admit(

admit(

cond(true,

R

↳Dependency Pair Analysis

ADMIT(x, .(u, .(v, .(w,z)))) -> COND(=(sum(x,u,v), w), .(u, .(v, .(w, admit(carry(x,u,v),z)))))

ADMIT(x, .(u, .(v, .(w,z)))) -> ADMIT(carry(x,u,v),z)

Furthermore,

R

↳DPs

→DP Problem 1

↳Argument Filtering and Ordering

**ADMIT( x, .(u, .(v, .(w, z)))) -> ADMIT(carry(x, u, v), z)**

admit(x, nil) -> nil

admit(x, .(u, .(v, .(w,z)))) -> cond(=(sum(x,u,v), w), .(u, .(v, .(w, admit(carry(x,u,v),z)))))

cond(true,y) ->y

innermost

The following dependency pair can be strictly oriented:

ADMIT(x, .(u, .(v, .(w,z)))) -> ADMIT(carry(x,u,v),z)

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(carry(x)_{1}, x_{2}, x_{3})= x _{1}+ x_{2}+ x_{3}_{ }^{ }_{ }^{ }POL(.(x)_{1}, x_{2})= x _{1}+ x_{2}_{ }^{ }_{ }^{ }POL(w)= 1 _{ }^{ }_{ }^{ }POL(ADMIT(x)_{1}, x_{2})= 1 + x _{1}+ x_{2}_{ }^{ }

resulting in one new DP problem.

Used Argument Filtering System:

ADMIT(x,_{1}x) -> ADMIT(_{2}x,_{1}x)_{2}

.(x,_{1}x) -> .(_{2}x,_{1}x)_{2}

carry(x,_{1}x,_{2}x) -> carry(_{3}x,_{1}x,_{2}x)_{3}

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳Dependency Graph

admit(x, nil) -> nil

admit(x, .(u, .(v, .(w,z)))) -> cond(=(sum(x,u,v), w), .(u, .(v, .(w, admit(carry(x,u,v),z)))))

cond(true,y) ->y

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes