f(0) -> 1

f(s(

g(0,

g(s(

g(s(

+(

+(

R

↳Dependency Pair Analysis

F(s(x)) -> G(x, s(x))

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

G(s(x),y) -> +'(y, s(x))

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

G(s(x),y) -> +'(y,x)

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Argument Filtering and Ordering

→DP Problem 2

↳AFS

**+'( x, s(y)) -> +'(x, y)**

f(0) -> 1

f(s(x)) -> g(x, s(x))

g(0,y) ->y

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

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

+(x, 0) ->x

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

The following dependency pair can be strictly oriented:

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

The following rules can be oriented:

f(0) -> 1

f(s(x)) -> g(x, s(x))

g(0,y) ->y

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

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

+(x, 0) ->x

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

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

{f, g, 1} > + > 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}

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

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

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

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 3

↳Dependency Graph

→DP Problem 2

↳AFS

f(0) -> 1

f(s(x)) -> g(x, s(x))

g(0,y) ->y

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

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

+(x, 0) ->x

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

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳Argument Filtering and Ordering

**G(s( x), y) -> G(x, s(+(y, x)))**

f(0) -> 1

f(s(x)) -> g(x, s(x))

g(0,y) ->y

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

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

+(x, 0) ->x

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

The following dependency pairs can be strictly oriented:

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

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

The following rules can be oriented:

+(x, 0) ->x

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

f(0) -> 1

f(s(x)) -> g(x, s(x))

g(0,y) ->y

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

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

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

{f, g, 1} > + > s

G > + > s

resulting in one new DP problem.

Used Argument Filtering System:

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

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

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

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

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

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 4

↳Dependency Graph

f(0) -> 1

f(s(x)) -> g(x, s(x))

g(0,y) ->y

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

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

+(x, 0) ->x

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

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes