+(

+(

f(0, s(0),

g(

g(

R

↳Dependency Pair Analysis

+'(X, s(Y)) -> +'(X,Y)

F(0, s(0),X) -> F(X, +(X,X),X)

F(0, s(0),X) -> +'(X,X)

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Remaining

**+'( X, s(Y)) -> +'(X, Y)**

+(X, 0) ->X

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

f(0, s(0),X) -> f(X, +(X,X),X)

g(X,Y) ->X

g(X,Y) ->Y

The following dependency pair can be strictly oriented:

+'(X, s(Y)) -> +'(X,Y)

Additionally, the following rules can be oriented:

+(X, 0) ->X

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

f(0, s(0),X) -> f(X, +(X,X),X)

g(X,Y) ->X

g(X,Y) ->Y

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(g(x)_{1}, x_{2})= x _{1}+ x_{2}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(+(x)_{1}, x_{2})= x _{1}+ x_{2}_{ }^{ }_{ }^{ }POL(f(x)_{1}, x_{2}, x_{3})= 0 _{ }^{ }_{ }^{ }POL(+'(x)_{1}, x_{2})= x _{2}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 3

↳Dependency Graph

→DP Problem 2

↳Remaining

+(X, 0) ->X

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

f(0, s(0),X) -> f(X, +(X,X),X)

g(X,Y) ->X

g(X,Y) ->Y

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Remaining Obligation(s)

The following remains to be proven:

**F(0, s(0), X) -> F(X, +(X, X), X)**

+(X, 0) ->X

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

f(0, s(0),X) -> f(X, +(X,X),X)

g(X,Y) ->X

g(X,Y) ->Y

Duration:

0:00 minutes