plus(

plus(0,

plus(s(

times(0,

times(s(0),

times(s(

div(0,

div(

div(div(

quot(0, s(

quot(s(

quot(

R

↳Dependency Pair Analysis

PLUS(s(x),y) -> PLUS(x,y)

TIMES(s(x),y) -> PLUS(y, times(x,y))

TIMES(s(x),y) -> TIMES(x,y)

DIV(x,y) -> QUOT(x,y,y)

DIV(div(x,y),z) -> DIV(x, times(y,z))

DIV(div(x,y),z) -> TIMES(y,z)

QUOT(s(x), s(y),z) -> QUOT(x,y,z)

QUOT(x, 0, s(z)) -> DIV(x, s(z))

Furthermore,

R

↳DPs

→DP Problem 1

↳Argument Filtering and Ordering

→DP Problem 2

↳AFS

→DP Problem 3

↳AFS

**PLUS(s( x), y) -> PLUS(x, y)**

plus(x, 0) ->x

plus(0,y) ->y

plus(s(x),y) -> s(plus(x,y))

times(0,y) -> 0

times(s(0),y) ->y

times(s(x),y) -> plus(y, times(x,y))

div(0,y) -> 0

div(x,y) -> quot(x,y,y)

div(div(x,y),z) -> div(x, times(y,z))

quot(0, s(y),z) -> 0

quot(s(x), s(y),z) -> quot(x,y,z)

quot(x, 0, s(z)) -> s(div(x, s(z)))

The following dependency pair can be strictly oriented:

PLUS(s(x),y) -> PLUS(x,y)

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

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

trivial

resulting in one new DP problem.

Used Argument Filtering System:

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

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

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 4

↳Dependency Graph

→DP Problem 2

↳AFS

→DP Problem 3

↳AFS

plus(x, 0) ->x

plus(0,y) ->y

plus(s(x),y) -> s(plus(x,y))

times(0,y) -> 0

times(s(0),y) ->y

times(s(x),y) -> plus(y, times(x,y))

div(0,y) -> 0

div(x,y) -> quot(x,y,y)

div(div(x,y),z) -> div(x, times(y,z))

quot(0, s(y),z) -> 0

quot(s(x), s(y),z) -> quot(x,y,z)

quot(x, 0, s(z)) -> s(div(x, s(z)))

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

↳AFS

**TIMES(s( x), y) -> TIMES(x, y)**

plus(x, 0) ->x

plus(0,y) ->y

plus(s(x),y) -> s(plus(x,y))

times(0,y) -> 0

times(s(0),y) ->y

times(s(x),y) -> plus(y, times(x,y))

div(0,y) -> 0

div(x,y) -> quot(x,y,y)

div(div(x,y),z) -> div(x, times(y,z))

quot(0, s(y),z) -> 0

quot(s(x), s(y),z) -> quot(x,y,z)

quot(x, 0, s(z)) -> s(div(x, s(z)))

The following dependency pair can be strictly oriented:

TIMES(s(x),y) -> TIMES(x,y)

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

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

trivial

resulting in one new DP problem.

Used Argument Filtering System:

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

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

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 5

↳Dependency Graph

→DP Problem 3

↳AFS

plus(x, 0) ->x

plus(0,y) ->y

plus(s(x),y) -> s(plus(x,y))

times(0,y) -> 0

times(s(0),y) ->y

times(s(x),y) -> plus(y, times(x,y))

div(0,y) -> 0

div(x,y) -> quot(x,y,y)

div(div(x,y),z) -> div(x, times(y,z))

quot(0, s(y),z) -> 0

quot(s(x), s(y),z) -> quot(x,y,z)

quot(x, 0, s(z)) -> s(div(x, s(z)))

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 3

↳Argument Filtering and Ordering

**DIV(div( x, y), z) -> DIV(x, times(y, z))**

plus(x, 0) ->x

plus(0,y) ->y

plus(s(x),y) -> s(plus(x,y))

times(0,y) -> 0

times(s(0),y) ->y

times(s(x),y) -> plus(y, times(x,y))

div(0,y) -> 0

div(x,y) -> quot(x,y,y)

div(div(x,y),z) -> div(x, times(y,z))

quot(0, s(y),z) -> 0

quot(s(x), s(y),z) -> quot(x,y,z)

quot(x, 0, s(z)) -> s(div(x, s(z)))

The following dependency pairs can be strictly oriented:

DIV(div(x,y),z) -> DIV(x, times(y,z))

QUOT(s(x), s(y),z) -> QUOT(x,y,z)

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

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

trivial

resulting in one new DP problem.

Used Argument Filtering System:

QUOT(x,_{1}x,_{2}x) ->_{3}x_{1}

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

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

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

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 3

↳AFS

→DP Problem 6

↳Instantiation Transformation

**QUOT( x, 0, s(z)) -> DIV(x, s(z))**

plus(x, 0) ->x

plus(0,y) ->y

plus(s(x),y) -> s(plus(x,y))

times(0,y) -> 0

times(s(0),y) ->y

times(s(x),y) -> plus(y, times(x,y))

div(0,y) -> 0

div(x,y) -> quot(x,y,y)

div(div(x,y),z) -> div(x, times(y,z))

quot(0, s(y),z) -> 0

quot(s(x), s(y),z) -> quot(x,y,z)

quot(x, 0, s(z)) -> s(div(x, s(z)))

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

As a result of transforming the rule

one new Dependency Pair is created:

DIV(x,y) -> QUOT(x,y,y)

DIV(x'', s(z'')) -> QUOT(x'', s(z''), s(z''))

The transformation is resulting in no new DP problems.

Duration:

0:00 minutes