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))))))

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
Polynomial Ordering
       →DP Problem 2
Polo
       →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))))))





The following dependency pair can be strictly oriented:

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


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

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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 4
Dependency Graph
       →DP Problem 2
Polo
       →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))))))





Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polynomial Ordering
       →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))))))





The following dependency pair can be strictly oriented:

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


Additionally, the following usable rules 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))=  x1  
  POL(0)=  1  
  POL(min(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
           →DP Problem 5
Dependency Graph
       →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))))))





Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →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))))))





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
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
           →DP Problem 6
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))))))





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
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
           →DP Problem 6
Nar
             ...
               →DP Problem 7
Polynomial Ordering


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))))))





The following dependency pair can be strictly oriented:

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


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

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)))


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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
           →DP Problem 6
Nar
             ...
               →DP Problem 8
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))))))





Using the Dependency Graph resulted in no new DP problems.

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