pred(s(

minus(

minus(

quot(0, s(

quot(s(

log(s(0)) -> 0

log(s(s(

R

↳Dependency Pair Analysis

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

MINUS(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)

LOG(s(s(x))) -> LOG(s(quot(x, s(s(0)))))

LOG(s(s(x))) -> QUOT(x, s(s(0)))

Furthermore,

R

↳DPs

→DP Problem 1

↳Argument Filtering and Ordering

→DP Problem 2

↳AFS

→DP Problem 3

↳AFS

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

pred(s(x)) ->x

minus(x, 0) ->x

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

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

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

log(s(0)) -> 0

log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

The following dependency pair can be strictly oriented:

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

The following rules can be oriented:

pred(s(x)) ->x

minus(x, 0) ->x

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

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

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

log(s(0)) -> 0

log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

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

{log, 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}

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

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

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

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

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 4

↳Dependency Graph

→DP Problem 2

↳AFS

→DP Problem 3

↳AFS

pred(s(x)) ->x

minus(x, 0) ->x

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

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

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

log(s(0)) -> 0

log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

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

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

pred(s(x)) ->x

minus(x, 0) ->x

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

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

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

log(s(0)) -> 0

log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

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

pred(s(x)) ->x

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

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

log(s(0)) -> 0

log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

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

{log, 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}

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

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

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

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 5

↳Dependency Graph

→DP Problem 3

↳AFS

pred(s(x)) ->x

minus(x, 0) ->x

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

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

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

log(s(0)) -> 0

log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

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

**LOG(s(s( x))) -> LOG(s(quot(x, s(s(0)))))**

pred(s(x)) ->x

minus(x, 0) ->x

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

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

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

log(s(0)) -> 0

log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

The following dependency pair can be strictly oriented:

LOG(s(s(x))) -> LOG(s(quot(x, s(s(0)))))

The following rules can be oriented:

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

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

minus(x, 0) ->x

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

pred(s(x)) ->x

log(s(0)) -> 0

log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

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

log > s

resulting in one new DP problem.

Used Argument Filtering System:

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

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

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

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

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

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

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 3

↳AFS

→DP Problem 6

↳Dependency Graph

pred(s(x)) ->x

minus(x, 0) ->x

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

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

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

log(s(0)) -> 0

log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes