Term Rewriting System R:
[x, y, z]
plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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)
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`
`         ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Inst`
`       →DP Problem 5`
`         ↳Nar`

Dependency Pair:

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

Rules:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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

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

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 6`
`             ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Inst`
`       →DP Problem 5`
`         ↳Nar`

Dependency Pair:

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

Rules:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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

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

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 6`
`             ↳FwdInst`
`             ...`
`               →DP Problem 7`
`                 ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Inst`
`       →DP Problem 5`
`         ↳Nar`

Dependency Pair:

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

Rules:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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(s(s(x''''))), y'''') -> PLUS(s(s(x'''')), y'''')

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

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

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 6`
`             ↳FwdInst`
`             ...`
`               →DP Problem 8`
`                 ↳Dependency Graph`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Inst`
`       →DP Problem 5`
`         ↳Nar`

Dependency Pair:

Rules:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Forward Instantiation Transformation`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Inst`
`       →DP Problem 5`
`         ↳Nar`

Dependency Pair:

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

Rules:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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

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

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 9`
`             ↳Forward Instantiation Transformation`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Inst`
`       →DP Problem 5`
`         ↳Nar`

Dependency Pair:

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

Rules:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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

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

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 9`
`             ↳FwdInst`
`             ...`
`               →DP Problem 10`
`                 ↳Polynomial Ordering`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Inst`
`       →DP Problem 5`
`         ↳Nar`

Dependency Pair:

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

Rules:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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(s(s(x''''))), s(s(s(y'''')))) -> EQ(s(s(x'''')), s(s(y'''')))

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

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

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 9`
`             ↳FwdInst`
`             ...`
`               →DP Problem 11`
`                 ↳Dependency Graph`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Inst`
`       →DP Problem 5`
`         ↳Nar`

Dependency Pair:

Rules:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳Forward Instantiation Transformation`
`       →DP Problem 4`
`         ↳Inst`
`       →DP Problem 5`
`         ↳Nar`

Dependency Pair:

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

Rules:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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

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

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`           →DP Problem 12`
`             ↳Forward Instantiation Transformation`
`       →DP Problem 4`
`         ↳Inst`
`       →DP Problem 5`
`         ↳Nar`

Dependency Pair:

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

Rules:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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

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

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`           →DP Problem 12`
`             ↳FwdInst`
`             ...`
`               →DP Problem 13`
`                 ↳Polynomial Ordering`
`       →DP Problem 4`
`         ↳Inst`
`       →DP Problem 5`
`         ↳Nar`

Dependency Pair:

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

Rules:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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(s(s(x''''))), y'''') -> TIMES(s(s(x'''')), y'''')

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

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

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`           →DP Problem 12`
`             ↳FwdInst`
`             ...`
`               →DP Problem 14`
`                 ↳Dependency Graph`
`       →DP Problem 4`
`         ↳Inst`
`       →DP Problem 5`
`         ↳Nar`

Dependency Pair:

Rules:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Instantiation Transformation`
`       →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:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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

DIV(x, y) -> QUOT(x, y, y)
one new Dependency Pair is created:

DIV(x'', s(z'')) -> QUOT(x'', s(z''), s(z''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Inst`
`           →DP Problem 15`
`             ↳Forward Instantiation Transformation`
`       →DP Problem 5`
`         ↳Nar`

Dependency Pairs:

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

Rules:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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

QUOT(s(x), s(y), z) -> QUOT(x, y, z)
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Inst`
`           →DP Problem 15`
`             ↳FwdInst`
`             ...`
`               →DP Problem 16`
`                 ↳Forward Instantiation Transformation`
`       →DP Problem 5`
`         ↳Nar`

Dependency Pairs:

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

Rules:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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

DIV(x'', s(z'')) -> QUOT(x'', s(z''), s(z''))
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Inst`
`           →DP Problem 15`
`             ↳FwdInst`
`             ...`
`               →DP Problem 17`
`                 ↳Forward Instantiation Transformation`
`       →DP Problem 5`
`         ↳Nar`

Dependency Pairs:

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

Rules:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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

QUOT(x, 0, s(z)) -> DIV(x, s(z))
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Inst`
`           →DP Problem 15`
`             ↳FwdInst`
`             ...`
`               →DP Problem 18`
`                 ↳Forward Instantiation Transformation`
`       →DP Problem 5`
`         ↳Nar`

Dependency Pairs:

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

Rules:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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

QUOT(s(s(x'')), s(s(y'')), z'') -> QUOT(s(x''), s(y''), z'')
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Inst`
`           →DP Problem 15`
`             ↳FwdInst`
`             ...`
`               →DP Problem 19`
`                 ↳Forward Instantiation Transformation`
`       →DP Problem 5`
`         ↳Nar`

Dependency Pairs:

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

Rules:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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

QUOT(s(x''), s(0), s(z'')) -> QUOT(x'', 0, s(z''))
two new Dependency Pairs are created:

QUOT(s(s(s(x''''''''))), s(0), s(s(y''''''''))) -> QUOT(s(s(x'''''''')), 0, s(s(y'''''''')))
QUOT(s(s(x'''''''')), s(0), s(0)) -> QUOT(s(x''''''''), 0, s(0))

The transformation is resulting in two new DP problems:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Inst`
`           →DP Problem 15`
`             ↳FwdInst`
`             ...`
`               →DP Problem 20`
`                 ↳Polynomial Ordering`
`       →DP Problem 5`
`         ↳Nar`

Dependency Pairs:

DIV(s(x''''), s(0)) -> QUOT(s(x''''), s(0), s(0))
QUOT(s(x''''''), 0, s(0)) -> DIV(s(x''''''), s(0))
QUOT(s(s(x'''''''')), s(0), s(0)) -> QUOT(s(x''''''''), 0, s(0))

Rules:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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(s(x'''''''')), s(0), s(0)) -> QUOT(s(x''''''''), 0, s(0))

There are no usable rules for innermost w.r.t. to the implicit AFS 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`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Inst`
`           →DP Problem 15`
`             ↳FwdInst`
`             ...`
`               →DP Problem 22`
`                 ↳Dependency Graph`
`       →DP Problem 5`
`         ↳Nar`

Dependency Pairs:

DIV(s(x''''), s(0)) -> QUOT(s(x''''), s(0), s(0))
QUOT(s(x''''''), 0, s(0)) -> DIV(s(x''''''), s(0))

Rules:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Inst`
`           →DP Problem 15`
`             ↳FwdInst`
`             ...`
`               →DP Problem 21`
`                 ↳Polynomial Ordering`
`       →DP Problem 5`
`         ↳Nar`

Dependency Pairs:

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

Rules:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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 pairs can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the implicit AFS 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`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Inst`
`           →DP Problem 15`
`             ↳FwdInst`
`             ...`
`               →DP Problem 23`
`                 ↳Dependency Graph`
`       →DP Problem 5`
`         ↳Nar`

Dependency Pairs:

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

Rules:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Inst`
`       →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:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Inst`
`       →DP Problem 5`
`         ↳Nar`
`           →DP Problem 24`
`             ↳Narrowing 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:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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(eq(x'', times(div(x'', s(s(y''))), s(s(y'')))), x'', s(y''))
two new Dependency Pairs are created:

PR(0, s(s(y'''))) -> IF(eq(0, times(0, s(s(y''')))), 0, s(y'''))
PR(x''', s(s(y'''))) -> IF(eq(x''', times(quot(x''', s(s(y''')), s(s(y'''))), s(s(y''')))), x''', s(y'''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Inst`
`       →DP Problem 5`
`         ↳Nar`
`           →DP Problem 24`
`             ↳Nar`
`             ...`
`               →DP Problem 25`
`                 ↳Instantiation Transformation`

Dependency Pairs:

PR(x''', s(s(y'''))) -> IF(eq(x''', times(quot(x''', s(s(y''')), s(s(y'''))), s(s(y''')))), x''', s(y'''))
PR(0, s(s(y'''))) -> IF(eq(0, times(0, s(s(y''')))), 0, s(y'''))
IF(false, x, y) -> PR(x, y)

Rules:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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)
two new Dependency Pairs are created:

IF(false, 0, s(y''''')) -> PR(0, s(y'''''))
IF(false, x', s(y''''')) -> PR(x', s(y'''''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Inst`
`       →DP Problem 5`
`         ↳Nar`
`           →DP Problem 24`
`             ↳Nar`
`             ...`
`               →DP Problem 26`
`                 ↳Forward Instantiation Transformation`

Dependency Pairs:

IF(false, x', s(y''''')) -> PR(x', s(y'''''))
PR(0, s(s(y'''))) -> IF(eq(0, times(0, s(s(y''')))), 0, s(y'''))
IF(false, 0, s(y''''')) -> PR(0, s(y'''''))
PR(x''', s(s(y'''))) -> IF(eq(x''', times(quot(x''', s(s(y''')), s(s(y'''))), s(s(y''')))), x''', s(y'''))

Rules:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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, 0, s(y''''')) -> PR(0, s(y'''''))
one new Dependency Pair is created:

IF(false, 0, s(s(y''''''))) -> PR(0, 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`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Inst`
`       →DP Problem 5`
`         ↳Nar`
`           →DP Problem 24`
`             ↳Nar`
`             ...`
`               →DP Problem 27`
`                 ↳Forward Instantiation Transformation`

Dependency Pairs:

PR(x''', s(s(y'''))) -> IF(eq(x''', times(quot(x''', s(s(y''')), s(s(y'''))), s(s(y''')))), x''', s(y'''))
IF(false, 0, s(s(y''''''))) -> PR(0, s(s(y'''''')))
PR(0, s(s(y'''))) -> IF(eq(0, times(0, s(s(y''')))), 0, s(y'''))
IF(false, x', s(y''''')) -> PR(x', s(y'''''))

Rules:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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'''''))
two new Dependency Pairs are created:

IF(false, 0, s(s(y''''''))) -> PR(0, s(s(y'''''')))
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`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Inst`
`       →DP Problem 5`
`         ↳Nar`
`           →DP Problem 24`
`             ↳Nar`
`             ...`
`               →DP Problem 28`
`                 ↳Polynomial Ordering`

Dependency Pairs:

IF(false, x'', s(s(y''''''))) -> PR(x'', s(s(y'''''')))
IF(false, 0, s(s(y''''''))) -> PR(0, s(s(y'''''')))
PR(0, s(s(y'''))) -> IF(eq(0, times(0, s(s(y''')))), 0, s(y'''))
IF(false, 0, s(s(y''''''))) -> PR(0, s(s(y'''''')))
PR(x''', s(s(y'''))) -> IF(eq(x''', times(quot(x''', s(s(y''')), s(s(y'''))), s(s(y''')))), x''', s(y'''))

Rules:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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 pairs can be strictly oriented:

PR(0, s(s(y'''))) -> IF(eq(0, times(0, s(s(y''')))), 0, s(y'''))
PR(x''', s(s(y'''))) -> IF(eq(x''', times(quot(x''', s(s(y''')), s(s(y'''))), s(s(y''')))), x''', s(y'''))

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(PR(x1, x2)) =  x2 POL(plus(x1, x2)) =  0 POL(eq(x1, x2)) =  0 POL(0) =  0 POL(false) =  0 POL(times(x1, x2)) =  0 POL(true) =  0 POL(quot(x1, x2, x3)) =  0 POL(s(x1)) =  1 + x1 POL(div(x1, x2)) =  0 POL(IF(x1, x2, x3)) =  x3

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Inst`
`       →DP Problem 5`
`         ↳Nar`
`           →DP Problem 24`
`             ↳Nar`
`             ...`
`               →DP Problem 29`
`                 ↳Dependency Graph`

Dependency Pairs:

IF(false, x'', s(s(y''''''))) -> PR(x'', s(s(y'''''')))
IF(false, 0, s(s(y''''''))) -> PR(0, s(s(y'''''')))
IF(false, 0, s(s(y''''''))) -> PR(0, s(s(y'''''')))

Rules:

plus(x, 0) -> x
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, 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.

Innermost Termination of R successfully shown.
Duration:
0:02 minutes