+(a, b) -> +(b, a)

+(a, +(b,

+(+(

f(a,

f(b,

f(+(

R

↳Dependency Pair Analysis

+'(a, b) -> +'(b, a)

+'(a, +(b,z)) -> +'(b, +(a,z))

+'(a, +(b,z)) -> +'(a,z)

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

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

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

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

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Polo

→DP Problem 3

↳Polo

**+'(a, +(b, z)) -> +'(a, z)**

+(a, b) -> +(b, a)

+(a, +(b,z)) -> +(b, +(a,z))

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

f(a,y) -> a

f(b,y) -> b

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

The following dependency pair can be strictly oriented:

+'(a, +(b,z)) -> +'(a,z)

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

Used ordering: Polynomial ordering with Polynomial interpretation:

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 4

↳Dependency Graph

→DP Problem 2

↳Polo

→DP Problem 3

↳Polo

+(a, b) -> +(b, a)

+(a, +(b,z)) -> +(b, +(a,z))

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

f(a,y) -> a

f(b,y) -> b

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

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polynomial Ordering

→DP Problem 3

↳Polo

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

+(a, b) -> +(b, a)

+(a, +(b,z)) -> +(b, +(a,z))

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

f(a,y) -> a

f(b,y) -> b

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

The following dependency pairs can be strictly oriented:

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

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 5

↳Dependency Graph

→DP Problem 3

↳Polo

+(a, b) -> +(b, a)

+(a, +(b,z)) -> +(b, +(a,z))

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

f(a,y) -> a

f(b,y) -> b

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

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 3

↳Polynomial Ordering

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

+(a, b) -> +(b, a)

+(a, +(b,z)) -> +(b, +(a,z))

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

f(a,y) -> a

f(b,y) -> b

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

The following dependency pairs can be strictly oriented:

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

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(F(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(+(x)_{1}, x_{2})= 1 + x _{1}+ x_{2}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 3

↳Polo

→DP Problem 6

↳Dependency Graph

+(a, b) -> +(b, a)

+(a, +(b,z)) -> +(b, +(a,z))

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

f(a,y) -> a

f(b,y) -> b

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

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes