Term Rewriting System R:
[y, z, x]
quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

QUOT(s(x), s(y), z) -> QUOT(x, y, z)
QUOT(x, 0, s(z)) -> QUOT(x, plus(z, s(0)), s(z))
QUOT(x, 0, s(z)) -> PLUS(z, s(0))
PLUS(s(x), y) -> PLUS(x, y)

Furthermore, R contains two SCCs. 


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


Dependency Pair:

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


Rules:


quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 3
             ↳Dependency Graph
       →DP Problem 2
         ↳Polo


Dependency Pair:


Rules:


quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pairs:

QUOT(x, 0, s(z)) -> QUOT(x, plus(z, s(0)), s(z))
QUOT(s(x), s(y), z) -> QUOT(x, y, z)


Rules:


quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

QUOT(s(x), s(y), z) -> QUOT(x, y, z)


Additionally, 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(plus(x1, x2))=  x1 + x2  
  POL(QUOT(x1, x2, x3))=  1 + x1  
  POL(0)=  0  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 4
             ↳Narrowing Transformation


Dependency Pair:

QUOT(x, 0, s(z)) -> QUOT(x, plus(z, s(0)), s(z))


Rules:


quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))


Strategy:

innermost




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

QUOT(x, 0, s(z)) -> QUOT(x, plus(z, s(0)), s(z))
two new Dependency Pairs are created:

QUOT(x, 0, s(0)) -> QUOT(x, s(0), s(0))
QUOT(x, 0, s(s(x''))) -> QUOT(x, s(plus(x'', s(0))), s(s(x'')))

The transformation is resulting in no new DP problems.


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