bin(

bin(0, s(

bin(s(

R

↳Dependency Pair Analysis

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

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

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

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

bin(0, s(y)) -> 0

bin(s(x), s(y)) -> +(bin(x, s(y)), bin(x,y))

The following dependency pairs can be strictly oriented:

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

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

Additionally, the following rules can be oriented:

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

bin(0, s(y)) -> 0

bin(s(x), s(y)) -> +(bin(x, s(y)), bin(x,y))

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(bin(x)_{1}, x_{2})= 1 _{ }^{ }_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(+(x)_{1}, x_{2})= 0 _{ }^{ }_{ }^{ }POL(BIN(x)_{1}, x_{2})= x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Dependency Graph

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

bin(0, s(y)) -> 0

bin(s(x), s(y)) -> +(bin(x, s(y)), bin(x,y))

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes