min(

min(s(

quot(0, s(

quot(s(

log(s(0)) -> 0

log(s(s(

R

↳Dependency Pair Analysis

MIN(s(X), s(Y)) -> MIN(X,Y)

QUOT(s(X), s(Y)) -> QUOT(min(X,Y), s(Y))

QUOT(s(X), s(Y)) -> MIN(X,Y)

LOG(s(s(X))) -> LOG(s(quot(X, s(s(0)))))

LOG(s(s(X))) -> QUOT(X, s(s(0)))

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Polo

→DP Problem 3

↳Polo

**MIN(s( X), s(Y)) -> MIN(X, Y)**

min(X, 0) ->X

min(s(X), s(Y)) -> min(X,Y)

quot(0, s(Y)) -> 0

quot(s(X), s(Y)) -> s(quot(min(X,Y), s(Y)))

log(s(0)) -> 0

log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

The following dependency pair can be strictly oriented:

MIN(s(X), s(Y)) -> MIN(X,Y)

Additionally, the following rules can be oriented:

min(X, 0) ->X

min(s(X), s(Y)) -> min(X,Y)

quot(0, s(Y)) -> 0

quot(s(X), s(Y)) -> s(quot(min(X,Y), s(Y)))

log(s(0)) -> 0

log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(log(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(MIN(x)_{1}, x_{2})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(min(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(quot(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

min(X, 0) ->X

min(s(X), s(Y)) -> min(X,Y)

quot(0, s(Y)) -> 0

quot(s(X), s(Y)) -> s(quot(min(X,Y), s(Y)))

log(s(0)) -> 0

log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

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

**QUOT(s( X), s(Y)) -> QUOT(min(X, Y), s(Y))**

min(X, 0) ->X

min(s(X), s(Y)) -> min(X,Y)

quot(0, s(Y)) -> 0

quot(s(X), s(Y)) -> s(quot(min(X,Y), s(Y)))

log(s(0)) -> 0

log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

The following dependency pair can be strictly oriented:

QUOT(s(X), s(Y)) -> QUOT(min(X,Y), s(Y))

Additionally, the following rules can be oriented:

min(X, 0) ->X

min(s(X), s(Y)) -> min(X,Y)

quot(0, s(Y)) -> 0

quot(s(X), s(Y)) -> s(quot(min(X,Y), s(Y)))

log(s(0)) -> 0

log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(QUOT(x)_{1}, x_{2})= 1 + x _{1}+ x_{2}_{ }^{ }_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(log(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(min(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(quot(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 2

↳Polo

→DP Problem 5

↳Dependency Graph

→DP Problem 3

↳Polo

min(X, 0) ->X

min(s(X), s(Y)) -> min(X,Y)

quot(0, s(Y)) -> 0

quot(s(X), s(Y)) -> s(quot(min(X,Y), s(Y)))

log(s(0)) -> 0

log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

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

**LOG(s(s( X))) -> LOG(s(quot(X, s(s(0)))))**

min(X, 0) ->X

min(s(X), s(Y)) -> min(X,Y)

quot(0, s(Y)) -> 0

quot(s(X), s(Y)) -> s(quot(min(X,Y), s(Y)))

log(s(0)) -> 0

log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

The following dependency pair can be strictly oriented:

LOG(s(s(X))) -> LOG(s(quot(X, s(s(0)))))

Additionally, the following rules can be oriented:

min(X, 0) ->X

min(s(X), s(Y)) -> min(X,Y)

quot(0, s(Y)) -> 0

quot(s(X), s(Y)) -> s(quot(min(X,Y), s(Y)))

log(s(0)) -> 0

log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(log(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(min(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(quot(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(LOG(x)_{1})= 1 + x _{1}_{ }^{ }

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

min(X, 0) ->X

min(s(X), s(Y)) -> min(X,Y)

quot(0, s(Y)) -> 0

quot(s(X), s(Y)) -> s(quot(min(X,Y), s(Y)))

log(s(0)) -> 0

log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes