Term Rewriting System R:
[x, y, z]
p(0) -> 0
p(s(x)) -> x
plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> s(plus(p(s(x)), y))
plus(x, s(y)) -> s(plus(x, p(s(y))))
times(0, y) -> 0
times(s(0), y) -> y
times(s(x), y) -> plus(y, times(x, y))
div(0, y) -> 0
div(x, y) -> quot(x, y, y)
div(div(x, y), z) -> div(x, times(y, z))
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(z)))
eq(0, 0) -> true
eq(s(x), 0) -> false
eq(0, s(y)) -> false
eq(s(x), s(y)) -> eq(x, y)
divides(y, x) -> eq(x, times(div(x, y), y))
prime(s(s(x))) -> pr(s(s(x)), s(x))
pr(x, s(0)) -> true
pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
if(true, x, y) -> false
if(false, x, y) -> pr(x, y)

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

PLUS(s(x), y) -> PLUS(x, y)
PLUS(s(x), y) -> PLUS(p(s(x)), y)
PLUS(s(x), y) -> P(s(x))
PLUS(x, s(y)) -> PLUS(x, p(s(y)))
PLUS(x, s(y)) -> P(s(y))
TIMES(s(x), y) -> PLUS(y, times(x, y))
TIMES(s(x), y) -> TIMES(x, y)
DIV(x, y) -> QUOT(x, y, y)
DIV(div(x, y), z) -> DIV(x, times(y, z))
DIV(div(x, y), z) -> TIMES(y, z)
QUOT(s(x), s(y), z) -> QUOT(x, y, z)
QUOT(x, 0, s(z)) -> DIV(x, s(z))
EQ(s(x), s(y)) -> EQ(x, y)
DIVIDES(y, x) -> EQ(x, times(div(x, y), y))
DIVIDES(y, x) -> TIMES(div(x, y), y)
DIVIDES(y, x) -> DIV(x, y)
PRIME(s(s(x))) -> PR(s(s(x)), s(x))
PR(x, s(s(y))) -> IF(divides(s(s(y)), x), x, s(y))
PR(x, s(s(y))) -> DIVIDES(s(s(y)), x)
IF(false, x, y) -> PR(x, y)

Furthermore, R contains five SCCs.


   R
DPs
       →DP Problem 1
Rewriting Transformation
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Nar


Dependency Pairs:

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


Rules:


p(0) -> 0
p(s(x)) -> x
plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> s(plus(p(s(x)), y))
plus(x, s(y)) -> s(plus(x, p(s(y))))
times(0, y) -> 0
times(s(0), y) -> y
times(s(x), y) -> plus(y, times(x, y))
div(0, y) -> 0
div(x, y) -> quot(x, y, y)
div(div(x, y), z) -> div(x, times(y, z))
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(z)))
eq(0, 0) -> true
eq(s(x), 0) -> false
eq(0, s(y)) -> false
eq(s(x), s(y)) -> eq(x, y)
divides(y, x) -> eq(x, times(div(x, y), y))
prime(s(s(x))) -> pr(s(s(x)), s(x))
pr(x, s(0)) -> true
pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
if(true, x, y) -> false
if(false, x, y) -> pr(x, y)


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Rw
           →DP Problem 6
Rewriting Transformation
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Nar


Dependency Pairs:

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


Rules:


p(0) -> 0
p(s(x)) -> x
plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> s(plus(p(s(x)), y))
plus(x, s(y)) -> s(plus(x, p(s(y))))
times(0, y) -> 0
times(s(0), y) -> y
times(s(x), y) -> plus(y, times(x, y))
div(0, y) -> 0
div(x, y) -> quot(x, y, y)
div(div(x, y), z) -> div(x, times(y, z))
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(z)))
eq(0, 0) -> true
eq(s(x), 0) -> false
eq(0, s(y)) -> false
eq(s(x), s(y)) -> eq(x, y)
divides(y, x) -> eq(x, times(div(x, y), y))
prime(s(s(x))) -> pr(s(s(x)), s(x))
pr(x, s(0)) -> true
pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
if(true, x, y) -> false
if(false, x, y) -> pr(x, y)


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Rw
           →DP Problem 6
Rw
             ...
               →DP Problem 7
Polynomial Ordering
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Nar


Dependency Pairs:

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


Rules:


p(0) -> 0
p(s(x)) -> x
plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> s(plus(p(s(x)), y))
plus(x, s(y)) -> s(plus(x, p(s(y))))
times(0, y) -> 0
times(s(0), y) -> y
times(s(x), y) -> plus(y, times(x, y))
div(0, y) -> 0
div(x, y) -> quot(x, y, y)
div(div(x, y), z) -> div(x, times(y, z))
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(z)))
eq(0, 0) -> true
eq(s(x), 0) -> false
eq(0, s(y)) -> false
eq(s(x), s(y)) -> eq(x, y)
divides(y, x) -> eq(x, times(div(x, y), y))
prime(s(s(x))) -> pr(s(s(x)), s(x))
pr(x, s(0)) -> true
pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
if(true, x, y) -> false
if(false, x, y) -> pr(x, y)


