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
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳Remaining


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 that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
F(x1) -> F(x1)
s(x1) -> s(x1)


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


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
         ↳AFS
       →DP Problem 2
         ↳Remaining Obligation(s)




The following remains to be proven:
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



Innermost Termination of R could not be shown.
Duration:
0:00 minutes