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,

R

↳DPs

→DP Problem 1

↳Argument Filtering and Ordering

→DP Problem 2

↳AFS

→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))

The following dependency pair can be strictly oriented:

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

The following rules can be oriented:

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))

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

{i, g}

resulting in one new DP problem.

Used Argument Filtering System:

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

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

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

h(x) ->_{1}x_{1}

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

j(x,_{1}x) ->_{2}x_{1}

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 4

↳Dependency Graph

→DP Problem 2

↳AFS

→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))

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳Argument Filtering and 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))

The following dependency pairs can be strictly oriented:

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

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

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

The following rules can be oriented:

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

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

i > j > g

resulting in one new DP problem.

Used Argument Filtering System:

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

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

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

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

f(x,_{1}x) -> f_{2}

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

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 5

↳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))

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→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))

Duration:

0:00 minutes