quot(0, s(

quot(s(

quot(

plus(0,

plus(s(

R

↳Dependency Pair Analysis

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

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

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

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Argument Filtering and Ordering

→DP Problem 2

↳Remaining

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

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

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

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

plus(0,y) ->y

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

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.

Used ordering: Homeomorphic Embedding Order with EMB

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 3

↳Dependency Graph

→DP Problem 2

↳Remaining

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

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

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

plus(0,y) ->y

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

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳Remaining Obligation(s)

The following remains to be proven:

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

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

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

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

plus(0,y) ->y

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

innermost

Duration:

0:00 minutes