Termination Proof using AProVE

Term Rewriting System R:
[X, XS, X1, X2]
active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

ACTIVE(zeros) -> CONS(0, zeros)
ACTIVE(cons(X1, X2)) -> CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)
ACTIVE(tail(X)) -> TAIL(active(X))
ACTIVE(tail(X)) -> ACTIVE(X)
CONS(mark(X1), X2) -> CONS(X1, X2)
CONS(ok(X1), ok(X2)) -> CONS(X1, X2)
TAIL(mark(X)) -> TAIL(X)
TAIL(ok(X)) -> TAIL(X)
PROPER(cons(X1, X2)) -> CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) -> PROPER(X1)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(tail(X)) -> TAIL(proper(X))
PROPER(tail(X)) -> PROPER(X)
TOP(mark(X)) -> TOP(proper(X))
TOP(mark(X)) -> PROPER(X)
TOP(ok(X)) -> TOP(active(X))
TOP(ok(X)) -> ACTIVE(X)

Furthermore, R contains five SCCs.

     ↳DPs

       →DP Problem 1

         ↳Usable Rules (Innermost)

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

Dependency Pairs:

CONS(ok(X1), ok(X2)) -> CONS(X1, X2)
CONS(mark(X1), X2) -> CONS(X1, X2)

Rules:

active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

As we are in the innermost case, we can delete all 14 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 6

             ↳Size-Change Principle

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

Dependency Pairs:

CONS(ok(X1), ok(X2)) -> CONS(X1, X2)
CONS(mark(X1), X2) -> CONS(X1, X2)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

CONS(ok(X1), ok(X2)) -> CONS(X1, X2)
CONS(mark(X1), X2) -> CONS(X1, X2)

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

{1, 2}	,	{1, 2}
1	>	1
2	>	2

{1, 2}	,	{1, 2}
1	>	1
2	=	2

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

{1, 2}	,	{1, 2}
1	>	1
2	>	2

{1, 2}	,	{1, 2}
1	>	1
2	=	2

D_P: empty set
Oriented Rules: none

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

trivial

with Argument Filtering System:

mark(x₁) -> mark(x₁)
ok(x₁) -> ok(x₁)

We obtain no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳Usable Rules (Innermost)

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

Dependency Pairs:

TAIL(ok(X)) -> TAIL(X)
TAIL(mark(X)) -> TAIL(X)

Rules:

active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

As we are in the innermost case, we can delete all 14 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 7

             ↳Size-Change Principle

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

Dependency Pairs:

TAIL(ok(X)) -> TAIL(X)
TAIL(mark(X)) -> TAIL(X)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

TAIL(ok(X)) -> TAIL(X)
TAIL(mark(X)) -> TAIL(X)

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

{1, 2}	,	{1, 2}
1	>	1

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

{1, 2}	,	{1, 2}
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:

mark(x₁) -> mark(x₁)
ok(x₁) -> ok(x₁)

We obtain no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳Usable Rules (Innermost)

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

Dependency Pairs:

ACTIVE(tail(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)

Rules:

active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

As we are in the innermost case, we can delete all 14 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 8

             ↳Size-Change Principle

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

Dependency Pairs:

ACTIVE(tail(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

ACTIVE(tail(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)

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

{1, 2}	,	{1, 2}
1	>	1

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

{1, 2}	,	{1, 2}
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:

cons(x₁, x₂) -> cons(x₁, x₂)
tail(x₁) -> tail(x₁)

We obtain no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳Usable Rules (Innermost)

       →DP Problem 5

         ↳UsableRules

Dependency Pairs:

PROPER(tail(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)

Rules:

active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

As we are in the innermost case, we can delete all 14 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

           →DP Problem 9

             ↳Size-Change Principle

       →DP Problem 5

         ↳UsableRules

Dependency Pairs:

PROPER(tail(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

PROPER(tail(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)

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

{1, 2, 3}	,	{1, 2, 3}
1	>	1

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

{1, 2, 3}	,	{1, 2, 3}
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:

cons(x₁, x₂) -> cons(x₁, x₂)
tail(x₁) -> tail(x₁)

We obtain no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳Usable Rules (Innermost)

Dependency Pairs:

TOP(ok(X)) -> TOP(active(X))
TOP(mark(X)) -> TOP(proper(X))

Rules:

active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(cons(X1, X2)) -> cons(active(X1), X2)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
proper(tail(X)) -> tail(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

As we are in the innermost case, we can delete all 2 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

