Term Rewriting System R:
[Y, X]
minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

MINUS(n0, Y) -> 0'
MINUS(ns(X), ns(Y)) -> MINUS(activate(X), activate(Y))
MINUS(ns(X), ns(Y)) -> ACTIVATE(X)
MINUS(ns(X), ns(Y)) -> ACTIVATE(Y)
GEQ(ns(X), ns(Y)) -> GEQ(activate(X), activate(Y))
GEQ(ns(X), ns(Y)) -> ACTIVATE(X)
GEQ(ns(X), ns(Y)) -> ACTIVATE(Y)
DIV(s(X), ns(Y)) -> IF(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
DIV(s(X), ns(Y)) -> GEQ(X, activate(Y))
DIV(s(X), ns(Y)) -> ACTIVATE(Y)
DIV(s(X), ns(Y)) -> DIV(minus(X, activate(Y)), ns(activate(Y)))
DIV(s(X), ns(Y)) -> MINUS(X, activate(Y))
IF(true, X, Y) -> ACTIVATE(X)
IF(false, X, Y) -> ACTIVATE(Y)
ACTIVATE(n0) -> 0'
ACTIVATE(ns(X)) -> S(X)

Furthermore, R contains two SCCs. 


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


Dependency Pair:

MINUS(ns(X), ns(Y)) -> MINUS(activate(X), activate(Y))


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




The following dependency pair can be strictly oriented:

MINUS(ns(X), ns(Y)) -> MINUS(activate(X), activate(Y))


Additionally, the following usable rules for innermost can be oriented:

activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X
0 -> n0
s(X) -> ns(X)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(activate(x1))=  x1  
  POL(0)=  0  
  POL(MINUS(x1, x2))=  x1  
  POL(n__s(x1))=  1 + x1  
  POL(n__0)=  0  
  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:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

GEQ(ns(X), ns(Y)) -> GEQ(activate(X), activate(Y))


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




The following dependency pair can be strictly oriented:

GEQ(ns(X), ns(Y)) -> GEQ(activate(X), activate(Y))


Additionally, the following usable rules for innermost can be oriented:

activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X
0 -> n0
s(X) -> ns(X)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(GEQ(x1, x2))=  x1  
  POL(activate(x1))=  x1  
  POL(0)=  0  
  POL(n__s(x1))=  1 + x1  
  POL(n__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
             ↳Dependency Graph


Dependency Pair:


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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