Term Rewriting System R:
[Y, X]
le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

LE(s(X), s(Y)) -> LE(X, Y)
MINUS(s(X), Y) -> IFMINUS(le(s(X), Y), s(X), Y)
MINUS(s(X), Y) -> LE(s(X), Y)
IFMINUS(false, s(X), Y) -> MINUS(X, Y)
QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))
QUOT(s(X), s(Y)) -> MINUS(X, Y)

Furthermore, R contains three SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo


Dependency Pair:

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


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))





The following dependency pair can be strictly oriented:

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


Additionally, the following rules can be oriented:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(LE(x1, x2))=  1 + x1  
  POL(0)=  0  
  POL(false)=  0  
  POL(minus(x1, x2))=  x1  
  POL(true)=  0  
  POL(quot(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  
  POL(le(x1, x2))=  0  
  POL(ifMinus(x1, x2, x3))=  x2  

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:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))





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:

IFMINUS(false, s(X), Y) -> MINUS(X, Y)
MINUS(s(X), Y) -> IFMINUS(le(s(X), Y), s(X), Y)


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))





The following dependency pair can be strictly oriented:

IFMINUS(false, s(X), Y) -> MINUS(X, Y)


Additionally, the following rules can be oriented:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(IFMINUS(x1, x2, x3))=  x2  
  POL(false)=  0  
  POL(minus(x1, x2))=  x1  
  POL(MINUS(x1, x2))=  x1  
  POL(true)=  0  
  POL(quot(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  
  POL(le(x1, x2))=  0  
  POL(ifMinus(x1, x2, x3))=  x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 5
             ↳Dependency Graph
       →DP Problem 3
         ↳Polo


Dependency Pair:

MINUS(s(X), Y) -> IFMINUS(le(s(X), Y), s(X), Y)


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))





Using the Dependency Graph resulted 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(minus(X, Y), s(Y))


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))





The following dependency pair can be strictly oriented:

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


Additionally, the following rules can be oriented:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(QUOT(x1, x2))=  1 + x1 + x2  
  POL(0)=  0  
  POL(false)=  0  
  POL(minus(x1, x2))=  x1  
  POL(true)=  0  
  POL(quot(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  
  POL(le(x1, x2))=  0  
  POL(ifMinus(x1, x2, x3))=  x2  

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:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))





Using the Dependency Graph resulted in no new DP problems.

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