Term Rewriting System R:
[y, x]
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

LE(s(x), s(y)) -> LE(x, y)
MINUS(x, s(y)) -> PRED(minus(x, y))
MINUS(x, s(y)) -> MINUS(x, y)
GCD(s(x), s(y)) -> IFGCD(le(y, x), s(x), s(y))
GCD(s(x), s(y)) -> LE(y, x)
IFGCD(true, s(x), s(y)) -> GCD(minus(x, y), s(y))
IFGCD(true, s(x), s(y)) -> MINUS(x, y)
IFGCD(false, s(x), s(y)) -> GCD(minus(y, x), s(x))
IFGCD(false, s(x), s(y)) -> MINUS(y, x)

Furthermore, R contains three SCCs.


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


Dependency Pair:

LE(s(x), s(y)) -> LE(x, y)


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))


Strategy:

innermost




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

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

LE(s(s(x'')), s(s(y''))) -> LE(s(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
Nar


Dependency Pair:

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


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))


Strategy:

innermost




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

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

LE(s(s(s(x''''))), s(s(s(y'''')))) -> LE(s(s(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
Nar


Dependency Pair:

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


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost that need to be oriented.

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

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
Nar


Dependency Pair:


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))


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
Nar


Dependency Pair:

MINUS(x, s(y)) -> MINUS(x, y)


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))


Strategy:

innermost




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

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

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

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
Nar


Dependency Pair:

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


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))


Strategy:

innermost




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

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

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

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
Nar


Dependency Pair:

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


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(MINUS(x1, x2))=  1 + x2  
  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
Nar


Dependency Pair:


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))


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
Narrowing Transformation


Dependency Pairs:

IFGCD(false, s(x), s(y)) -> GCD(minus(y, x), s(x))
IFGCD(true, s(x), s(y)) -> GCD(minus(x, y), s(y))
GCD(s(x), s(y)) -> IFGCD(le(y, x), s(x), s(y))


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))


Strategy:

innermost




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

GCD(s(x), s(y)) -> IFGCD(le(y, x), s(x), s(y))
three new Dependency Pairs are created:

GCD(s(x'), s(0)) -> IFGCD(true, s(x'), s(0))
GCD(s(0), s(s(x''))) -> IFGCD(false, s(0), s(s(x'')))
GCD(s(s(y'')), s(s(x''))) -> IFGCD(le(x'', y''), s(s(y'')), 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
Nar
           →DP Problem 10
Narrowing Transformation


Dependency Pairs:

GCD(s(s(y'')), s(s(x''))) -> IFGCD(le(x'', y''), s(s(y'')), s(s(x'')))
GCD(s(0), s(s(x''))) -> IFGCD(false, s(0), s(s(x'')))
IFGCD(true, s(x), s(y)) -> GCD(minus(x, y), s(y))
GCD(s(x'), s(0)) -> IFGCD(true, s(x'), s(0))
IFGCD(false, s(x), s(y)) -> GCD(minus(y, x), s(x))


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))


Strategy:

innermost




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

IFGCD(true, s(x), s(y)) -> GCD(minus(x, y), s(y))
two new Dependency Pairs are created:

IFGCD(true, s(x''), s(0)) -> GCD(x'', s(0))
IFGCD(true, s(x''), s(s(y''))) -> GCD(pred(minus(x'', y'')), s(s(y'')))

The transformation is resulting in two new DP problems:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
Nar
             ...
               →DP Problem 11
Forward Instantiation Transformation


Dependency Pairs:

IFGCD(true, s(x''), s(0)) -> GCD(x'', s(0))
GCD(s(x'), s(0)) -> IFGCD(true, s(x'), s(0))


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))


Strategy:

innermost




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

IFGCD(true, s(x''), s(0)) -> GCD(x'', s(0))
one new Dependency Pair is created:

IFGCD(true, s(s(x'''')), s(0)) -> GCD(s(x''''), s(0))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
Nar
             ...
               →DP Problem 13
Forward Instantiation Transformation


Dependency Pairs:

IFGCD(true, s(s(x'''')), s(0)) -> GCD(s(x''''), s(0))
GCD(s(x'), s(0)) -> IFGCD(true, s(x'), s(0))


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))


Strategy:

innermost




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

GCD(s(x'), s(0)) -> IFGCD(true, s(x'), s(0))
one new Dependency Pair is created:

GCD(s(s(x'''''')), s(0)) -> IFGCD(true, s(s(x'''''')), s(0))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
Nar
             ...
               →DP Problem 15
Polynomial Ordering


Dependency Pairs:

GCD(s(s(x'''''')), s(0)) -> IFGCD(true, s(s(x'''''')), s(0))
IFGCD(true, s(s(x'''')), s(0)) -> GCD(s(x''''), s(0))


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))


Strategy:

innermost




The following dependency pair can be strictly oriented:

IFGCD(true, s(s(x'''')), s(0)) -> GCD(s(x''''), s(0))


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(GCD(x1, x2))=  x1  
  POL(IF_GCD(x1, x2, x3))=  x2  
  POL(true)=  0  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
Nar
             ...
               →DP Problem 20
Dependency Graph


Dependency Pair:

GCD(s(s(x'''''')), s(0)) -> IFGCD(true, s(s(x'''''')), s(0))


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))


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
Nar
           →DP Problem 10
Nar
             ...
               →DP Problem 12
Narrowing Transformation


Dependency Pairs:

IFGCD(true, s(x''), s(s(y''))) -> GCD(pred(minus(x'', y'')), s(s(y'')))
GCD(s(0), s(s(x''))) -> IFGCD(false, s(0), s(s(x'')))
IFGCD(false, s(x), s(y)) -> GCD(minus(y, x), s(x))
GCD(s(s(y'')), s(s(x''))) -> IFGCD(le(x'', y''), s(s(y'')), s(s(x'')))


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))


Strategy:

innermost




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

IFGCD(false, s(x), s(y)) -> GCD(minus(y, x), s(x))
two new Dependency Pairs are created:

IFGCD(false, s(0), s(y')) -> GCD(y', s(0))
IFGCD(false, s(s(y'')), s(y0)) -> GCD(pred(minus(y0, y'')), s(s(y'')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
Nar
             ...
               →DP Problem 14
Narrowing Transformation


Dependency Pairs:

IFGCD(false, s(s(y'')), s(y0)) -> GCD(pred(minus(y0, y'')), s(s(y'')))
GCD(s(s(y'')), s(s(x''))) -> IFGCD(le(x'', y''), s(s(y'')), s(s(x'')))
IFGCD(true, s(x''), s(s(y''))) -> GCD(pred(minus(x'', y'')), s(s(y'')))


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))


Strategy:

innermost




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

GCD(s(s(y'')), s(s(x''))) -> IFGCD(le(x'', y''), s(s(y'')), s(s(x'')))
three new Dependency Pairs are created:

GCD(s(s(y''')), s(s(0))) -> IFGCD(true, s(s(y''')), s(s(0)))
GCD(s(s(0)), s(s(s(x')))) -> IFGCD(false, s(s(0)), s(s(s(x'))))
GCD(s(s(s(y'))), s(s(s(x')))) -> IFGCD(le(x', y'), s(s(s(y'))), s(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
Nar
           →DP Problem 10
Nar
             ...
               →DP Problem 16
Narrowing Transformation


Dependency Pairs:

GCD(s(s(s(y'))), s(s(s(x')))) -> IFGCD(le(x', y'), s(s(s(y'))), s(s(s(x'))))
GCD(s(s(0)), s(s(s(x')))) -> IFGCD(false, s(s(0)), s(s(s(x'))))
IFGCD(true, s(x''), s(s(y''))) -> GCD(pred(minus(x'', y'')), s(s(y'')))
GCD(s(s(y''')), s(s(0))) -> IFGCD(true, s(s(y''')), s(s(0)))
IFGCD(false, s(s(y'')), s(y0)) -> GCD(pred(minus(y0, y'')), s(s(y'')))


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))


