*(

*(

*(

*(

R

↳Dependency Pair Analysis

*'(X, +(Y, 1)) -> *'(X, +(Y, *(1, 0)))

*'(X, +(Y, 1)) -> *'(1, 0)

Furthermore,

R

↳DPs

→DP Problem 1

↳Modular Removal of Rules

***'( X, +(Y, 1)) -> *'(X, +(Y, *(1, 0)))**

*(X, +(Y, 1)) -> +(*(X, +(Y, *(1, 0))),X)

*(X, 1) ->X

*(X, 0) ->X

*(X, 0) -> 0

We have the following set of usable rules:

To remove rules and DPs from this DP problem we used the following monotonic and C

*(X, 0) ->X

*(X, 0) -> 0

Polynomial interpretation:

_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(*'(x)_{1}, x_{2})= 1 + x _{1}+ x_{2}_{ }^{ }_{ }^{ }POL(1)= 0 _{ }^{ }_{ }^{ }POL(*(x)_{1}, x_{2})= x _{1}+ x_{2}_{ }^{ }_{ }^{ }POL(+(x)_{1}, x_{2})= x _{1}+ x_{2}_{ }^{ }

We have the following set D of usable symbols: {0, *', 1, *, +}

No Dependency Pairs can be deleted.

2 non usable rules have been deleted.

The result of this processor delivers one new DP problem.

R

↳DPs

→DP Problem 1

↳MRR

→DP Problem 2

↳Non Termination

***'( X, +(Y, 1)) -> *'(X, +(Y, *(1, 0)))**

*(X, 0) ->X

*(X, 0) -> 0

Found an infinite P-chain over R:

P =

*'(X, +(Y, 1)) -> *'(X, +(Y, *(1, 0)))

R =

*(X, 0) ->X

*(X, 0) -> 0

s = *'(

evaluates to t =*'(

Thus, s starts an infinite chain.

Duration:

0:00 minutes