Term Rewriting System R:
[x, y]
sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

SQR(s(x)) -> +'(sqr(x), s(double(x)))
SQR(s(x)) -> SQR(x)
SQR(s(x)) -> DOUBLE(x)
SQR(s(x)) -> +'(sqr(x), double(x))
DOUBLE(s(x)) -> DOUBLE(x)
+'(x, s(y)) -> +'(x, y)

Furthermore, R contains three SCCs.


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


Dependency Pair:

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


Rules:


sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem:



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


Dependency Pair:

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


Rules:


sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem:



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


Dependency Pair:

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


Rules:


sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 4
FwdInst
             ...
               →DP Problem 6
Dependency Graph
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst


Dependency Pair:


Rules:


sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, 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


Dependency Pair:

DOUBLE(s(x)) -> DOUBLE(x)


Rules:


sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))


Strategy:

innermost




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

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

DOUBLE(s(s(x''))) -> DOUBLE(s(x''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 7
Forward Instantiation Transformation
       →DP Problem 3
FwdInst


Dependency Pair:

DOUBLE(s(s(x''))) -> DOUBLE(s(x''))


Rules:


sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))


Strategy:

innermost




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

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

DOUBLE(s(s(s(x'''')))) -> DOUBLE(s(s(x'''')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 7
FwdInst
             ...
               →DP Problem 8
Polynomial Ordering
       →DP Problem 3
FwdInst


Dependency Pair:

DOUBLE(s(s(s(x'''')))) -> DOUBLE(s(s(x'''')))


Rules:


sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

DOUBLE(s(s(s(x'''')))) -> DOUBLE(s(s(x'''')))


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 7
FwdInst
             ...
               →DP Problem 9
Dependency Graph
       →DP Problem 3
FwdInst


Dependency Pair:


Rules:


sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, 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


Dependency Pair:

SQR(s(x)) -> SQR(x)


Rules:


sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))


Strategy:

innermost




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

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

SQR(s(s(x''))) -> SQR(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 10
Forward Instantiation Transformation


Dependency Pair:

SQR(s(s(x''))) -> SQR(s(x''))


Rules:


sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))


Strategy:

innermost




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

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

SQR(s(s(s(x'''')))) -> SQR(s(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 10
FwdInst
             ...
               →DP Problem 11
Polynomial Ordering


Dependency Pair:

SQR(s(s(s(x'''')))) -> SQR(s(s(x'''')))


Rules:


sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

SQR(s(s(s(x'''')))) -> SQR(s(s(x'''')))


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(s(x1))=  1 + x1  
  POL(SQR(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 10
FwdInst
             ...
               →DP Problem 12
Dependency Graph


Dependency Pair:


Rules:


sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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