Strategy:

innermost




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

IFGCD(true, s(x''), s(s(y''))) -> GCD(pred(minus(x'', y'')), s(s(y'')))
two new Dependency Pairs are created:

IFGCD(true, s(x'''), s(s(0))) -> GCD(pred(x'''), s(s(0)))
IFGCD(true, s(x'''), s(s(s(y')))) -> GCD(pred(pred(minus(x''', y'))), s(s(s(y'))))

The transformation is resulting in two new DP problems:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
Nar
             ...
               →DP Problem 17
Polynomial Ordering


Dependency Pairs:

IFGCD(true, s(x'''), s(s(0))) -> GCD(pred(x'''), s(s(0)))
GCD(s(s(y''')), s(s(0))) -> IFGCD(true, s(s(y''')), s(s(0)))


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))


Strategy:

innermost




The following dependency pair can be strictly oriented:

IFGCD(true, s(x'''), s(s(0))) -> GCD(pred(x'''), s(s(0)))


Additionally, the following usable rule for innermost can be oriented:

pred(s(x)) -> x


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(GCD(x1, x2))=  x1  
  POL(pred(x1))=  x1  
  POL(IF_GCD(x1, x2, x3))=  x2  
  POL(true)=  0  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
Nar
             ...
               →DP Problem 21
Dependency Graph


Dependency Pair:

GCD(s(s(y''')), s(s(0))) -> IFGCD(true, s(s(y''')), s(s(0)))


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))


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
Nar
           →DP Problem 10
Nar
             ...
               →DP Problem 18
Narrowing Transformation


Dependency Pairs:

IFGCD(true, s(x'''), s(s(s(y')))) -> GCD(pred(pred(minus(x''', y'))), s(s(s(y'))))
GCD(s(s(0)), s(s(s(x')))) -> IFGCD(false, s(s(0)), s(s(s(x'))))
IFGCD(false, s(s(y'')), s(y0)) -> GCD(pred(minus(y0, y'')), s(s(y'')))
GCD(s(s(s(y'))), s(s(s(x')))) -> IFGCD(le(x', y'), s(s(s(y'))), s(s(s(x'))))


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))


Strategy:

innermost




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

IFGCD(false, s(s(y'')), s(y0)) -> GCD(pred(minus(y0, y'')), s(s(y'')))
two new Dependency Pairs are created:

IFGCD(false, s(s(0)), s(y0')) -> GCD(pred(y0'), s(s(0)))
IFGCD(false, s(s(s(y'))), s(y0')) -> GCD(pred(pred(minus(y0', y'))), s(s(s(y'))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
Nar
             ...
               →DP Problem 19
Polynomial Ordering


Dependency Pairs:

IFGCD(false, s(s(s(y'))), s(y0')) -> GCD(pred(pred(minus(y0', y'))), s(s(s(y'))))
GCD(s(s(s(y'))), s(s(s(x')))) -> IFGCD(le(x', y'), s(s(s(y'))), s(s(s(x'))))
IFGCD(true, s(x'''), s(s(s(y')))) -> GCD(pred(pred(minus(x''', y'))), s(s(s(y'))))


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

IFGCD(false, s(s(s(y'))), s(y0')) -> GCD(pred(pred(minus(y0', y'))), s(s(s(y'))))
IFGCD(true, s(x'''), s(s(s(y')))) -> GCD(pred(pred(minus(x''', y'))), s(s(s(y'))))


Additionally, the following usable rules for innermost can be oriented:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(GCD(x1, x2))=  x1 + x2  
  POL(false)=  0  
  POL(pred(x1))=  x1  
  POL(minus(x1, x2))=  x1  
  POL(true)=  0  
  POL(IF_GCD(x1, x2, x3))=  x2 + x3  
  POL(s(x1))=  1 + x1  
  POL(le(x1, x2))=  0  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
Nar
             ...
               →DP Problem 22
Dependency Graph


Dependency Pair:

GCD(s(s(s(y'))), s(s(s(x')))) -> IFGCD(le(x', y'), s(s(s(y'))), s(s(s(x'))))


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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