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

MINUS(X, s(Y)) -> PRED(minus(X, Y))
MINUS(X, s(Y)) -> MINUS(X, Y)
LE(s(X), s(Y)) -> LE(X, Y)
GCD(s(X), s(Y)) -> IF(le(Y, X), s(X), s(Y))
GCD(s(X), s(Y)) -> LE(Y, X)
IF(true, s(X), s(Y)) -> GCD(minus(X, Y), s(Y))
IF(true, s(X), s(Y)) -> MINUS(X, Y)
IF(false, s(X), s(Y)) -> GCD(minus(Y, X), s(X))
IF(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:

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


Rules:


minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(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:

minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) -> gcd(minus(Y, X), s(X))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(if(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 4
             ↳Dependency Graph
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo


Dependency Pair:


Rules:


minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(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:

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


Rules:


minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(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:

minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) -> gcd(minus(Y, X), s(X))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(if(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 2
         ↳Polo
           →DP Problem 5
             ↳Dependency Graph
       →DP Problem 3
         ↳Polo


Dependency Pair:


Rules:


minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(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:

IF(false, s(X), s(Y)) -> GCD(minus(Y, X), s(X))
IF(true, s(X), s(Y)) -> GCD(minus(X, Y), s(Y))
GCD(s(X), s(Y)) -> IF(le(Y, X), s(X), s(Y))


Rules:


minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) -> gcd(minus(Y, X), s(X))





The following dependency pairs can be strictly oriented:

IF(false, s(X), s(Y)) -> GCD(minus(Y, X), s(X))
IF(true, s(X), s(Y)) -> GCD(minus(X, Y), s(Y))


Additionally, the following rules can be oriented:

minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) -> gcd(minus(Y, X), s(X))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(if(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(s(x1))=  1 + x1  
  POL(le(x1, x2))=  0  
  POL(IF(x1, x2, x3))=  x2 + x3  
  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)) -> IF(le(Y, X), s(X), s(Y))


Rules:


minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(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