minus(

minus(s(

le(0,

le(s(

le(s(

quot(0, s(

quot(s(

R

↳Dependency Pair Analysis

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

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

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

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Polo

→DP Problem 3

↳Remaining

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

minus(x, 0) ->x

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

le(0,y) -> true

le(s(x), 0) -> false

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

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

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

The following dependency pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(MINUS(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 4

↳Dependency Graph

→DP Problem 2

↳Polo

→DP Problem 3

↳Remaining

minus(x, 0) ->x

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

le(0,y) -> true

le(s(x), 0) -> false

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

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

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

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polynomial Ordering

→DP Problem 3

↳Remaining

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

minus(x, 0) ->x

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

le(0,y) -> true

le(s(x), 0) -> false

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

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

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

The following dependency pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(LE(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 5

↳Dependency Graph

→DP Problem 3

↳Remaining

minus(x, 0) ->x

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

le(0,y) -> true

le(s(x), 0) -> false

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

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

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

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 3

↳Remaining Obligation(s)

The following remains to be proven:

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

minus(x, 0) ->x

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

le(0,y) -> true

le(s(x), 0) -> false

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

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

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

Duration:

0:00 minutes