Term Rewriting System R:
[x, y]
prime(0) -> false
prime(s(0)) -> false
prime(s(s(x))) -> prime1(s(s(x)), s(x))
prime1(x, 0) -> false
prime1(x, s(0)) -> true
prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
divp(x, y) -> =(rem(x, y), 0)

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

PRIME(s(s(x))) -> PRIME1(s(s(x)), s(x))
PRIME1(x, s(s(y))) -> DIVP(s(s(y)), x)
PRIME1(x, s(s(y))) -> PRIME1(x, s(y))

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Forward Instantiation Transformation


Dependency Pair:

PRIME1(x, s(s(y))) -> PRIME1(x, s(y))


Rules:


prime(0) -> false
prime(s(0)) -> false
prime(s(s(x))) -> prime1(s(s(x)), s(x))
prime1(x, 0) -> false
prime1(x, s(0)) -> true
prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
divp(x, y) -> =(rem(x, y), 0)


Strategy:

innermost




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

PRIME1(x, s(s(y))) -> PRIME1(x, s(y))
one new Dependency Pair is created:

PRIME1(x'', s(s(s(y'')))) -> PRIME1(x'', s(s(y'')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳Forward Instantiation Transformation


Dependency Pair:

PRIME1(x'', s(s(s(y'')))) -> PRIME1(x'', s(s(y'')))


Rules:


prime(0) -> false
prime(s(0)) -> false
prime(s(s(x))) -> prime1(s(s(x)), s(x))
prime1(x, 0) -> false
prime1(x, s(0)) -> true
prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
divp(x, y) -> =(rem(x, y), 0)


Strategy:

innermost




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

PRIME1(x'', s(s(s(y'')))) -> PRIME1(x'', s(s(y'')))
one new Dependency Pair is created:

PRIME1(x'''', s(s(s(s(y''''))))) -> PRIME1(x'''', s(s(s(y''''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳FwdInst
             ...
               →DP Problem 3
                 ↳Polynomial Ordering


Dependency Pair:

PRIME1(x'''', s(s(s(s(y''''))))) -> PRIME1(x'''', s(s(s(y''''))))


Rules:


prime(0) -> false
prime(s(0)) -> false
prime(s(s(x))) -> prime1(s(s(x)), s(x))
prime1(x, 0) -> false
prime1(x, s(0)) -> true
prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
divp(x, y) -> =(rem(x, y), 0)


Strategy:

innermost




The following dependency pair can be strictly oriented:

PRIME1(x'''', s(s(s(s(y''''))))) -> PRIME1(x'''', s(s(s(y''''))))


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(PRIME1(x1, x2))=  1 + x1 + x2  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳FwdInst
             ...
               →DP Problem 4
                 ↳Dependency Graph


Dependency Pair:


Rules:


prime(0) -> false
prime(s(0)) -> false
prime(s(s(x))) -> prime1(s(s(x)), s(x))
prime1(x, 0) -> false
prime1(x, s(0)) -> true
prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
divp(x, y) -> =(rem(x, y), 0)


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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