le(0,

le(s(

le(s(

pred(s(

minus(

minus(

gcd(0,

gcd(s(

gcd(s(

if

if

R

↳Dependency Pair Analysis

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

MINUS(x, s(y)) -> PRED(minus(x,y))

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

GCD(s(x), s(y)) -> IF_{GCD}(le(y,x), s(x), s(y))

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

IF_{GCD}(true, s(x), s(y)) -> GCD(minus(x,y), s(y))

IF_{GCD}(true, s(x), s(y)) -> MINUS(x,y)

IF_{GCD}(false, s(x), s(y)) -> GCD(minus(y,x), s(x))

IF_{GCD}(false, s(x), s(y)) -> MINUS(y,x)

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Polo

→DP Problem 3

↳Polo

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

le(0,y) -> true

le(s(x), 0) -> false

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

pred(s(x)) ->x

minus(x, 0) ->x

minus(x, s(y)) -> pred(minus(x,y))

gcd(0,y) ->y

gcd(s(x), 0) -> s(x)

gcd(s(x), s(y)) -> if_{gcd}(le(y,x), s(x), s(y))

if_{gcd}(true, s(x), s(y)) -> gcd(minus(x,y), s(y))

if_{gcd}(false, s(x), s(y)) -> gcd(minus(y,x), s(x))

innermost

The following dependency pair can be strictly oriented:

LE(s(x), s(y)) -> LE(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(LE(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 4

↳Dependency Graph

→DP Problem 2

↳Polo

→DP Problem 3

↳Polo

le(0,y) -> true

le(s(x), 0) -> false

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

pred(s(x)) ->x

minus(x, 0) ->x

minus(x, s(y)) -> pred(minus(x,y))

gcd(0,y) ->y

gcd(s(x), 0) -> s(x)

gcd(s(x), s(y)) -> if_{gcd}(le(y,x), s(x), s(y))

if_{gcd}(true, s(x), s(y)) -> gcd(minus(x,y), s(y))

if_{gcd}(false, s(x), s(y)) -> gcd(minus(y,x), s(x))

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polynomial Ordering

→DP Problem 3

↳Polo

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

le(0,y) -> true

le(s(x), 0) -> false

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

pred(s(x)) ->x

minus(x, 0) ->x

minus(x, s(y)) -> pred(minus(x,y))

gcd(0,y) ->y

gcd(s(x), 0) -> s(x)

gcd(s(x), s(y)) -> if_{gcd}(le(y,x), s(x), s(y))

if_{gcd}(true, s(x), s(y)) -> gcd(minus(x,y), s(y))

if_{gcd}(false, s(x), s(y)) -> gcd(minus(y,x), s(x))

innermost

The following dependency pair can be strictly oriented:

MINUS(x, s(y)) -> MINUS(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(MINUS(x)_{1}, x_{2})= x _{2}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 5

↳Dependency Graph

→DP Problem 3

↳Polo

le(0,y) -> true

le(s(x), 0) -> false

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

pred(s(x)) ->x

minus(x, 0) ->x

minus(x, s(y)) -> pred(minus(x,y))

gcd(0,y) ->y

gcd(s(x), 0) -> s(x)

gcd(s(x), s(y)) -> if_{gcd}(le(y,x), s(x), s(y))

if_{gcd}(true, s(x), s(y)) -> gcd(minus(x,y), s(y))

if_{gcd}(false, s(x), s(y)) -> gcd(minus(y,x), s(x))

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 3

↳Polynomial Ordering

**IF _{GCD}(false, s(x), s(y)) -> GCD(minus(y, x), s(x))**

le(0,y) -> true

le(s(x), 0) -> false

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

pred(s(x)) ->x

minus(x, 0) ->x

minus(x, s(y)) -> pred(minus(x,y))

gcd(0,y) ->y

gcd(s(x), 0) -> s(x)

gcd(s(x), s(y)) -> if_{gcd}(le(y,x), s(x), s(y))

if_{gcd}(true, s(x), s(y)) -> gcd(minus(x,y), s(y))

if_{gcd}(false, s(x), s(y)) -> gcd(minus(y,x), s(x))

innermost

The following dependency pairs can be strictly oriented:

IF_{GCD}(false, s(x), s(y)) -> GCD(minus(y,x), s(x))

IF_{GCD}(true, s(x), s(y)) -> GCD(minus(x,y), s(y))

Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

minus(x, 0) ->x

minus(x, s(y)) -> pred(minus(x,y))

pred(s(x)) ->x

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(GCD(x)_{1}, x_{2})= x _{1}+ x_{2}_{ }^{ }_{ }^{ }POL(false)= 0 _{ }^{ }_{ }^{ }POL(pred(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(minus(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(true)= 0 _{ }^{ }_{ }^{ }POL(IF_GCD(x)_{1}, x_{2}, x_{3})= x _{2}+ x_{3}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(le(x)_{1}, x_{2})= 0 _{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 3

↳Polo

→DP Problem 6

↳Dependency Graph

**GCD(s( x), s(y)) -> IF_{GCD}(le(y, x), s(x), s(y))**

le(0,y) -> true

le(s(x), 0) -> false

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

pred(s(x)) ->x

minus(x, 0) ->x

minus(x, s(y)) -> pred(minus(x,y))

gcd(0,y) ->y

gcd(s(x), 0) -> s(x)

gcd(s(x), s(y)) -> if_{gcd}(le(y,x), s(x), s(y))

if_{gcd}(true, s(x), s(y)) -> gcd(minus(x,y), s(y))

if_{gcd}(false, s(x), s(y)) -> gcd(minus(y,x), s(x))

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes