minus(

minus(s(

quot(0, s(

quot(s(

R

↳Dependency Pair Analysis

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

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

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Argument Filtering and Ordering

→DP Problem 2

↳AFS

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

minus(x, 0) ->x

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

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

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

The following dependency pair can be strictly oriented:

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

The following rules can be oriented:

minus(x, 0) ->x

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

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

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

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

quot > s

resulting in one new DP problem.

Used Argument Filtering System:

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

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

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

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

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 3

↳Dependency Graph

→DP Problem 2

↳AFS

minus(x, 0) ->x

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

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

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

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳Argument Filtering and Ordering

**QUOT(s( x), s(y)) -> QUOT(minus(x, y), s(y))**

minus(x, 0) ->x

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

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

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

The following dependency pair can be strictly oriented:

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

The following rules can be oriented:

minus(x, 0) ->x

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

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

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

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

quot > s

resulting in one new DP problem.

Used Argument Filtering System:

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

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

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

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

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 4

↳Dependency Graph

minus(x, 0) ->x

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

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

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

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes