+(

+(

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

↳Argument Filtering and 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)

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: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

+ > s

resulting in one new DP problem.

Used Argument Filtering System:

+'(x,_{1}x) -> +'(_{2}x,_{1}x)_{2}

s(x) -> s(_{1}x)_{1}

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

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

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

R

↳DPs

→DP Problem 1

↳AFS

→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

↳AFS

→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