times(

times(

plus(

plus(0,

plus(

plus(s(

R

↳Dependency Pair Analysis

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

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

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

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Argument Filtering and Ordering

→DP Problem 2

↳AFS

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

times(x, 0) -> 0

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

plus(x, 0) ->x

plus(0,x) ->x

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

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

innermost

The following dependency pairs can be strictly oriented:

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

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

There are no usable rules for innermost 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:

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

↳AFS

times(x, 0) -> 0

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

plus(x, 0) ->x

plus(0,x) ->x

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

↳Argument Filtering and Ordering

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

times(x, 0) -> 0

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

plus(x, 0) ->x

plus(0,x) ->x

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

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

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost 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:

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

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

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 4

↳Dependency Graph

times(x, 0) -> 0

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

plus(x, 0) ->x

plus(0,x) ->x

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

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

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes