f(

i(

i(

i(h(

j(g(

R

↳Dependency Pair Analysis

F(x, g(y)) -> F(h(x), i(x,y))

F(x, g(y)) -> I(x,y)

I(x, j(y,z)) -> J(g(y), i(x,z))

I(x, j(y,z)) -> I(x,z)

I(h(x), j(j(y,z), 0)) -> J(i(h(x), j(y,z)), i(x, j(y,z)))

I(h(x), j(j(y,z), 0)) -> I(h(x), j(y,z))

I(h(x), j(j(y,z), 0)) -> I(x, j(y,z))

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

Furthermore,

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↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Polo

→DP Problem 3

↳Remaining

**J(g( x), g(y)) -> J(x, y)**

f(x, g(y)) -> f(h(x), i(x,y))

i(x, j(0, 0)) -> g(0)

i(x, j(y,z)) -> j(g(y), i(x,z))

i(h(x), j(j(y,z), 0)) -> j(i(h(x), j(y,z)), i(x, j(y,z)))

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

innermost

The following dependency pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

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

f(x, g(y)) -> f(h(x), i(x,y))

i(x, j(0, 0)) -> g(0)

i(x, j(y,z)) -> j(g(y), i(x,z))

i(h(x), j(j(y,z), 0)) -> j(i(h(x), j(y,z)), i(x, j(y,z)))

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

innermost

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

**I(h( x), j(j(y, z), 0)) -> I(x, j(y, z))**

f(x, g(y)) -> f(h(x), i(x,y))

i(x, j(0, 0)) -> g(0)

i(x, j(y,z)) -> j(g(y), i(x,z))

i(h(x), j(j(y,z), 0)) -> j(i(h(x), j(y,z)), i(x, j(y,z)))

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

innermost

The following dependency pair can be strictly oriented:

I(h(x), j(j(y,z), 0)) -> I(x, j(y,z))

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(I(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(g(x)_{1})= 0 _{ }^{ }_{ }^{ }POL(h(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(j(x)_{1}, x_{2})= 0 _{ }^{ }

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

**I(h( x), j(j(y, z), 0)) -> I(h(x), j(y, z))**

f(x, g(y)) -> f(h(x), i(x,y))

i(x, j(0, 0)) -> g(0)

i(x, j(y,z)) -> j(g(y), i(x,z))

i(h(x), j(j(y,z), 0)) -> j(i(h(x), j(y,z)), i(x, j(y,z)))

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

innermost

The following dependency pair can be strictly oriented:

I(h(x), j(j(y,z), 0)) -> I(h(x), j(y,z))

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(I(x)_{1}, x_{2})= 1 + x _{1}+ x_{2}_{ }^{ }_{ }^{ }POL(0)= 1 _{ }^{ }_{ }^{ }POL(g(x)_{1})= 0 _{ }^{ }_{ }^{ }POL(h(x)_{1})= 0 _{ }^{ }_{ }^{ }POL(j(x)_{1}, x_{2})= 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

**I( x, j(y, z)) -> I(x, z)**

f(x, g(y)) -> f(h(x), i(x,y))

i(x, j(0, 0)) -> g(0)

i(x, j(y,z)) -> j(g(y), i(x,z))

i(h(x), j(j(y,z), 0)) -> j(i(h(x), j(y,z)), i(x, j(y,z)))

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

innermost

The following dependency pair can be strictly oriented:

I(x, j(y,z)) -> I(x,z)

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(I(x)_{1}, x_{2})= x _{2}_{ }^{ }_{ }^{ }POL(j(x)_{1}, x_{2})= 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 7

↳Dependency Graph

→DP Problem 3

↳Remaining

f(x, g(y)) -> f(h(x), i(x,y))

i(x, j(0, 0)) -> g(0)

i(x, j(y,z)) -> j(g(y), i(x,z))

i(h(x), j(j(y,z), 0)) -> j(i(h(x), j(y,z)), i(x, j(y,z)))

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

innermost

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:

**F( x, g(y)) -> F(h(x), i(x, y))**

f(x, g(y)) -> f(h(x), i(x,y))

i(x, j(0, 0)) -> g(0)

i(x, j(y,z)) -> j(g(y), i(x,z))

i(h(x), j(j(y,z), 0)) -> j(i(h(x), j(y,z)), i(x, j(y,z)))

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

innermost

Duration:

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