TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60001 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (47ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – Problem 3 was processed with processor BackwardInstantiation (4ms).
 |    | – Problem 6 was processed with processor BackwardInstantiation (1ms).
 |    |    | – Problem 7 was processed with processor Propagation (2ms).
 |    |    |    | – Problem 8 remains open; application of the following processors failed [ForwardNarrowing (0ms), BackwardInstantiation (1ms), ForwardInstantiation (1ms), Propagation (1ms)].
 | – Problem 4 was processed with processor PolynomialLinearRange4iUR (73ms).
 |    | – Problem 5 was processed with processor PolynomialLinearRange4iUR (23ms).

The following open problems remain:



Open Dependency Pair Problem 3

Dependency Pairs

if#(false, x, y, z)gen#(x, y, s(z))gen#(x, y, z)if#(ge(z, x), x, y, z)

Rewrite Rules

times(x, y)sum(generate(x, y))generate(x, y)gen(x, y, 0)
gen(x, y, z)if(ge(z, x), x, y, z)if(true, x, y, z)nil
if(false, x, y, z)cons(y, gen(x, y, s(z)))sum(nil)0
sum(cons(0, xs))sum(xs)sum(cons(s(x), xs))s(sum(cons(x, xs)))
ge(x, 0)truege(0, s(y))false
ge(s(x), s(y))ge(x, y)

Original Signature

Termination of terms over the following signature is verified: gen, 0, s, generate, times, if, false, true, sum, ge, cons, nil


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

if#(false, x, y, z)gen#(x, y, s(z))gen#(x, y, z)if#(ge(z, x), x, y, z)
sum#(cons(0, xs))sum#(xs)gen#(x, y, z)ge#(z, x)
generate#(x, y)gen#(x, y, 0)sum#(cons(s(x), xs))sum#(cons(x, xs))
ge#(s(x), s(y))ge#(x, y)times#(x, y)generate#(x, y)
times#(x, y)sum#(generate(x, y))

Rewrite Rules

times(x, y)sum(generate(x, y))generate(x, y)gen(x, y, 0)
gen(x, y, z)if(ge(z, x), x, y, z)if(true, x, y, z)nil
if(false, x, y, z)cons(y, gen(x, y, s(z)))sum(nil)0
sum(cons(0, xs))sum(xs)sum(cons(s(x), xs))s(sum(cons(x, xs)))
ge(x, 0)truege(0, s(y))false
ge(s(x), s(y))ge(x, y)

Original Signature

Termination of terms over the following signature is verified: gen, 0, generate, s, times, if, sum, true, false, ge, nil, cons

Strategy


The following SCCs where found

sum#(cons(0, xs)) → sum#(xs)sum#(cons(s(x), xs)) → sum#(cons(x, xs))

if#(false, x, y, z) → gen#(x, y, s(z))gen#(x, y, z) → if#(ge(z, x), x, y, z)

ge#(s(x), s(y)) → ge#(x, y)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

ge#(s(x), s(y))ge#(x, y)

Rewrite Rules

times(x, y)sum(generate(x, y))generate(x, y)gen(x, y, 0)
gen(x, y, z)if(ge(z, x), x, y, z)if(true, x, y, z)nil
if(false, x, y, z)cons(y, gen(x, y, s(z)))sum(nil)0
sum(cons(0, xs))sum(xs)sum(cons(s(x), xs))s(sum(cons(x, xs)))
ge(x, 0)truege(0, s(y))false
ge(s(x), s(y))ge(x, y)

Original Signature

Termination of terms over the following signature is verified: gen, 0, generate, s, times, if, sum, true, false, ge, nil, cons

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

ge#(s(x), s(y))ge#(x, y)

Problem 3: BackwardInstantiation



Dependency Pair Problem

Dependency Pairs

if#(false, x, y, z)gen#(x, y, s(z))gen#(x, y, z)if#(ge(z, x), x, y, z)

Rewrite Rules

times(x, y)sum(generate(x, y))generate(x, y)gen(x, y, 0)
gen(x, y, z)if(ge(z, x), x, y, z)if(true, x, y, z)nil
if(false, x, y, z)cons(y, gen(x, y, s(z)))sum(nil)0
sum(cons(0, xs))sum(xs)sum(cons(s(x), xs))s(sum(cons(x, xs)))
ge(x, 0)truege(0, s(y))false
ge(s(x), s(y))ge(x, y)

Original Signature

Termination of terms over the following signature is verified: gen, 0, generate, s, times, if, sum, true, false, ge, nil, cons

Strategy


Instantiation

For all potential predecessors l → r of the rule gen#(x, y, z) → if#(ge(z, x), x, y, z) on dependency pair chains it holds that: Thus, gen#(x, y, z) → if#(ge(z, x), x, y, z) is replaced by instances determined through the above matching. These instances are:
gen#(_x, _y, s(_z)) → if#(ge(s(_z), _x), _x, _y, s(_z))

