Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
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.

     ↳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.

     ↳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(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:

     ↳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(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:

     ↳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(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.

     ↳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(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.

     ↳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(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.

     ↳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(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.

     ↳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(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.

     ↳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(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.

     ↳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(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.

     ↳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(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.

     ↳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(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.

     ↳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(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:

     ↳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(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:

     ↳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

                 ↳Remaining Obligation(s)

The following remains to be proven:
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(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:
0:00 minutes

POL(QUOT(x₁, x₂, x₃))	= x₁
POL(0)	= 0
POL(DIV(x₁, x₂))	= x₁
POL(s(x₁))	= 1 + x₁