Term Rewriting System R:
[x, y]
p(s(x)) -> x
s(p(x)) -> x
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

+'(s(x), y) -> S(+(x, y))
+'(s(x), y) -> +'(x, y)
+'(p(x), y) -> P(+(x, y))
+'(p(x), y) -> +'(x, y)
MINUS(s(x)) -> P(minus(x))
MINUS(s(x)) -> MINUS(x)
MINUS(p(x)) -> S(minus(x))
MINUS(p(x)) -> MINUS(x)
*'(s(x), y) -> +'(*(x, y), y)
*'(s(x), y) -> *'(x, y)
*'(p(x), y) -> +'(*(x, y), minus(y))
*'(p(x), y) -> *'(x, y)
*'(p(x), y) -> MINUS(y)

Furthermore, R contains six SCCs.


   R
DPs
       →DP Problem 1
Forward Instantiation Transformation
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
FwdInst
       →DP Problem 5
FwdInst
       →DP Problem 6
FwdInst


Dependency Pair:

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


Rules:


p(s(x)) -> x
s(p(x)) -> x
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




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

+'(p(x), y) -> +'(x, y)
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 7
Polynomial Ordering
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
FwdInst
       →DP Problem 5
FwdInst
       →DP Problem 6
FwdInst


Dependency Pair:

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


Rules:


p(s(x)) -> x
s(p(x)) -> x
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


Additionally, the following usable rule for innermost w.r.t. to the implicit AFS can be oriented:

p(s(x)) -> x


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(s(x1))=  1 + x1  
  POL(p(x1))=  1 + x1  
  POL(+'(x1, x2))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 7
Polo
             ...
               →DP Problem 8
Dependency Graph
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
FwdInst
       →DP Problem 5
FwdInst
       →DP Problem 6
FwdInst


Dependency Pair:


Rules:


p(s(x)) -> x
s(p(x)) -> x
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Forward Instantiation Transformation
       →DP Problem 3
FwdInst
       →DP Problem 4
FwdInst
       →DP Problem 5
FwdInst
       →DP Problem 6
FwdInst


Dependency Pair:

MINUS(p(x)) -> MINUS(x)


Rules:


p(s(x)) -> x
s(p(x)) -> x
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




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

MINUS(p(x)) -> MINUS(x)
one new Dependency Pair is created:

MINUS(p(p(x''))) -> MINUS(p(x''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 9
Polynomial Ordering
       →DP Problem 3
FwdInst
       →DP Problem 4
FwdInst
       →DP Problem 5
FwdInst
       →DP Problem 6
FwdInst


Dependency Pair:

MINUS(p(p(x''))) -> MINUS(p(x''))


Rules:


p(s(x)) -> x
s(p(x)) -> x
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

MINUS(p(p(x''))) -> MINUS(p(x''))


Additionally, the following usable rule for innermost w.r.t. to the implicit AFS can be oriented:

p(s(x)) -> x


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(MINUS(x1))=  x1  
  POL(s(x1))=  1 + x1  
  POL(p(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 9
Polo
             ...
               →DP Problem 10
Dependency Graph
       →DP Problem 3
FwdInst
       →DP Problem 4
FwdInst
       →DP Problem 5
FwdInst
       →DP Problem 6
FwdInst


Dependency Pair:


Rules:


p(s(x)) -> x
s(p(x)) -> x
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Forward Instantiation Transformation
       →DP Problem 4
FwdInst
       →DP Problem 5
FwdInst
       →DP Problem 6
FwdInst


Dependency Pair:

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


Rules:


p(s(x)) -> x
s(p(x)) -> x
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




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

+'(s(x), y) -> +'(x, y)
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
           →DP Problem 11
Polynomial Ordering
       →DP Problem 4
FwdInst
       →DP Problem 5
FwdInst
       →DP Problem 6
FwdInst


Dependency Pair:

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


Rules:


p(s(x)) -> x
s(p(x)) -> x
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


Additionally, the following usable rule for innermost w.r.t. to the implicit AFS can be oriented:

s(p(x)) -> x


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(s(x1))=  1 + x1  
  POL(p(x1))=  1 + x1  
  POL(+'(x1, x2))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
           →DP Problem 11
Polo
             ...
               →DP Problem 12
Dependency Graph
       →DP Problem 4
FwdInst
       →DP Problem 5
FwdInst
       →DP Problem 6
FwdInst


Dependency Pair:


Rules:


p(s(x)) -> x
s(p(x)) -> x
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
Forward Instantiation Transformation
       →DP Problem 5
FwdInst
       →DP Problem 6
FwdInst


Dependency Pair:

MINUS(s(x)) -> MINUS(x)


Rules:


p(s(x)) -> x
s(p(x)) -> x
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




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

MINUS(s(x)) -> MINUS(x)
one new Dependency Pair is created:

MINUS(s(s(x''))) -> MINUS(s(x''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
FwdInst
           →DP Problem 13
Polynomial Ordering
       →DP Problem 5
FwdInst
       →DP Problem 6
FwdInst


Dependency Pair:

MINUS(s(s(x''))) -> MINUS(s(x''))


Rules:


p(s(x)) -> x
s(p(x)) -> x
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

MINUS(s(s(x''))) -> MINUS(s(x''))


Additionally, the following usable rule for innermost w.r.t. to the implicit AFS can be oriented:

s(p(x)) -> x


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(MINUS(x1))=  x1  
  POL(s(x1))=  1 + x1  
  POL(p(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
FwdInst
           →DP Problem 13
Polo
             ...
               →DP Problem 14
Dependency Graph
       →DP Problem 5
FwdInst
       →DP Problem 6
FwdInst


Dependency Pair:


Rules:


p(s(x)) -> x
s(p(x)) -> x
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
FwdInst
       →DP Problem 5
Forward Instantiation Transformation
       →DP Problem 6
FwdInst


Dependency Pair:

*'(p(x), y) -> *'(x, y)


Rules:


p(s(x)) -> x
s(p(x)) -> x
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




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

*'(p(x), y) -> *'(x, y)
one new Dependency Pair is created:

*'(p(p(x'')), y'') -> *'(p(x''), y'')

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
FwdInst
       →DP Problem 5
FwdInst
           →DP Problem 15
Polynomial Ordering
       →DP Problem 6
FwdInst


Dependency Pair:

*'(p(p(x'')), y'') -> *'(p(x''), y'')


Rules:


p(s(x)) -> x
s(p(x)) -> x
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

*'(p(p(x'')), y'') -> *'(p(x''), y'')


Additionally, the following usable rule for innermost w.r.t. to the implicit AFS can be oriented:

p(s(x)) -> x


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(*'(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  
  POL(p(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
FwdInst
       →DP Problem 5
FwdInst
           →DP Problem 15
Polo
             ...
               →DP Problem 16
Dependency Graph
       →DP Problem 6
FwdInst


Dependency Pair:


Rules:


p(s(x)) -> x
s(p(x)) -> x
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
FwdInst
       →DP Problem 5
FwdInst
       →DP Problem 6
Forward Instantiation Transformation


Dependency Pair:

*'(s(x), y) -> *'(x, y)


Rules:


p(s(x)) -> x
s(p(x)) -> x
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




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

*'(s(x), y) -> *'(x, y)
one new Dependency Pair is created:

*'(s(s(x'')), y'') -> *'(s(x''), y'')

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
FwdInst
       →DP Problem 5
FwdInst
       →DP Problem 6
FwdInst
           →DP Problem 17
Polynomial Ordering


Dependency Pair:

*'(s(s(x'')), y'') -> *'(s(x''), y'')


Rules:


p(s(x)) -> x
s(p(x)) -> x
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

*'(s(s(x'')), y'') -> *'(s(x''), y'')


Additionally, the following usable rule for innermost w.r.t. to the implicit AFS can be oriented:

s(p(x)) -> x


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(*'(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  
  POL(p(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
FwdInst
       →DP Problem 5
FwdInst
       →DP Problem 6
FwdInst
           →DP Problem 17
Polo
             ...
               →DP Problem 18
Dependency Graph


Dependency Pair:


Rules:


p(s(x)) -> x
s(p(x)) -> x
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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