Term Rewriting System R:
[Y, X, X1, X2]
minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
minus(X1, X2) -> nminus(X1, X2)
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(ndiv(nminus(X, activate(Y)), ns(activate(Y)))), n0)
div(X1, X2) -> ndiv(X1, X2)
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(activate(X))
activate(ndiv(X1, X2)) -> div(activate(X1), X2)
activate(nminus(X1, X2)) -> minus(X1, X2)
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(ndiv(nminus(X, activate(Y)), ns(activate(Y)))), n0)
DIV(s(X), ns(Y)) -> GEQ(X, activate(Y))
DIV(s(X), ns(Y)) -> ACTIVATE(Y)
IF(true, X, Y) -> ACTIVATE(X)
IF(false, X, Y) -> ACTIVATE(Y)
ACTIVATE(n0) -> 0'
ACTIVATE(ns(X)) -> S(activate(X))
ACTIVATE(ns(X)) -> ACTIVATE(X)
ACTIVATE(ndiv(X1, X2)) -> DIV(activate(X1), X2)
ACTIVATE(ndiv(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(nminus(X1, X2)) -> MINUS(X1, X2)

Furthermore, R contains two SCCs. 


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


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))
minus(X1, X2) -> nminus(X1, X2)
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(ndiv(nminus(X, activate(Y)), ns(activate(Y)))), n0)
div(X1, X2) -> ndiv(X1, X2)
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(activate(X))
activate(ndiv(X1, X2)) -> div(activate(X1), X2)
activate(nminus(X1, X2)) -> minus(X1, X2)
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 w.r.t. to the implicit AFS can be oriented:

activate(n0) -> 0
activate(ns(X)) -> s(activate(X))
activate(ndiv(X1, X2)) -> div(activate(X1), X2)
activate(nminus(X1, X2)) -> minus(X1, X2)
activate(X) -> X
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
minus(X1, X2) -> nminus(X1, X2)
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(ndiv(nminus(X, activate(Y)), ns(activate(Y)))), n0)
div(X1, X2) -> ndiv(X1, X2)
0 -> n0
s(X) -> ns(X)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(GEQ(x1, x2))=  1 + x1  
  POL(activate(x1))=  x1  
  POL(geq(x1, x2))=  0  
  POL(false)=  0  
  POL(minus(x1, x2))=  0  
  POL(n__s(x1))=  1 + x1  
  POL(n__minus(x1, x2))=  0  
  POL(true)=  0  
  POL(if(x1, x2, x3))=  x2 + x3  
  POL(0)=  0  
  POL(n__div(x1, x2))=  x1  
  POL(n__0)=  0  
  POL(s(x1))=  1 + x1  
  POL(div(x1, x2))=  x1  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
minus(X1, X2) -> nminus(X1, X2)
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(ndiv(nminus(X, activate(Y)), ns(activate(Y)))), n0)
div(X1, X2) -> ndiv(X1, X2)
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(activate(X))
activate(ndiv(X1, X2)) -> div(activate(X1), X2)
activate(nminus(X1, X2)) -> minus(X1, X2)
activate(X) -> X


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

MINUS(ns(X), ns(Y)) -> ACTIVATE(Y)
ACTIVATE(nminus(X1, X2)) -> MINUS(X1, X2)
ACTIVATE(ndiv(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(ns(X)) -> ACTIVATE(X)
MINUS(ns(X), ns(Y)) -> ACTIVATE(X)
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))
minus(X1, X2) -> nminus(X1, X2)
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(ndiv(nminus(X, activate(Y)), ns(activate(Y)))), n0)
div(X1, X2) -> ndiv(X1, X2)
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(activate(X))
activate(ndiv(X1, X2)) -> div(activate(X1), X2)
activate(nminus(X1, X2)) -> minus(X1, X2)
activate(X) -> X


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:02 minutes