minus(

minus(s(

p(s(

div(0, s(

div(s(

R

↳Dependency Pair Analysis

MINUS(s(X), s(Y)) -> P(minus(X,Y))

MINUS(s(X), s(Y)) -> MINUS(X,Y)

DIV(s(X), s(Y)) -> DIV(minus(X,Y), s(Y))

DIV(s(X), s(Y)) -> MINUS(X,Y)

Furthermore,

R

↳DPs

→DP Problem 1

↳Argument Filtering and Ordering

→DP Problem 2

↳AFS

**MINUS(s( X), s(Y)) -> MINUS(X, Y)**

minus(X, 0) ->X

minus(s(X), s(Y)) -> p(minus(X,Y))

p(s(X)) ->X

div(0, s(Y)) -> 0

div(s(X), s(Y)) -> s(div(minus(X,Y), s(Y)))

The following dependency pair can be strictly oriented:

MINUS(s(X), s(Y)) -> MINUS(X,Y)

The following rules can be oriented:

minus(X, 0) ->X

minus(s(X), s(Y)) -> p(minus(X,Y))

p(s(X)) ->X

div(0, s(Y)) -> 0

div(s(X), s(Y)) -> s(div(minus(X,Y), s(Y)))

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

div > s > p

resulting in one new DP problem.

Used Argument Filtering System:

MINUS(x,_{1}x) -> MINUS(_{2}x,_{1}x)_{2}

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

minus(x,_{1}x) ->_{2}x_{1}

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

div(x,_{1}x) -> div(_{2}x,_{1}x)_{2}

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 3

↳Dependency Graph

→DP Problem 2

↳AFS

minus(X, 0) ->X

minus(s(X), s(Y)) -> p(minus(X,Y))

p(s(X)) ->X

div(0, s(Y)) -> 0

div(s(X), s(Y)) -> s(div(minus(X,Y), s(Y)))

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳Argument Filtering and Ordering

**DIV(s( X), s(Y)) -> DIV(minus(X, Y), s(Y))**

minus(X, 0) ->X

minus(s(X), s(Y)) -> p(minus(X,Y))

p(s(X)) ->X

div(0, s(Y)) -> 0

div(s(X), s(Y)) -> s(div(minus(X,Y), s(Y)))

The following dependency pair can be strictly oriented:

DIV(s(X), s(Y)) -> DIV(minus(X,Y), s(Y))

The following rules can be oriented:

minus(X, 0) ->X

minus(s(X), s(Y)) -> p(minus(X,Y))

p(s(X)) ->X

div(0, s(Y)) -> 0

div(s(X), s(Y)) -> s(div(minus(X,Y), s(Y)))

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

div > s > p

resulting in one new DP problem.

Used Argument Filtering System:

DIV(x,_{1}x) -> DIV(_{2}x,_{1}x)_{2}

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

minus(x,_{1}x) ->_{2}x_{1}

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

div(x,_{1}x) -> div(_{2}x,_{1}x)_{2}

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 4

↳Dependency Graph

minus(X, 0) ->X

minus(s(X), s(Y)) -> p(minus(X,Y))

p(s(X)) ->X

div(0, s(Y)) -> 0

div(s(X), s(Y)) -> s(div(minus(X,Y), s(Y)))

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes