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

         ↳Usable Rules (Innermost)

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

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

As we are in the innermost case, we can delete all 8 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 4

             ↳Size-Change Principle

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

Dependency Pair:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:

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

and get the following Size-Change Graph(s):

{1}	,	{1}
1	>	1

which lead(s) to this/these maximal multigraph(s):

{1}	,	{1}
1	>	1

D_P: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.

trivial

with Argument Filtering System:

s(x₁) -> s(x₁)

We obtain no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳Usable Rules (Innermost)

       →DP Problem 3

         ↳UsableRules

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

As we are in the innermost case, we can delete all 8 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 5

             ↳Size-Change Principle

       →DP Problem 3

         ↳UsableRules

Dependency Pair:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:

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

and get the following Size-Change Graph(s):

{1}	,	{1}
1	>	1

which lead(s) to this/these maximal multigraph(s):

{1}	,	{1}
1	>	1

D_P: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.

trivial

with Argument Filtering System:

s(x₁) -> s(x₁)

We obtain no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳Usable Rules (Innermost)

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

As we are in the innermost case, we can delete all 5 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 6

             ↳Modular Removal of Rules

Dependency Pair:

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

Rules:

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

Strategy:

innermost

We have the following set of usable rules:

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

To remove rules and DPs from this DP problem we used the following monotonic and C_E-compatible order: Polynomial ordering.
Polynomial interpretation:

POL(0) = 0

POL(CONV(x₁)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

POL(half(x₁)) = x₁

We have the following set D of usable symbols: {0, CONV, s, half}
No Dependency Pairs can be deleted.
The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

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

The result of this processor delivers one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 6

             ↳MRR

...

               →DP Problem 7

                 ↳Modular Removal of Rules

Dependency Pair:

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

Rule:

half(0) -> 0

Strategy:

innermost

We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and C_E-compatible order: Polynomial ordering.
Polynomial interpretation:

POL(CONV(x₁)) = x₁

POL(s(x₁)) = x₁

POL(half(x₁)) = x₁

We have the following set D of usable symbols: {CONV, s, half}
No Dependency Pairs can be deleted.
1 non usable rules have been deleted.

The result of this processor delivers one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 6

             ↳MRR

...

               →DP Problem 8

                 ↳Dependency Graph

Dependency Pair:

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

Rule:

none

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes

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