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. 



   R
     ↳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. 


   R
     ↳DPs
       →DP Problem 1
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS


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




The following dependency pairs can be strictly oriented:

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


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:
I > 0
O > 0
+ > 0
+' > 0

resulting in one new DP problem.
Used Argument Filtering System:
+'(x1, x2) -> x2
O(x1) -> O(x1)
I(x1) -> I(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 4
             ↳Narrowing Transformation
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS


Dependency Pair:

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


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




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

+'(I(x), I(y)) -> +'(+(x, y), I(0))
six new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 4
             ↳Nar
             ...
               →DP Problem 5
                 ↳Instantiation Transformation
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS


Dependency Pairs:

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


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




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

+'(I(0), I(y')) -> +'(y', I(0))
one new Dependency Pair is created:

+'(I(0), I(0)) -> +'(0, I(0))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 4
             ↳Nar
             ...
               →DP Problem 6
                 ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS


Dependency Pair:

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


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




The following dependency pair can be strictly oriented:

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


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:
+'(x1, x2) -> +'(x1, x2)
I(x1) -> I(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 4
             ↳Nar
             ...
               →DP Problem 7
                 ↳Dependency Graph
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS


Dependency Pair:


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




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Argument Filtering and Ordering
       →DP Problem 3
         ↳AFS


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




The following dependency pairs can be strictly oriented:

-'(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)


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

-(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))
O(0) -> 0


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{-', I, O, 1}

resulting in one new DP problem.
Used Argument Filtering System:
-'(x1, x2) -> -'(x1, x2)
I(x1) -> I(x1)
O(x1) -> O(x1)
-(x1, x2) -> x1


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
           →DP Problem 8
             ↳Dependency Graph
       →DP Problem 3
         ↳AFS


Dependency Pair:


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




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Argument Filtering and Ordering


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




The following dependency pairs can be strictly oriented:

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


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:
*'(x1, x2) -> *'(x1, x2)
I(x1) -> I(x1)
O(x1) -> O(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
           →DP Problem 9
             ↳Dependency Graph


Dependency Pair:


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




Using the Dependency Graph resulted in no new DP problems.

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