Termination Proof using AProVE

Term Rewriting System R:
[x]
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
lastbit(0) -> 0
lastbit(s(0)) -> s(0)
lastbit(s(s(x))) -> lastbit(x)
conv(0) -> cons(nil, 0)
conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x)))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

HALF(s(s(x))) -> HALF(x)
LASTBIT(s(s(x))) -> LASTBIT(x)
CONV(s(x)) -> CONV(half(s(x)))
CONV(s(x)) -> HALF(s(x))
CONV(s(x)) -> LASTBIT(s(x))

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

Rules:

half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
lastbit(0) -> 0
lastbit(s(0)) -> s(0)
lastbit(s(s(x))) -> lastbit(x)
conv(0) -> cons(nil, 0)
conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(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 4

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

Rules:

half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
lastbit(0) -> 0
lastbit(s(0)) -> s(0)
lastbit(s(s(x))) -> lastbit(x)
conv(0) -> cons(nil, 0)
conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(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 4

             ↳FwdInst

...

               →DP Problem 5

                 ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

Rules:

half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
lastbit(0) -> 0
lastbit(s(0)) -> s(0)
lastbit(s(s(x))) -> lastbit(x)
conv(0) -> cons(nil, 0)
conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(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 that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(HALF(x₁)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

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 4

             ↳FwdInst

...

               →DP Problem 6

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

Dependency Pair:

Rules:

half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
lastbit(0) -> 0
lastbit(s(0)) -> s(0)
lastbit(s(s(x))) -> lastbit(x)
conv(0) -> cons(nil, 0)
conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x)))

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

         ↳Nar

Dependency Pair:

LASTBIT(s(s(x))) -> LASTBIT(x)

Rules:

half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
lastbit(0) -> 0
lastbit(s(0)) -> s(0)
lastbit(s(s(x))) -> lastbit(x)
conv(0) -> cons(nil, 0)
conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x)))

Strategy:

innermost

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

LASTBIT(s(s(x))) -> LASTBIT(x)

one new Dependency Pair is created:

LASTBIT(s(s(s(s(x''))))) -> LASTBIT(s(s(x'')))

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

         ↳Nar

Dependency Pair:

LASTBIT(s(s(s(s(x''))))) -> LASTBIT(s(s(x'')))

Rules:

half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
lastbit(0) -> 0
lastbit(s(0)) -> s(0)
lastbit(s(s(x))) -> lastbit(x)
conv(0) -> cons(nil, 0)
conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x)))

Strategy:

innermost

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

LASTBIT(s(s(s(s(x''))))) -> LASTBIT(s(s(x'')))

one new Dependency Pair is created:

LASTBIT(s(s(s(s(s(s(x''''))))))) -> LASTBIT(s(s(s(s(x'''')))))

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

                 ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳Nar

Dependency Pair:

LASTBIT(s(s(s(s(s(s(x''''))))))) -> LASTBIT(s(s(s(s(x'''')))))

Rules:

half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
lastbit(0) -> 0
lastbit(s(0)) -> s(0)
lastbit(s(s(x))) -> lastbit(x)
conv(0) -> cons(nil, 0)
conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

LASTBIT(s(s(s(s(s(s(x''''))))))) -> LASTBIT(s(s(s(s(x'''')))))

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(LASTBIT(x₁)) = 1 + x₁

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

LASTBIT(x₁) -> LASTBIT(x₁)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 9

                 ↳Dependency Graph

       →DP Problem 3

         ↳Nar

Dependency Pair:

Rules:

half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
lastbit(0) -> 0
lastbit(s(0)) -> s(0)
lastbit(s(s(x))) -> lastbit(x)
conv(0) -> cons(nil, 0)
conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Narrowing Transformation

Dependency Pair:

CONV(s(x)) -> CONV(half(s(x)))

Rules:

half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
lastbit(0) -> 0
lastbit(s(0)) -> s(0)
lastbit(s(s(x))) -> lastbit(x)
conv(0) -> cons(nil, 0)
conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x)))

Strategy:

innermost

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

CONV(s(x)) -> CONV(half(s(x)))

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Narrowing Transformation

Dependency Pair:

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

Rules:

half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
lastbit(0) -> 0
lastbit(s(0)) -> s(0)
lastbit(s(s(x))) -> lastbit(x)
conv(0) -> cons(nil, 0)
conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x)))

Strategy:

innermost

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 11

                 ↳Argument Filtering and Ordering

Dependency Pair:

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

Rules:

half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
lastbit(0) -> 0
lastbit(s(0)) -> s(0)
lastbit(s(s(x))) -> lastbit(x)
conv(0) -> cons(nil, 0)
conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

The following usable rules for innermost can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(CONV(x₁)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

POL(half(x₁)) = x₁

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

CONV(x₁) -> CONV(x₁)
s(x₁) -> s(x₁)
half(x₁) -> half(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 12

                 ↳Dependency Graph

Dependency Pair:

Rules:

half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
lastbit(0) -> 0
lastbit(s(0)) -> s(0)
lastbit(s(s(x))) -> lastbit(x)
conv(0) -> cons(nil, 0)
conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

POL(0)	= 0
POL(CONV(x₁))	= 1 + x₁
POL(s(x₁))	= 1 + x₁
POL(half(x₁))	= x₁