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

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳Nar


Dependency Pair:

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


Rules:


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


Dependency Pair:


Rules:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Narrowing Transformation


Dependency Pairs:

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


Rules:


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


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

G(x, c(y)) -> G(x, if(f(x), c(g(s(x), y)), c(y)))
five new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

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


Rules:


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


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:01 minutes