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


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

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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Rw
           →DP Problem 6
Rw
             ...
               →DP Problem 8
Polynomial Ordering
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Nar


Dependency Pair:

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


Rules:


p(0) -> 0
p(s(x)) -> x
plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> s(plus(p(s(x)), y))
plus(x, s(y)) -> s(plus(x, p(s(y))))
times(0, y) -> 0
times(s(0), y) -> y
times(s(x), y) -> plus(y, times(x, y))
div(0, y) -> 0
div(x, y) -> quot(x, y, y)
div(div(x, y), z) -> div(x, times(y, z))
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(z)))
eq(0, 0) -> true
eq(s(x), 0) -> false
eq(0, s(y)) -> false
eq(s(x), s(y)) -> eq(x, y)
divides(y, x) -> eq(x, times(div(x, y), y))
prime(s(s(x))) -> pr(s(s(x)), s(x))
pr(x, s(0)) -> true
pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
if(true, x, y) -> false
if(false, x, y) -> pr(x, y)


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


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

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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Rw
           →DP Problem 6
Rw
             ...
               →DP Problem 9
Dependency Graph
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Nar


Dependency Pair:


Rules:


p(0) -> 0
p(s(x)) -> x
plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> s(plus(p(s(x)), y))
plus(x, s(y)) -> s(plus(x, p(s(y))))
times(0, y) -> 0
times(s(0), y) -> y
times(s(x), y) -> plus(y, times(x, y))
div(0, y) -> 0
div(x, y) -> quot(x, y, y)
div(div(x, y), z) -> div(x, times(y, z))
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(z)))
eq(0, 0) -> true
eq(s(x), 0) -> false
eq(0, s(y)) -> false
eq(s(x), s(y)) -> eq(x, y)
divides(y, x) -> eq(x, times(div(x, y), y))
prime(s(s(x))) -> pr(s(s(x)), s(x))
pr(x, s(0)) -> true
pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
if(true, x, y) -> false
if(false, x, y) -> pr(x, y)


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Rw
       →DP Problem 2
Polynomial Ordering
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Nar


Dependency Pair:

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


Rules:


p(0) -> 0
p(s(x)) -> x
plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> s(plus(p(s(x)), y))
plus(x, s(y)) -> s(plus(x, p(s(y))))
times(0, y) -> 0
times(s(0), y) -> y
times(s(x), y) -> plus(y, times(x, y))
div(0, y) -> 0
div(x, y) -> quot(x, y, y)
div(div(x, y), z) -> div(x, times(y, z))
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(z)))
eq(0, 0) -> true
eq(s(x), 0) -> false
eq(0, s(y)) -> false
eq(s(x), s(y)) -> eq(x, y)
divides(y, x) -> eq(x, times(div(x, y), y))
prime(s(s(x))) -> pr(s(s(x)), s(x))
pr(x, s(0)) -> true
pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
if(true, x, y) -> false
if(false, x, y) -> pr(x, y)


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


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

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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Rw
       →DP Problem 2
Polo
           →DP Problem 10
Dependency Graph
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Nar


Dependency Pair:


Rules:


p(0) -> 0
p(s(x)) -> x
plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> s(plus(p(s(x)), y))
plus(x, s(y)) -> s(plus(x, p(s(y))))
times(0, y) -> 0
times(s(0), y) -> y
times(s(x), y) -> plus(y, times(x, y))
div(0, y) -> 0
div(x, y) -> quot(x, y, y)
div(div(x, y), z) -> div(x, times(y, z))
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(z)))
eq(0, 0) -> true
eq(s(x), 0) -> false
eq(0, s(y)) -> false
eq(s(x), s(y)) -> eq(x, y)
divides(y, x) -> eq(x, times(div(x, y), y))
prime(s(s(x))) -> pr(s(s(x)), s(x))
pr(x, s(0)) -> true
pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
if(true, x, y) -> false
if(false, x, y) -> pr(x, y)


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Rw
       →DP Problem 2
Polo
       →DP Problem 3
Polynomial Ordering
       →DP Problem 4
Polo
       →DP Problem 5
Nar


Dependency Pair:

