-(

-(s(

p(s(

f(s(

f(

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↳Dependency Pair Analysis

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

F(s(x),y) -> F(p(-(s(x),y)), p(-(y, s(x))))

F(s(x),y) -> P(-(s(x),y))

F(s(x),y) -> -'(s(x),y)

F(s(x),y) -> P(-(y, s(x)))

F(s(x),y) -> -'(y, s(x))

F(x, s(y)) -> F(p(-(x, s(y))), p(-(s(y),x)))

F(x, s(y)) -> P(-(x, s(y)))

F(x, s(y)) -> -'(x, s(y))

F(x, s(y)) -> P(-(s(y),x))

F(x, s(y)) -> -'(s(y),x)

Furthermore,

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↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Remaining

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

-(x, 0) ->x

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

p(s(x)) ->x

f(s(x),y) -> f(p(-(s(x),y)), p(-(y, s(x))))

f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y),x)))

The following dependency pair can be strictly oriented:

-'(s(x), s(y)) -> -'(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(-'(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

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↳DPs

→DP Problem 1

↳Polo

→DP Problem 3

↳Dependency Graph

→DP Problem 2

↳Remaining

-(x, 0) ->x

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

p(s(x)) ->x

f(s(x),y) -> f(p(-(s(x),y)), p(-(y, s(x))))

f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y),x)))

Using the Dependency Graph resulted in no new DP problems.

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↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Remaining Obligation(s)

The following remains to be proven:

**F( x, s(y)) -> F(p(-(x, s(y))), p(-(s(y), x)))**

-(x, 0) ->x

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

p(s(x)) ->x

f(s(x),y) -> f(p(-(s(x),y)), p(-(y, s(x))))

f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y),x)))

Duration:

0:00 minutes