f(

f(s(

f(c(

g(

g(

R

↳Dependency Pair Analysis

F(x, c(x), c(y)) -> F(y,y, f(y,x,y))

F(x, c(x), c(y)) -> F(y,x,y)

F(s(x),y,z) -> F(x, s(c(y)), c(z))

Furthermore,

R

↳DPs

→DP Problem 1

↳Instantiation Transformation

→DP Problem 2

↳Inst

**F(s( x), y, z) -> F(x, s(c(y)), c(z))**

f(x, c(x), c(y)) -> f(y,y, f(y,x,y))

f(s(x),y,z) -> f(x, s(c(y)), c(z))

f(c(x),x,y) -> c(y)

g(x,y) ->x

g(x,y) ->y

innermost

On this DP problem, an Instantiation SCC transformation can be performed.

As a result of transforming the rule

one new Dependency Pair is created:

F(s(x),y,z) -> F(x, s(c(y)), c(z))

F(s(x''), s(c(y'')), c(z'')) -> F(x'', s(c(s(c(y'')))), c(c(z'')))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Inst

→DP Problem 3

↳Instantiation Transformation

→DP Problem 2

↳Inst

**F(s( x''), s(c(y'')), c(z'')) -> F(x'', s(c(s(c(y'')))), c(c(z'')))**

f(x, c(x), c(y)) -> f(y,y, f(y,x,y))

f(s(x),y,z) -> f(x, s(c(y)), c(z))

f(c(x),x,y) -> c(y)

g(x,y) ->x

g(x,y) ->y

innermost

On this DP problem, an Instantiation SCC transformation can be performed.

As a result of transforming the rule

one new Dependency Pair is created:

F(s(x''), s(c(y'')), c(z'')) -> F(x'', s(c(s(c(y'')))), c(c(z'')))

F(s(x''''), s(c(s(c(y'''')))), c(c(z''''))) -> F(x'''', s(c(s(c(s(c(y'''')))))), c(c(c(z''''))))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Inst

→DP Problem 3

↳Inst

...

→DP Problem 4

↳Polynomial Ordering

→DP Problem 2

↳Inst

**F(s( x''''), s(c(s(c(y'''')))), c(c(z''''))) -> F(x'''', s(c(s(c(s(c(y'''')))))), c(c(c(z''''))))**

f(x, c(x), c(y)) -> f(y,y, f(y,x,y))

f(s(x),y,z) -> f(x, s(c(y)), c(z))

f(c(x),x,y) -> c(y)

g(x,y) ->x

g(x,y) ->y

innermost

The following dependency pair can be strictly oriented:

F(s(x''''), s(c(s(c(y'''')))), c(c(z''''))) -> F(x'''', s(c(s(c(s(c(y'''')))))), c(c(c(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(c(x)_{1})= 0 _{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(F(x)_{1}, x_{2}, x_{3})= x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Inst

→DP Problem 3

↳Inst

...

→DP Problem 5

↳Dependency Graph

→DP Problem 2

↳Inst

f(x, c(x), c(y)) -> f(y,y, f(y,x,y))

f(s(x),y,z) -> f(x, s(c(y)), c(z))

f(c(x),x,y) -> c(y)

g(x,y) ->x

g(x,y) ->y

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Inst

→DP Problem 2

↳Instantiation Transformation

**F( x, c(x), c(y)) -> F(y, x, y)**

f(x, c(x), c(y)) -> f(y,y, f(y,x,y))

f(s(x),y,z) -> f(x, s(c(y)), c(z))

f(c(x),x,y) -> c(y)

g(x,y) ->x

g(x,y) ->y

innermost

On this DP problem, an Instantiation SCC transformation can be performed.

As a result of transforming the rule

one new Dependency Pair is created:

F(x, c(x), c(y)) -> F(y,x,y)

F(c(y''), c(c(y'')), c(y'')) -> F(y'', c(y''),y'')

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Inst

→DP Problem 2

↳Inst

→DP Problem 6

↳Forward Instantiation Transformation

**F(c( y''), c(c(y'')), c(y'')) -> F(y'', c(y''), y'')**

f(x, c(x), c(y)) -> f(y,y, f(y,x,y))

f(s(x),y,z) -> f(x, s(c(y)), c(z))

f(c(x),x,y) -> c(y)

g(x,y) ->x

g(x,y) ->y

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.

As a result of transforming the rule

one new Dependency Pair is created:

F(c(y''), c(c(y'')), c(y'')) -> F(y'', c(y''),y'')

F(c(c(y''''')), c(c(c(y'''''))), c(c(y'''''))) -> F(c(y'''''), c(c(y''''')), c(y'''''))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Inst

→DP Problem 2

↳Inst

→DP Problem 6

↳FwdInst

...

→DP Problem 7

↳Polynomial Ordering

**F(c(c( y''''')), c(c(c(y'''''))), c(c(y'''''))) -> F(c(y'''''), c(c(y''''')), c(y'''''))**

f(x, c(x), c(y)) -> f(y,y, f(y,x,y))

f(s(x),y,z) -> f(x, s(c(y)), c(z))

f(c(x),x,y) -> c(y)

g(x,y) ->x

g(x,y) ->y

innermost

The following dependency pair can be strictly oriented:

F(c(c(y''''')), c(c(c(y'''''))), c(c(y'''''))) -> F(c(y'''''), c(c(y''''')), c(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(c(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(F(x)_{1}, x_{2}, x_{3})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Inst

→DP Problem 2

↳Inst

→DP Problem 6

↳FwdInst

...

→DP Problem 8

↳Dependency Graph

f(x, c(x), c(y)) -> f(y,y, f(y,x,y))

f(s(x),y,z) -> f(x, s(c(y)), c(z))

f(c(x),x,y) -> c(y)

g(x,y) ->x

g(x,y) ->y

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes