times(

times(

times(

plus(

plus(

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↳Dependency Pair Analysis

TIMES(x, plus(y, s(z))) -> PLUS(times(x, plus(y, times(s(z), 0))), times(x, s(z)))

TIMES(x, plus(y, s(z))) -> TIMES(x, plus(y, times(s(z), 0)))

TIMES(x, plus(y, s(z))) -> PLUS(y, times(s(z), 0))

TIMES(x, plus(y, s(z))) -> TIMES(s(z), 0)

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

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

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

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

Furthermore,

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↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Remaining

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

times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))

times(x, 0) -> 0

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

plus(x, 0) ->x

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

The following dependency pair can be strictly oriented:

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

There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(PLUS(x)_{1}, x_{2})= x _{2}_{ }^{ }_{ }^{ }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

times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))

times(x, 0) -> 0

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

plus(x, 0) ->x

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

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:

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

times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))

times(x, 0) -> 0

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

plus(x, 0) ->x

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

Duration:

0:00 minutes