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


Rules:


p(0) -> 0
p(s(x)) -> x
plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> s(plus(p(s(x)), y))
plus(x, s(y)) -> s(plus(x, p(s(y))))
times(0, y) -> 0
times(s(0), y) -> y
times(s(x), y) -> plus(y, times(x, y))
div(0, y) -> 0
div(x, y) -> quot(x, y, y)
div(div(x, y), z) -> div(x, times(y, z))
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(z)))
eq(0, 0) -> true
eq(s(x), 0) -> false
eq(0, s(y)) -> false
eq(s(x), s(y)) -> eq(x, y)
divides(y, x) -> eq(x, times(div(x, y), y))
prime(s(s(x))) -> pr(s(s(x)), s(x))
pr(x, s(0)) -> true
pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
if(true, x, y) -> false
if(false, x, y) -> pr(x, y)


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


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

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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Rw
       →DP Problem 2
Polo
       →DP Problem 3
Polo
           →DP Problem 11
Dependency Graph
       →DP Problem 4
Polo
       →DP Problem 5
Nar


Dependency Pair:


Rules:


p(0) -> 0
p(s(x)) -> x
plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> s(plus(p(s(x)), y))
plus(x, s(y)) -> s(plus(x, p(s(y))))
times(0, y) -> 0
times(s(0), y) -> y
times(s(x), y) -> plus(y, times(x, y))
div(0, y) -> 0
div(x, y) -> quot(x, y, y)
div(div(x, y), z) -> div(x, times(y, z))
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(z)))
eq(0, 0) -> true
eq(s(x), 0) -> false
eq(0, s(y)) -> false
eq(s(x), s(y)) -> eq(x, y)
divides(y, x) -> eq(x, times(div(x, y), y))
prime(s(s(x))) -> pr(s(s(x)), s(x))
pr(x, s(0)) -> true
pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
if(true, x, y) -> false
if(false, x, y) -> pr(x, y)


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Rw
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polynomial Ordering
       →DP Problem 5
Nar


Dependency Pairs:

QUOT(x, 0, s(z)) -> DIV(x, s(z))
QUOT(s(x), s(y), z) -> QUOT(x, y, z)
DIV(x, y) -> QUOT(x, y, y)


Rules:


p(0) -> 0
p(s(x)) -> x
plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> s(plus(p(s(x)), y))
plus(x, s(y)) -> s(plus(x, p(s(y))))
times(0, y) -> 0
times(s(0), y) -> y
times(s(x), y) -> plus(y, times(x, y))
div(0, y) -> 0
div(x, y) -> quot(x, y, y)
div(div(x, y), z) -> div(x, times(y, z))
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(z)))
eq(0, 0) -> true
eq(s(x), 0) -> false
eq(0, s(y)) -> false
eq(s(x), s(y)) -> eq(x, y)
divides(y, x) -> eq(x, times(div(x, y), y))
prime(s(s(x))) -> pr(s(s(x)), s(x))
pr(x, s(0)) -> true
pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
if(true, x, y) -> false
if(false, x, y) -> pr(x, y)


Strategy:

innermost




The following dependency pair can be strictly oriented:

QUOT(s(x), s(y), z) -> QUOT(x, y, z)


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(QUOT(x1, x2, x3))=  x1  
  POL(0)=  0  
  POL(DIV(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Rw
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
           →DP Problem 12
Dependency Graph
       →DP Problem 5
Nar


Dependency Pairs:

QUOT(x, 0, s(z)) -> DIV(x, s(z))
DIV(x, y) -> QUOT(x, y, y)


Rules:


p(0) -> 0
p(s(x)) -> x
plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> s(plus(p(s(x)), y))
plus(x, s(y)) -> s(plus(x, p(s(y))))
times(0, y) -> 0
times(s(0), y) -> y
times(s(x), y) -> plus(y, times(x, y))
div(0, y) -> 0
div(x, y) -> quot(x, y, y)
div(div(x, y), z) -> div(x, times(y, z))
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(z)))
eq(0, 0) -> true
eq(s(x), 0) -> false
eq(0, s(y)) -> false
eq(s(x), s(y)) -> eq(x, y)
divides(y, x) -> eq(x, times(div(x, y), y))
prime(s(s(x))) -> pr(s(s(x)), s(x))
pr(x, s(0)) -> true
pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
if(true, x, y) -> false
if(false, x, y) -> pr(x, y)


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Rw
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Narrowing Transformation


Dependency Pairs:

IF(false, x, y) -> PR(x, y)
PR(x, s(s(y))) -> IF(divides(s(s(y)), x), x, s(y))


