p(s(

fact(0) -> s(0)

fact(s(

*(0,

*(s(

+(

+(

R

↳Dependency Pair Analysis

FACT(s(x)) -> *'(s(x), fact(p(s(x))))

FACT(s(x)) -> FACT(p(s(x)))

FACT(s(x)) -> P(s(x))

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

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

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Argument Filtering and Ordering

→DP Problem 2

↳AFS

→DP Problem 3

↳Remaining

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

p(s(x)) ->x

fact(0) -> s(0)

fact(s(x)) -> *(s(x), fact(p(s(x))))

*(0,y) -> 0

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

+(x, 0) ->x

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

The following dependency pair can be strictly oriented:

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

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

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

trivial

resulting in one new DP problem.

Used Argument Filtering System:

+'(x,_{1}x) -> +'(_{2}x,_{1}x)_{2}

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

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 4

↳Dependency Graph

→DP Problem 2

↳AFS

→DP Problem 3

↳Remaining

p(s(x)) ->x

fact(0) -> s(0)

fact(s(x)) -> *(s(x), fact(p(s(x))))

*(0,y) -> 0

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

+(x, 0) ->x

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

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

↳Remaining

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

p(s(x)) ->x

fact(0) -> s(0)

fact(s(x)) -> *(s(x), fact(p(s(x))))

*(0,y) -> 0

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

+(x, 0) ->x

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

The following dependency pair can be strictly oriented:

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

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

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

trivial

resulting in one new DP problem.

Used Argument Filtering System:

*'(x,_{1}x) -> *'(_{2}x,_{1}x)_{2}

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

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 5

↳Dependency Graph

→DP Problem 3

↳Remaining

p(s(x)) ->x

fact(0) -> s(0)

fact(s(x)) -> *(s(x), fact(p(s(x))))

*(0,y) -> 0

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

+(x, 0) ->x

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

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 3

↳Remaining Obligation(s)

The following remains to be proven:

**FACT(s( x)) -> FACT(p(s(x)))**

p(s(x)) ->x

fact(0) -> s(0)

fact(s(x)) -> *(s(x), fact(p(s(x))))

*(0,y) -> 0

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

+(x, 0) ->x

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

Duration:

0:00 minutes