minus(0) -> 0

minus(minus(

+(

+(0,

+(minus(1), 1) -> 0

+(

+(

+(minus(+(

R

↳Dependency Pair Analysis

+'(x, minus(y)) -> MINUS(+(minus(x),y))

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

+'(x, minus(y)) -> MINUS(x)

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

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

+'(minus(+(x, 1)), 1) -> MINUS(x)

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

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

minus(0) -> 0

minus(minus(x)) ->x

+(x, 0) ->x

+(0,y) ->y

+(minus(1), 1) -> 0

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

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

+(minus(+(x, 1)), 1) -> minus(x)

The following dependency pairs can be strictly oriented:

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

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

Additionally, the following usable rules using the Ce-refinement can be oriented:

+(x, 0) ->x

+(0,y) ->y

+(minus(1), 1) -> 0

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

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

+(minus(+(x, 1)), 1) -> minus(x)

minus(0) -> 0

minus(minus(x)) ->x

Used ordering: Polynomial ordering with Polynomial interpretation:

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polynomial Ordering

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

minus(0) -> 0

minus(minus(x)) ->x

+(x, 0) ->x

+(0,y) ->y

+(minus(1), 1) -> 0

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

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

+(minus(+(x, 1)), 1) -> minus(x)

The following dependency pair can be strictly oriented:

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

Additionally, the following usable rules using the Ce-refinement can be oriented:

minus(0) -> 0

minus(minus(x)) ->x

Used ordering: Polynomial ordering with Polynomial interpretation:

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

...

→DP Problem 3

↳Dependency Graph

minus(0) -> 0

minus(minus(x)) ->x

+(x, 0) ->x

+(0,y) ->y

+(minus(1), 1) -> 0

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

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

+(minus(+(x, 1)), 1) -> minus(x)

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes