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
Polynomial Ordering
       →DP Problem 2
Polo
       →DP Problem 3
Polo


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 w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(PLUS(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
Polo


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
Polo
       →DP Problem 2
Polynomial Ordering
       →DP Problem 3
Polo


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)


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(plus(x1, x2))=  0  
  POL(Z)=  0  
  POL(0)=  0  
  POL(MIN(x1, x2))=  x1  
  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
Narrowing Transformation
       →DP Problem 3
Polo


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




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

MIN(min(X, Y), Z) -> MIN(X, plus(Y, Z))
two new Dependency Pairs are created:

MIN(min(X, 0), Z) -> MIN(X, Z)
MIN(min(X, s(X'')), Z) -> MIN(X, s(plus(X'', Z)))

The transformation is resulting in no new DP problems.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polynomial 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))


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


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(Z)=  0  
  POL(plus(x1, x2))=  0  
  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 3
Polo
           →DP Problem 6
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