Term Rewriting System R:
[y, x]
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))

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(x, s(y)) -> PRED(minus(x, y))
MINUS(x, s(y)) -> MINUS(x, y)
GCD(s(x), s(y)) -> IFGCD(le(y, x), s(x), s(y))
GCD(s(x), s(y)) -> LE(y, x)
IFGCD(true, s(x), s(y)) -> GCD(minus(x, y), s(y))
IFGCD(true, s(x), s(y)) -> MINUS(x, y)
IFGCD(false, s(x), s(y)) -> GCD(minus(y, x), s(x))
IFGCD(false, s(x), s(y)) -> MINUS(y, x)

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)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))





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)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(if_gcd(x1, x2, x3))=  x2 + x3  
  POL(LE(x1, x2))=  1 + x1  
  POL(0)=  0  
  POL(false)=  0  
  POL(pred(x1))=  x1  
  POL(minus(x1, x2))=  x1  
  POL(true)=  0  
  POL(s(x1))=  1 + x1  
  POL(le(x1, x2))=  0  
  POL(gcd(x1, x2))=  x1 + 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)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))





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:

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


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))





The following dependency pair can be strictly oriented:

MINUS(x, s(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)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(if_gcd(x1, x2, x3))=  x2 + x3  
  POL(0)=  0  
  POL(false)=  0  
  POL(pred(x1))=  x1  
  POL(minus(x1, x2))=  x1  
  POL(MINUS(x1, x2))=  1 + x1 + x2  
  POL(true)=  0  
  POL(s(x1))=  1 + x1  
  POL(le(x1, x2))=  0  
  POL(gcd(x1, x2))=  x1 + 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:


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))





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:

IFGCD(false, s(x), s(y)) -> GCD(minus(y, x), s(x))
IFGCD(true, s(x), s(y)) -> GCD(minus(x, y), s(y))
GCD(s(x), s(y)) -> IFGCD(le(y, x), s(x), s(y))


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))





The following dependency pairs can be strictly oriented:

IFGCD(false, s(x), s(y)) -> GCD(minus(y, x), s(x))
IFGCD(true, s(x), s(y)) -> GCD(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)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(if_gcd(x1, x2, x3))=  x2 + x3  
  POL(0)=  0  
  POL(GCD(x1, x2))=  x1 + x2  
  POL(false)=  0  
  POL(pred(x1))=  x1  
  POL(minus(x1, x2))=  x1  
  POL(true)=  0  
  POL(IF_GCD(x1, x2, x3))=  x2 + x3  
  POL(s(x1))=  1 + x1  
  POL(le(x1, x2))=  0  
  POL(gcd(x1, x2))=  x1 + 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:

GCD(s(x), s(y)) -> IFGCD(le(y, x), s(x), s(y))


Rules:


le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> ifgcd(le(y, x), s(x), s(y))
ifgcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
ifgcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))





Using the Dependency Graph resulted in no new DP problems.

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