Term Rewriting System R:
[y, x, u, z]
minus(0, y) -> 0
minus(s(x), 0) -> s(x)
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

MINUS(s(x), s(y)) -> MINUS(x, y)
LE(s(x), s(y)) -> LE(x, y)
PERFECTP(s(x)) -> F(x, s(0), s(x), s(x))
F(s(x), 0, z, u) -> F(x, u, minus(z, s(x)), u)
F(s(x), 0, z, u) -> MINUS(z, s(x))
F(s(x), s(y), z, u) -> IF(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))
F(s(x), s(y), z, u) -> LE(x, y)
F(s(x), s(y), z, u) -> F(s(x), minus(y, x), z, u)
F(s(x), s(y), z, u) -> MINUS(y, x)
F(s(x), s(y), z, u) -> F(x, u, z, u)

Furthermore, R contains three SCCs. 


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


Dependency Pair:

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


Rules:


minus(0, y) -> 0
minus(s(x), 0) -> s(x)
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))





The following dependency pair can be strictly oriented:

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


Additionally, the following rules can be oriented:

minus(0, y) -> 0
minus(s(x), 0) -> s(x)
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(if(x1, x2, x3))=  x2 + x3  
  POL(0)=  0  
  POL(false)=  0  
  POL(perfectp(x1))=  0  
  POL(minus(x1, x2))=  x1  
  POL(MINUS(x1, x2))=  x1  
  POL(true)=  0  
  POL(s(x1))=  1 + x1  
  POL(le(x1, x2))=  0  
  POL(f(x1, x2, x3, x4))=  0  

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:


minus(0, y) -> 0
minus(s(x), 0) -> s(x)
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))





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 Pair:

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


Rules:


minus(0, y) -> 0
minus(s(x), 0) -> s(x)
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))





The following dependency pair can be strictly oriented:

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


Additionally, the following rules can be oriented:

minus(0, y) -> 0
minus(s(x), 0) -> s(x)
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(if(x1, x2, x3))=  x2 + x3  
  POL(LE(x1, x2))=  x1  
  POL(0)=  0  
  POL(false)=  0  
  POL(perfectp(x1))=  0  
  POL(minus(x1, x2))=  x1  
  POL(true)=  0  
  POL(s(x1))=  1 + x1  
  POL(le(x1, x2))=  0  
  POL(f(x1, x2, x3, x4))=  0  

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:


Rules:


minus(0, y) -> 0
minus(s(x), 0) -> s(x)
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))





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 Pairs:

F(s(x), s(y), z, u) -> F(x, u, z, u)
F(s(x), 0, z, u) -> F(x, u, minus(z, s(x)), u)
F(s(x), s(y), z, u) -> F(s(x), minus(y, x), z, u)


Rules:


minus(0, y) -> 0
minus(s(x), 0) -> s(x)
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))





The following dependency pairs can be strictly oriented:

F(s(x), s(y), z, u) -> F(x, u, z, u)
F(s(x), 0, z, u) -> F(x, u, minus(z, s(x)), u)


Additionally, the following rules can be oriented:

minus(0, y) -> 0
minus(s(x), 0) -> s(x)
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(if(x1, x2, x3))=  x2 + x3  
  POL(0)=  0  
  POL(false)=  0  
  POL(perfectp(x1))=  0  
  POL(minus(x1, x2))=  x1  
  POL(true)=  0  
  POL(s(x1))=  1 + x1  
  POL(le(x1, x2))=  0  
  POL(f(x1, x2, x3, x4))=  0  
  POL(F(x1, x2, x3, x4))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
           →DP Problem 6
             ↳Polynomial Ordering


Dependency Pair:

F(s(x), s(y), z, u) -> F(s(x), minus(y, x), z, u)


Rules:


minus(0, y) -> 0
minus(s(x), 0) -> s(x)
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))





The following dependency pair can be strictly oriented:

F(s(x), s(y), z, u) -> F(s(x), minus(y, x), z, u)


Additionally, the following rules can be oriented:

minus(0, y) -> 0
minus(s(x), 0) -> s(x)
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(if(x1, x2, x3))=  x2 + x3  
  POL(0)=  0  
  POL(false)=  0  
  POL(perfectp(x1))=  0  
  POL(minus(x1, x2))=  x1  
  POL(true)=  0  
  POL(s(x1))=  1 + x1  
  POL(le(x1, x2))=  0  
  POL(f(x1, x2, x3, x4))=  0  
  POL(F(x1, x2, x3, x4))=  x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
           →DP Problem 6
             ↳Polo
             ...
               →DP Problem 7
                 ↳Dependency Graph


Dependency Pair:


Rules:


minus(0, y) -> 0
minus(s(x), 0) -> s(x)
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))





Using the Dependency Graph resulted in no new DP problems.

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