w(r(

b(r(

b(w(

R

↳Dependency Pair Analysis

W(r(x)) -> W(x)

B(r(x)) -> B(x)

B(w(x)) -> W(b(x))

B(w(x)) -> B(x)

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Polo

→DP Problem 3

↳Polo

**W(r( x)) -> W(x)**

w(r(x)) -> r(w(x))

b(r(x)) -> r(b(x))

b(w(x)) -> w(b(x))

innermost

The following dependency pair can be strictly oriented:

W(r(x)) -> W(x)

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(W(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(r(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

w(r(x)) -> r(w(x))

b(r(x)) -> r(b(x))

b(w(x)) -> w(b(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

**B(r( x)) -> B(x)**

w(r(x)) -> r(w(x))

b(r(x)) -> r(b(x))

b(w(x)) -> w(b(x))

innermost

The following dependency pair can be strictly oriented:

B(r(x)) -> B(x)

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(B(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(r(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

w(r(x)) -> r(w(x))

b(r(x)) -> r(b(x))

b(w(x)) -> w(b(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

**B(w( x)) -> B(x)**

w(r(x)) -> r(w(x))

b(r(x)) -> r(b(x))

b(w(x)) -> w(b(x))

innermost

The following dependency pair can be strictly oriented:

B(w(x)) -> B(x)

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(B(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(w(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

w(r(x)) -> r(w(x))

b(r(x)) -> r(b(x))

b(w(x)) -> w(b(x))

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes