minus(

minus(s(

le(0,

le(s(

le(s(

quot(

if

if

R

↳Dependency Pair Analysis

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

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

QUOT(x, s(y)) -> IF_{QUOT}(le(s(y),x),x, s(y))

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

IF_{QUOT}(true,x,y) -> QUOT(minus(x,y),y)

IF_{QUOT}(true,x,y) -> MINUS(x,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(x, s(y)) -> if_{quot}(le(s(y),x),x, s(y))

if_{quot}(true,x,y) -> s(quot(minus(x,y),y))

if_{quot}(false,x,y) -> 0

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost 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(x, s(y)) -> if_{quot}(le(s(y),x),x, s(y))

if_{quot}(true,x,y) -> s(quot(minus(x,y),y))

if_{quot}(false,x,y) -> 0

innermost

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(x, s(y)) -> if_{quot}(le(s(y),x),x, s(y))

if_{quot}(true,x,y) -> s(quot(minus(x,y),y))

if_{quot}(false,x,y) -> 0

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost 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(x, s(y)) -> if_{quot}(le(s(y),x),x, s(y))

if_{quot}(true,x,y) -> s(quot(minus(x,y),y))

if_{quot}(false,x,y) -> 0

innermost

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:

**IF _{QUOT}(true, x, y) -> QUOT(minus(x, y), 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(x, s(y)) -> if_{quot}(le(s(y),x),x, s(y))

if_{quot}(true,x,y) -> s(quot(minus(x,y),y))

if_{quot}(false,x,y) -> 0

innermost

Duration:

0:00 minutes