i(0) -> 0

i(i(

i(+(

+(0,

+(

+(i(

+(

+(

+(+(

+(+(

R

↳Dependency Pair Analysis

I(+(x,y)) -> +'(i(x), i(y))

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

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

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

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Argument Filtering and Ordering

→DP Problem 2

↳AFS

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

i(0) -> 0

i(i(x)) ->x

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

+(0,y) ->y

+(x, 0) ->x

+(i(x),x) -> 0

+(x, i(x)) -> 0

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

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

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

The following dependency pairs can be strictly oriented:

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

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

There are no usable rules w.r.t. to the AFS that need to be oriented.

Used ordering: Homeomorphic Embedding Order with EMB

resulting in one new DP problem.

Used Argument Filtering System:

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

i(0) -> 0

i(i(x)) ->x

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

+(0,y) ->y

+(x, 0) ->x

+(i(x),x) -> 0

+(x, i(x)) -> 0

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

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

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

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳Argument Filtering and Ordering

**I(+( x, y)) -> I(y)**

i(0) -> 0

i(i(x)) ->x

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

+(0,y) ->y

+(x, 0) ->x

+(i(x),x) -> 0

+(x, i(x)) -> 0

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

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

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

The following dependency pairs can be strictly oriented:

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

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

There are no usable rules w.r.t. to the AFS that need to be oriented.

Used ordering: Homeomorphic Embedding Order with EMB

resulting in one new DP problem.

Used Argument Filtering System:

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

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

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 4

↳Dependency Graph

i(0) -> 0

i(i(x)) ->x

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

+(0,y) ->y

+(x, 0) ->x

+(i(x),x) -> 0

+(x, i(x)) -> 0

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

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

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

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes