h(

d(

d(

d(c(

g(e(

R

↳Dependency Pair Analysis

H(z, e(x)) -> H(c(z), d(z,x))

H(z, e(x)) -> D(z,x)

D(z, g(x,y)) -> G(e(x), d(z,y))

D(z, g(x,y)) -> D(z,y)

D(c(z), g(g(x,y), 0)) -> G(d(c(z), g(x,y)), d(z, g(x,y)))

D(c(z), g(g(x,y), 0)) -> D(c(z), g(x,y))

D(c(z), g(g(x,y), 0)) -> D(z, g(x,y))

G(e(x), e(y)) -> G(x,y)

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Polo

→DP Problem 3

↳Remaining

**G(e( x), e(y)) -> G(x, y)**

h(z, e(x)) -> h(c(z), d(z,x))

d(z, g(0, 0)) -> e(0)

d(z, g(x,y)) -> g(e(x), d(z,y))

d(c(z), g(g(x,y), 0)) -> g(d(c(z), g(x,y)), d(z, g(x,y)))

g(e(x), e(y)) -> e(g(x,y))

The following dependency pair can be strictly oriented:

G(e(x), e(y)) -> G(x,y)

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 4

↳Dependency Graph

→DP Problem 2

↳Polo

→DP Problem 3

↳Remaining

h(z, e(x)) -> h(c(z), d(z,x))

d(z, g(0, 0)) -> e(0)

d(z, g(x,y)) -> g(e(x), d(z,y))

d(c(z), g(g(x,y), 0)) -> g(d(c(z), g(x,y)), d(z, g(x,y)))

g(e(x), e(y)) -> e(g(x,y))

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polynomial Ordering

→DP Problem 3

↳Remaining

**D(c( z), g(g(x, y), 0)) -> D(z, g(x, y))**

h(z, e(x)) -> h(c(z), d(z,x))

d(z, g(0, 0)) -> e(0)

d(z, g(x,y)) -> g(e(x), d(z,y))

d(c(z), g(g(x,y), 0)) -> g(d(c(z), g(x,y)), d(z, g(x,y)))

g(e(x), e(y)) -> e(g(x,y))

The following dependency pair can be strictly oriented:

D(c(z), g(g(x,y), 0)) -> D(z, g(x,y))

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(c(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(g(x)_{1}, x_{2})= 0 _{ }^{ }_{ }^{ }POL(e(x)_{1})= 0 _{ }^{ }_{ }^{ }POL(D(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

↳Polynomial Ordering

→DP Problem 3

↳Remaining

**D(c( z), g(g(x, y), 0)) -> D(c(z), g(x, y))**

h(z, e(x)) -> h(c(z), d(z,x))

d(z, g(0, 0)) -> e(0)

d(z, g(x,y)) -> g(e(x), d(z,y))

d(c(z), g(g(x,y), 0)) -> g(d(c(z), g(x,y)), d(z, g(x,y)))

g(e(x), e(y)) -> e(g(x,y))

The following dependency pair can be strictly oriented:

D(c(z), g(g(x,y), 0)) -> D(c(z), g(x,y))

Additionally, the following usable rule w.r.t. to the implicit AFS can be oriented:

g(e(x), e(y)) -> e(g(x,y))

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(c(x)_{1})= 0 _{ }^{ }_{ }^{ }POL(0)= 1 _{ }^{ }_{ }^{ }POL(g(x)_{1}, x_{2})= x _{1}+ x_{2}_{ }^{ }_{ }^{ }POL(e(x)_{1})= 0 _{ }^{ }_{ }^{ }POL(D(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 5

↳Polo

...

→DP Problem 6

↳Polynomial Ordering

→DP Problem 3

↳Remaining

**D( z, g(x, y)) -> D(z, y)**

h(z, e(x)) -> h(c(z), d(z,x))

d(z, g(0, 0)) -> e(0)

d(z, g(x,y)) -> g(e(x), d(z,y))

d(c(z), g(g(x,y), 0)) -> g(d(c(z), g(x,y)), d(z, g(x,y)))

g(e(x), e(y)) -> e(g(x,y))

The following dependency pair can be strictly oriented:

D(z, g(x,y)) -> D(z,y)

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 5

↳Polo

...

→DP Problem 7

↳Dependency Graph

→DP Problem 3

↳Remaining

h(z, e(x)) -> h(c(z), d(z,x))

d(z, g(0, 0)) -> e(0)

d(z, g(x,y)) -> g(e(x), d(z,y))

d(c(z), g(g(x,y), 0)) -> g(d(c(z), g(x,y)), d(z, g(x,y)))

g(e(x), e(y)) -> e(g(x,y))

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 3

↳Remaining Obligation(s)

The following remains to be proven:

**H( z, e(x)) -> H(c(z), d(z, x))**

h(z, e(x)) -> h(c(z), d(z,x))

d(z, g(0, 0)) -> e(0)

d(z, g(x,y)) -> g(e(x), d(z,y))

d(c(z), g(g(x,y), 0)) -> g(d(c(z), g(x,y)), d(z, g(x,y)))

g(e(x), e(y)) -> e(g(x,y))

Duration:

0:00 minutes