Problem 6: BackwardInstantiation



Dependency Pair Problem

Dependency Pairs

if#(false, x, y, z)gen#(x, y, s(z))gen#(_x, _y, s(_z))if#(ge(s(_z), _x), _x, _y, s(_z))

Rewrite Rules

times(x, y)sum(generate(x, y))generate(x, y)gen(x, y, 0)
gen(x, y, z)if(ge(z, x), x, y, z)if(true, x, y, z)nil
if(false, x, y, z)cons(y, gen(x, y, s(z)))sum(nil)0
sum(cons(0, xs))sum(xs)sum(cons(s(x), xs))s(sum(cons(x, xs)))
ge(x, 0)truege(0, s(y))false
ge(s(x), s(y))ge(x, y)

Original Signature

Termination of terms over the following signature is verified: gen, 0, s, generate, times, if, false, true, sum, ge, cons, nil

Strategy


Instantiation

For all potential predecessors l → r of the rule gen#(_x, _y, s(_z)) → if#(ge(s(_z), _x), _x, _y, s(_z)) on dependency pair chains it holds that: Thus, gen#(_x, _y, s(_z)) → if#(ge(s(_z), _x), _x, _y, s(_z)) is replaced by instances determined through the above matching. These instances are:
gen#(x, y, s(z)) → if#(ge(s(z), x), x, y, s(z))

Problem 7: Propagation



Dependency Pair Problem

Dependency Pairs

if#(false, x, y, z)gen#(x, y, s(z))gen#(x, y, s(z))if#(ge(s(z), x), x, y, s(z))

Rewrite Rules

times(x, y)sum(generate(x, y))generate(x, y)gen(x, y, 0)
gen(x, y, z)if(ge(z, x), x, y, z)if(true, x, y, z)nil
if(false, x, y, z)cons(y, gen(x, y, s(z)))sum(nil)0
sum(cons(0, xs))sum(xs)sum(cons(s(x), xs))s(sum(cons(x, xs)))
ge(x, 0)truege(0, s(y))false
ge(s(x), s(y))ge(x, y)

Original Signature

Termination of terms over the following signature is verified: gen, 0, generate, s, times, if, sum, true, false, ge, nil, cons

Strategy


The dependency pairs if#(false, x, y, z) → gen#(x, y, s(z)) and gen#(x, y, s(z)) → if#(ge(s(z), x), x, y, s(z)) are consolidated into the rule if#(false, x, y, z) → if#(ge(s(z), x), x, y, s(z)) .

This is possible as

The dependency pairs if#(false, x, y, z) → gen#(x, y, s(z)) and gen#(x, y, s(z)) → if#(ge(s(z), x), x, y, s(z)) are consolidated into the rule if#(false, x, y, z) → if#(ge(s(z), x), x, y, s(z)) .

This is possible as


Summary

Removed Dependency PairsAdded Dependency Pairs
if#(false, x, y, z) → gen#(x, y, s(z))if#(false, x, y, z) → if#(ge(s(z), x), x, y, s(z))
gen#(x, y, s(z)) → if#(ge(s(z), x), x, y, s(z)) 

Problem 4: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

sum#(cons(0, xs))sum#(xs)sum#(cons(s(x), xs))sum#(cons(x, xs))

Rewrite Rules

times(x, y)sum(generate(x, y))generate(x, y)gen(x, y, 0)
gen(x, y, z)if(ge(z, x), x, y, z)if(true, x, y, z)nil
if(false, x, y, z)cons(y, gen(x, y, s(z)))sum(nil)0
sum(cons(0, xs))sum(xs)sum(cons(s(x), xs))s(sum(cons(x, xs)))
ge(x, 0)truege(0, s(y))false
ge(s(x), s(y))ge(x, y)

Original Signature

Termination of terms over the following signature is verified: gen, 0, generate, s, times, if, sum, true, false, ge, nil, cons

Strategy


Polynomial Interpretation

There are no usable rules

The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:

sum#(cons(s(x), xs))sum#(cons(x, xs))

Problem 5: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

sum#(cons(0, xs))sum#(xs)

Rewrite Rules

times(x, y)sum(generate(x, y))generate(x, y)gen(x, y, 0)
gen(x, y, z)if(ge(z, x), x, y, z)if(true, x, y, z)nil
if(false, x, y, z)cons(y, gen(x, y, s(z)))sum(nil)0
sum(cons(0, xs))sum(xs)sum(cons(s(x), xs))s(sum(cons(x, xs)))
ge(x, 0)truege(0, s(y))false
ge(s(x), s(y))ge(x, y)

Original Signature

Termination of terms over the following signature is verified: gen, 0, s, generate, times, if, false, true, sum, ge, cons, nil

Strategy


Polynomial Interpretation

There are no usable rules

The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:

sum#(cons(0, xs))sum#(xs)