(s(s(s(s(x''))))) -> LOG(s(s(half(x''))))

two new Dependency Pairs are created:

LOG(s(s(s(s(0))))) -> LOG(s(s(0)))
LOG(s(s(s(s(s(s(x'))))))) -> LOG(s(s(s(half(x')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 7

                 ↳Argument Filtering and Ordering

Dependency Pair:

LOG(s(s(s(s(s(s(x'))))))) -> LOG(s(s(s(half(x')))))

Rules:

half(0) -> 0
half(s(s(x))) -> s(half(x))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(half(x))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

LOG(s(s(s(s(s(s(x'))))))) -> LOG(s(s(s(half(x')))))

The following usable rules for innermost w.r.t. to the AFS can be oriented:

half(0) -> 0
half(s(s(x))) -> s(half(x))

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

{half, s}

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

LOG(x₁) -> LOG(x₁)
s(x₁) -> s(x₁)
half(x₁) -> half(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 8

                 ↳Dependency Graph

Dependency Pair:

Rules:

half(0) -> 0
half(s(s(x))) -> s(half(x))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(half(x))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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