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
         ↳AFS


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: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

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
         ↳AFS


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
         ↳Argument Filtering and Ordering


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




The following dependency pairs can be strictly oriented:

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)


The following usable rules for innermost can be oriented:

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


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
g > c > {f, true, false} > s
g > c > G

resulting in one new DP problem.
Used Argument Filtering System:
G(x1, x2) -> G(x1, x2)
c(x1) -> c(x1)
if(x1, x2, x3) -> x1
f(x1) -> f
s(x1) -> s
g(x1, x2) -> g(x1, x2)


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


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.

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