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

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

PLUS(s(X), Y) -> PLUS(X, Y)
MIN(s(X), s(Y)) -> MIN(X, Y)
MIN(min(X, Y), Z) -> MIN(X, plus(Y, Z))
MIN(min(X, Y), Z) -> PLUS(Y, Z)
QUOT(s(X), s(Y)) -> QUOT(min(X, Y), s(Y))
QUOT(s(X), s(Y)) -> MIN(X, Y)

Furthermore, R contains three SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS


Dependency Pair:

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


Rules:


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


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(PLUS(x1, x2))=  x1 + x2  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.
Used Argument Filtering System:
PLUS(x1, x2) -> PLUS(x1, x2)
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 4
             ↳Dependency Graph
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS


Dependency Pair:


Rules:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Argument Filtering and Ordering
       →DP Problem 3
         ↳AFS


Dependency Pairs:

MIN(min(X, Y), Z) -> MIN(X, plus(Y, Z))
MIN(s(X), s(Y)) -> MIN(X, Y)


Rules:


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


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


The following usable rules for innermost can be oriented:

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


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

resulting in one new DP problem.
Used Argument Filtering System:
MIN(x1, x2) -> MIN(x1, x2)
s(x1) -> s(x1)
min(x1, x2) -> min(x1, x2)
plus(x1, x2) -> plus(x1, x2)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
           →DP Problem 5
             ↳Argument Filtering and Ordering
       →DP Problem 3
         ↳AFS


Dependency Pair:

MIN(min(X, Y), Z) -> MIN(X, plus(Y, Z))


Rules:


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


Strategy:

innermost




The following dependency pair can be strictly oriented:

MIN(min(X, Y), Z) -> MIN(X, plus(Y, Z))


The following usable rules for innermost can be oriented:

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


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

resulting in one new DP problem.
Used Argument Filtering System:
MIN(x1, x2) -> MIN(x1, x2)
min(x1, x2) -> min(x1, x2)
plus(x1, x2) -> plus(x1, x2)
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
           →DP Problem 5
             ↳AFS
             ...
               →DP Problem 6
                 ↳Dependency Graph
       →DP Problem 3
         ↳AFS


Dependency Pair:


Rules:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Argument Filtering and Ordering


Dependency Pair:

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


Rules:


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


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


The following usable rules for innermost can be oriented:

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


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

resulting in one new DP problem.
Used Argument Filtering System:
QUOT(x1, x2) -> QUOT(x1, x2)
s(x1) -> s(x1)
min(x1, x2) -> x1
plus(x1, x2) -> plus(x1, x2)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
           →DP Problem 7
             ↳Dependency Graph


Dependency Pair:


Rules:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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