Term Rewriting System R:
[X, Y]
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

MIN(s(X), s(Y)) -> MIN(X, Y)
QUOT(s(X), s(Y)) -> QUOT(min(X, Y), s(Y))
QUOT(s(X), s(Y)) -> MIN(X, Y)
LOG(s(s(X))) -> LOG(s(quot(X, s(s(0)))))
LOG(s(s(X))) -> QUOT(X, s(s(0)))

Furthermore, R contains three SCCs.


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


Dependency Pair:

MIN(s(X), s(Y)) -> MIN(X, Y)


Rules:


min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))


Strategy:

innermost




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

MIN(s(X), s(Y)) -> MIN(X, Y)
one new Dependency Pair is created:

MIN(s(s(X'')), s(s(Y''))) -> MIN(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
Nar
       →DP Problem 3
Nar


Dependency Pair:

MIN(s(s(X'')), s(s(Y''))) -> MIN(s(X''), s(Y''))


Rules:


min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))


Strategy:

innermost




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

MIN(s(s(X'')), s(s(Y''))) -> MIN(s(X''), s(Y''))
one new Dependency Pair is created:

MIN(s(s(s(X''''))), s(s(s(Y'''')))) -> MIN(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
Nar
       →DP Problem 3
Nar


Dependency Pair:

MIN(s(s(s(X''''))), s(s(s(Y'''')))) -> MIN(s(s(X'''')), s(s(Y'''')))


Rules:


min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))


Strategy:

innermost




The following dependency pair can be strictly oriented:

MIN(s(s(s(X''''))), s(s(s(Y'''')))) -> MIN(s(s(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(MIN(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
Nar
       →DP Problem 3
Nar


Dependency Pair:


Rules:


min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

QUOT(s(X), s(Y)) -> QUOT(min(X, Y), s(Y))


Rules:


min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))


Strategy:

innermost




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

QUOT(s(X), s(Y)) -> QUOT(min(X, Y), s(Y))
two new Dependency Pairs are created:

QUOT(s(X''), s(0)) -> QUOT(X'', s(0))
QUOT(s(s(X'')), s(s(Y''))) -> QUOT(min(X'', Y''), s(s(Y'')))

The transformation is resulting in two new DP problems:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
           →DP Problem 7
Forward Instantiation Transformation
           →DP Problem 8
Nar
       →DP Problem 3
Nar


Dependency Pair:

QUOT(s(X''), s(0)) -> QUOT(X'', s(0))


Rules:


min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))


Strategy:

innermost




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

QUOT(s(X''), s(0)) -> QUOT(X'', s(0))
one new Dependency Pair is created:

QUOT(s(s(X'''')), s(0)) -> QUOT(s(X''''), s(0))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
           →DP Problem 7
FwdInst
             ...
               →DP Problem 9
Polynomial Ordering
           →DP Problem 8
Nar
       →DP Problem 3
Nar


Dependency Pair:

QUOT(s(s(X'''')), s(0)) -> QUOT(s(X''''), s(0))


Rules:


min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))


Strategy:

innermost




The following dependency pair can be strictly oriented:

QUOT(s(s(X'''')), s(0)) -> QUOT(s(X''''), s(0))


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(QUOT(x1, x2))=  1 + x1  
  POL(0)=  0  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
           →DP Problem 7
FwdInst
             ...
               →DP Problem 12
Dependency Graph
           →DP Problem 8
Nar
       →DP Problem 3
Nar


Dependency Pair:


Rules:


min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
           →DP Problem 7
FwdInst
           →DP Problem 8
Narrowing Transformation
       →DP Problem 3
Nar


Dependency Pair:

QUOT(s(s(X'')), s(s(Y''))) -> QUOT(min(X'', Y''), s(s(Y'')))


Rules:


min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))


Strategy:

innermost




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

QUOT(s(s(X'')), s(s(Y''))) -> QUOT(min(X'', Y''), s(s(Y'')))
two new Dependency Pairs are created:

QUOT(s(s(X''')), s(s(0))) -> QUOT(X''', s(s(0)))
QUOT(s(s(s(X'))), s(s(s(Y')))) -> QUOT(min(X', Y'), s(s(s(Y'))))

The transformation is resulting in two new DP problems:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
           →DP Problem 7
FwdInst
           →DP Problem 8
Nar
             ...
               →DP Problem 10
Polynomial Ordering
       →DP Problem 3
Nar


Dependency Pair:

QUOT(s(s(X''')), s(s(0))) -> QUOT(X''', s(s(0)))


Rules:


min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))


Strategy:

innermost




The following dependency pair can be strictly oriented:

QUOT(s(s(X''')), s(s(0))) -> QUOT(X''', s(s(0)))


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(QUOT(x1, x2))=  1 + x1  
  POL(0)=  0  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
           →DP Problem 7
FwdInst
           →DP Problem 8
Nar
             ...
               →DP Problem 11
Polynomial Ordering
       →DP Problem 3
Nar


Dependency Pair:

QUOT(s(s(s(X'))), s(s(s(Y')))) -> QUOT(min(X', Y'), s(s(s(Y'))))


Rules:


min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))


Strategy:

innermost




The following dependency pair can be strictly oriented:

QUOT(s(s(s(X'))), s(s(s(Y')))) -> QUOT(min(X', Y'), s(s(s(Y'))))


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

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(QUOT(x1, x2))=  1 + x1 + x2  
  POL(0)=  1  
  POL(min(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



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


Dependency Pair:

LOG(s(s(X))) -> LOG(s(quot(X, s(s(0)))))


Rules:


min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))


Strategy:

innermost




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

LOG(s(s(X))) -> LOG(s(quot(X, s(s(0)))))
two new Dependency Pairs are created:

LOG(s(s(0))) -> LOG(s(0))
LOG(s(s(s(X'')))) -> LOG(s(s(quot(min(X'', s(0)), s(s(0))))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 15
Narrowing Transformation


Dependency Pair:

LOG(s(s(s(X'')))) -> LOG(s(s(quot(min(X'', s(0)), s(s(0))))))


Rules:


min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))


Strategy:

innermost




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

LOG(s(s(s(X'')))) -> LOG(s(s(quot(min(X'', s(0)), s(s(0))))))
one new Dependency Pair is created:

LOG(s(s(s(s(X'))))) -> LOG(s(s(quot(min(X', 0), s(s(0))))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 15
Nar
             ...
               →DP Problem 16
Rewriting Transformation


Dependency Pair:

LOG(s(s(s(s(X'))))) -> LOG(s(s(quot(min(X', 0), s(s(0))))))


Rules:


min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))


Strategy:

innermost




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

LOG(s(s(s(s(X'))))) -> LOG(s(s(quot(min(X', 0), s(s(0))))))
one new Dependency Pair is created:

LOG(s(s(s(s(X'))))) -> LOG(s(s(quot(X', s(s(0))))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 15
Nar
             ...
               →DP Problem 17
Polynomial Ordering


Dependency Pair:

LOG(s(s(s(s(X'))))) -> LOG(s(s(quot(X', s(s(0))))))


Rules:


min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))


Strategy:

innermost




The following dependency pair can be strictly oriented:

LOG(s(s(s(s(X'))))) -> LOG(s(s(quot(X', s(s(0))))))


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

quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(quot(x1, x2))=  x1  
  POL(min(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  
  POL(LOG(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 15
Nar
             ...
               →DP Problem 18
Dependency Graph


Dependency Pair:


Rules:


min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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