           →DP Problem 10

             ↳Modular Removal of Rules

Dependency Pairs:

TOP(ok(X)) -> TOP(active(X))
TOP(mark(X)) -> TOP(proper(X))

Rules:

active(cons(X1, X2)) -> cons(active(X1), X2)
active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(tail(X)) -> tail(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)

Strategy:

innermost

We have the following set of usable rules:

active(cons(X1, X2)) -> cons(active(X1), X2)
active(zeros) -> mark(cons(0, zeros))
active(tail(cons(X, XS))) -> mark(XS)
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(tail(X)) -> tail(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)

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

POL(active(x₁)) = x₁

POL(proper(x₁)) = x₁

POL(0) = 0

POL(cons(x₁, x₂)) = x₁ + x₂

POL(tail(x₁)) = 1 + x₁

POL(mark(x₁)) = x₁

POL(zeros) = 0

POL(TOP(x₁)) = x₁

POL(ok(x₁)) = x₁

We have the following set D of usable symbols: {proper, active, 0, cons, tail, zeros, mark, TOP, ok}
No Dependency Pairs can be deleted.
The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

active(tail(cons(X, XS))) -> mark(XS)

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 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

           →DP Problem 10

             ↳MRR

...

               →DP Problem 11

                 ↳Negative Polynomial Order

Dependency Pairs:

TOP(ok(X)) -> TOP(active(X))
TOP(mark(X)) -> TOP(proper(X))

Rules:

active(cons(X1, X2)) -> cons(active(X1), X2)
active(zeros) -> mark(cons(0, zeros))
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(tail(X)) -> tail(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)

Strategy:

innermost

The following Dependency Pair can be strictly oriented using the given order.

TOP(ok(X)) -> TOP(active(X))

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

active(cons(X1, X2)) -> cons(active(X1), X2)
cons(mark(X1), X2) -> mark(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
active(zeros) -> mark(cons(0, zeros))
active(tail(X)) -> tail(active(X))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
proper(tail(X)) -> tail(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)

Used ordering:
Polynomial Order with Interpretation:

POL( TOP(x₁) ) = x₁

POL( ok(x₁) ) = x₁ + 1

POL( active(x₁) ) = x₁

POL( mark(x₁) ) = x₁ + 1

POL( proper(x₁) ) = x₁ + 1

POL( cons(x₁, x₂) ) = x₁

POL( tail(x₁) ) = x₁

POL( zeros ) = 1

POL( 0 ) = 0

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

           →DP Problem 10

             ↳MRR

...

               →DP Problem 12

                 ↳Usable Rules (Innermost)

Dependency Pair:

TOP(mark(X)) -> TOP(proper(X))

Rules:

active(cons(X1, X2)) -> cons(active(X1), X2)
active(zeros) -> mark(cons(0, zeros))
active(tail(X)) -> tail(active(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
proper(tail(X)) -> tail(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)

Strategy:

innermost

As we are in the innermost case, we can delete all 3 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

           →DP Problem 10

             ↳MRR

...

               →DP Problem 13

                 ↳Modular Removal of Rules

Dependency Pair:

TOP(mark(X)) -> TOP(proper(X))

Rules:

proper(tail(X)) -> tail(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))

Strategy:

innermost

We have the following set of usable rules:

proper(tail(X)) -> tail(proper(X))
proper(zeros) -> ok(zeros)
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(0) -> ok(0)
tail(mark(X)) -> mark(tail(X))
tail(ok(X)) -> ok(tail(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))

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

POL(proper(x₁)) = x₁

POL(0) = 0

POL(cons(x₁, x₂)) = x₁ + x₂

POL(tail(x₁)) = x₁

POL(mark(x₁)) = 1 + x₁

POL(zeros) = 0

POL(TOP(x₁)) = x₁

POL(ok(x₁)) = x₁

We have the following set D of usable symbols: {proper, 0, cons, tail, mark, zeros, TOP, ok}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

TOP(mark(X)) -> TOP(proper(X))

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved.

Innermost Termination of R successfully shown.
Duration:
0:02 minutes

POL(active(x₁))	= x₁
POL(proper(x₁))	= x₁
POL(0)	= 0
POL(cons(x₁, x₂))	= x₁ + x₂
POL(tail(x₁))	= 1 + x₁
POL(mark(x₁))	= x₁
POL(zeros)	= 0
POL(TOP(x₁))	= x₁
POL(ok(x₁))	= x₁