Term Rewriting System R:
[x, y]
p(0) -> 0
p(s(x)) -> x
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, 0) -> x
minus(x, s(y)) -> if(le(x, s(y)), 0, p(minus(x, p(s(y)))))
if(true, x, y) -> x
if(false, x, y) -> y

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

LE(s(x), s(y)) -> LE(x, y)
MINUS(x, s(y)) -> IF(le(x, s(y)), 0, p(minus(x, p(s(y)))))
MINUS(x, s(y)) -> LE(x, s(y))
MINUS(x, s(y)) -> P(minus(x, p(s(y))))
MINUS(x, s(y)) -> MINUS(x, p(s(y)))
MINUS(x, s(y)) -> P(s(y))

Furthermore, R contains two SCCs. 


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


Dependency Pair:

LE(s(x), s(y)) -> LE(x, y)


Rules:


p(0) -> 0
p(s(x)) -> x
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, 0) -> x
minus(x, s(y)) -> if(le(x, s(y)), 0, p(minus(x, p(s(y)))))
if(true, x, y) -> x
if(false, x, y) -> y





The following dependency pair can be strictly oriented:

LE(s(x), s(y)) -> LE(x, y)


There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
LE(x1, x2) -> LE(x1, x2)
s(x1) -> s(x1)


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


Dependency Pair:


Rules:


p(0) -> 0
p(s(x)) -> x
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, 0) -> x
minus(x, s(y)) -> if(le(x, s(y)), 0, p(minus(x, p(s(y)))))
if(true, x, y) -> x
if(false, x, y) -> y





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 Pair:

MINUS(x, s(y)) -> MINUS(x, p(s(y)))


Rules:


p(0) -> 0
p(s(x)) -> x
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, 0) -> x
minus(x, s(y)) -> if(le(x, s(y)), 0, p(minus(x, p(s(y)))))
if(true, x, y) -> x
if(false, x, y) -> y




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