Rules:


p(0) -> 0
p(s(x)) -> x
plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> s(plus(p(s(x)), y))
plus(x, s(y)) -> s(plus(x, p(s(y))))
times(0, y) -> 0
times(s(0), y) -> y
times(s(x), y) -> plus(y, times(x, y))
div(0, y) -> 0
div(x, y) -> quot(x, y, y)
div(div(x, y), z) -> div(x, times(y, z))
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(z)))
eq(0, 0) -> true
eq(s(x), 0) -> false
eq(0, s(y)) -> false
eq(s(x), s(y)) -> eq(x, y)
divides(y, x) -> eq(x, times(div(x, y), y))
prime(s(s(x))) -> pr(s(s(x)), s(x))
pr(x, s(0)) -> true
pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
if(true, x, y) -> false
if(false, x, y) -> pr(x, y)


Strategy:

innermost




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

PR(x, s(s(y))) -> IF(divides(s(s(y)), x), x, s(y))
one new Dependency Pair is created:

PR(x'', s(s(y''))) -> IF(eq(x'', times(div(x'', s(s(y''))), s(s(y'')))), x'', s(y''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Rw
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Nar
           →DP Problem 13
Instantiation Transformation


Dependency Pairs:

PR(x'', s(s(y''))) -> IF(eq(x'', times(div(x'', s(s(y''))), s(s(y'')))), x'', s(y''))
IF(false, x, y) -> PR(x, y)


Rules:


p(0) -> 0
p(s(x)) -> x
plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> s(plus(p(s(x)), y))
plus(x, s(y)) -> s(plus(x, p(s(y))))
times(0, y) -> 0
times(s(0), y) -> y
times(s(x), y) -> plus(y, times(x, y))
div(0, y) -> 0
div(x, y) -> quot(x, y, y)
div(div(x, y), z) -> div(x, times(y, z))
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(z)))
eq(0, 0) -> true
eq(s(x), 0) -> false
eq(0, s(y)) -> false
eq(s(x), s(y)) -> eq(x, y)
divides(y, x) -> eq(x, times(div(x, y), y))
prime(s(s(x))) -> pr(s(s(x)), s(x))
pr(x, s(0)) -> true
pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
if(true, x, y) -> false
if(false, x, y) -> pr(x, y)


Strategy:

innermost




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

IF(false, x, y) -> PR(x, y)
one new Dependency Pair is created:

IF(false, x', s(y'''')) -> PR(x', s(y''''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Rw
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Nar
           →DP Problem 13
Inst
             ...
               →DP Problem 14
Forward Instantiation Transformation


Dependency Pairs:

IF(false, x', s(y'''')) -> PR(x', s(y''''))
PR(x'', s(s(y''))) -> IF(eq(x'', times(div(x'', s(s(y''))), s(s(y'')))), x'', s(y''))


Rules:


p(0) -> 0
p(s(x)) -> x
plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> s(plus(p(s(x)), y))
plus(x, s(y)) -> s(plus(x, p(s(y))))
times(0, y) -> 0
times(s(0), y) -> y
times(s(x), y) -> plus(y, times(x, y))
div(0, y) -> 0
div(x, y) -> quot(x, y, y)
div(div(x, y), z) -> div(x, times(y, z))
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(z)))
eq(0, 0) -> true
eq(s(x), 0) -> false
eq(0, s(y)) -> false
eq(s(x), s(y)) -> eq(x, y)
divides(y, x) -> eq(x, times(div(x, y), y))
prime(s(s(x))) -> pr(s(s(x)), s(x))
pr(x, s(0)) -> true
pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
if(true, x, y) -> false
if(false, x, y) -> pr(x, y)


Strategy:

innermost




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

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

IF(false, x'', s(s(y'''''))) -> PR(x'', s(s(y''''')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Rw
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Nar
           →DP Problem 13
Inst
             ...
               →DP Problem 15
Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

IF(false, x'', s(s(y'''''))) -> PR(x'', s(s(y''''')))
PR(x'', s(s(y''))) -> IF(eq(x'', times(div(x'', s(s(y''))), s(s(y'')))), x'', s(y''))


Rules:


p(0) -> 0
p(s(x)) -> x
plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(s(x), y) -> s(plus(p(s(x)), y))
plus(x, s(y)) -> s(plus(x, p(s(y))))
times(0, y) -> 0
times(s(0), y) -> y
times(s(x), y) -> plus(y, times(x, y))
div(0, y) -> 0
div(x, y) -> quot(x, y, y)
div(div(x, y), z) -> div(x, times(y, z))
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(z)))
eq(0, 0) -> true
eq(s(x), 0) -> false
eq(0, s(y)) -> false
eq(s(x), s(y)) -> eq(x, y)
divides(y, x) -> eq(x, times(div(x, y), y))
prime(s(s(x))) -> pr(s(s(x)), s(x))
pr(x, s(0)) -> true
pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
if(true, x, y) -> false
if(false, x, y) -> pr(x, y)


Strategy:

innermost



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