Termination Proof using AProVE

Term Rewriting System R:
[x, y]
O(0) -> 0
+(0, x) -> x
+(x, 0) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
+(I(x), O(y)) -> I(+(x, y))
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
*(0, x) -> 0
*(x, 0) -> 0
*(O(x), y) -> O(*(x, y))
*(I(x), y) -> +(O(*(x, y)), y)
-(x, 0) -> x
-(0, x) -> 0
-(O(x), O(y)) -> O(-(x, y))
-(O(x), I(y)) -> I(-(-(x, y), I(1)))
-(I(x), O(y)) -> I(-(x, y))
-(I(x), I(y)) -> O(-(x, y))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

+'(O(x), O(y)) -> O'(+(x, y))
+'(O(x), O(y)) -> +'(x, y)
+'(O(x), I(y)) -> +'(x, y)
+'(I(x), O(y)) -> +'(x, y)
+'(I(x), I(y)) -> O'(+(+(x, y), I(0)))
+'(I(x), I(y)) -> +'(+(x, y), I(0))
+'(I(x), I(y)) -> +'(x, y)
*'(O(x), y) -> O'(*(x, y))
*'(O(x), y) -> *'(x, y)
*'(I(x), y) -> +'(O(*(x, y)), y)
*'(I(x), y) -> O'(*(x, y))
*'(I(x), y) -> *'(x, y)
-'(O(x), O(y)) -> O'(-(x, y))
-'(O(x), O(y)) -> -'(x, y)
-'(O(x), I(y)) -> -'(-(x, y), I(1))
-'(O(x), I(y)) -> -'(x, y)
-'(I(x), O(y)) -> -'(x, y)
-'(I(x), I(y)) -> O'(-(x, y))
-'(I(x), I(y)) -> -'(x, y)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Usable Rules (Innermost)

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

Dependency Pairs:

+'(I(x), I(y)) -> +'(x, y)
+'(I(x), I(y)) -> +'(+(x, y), I(0))
+'(I(x), O(y)) -> +'(x, y)
+'(O(x), I(y)) -> +'(x, y)
+'(O(x), O(y)) -> +'(x, y)

Rules:

O(0) -> 0
+(0, x) -> x
+(x, 0) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
+(I(x), O(y)) -> I(+(x, y))
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
*(0, x) -> 0
*(x, 0) -> 0
*(O(x), y) -> O(*(x, y))
*(I(x), y) -> +(O(*(x, y)), y)
-(x, 0) -> x
-(0, x) -> 0
-(O(x), O(y)) -> O(-(x, y))
-(O(x), I(y)) -> I(-(-(x, y), I(1)))
-(I(x), O(y)) -> I(-(x, y))
-(I(x), I(y)) -> O(-(x, y))

Strategy:

innermost

As we are in the innermost case, we can delete all 10 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 4

             ↳Modular Removal of Rules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

Dependency Pairs:

+'(I(x), I(y)) -> +'(x, y)
+'(I(x), I(y)) -> +'(+(x, y), I(0))
+'(I(x), O(y)) -> +'(x, y)
+'(O(x), I(y)) -> +'(x, y)
+'(O(x), O(y)) -> +'(x, y)

Rules:

+(I(x), O(y)) -> I(+(x, y))
+(x, 0) -> x
+(0, x) -> x
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
O(0) -> 0

Strategy:

innermost

We have the following set of usable rules:

+(I(x), O(y)) -> I(+(x, y))
+(x, 0) -> x
+(0, x) -> x
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
O(0) -> 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(I(x₁)) = 1 + x₁

POL(0) = 0

POL(O(x₁)) = x₁

POL(+(x₁, x₂)) = x₁ + x₂

POL(+'(x₁, x₂)) = 1 + x₁ + x₂

We have the following set D of usable symbols: {I, 0, O, +, +'}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

+'(I(x), I(y)) -> +'(x, y)
+'(I(x), I(y)) -> +'(+(x, y), I(0))
+'(I(x), O(y)) -> +'(x, y)
+'(O(x), I(y)) -> +'(x, y)

The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

+(I(x), I(y)) -> O(+(+(x, y), I(0)))

The result of this processor delivers one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 4

             ↳MRR

...

               →DP Problem 5

                 ↳Modular Removal of Rules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

Dependency Pair:

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

Rules:

+(I(x), O(y)) -> I(+(x, y))
+(x, 0) -> x
+(0, x) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
O(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(O(x₁)) = x₁

POL(+'(x₁, x₂)) = x₁ + x₂

We have the following set D of usable symbols: {+'}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

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

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.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳Usable Rules (Innermost)

       →DP Problem 3

         ↳UsableRules

Dependency Pairs:

-'(I(x), I(y)) -> -'(x, y)
-'(I(x), O(y)) -> -'(x, y)
-'(O(x), I(y)) -> -'(x, y)
-'(O(x), I(y)) -> -'(-(x, y), I(1))
-'(O(x), O(y)) -> -'(x, y)

Rules:

O(0) -> 0
+(0, x) -> x
+(x, 0) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
+(I(x), O(y)) -> I(+(x, y))
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
*(0, x) -> 0
*(x, 0) -> 0
*(O(x), y) -> O(*(x, y))
*(I(x), y) -> +(O(*(x, y)), y)
-(x, 0) -> x
-(0, x) -> 0
-(O(x), O(y)) -> O(-(x, y))
-(O(x), I(y)) -> I(-(-(x, y), I(1)))
-(I(x), O(y)) -> I(-(x, y))
-(I(x), I(y)) -> O(-(x, y))

Strategy:

innermost

As we are in the innermost case, we can delete all 10 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 6

             ↳Modular Removal of Rules

       →DP Problem 3

         ↳UsableRules

Dependency Pairs:

-'(I(x), I(y)) -> -'(x, y)
-'(I(x), O(y)) -> -'(x, y)
-'(O(x), I(y)) -> -'(x, y)
-'(O(x), I(y)) -> -'(-(x, y), I(1))
-'(O(x), O(y)) -> -'(x, y)

Rules:

-(I(x), O(y)) -> I(-(x, y))
-(x, 0) -> x
-(I(x), I(y)) -> O(-(x, y))
-(O(x), I(y)) -> I(-(-(x, y), I(1)))
-(0, x) -> 0
-(O(x), O(y)) -> O(-(x, y))
O(0) -> 0

Strategy:

innermost

We have the following set of usable rules:

-(I(x), O(y)) -> I(-(x, y))
-(x, 0) -> x
-(I(x), I(y)) -> O(-(x, y))
-(O(x), I(y)) -> I(-(-(x, y), I(1)))
-(0, x) -> 0
-(O(x), O(y)) -> O(-(x, y))
O(0) -> 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(I(x₁)) = x₁

POL(-'(x₁, x₂)) = 1 + x₁ + x₂

POL(0) = 1

POL(1) = 0

POL(O(x₁)) = x₁

POL(-(x₁, x₂)) = x₁ + x₂

We have the following set D of usable symbols: {I, -', 0, 1, O, -}
No Dependency Pairs can be deleted.
The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

-(x, 0) -> x

The result of this processor delivers one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 6

             ↳MRR

...

               →DP Problem 7

                 ↳Modular Removal of Rules

       →DP Problem 3

         ↳UsableRules

Dependency Pairs:

-'(I(x), I(y)) -> -'(x, y)
-'(I(x), O(y)) -> -'(x, y)
-'(O(x), I(y)) -> -'(x, y)
-'(O(x), I(y)) -> -'(-(x, y), I(1))
-'(O(x), O(y)) -> -'(x, y)

Rules:

-(I(x), O(y)) -> I(-(x, y))
-(I(x), I(y)) -> O(-(x, y))
-(O(x), I(y)) -> I(-(-(x, y), I(1)))
-(0, x) -> 0
-(O(x), O(y)) -> O(-(x, y))
O(0) -> 0

Strategy:

innermost

We have the following set of usable rules:

-(I(x), O(y)) -> I(-(x, y))
-(I(x), I(y)) -> O(-(x, y))
-(O(x), I(y)) -> I(-(-(x, y), I(1)))
O(0) -> 0
-(0, x) -> 0
-(O(x), O(y)) -> O(-(x, y))

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

POL(I(x₁)) = 1 + x₁

POL(-'(x₁, x₂)) = 1 + x₁ + x₂

POL(0) = 0

POL(1) = 0

POL(O(x₁)) = 1 + x₁

POL(-(x₁, x₂)) = x₁ + x₂

We have the following set D of usable symbols: {I, -', 0, 1, O, -}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

-'(I(x), I(y)) -> -'(x, y)
-'(I(x), O(y)) -> -'(x, y)
-'(O(x), I(y)) -> -'(x, y)
-'(O(x), I(y)) -> -'(-(x, y), I(1))
-'(O(x), O(y)) -> -'(x, y)

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.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳Usable Rules (Innermost)

Dependency Pairs:

*'(I(x), y) -> *'(x, y)
*'(O(x), y) -> *'(x, y)

Rules:

O(0) -> 0
+(0, x) -> x
+(x, 0) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
+(I(x), O(y)) -> I(+(x, y))
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
*(0, x) -> 0
*(x, 0) -> 0
*(O(x), y) -> O(*(x, y))
*(I(x), y) -> +(O(*(x, y)), y)
-(x, 0) -> x
-(0, x) -> 0
-(O(x), O(y)) -> O(-(x, y))
-(O(x), I(y)) -> I(-(-(x, y), I(1)))
-(I(x), O(y)) -> I(-(x, y))
-(I(x), I(y)) -> O(-(x, y))

Strategy:

innermost

As we are in the innermost case, we can delete all 17 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 8

             ↳Size-Change Principle

Dependency Pairs:

*'(I(x), y) -> *'(x, y)
*'(O(x), y) -> *'(x, y)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

*'(I(x), y) -> *'(x, y)
*'(O(x), y) -> *'(x, y)

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

{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

D_P: empty set
Oriented Rules: none

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

trivial

with Argument Filtering System:

I(x₁) -> I(x₁)
O(x₁) -> O(x₁)

We obtain no new DP problems.

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

POL(I(x₁))	= 1 + x₁
POL(0)	= 0
POL(O(x₁))	= x₁
POL(+(x₁, x₂))	= x₁ + x₂
POL(+'(x₁, x₂))	= 1 + x₁ + x₂

POL(I(x₁))	= x₁
POL(-'(x₁, x₂))	= 1 + x₁ + x₂
POL(0)	= 1
POL(1)	= 0
POL(O(x₁))	= x₁
POL(-(x₁, x₂))	= x₁ + x₂