Term Rewriting System R:
[x, y]
f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, x, y) -> x
if(false, x, y) -> y
g(s(x), s(y)) -> if(f(x), s(x), s(y))
g(x, c(y)) -> g(x, g(s(c(y)), y))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(s(x)) -> F(x)
G(s(x), s(y)) -> IF(f(x), s(x), s(y))
G(s(x), s(y)) -> F(x)
G(x, c(y)) -> G(x, g(s(c(y)), y))
G(x, c(y)) -> G(s(c(y)), y)

Furthermore, R contains two SCCs. 


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


Dependency Pair:

F(s(x)) -> F(x)


Rules:


f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, x, y) -> x
if(false, x, y) -> y
g(s(x), s(y)) -> if(f(x), s(x), s(y))
g(x, c(y)) -> g(x, g(s(c(y)), y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

F(s(x)) -> F(x)


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(s(x1))=  1 + x1  
  POL(F(x1))=  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:


f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, x, y) -> x
if(false, x, y) -> y
g(s(x), s(y)) -> if(f(x), s(x), s(y))
g(x, c(y)) -> g(x, g(s(c(y)), y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pairs:

G(x, c(y)) -> G(s(c(y)), y)
G(x, c(y)) -> G(x, g(s(c(y)), y))


Rules:


f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, x, y) -> x
if(false, x, y) -> y
g(s(x), s(y)) -> if(f(x), s(x), s(y))
g(x, c(y)) -> g(x, g(s(c(y)), y))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

G(x, c(y)) -> G(s(c(y)), y)
G(x, c(y)) -> G(x, g(s(c(y)), y))


Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

g(s(x), s(y)) -> if(f(x), s(x), s(y))
g(x, c(y)) -> g(x, g(s(c(y)), y))
if(true, x, y) -> x
if(false, x, y) -> y


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(if(x1, x2, x3))=  x2 + x3  
  POL(c(x1))=  1 + x1  
  POL(0)=  0  
  POL(g(x1, x2))=  0  
  POL(false)=  0  
  POL(G(x1, x2))=  x2  
  POL(1)=  0  
  POL(true)=  0  
  POL(s(x1))=  0  
  POL(f(x1))=  0  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 4
             ↳Dependency Graph


Dependency Pair:


Rules:


f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, x, y) -> x
if(false, x, y) -> y
g(s(x), s(y)) -> if(f(x), s(x), s(y))
g(x, c(y)) -> g(x, g(s(c(y)), y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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