TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60000 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (75ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – Problem 3 was processed with processor BackwardInstantiation (5ms).
 |    | – Problem 5 was processed with processor ForwardInstantiation (5ms).
 |    |    | – Problem 6 was processed with processor BackwardInstantiation (5ms).
 |    |    |    | – Problem 7 was processed with processor ForwardInstantiation (12ms).
 |    |    |    |    | – Problem 8 was processed with processor BackwardInstantiation (9ms).
 |    |    |    |    |    | – Problem 9 was processed with processor Propagation (8ms).
 |    |    |    |    |    |    | – Problem 10 remains open; application of the following processors failed [].
 | – Problem 4 was processed with processor SubtermCriterion (1ms).

The following open problems remain:



Open Dependency Pair Problem 3

Dependency Pairs

sumList#(xs, y)if#(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))if#(false, false, y, xs, ys, x)sumList#(ys, x)
if#(false, true, y, xs, ys, x)sumList#(xs, y)

Rewrite Rules

isEmpty(cons(x, xs))falseisEmpty(nil)true
isZero(0)trueisZero(s(x))false
head(cons(x, xs))xtail(cons(x, xs))xs
tail(nil)nilp(s(s(x)))s(p(s(x)))
p(s(0))0p(0)0
inc(s(x))s(inc(x))inc(0)s(0)
sumList(xs, y)if(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))if(true, b, y, xs, ys, x)y
if(false, true, y, xs, ys, x)sumList(xs, y)if(false, false, y, xs, ys, x)sumList(ys, x)
sum(xs)sumList(xs, 0)

Original Signature

Termination of terms over the following signature is verified: sumList, sum, true, tail, isEmpty, 0, s, inc, if, p, false, head, cons, nil, isZero


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

sumList#(xs, y)head#(xs)sumList#(xs, y)p#(head(xs))
sum#(xs)sumList#(xs, 0)sumList#(xs, y)inc#(y)
sumList#(xs, y)if#(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))sumList#(xs, y)tail#(xs)
inc#(s(x))inc#(x)if#(false, true, y, xs, ys, x)sumList#(xs, y)
if#(false, false, y, xs, ys, x)sumList#(ys, x)p#(s(s(x)))p#(s(x))
sumList#(xs, y)isZero#(head(xs))sumList#(xs, y)isEmpty#(xs)

Rewrite Rules

isEmpty(cons(x, xs))falseisEmpty(nil)true
isZero(0)trueisZero(s(x))false
head(cons(x, xs))xtail(cons(x, xs))xs
tail(nil)nilp(s(s(x)))s(p(s(x)))
p(s(0))0p(0)0
inc(s(x))s(inc(x))inc(0)s(0)
sumList(xs, y)if(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))if(true, b, y, xs, ys, x)y
if(false, true, y, xs, ys, x)sumList(xs, y)if(false, false, y, xs, ys, x)sumList(ys, x)
sum(xs)sumList(xs, 0)

Original Signature

Termination of terms over the following signature is verified: sumList, sum, true, tail, isEmpty, 0, s, inc, if, p, false, head, isZero, nil, cons

Strategy


The following SCCs where found

inc#(s(x)) → inc#(x)

p#(s(s(x))) → p#(s(x))

sumList#(xs, y) → if#(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))if#(false, true, y, xs, ys, x) → sumList#(xs, y)
if#(false, false, y, xs, ys, x) → sumList#(ys, x)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

p#(s(s(x)))p#(s(x))

Rewrite Rules

isEmpty(cons(x, xs))falseisEmpty(nil)true
isZero(0)trueisZero(s(x))false
head(cons(x, xs))xtail(cons(x, xs))xs
tail(nil)nilp(s(s(x)))s(p(s(x)))
p(s(0))0p(0)0
inc(s(x))s(inc(x))inc(0)s(0)
sumList(xs, y)if(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))if(true, b, y, xs, ys, x)y
if(false, true, y, xs, ys, x)sumList(xs, y)if(false, false, y, xs, ys, x)sumList(ys, x)
sum(xs)sumList(xs, 0)

Original Signature

Termination of terms over the following signature is verified: sumList, sum, true, tail, isEmpty, 0, s, inc, if, p, false, head, isZero, nil, cons

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

p#(s(s(x)))p#(s(x))

Problem 3: BackwardInstantiation



Dependency Pair Problem

Dependency Pairs

sumList#(xs, y)if#(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))if#(false, true, y, xs, ys, x)sumList#(xs, y)
if#(false, false, y, xs, ys, x)sumList#(ys, x)

Rewrite Rules

isEmpty(cons(x, xs))falseisEmpty(nil)true
isZero(0)trueisZero(s(x))false
head(cons(x, xs))xtail(cons(x, xs))xs
tail(nil)nilp(s(s(x)))s(p(s(x)))
p(s(0))0p(0)0
inc(s(x))s(inc(x))inc(0)s(0)
sumList(xs, y)if(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))if(true, b, y, xs, ys, x)y
if(false, true, y, xs, ys, x)sumList(xs, y)if(false, false, y, xs, ys, x)sumList(ys, x)
sum(xs)sumList(xs, 0)

Original Signature

Termination of terms over the following signature is verified: sumList, sum, true, tail, isEmpty, 0, s, inc, if, p, false, head, isZero, nil, cons

Strategy


Instantiation

For all potential predecessors l → r of the rule sumList#(xs, y) → if#(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y)) on dependency pair chains it holds that: Thus, sumList#(xs, y) → if#(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y)) is replaced by instances determined through the above matching. These instances are:
sumList#(ys, x) → if#(isEmpty(ys), isZero(head(ys)), x, tail(ys), cons(p(head(ys)), tail(ys)), inc(x))sumList#(_xs, _y) → if#(isEmpty(_xs), isZero(head(_xs)), _y, tail(_xs), cons(p(head(_xs)), tail(_xs)), inc(_y))

Problem 5: ForwardInstantiation



Dependency Pair Problem

Dependency Pairs

if#(false, false, y, xs, ys, x)sumList#(ys, x)if#(false, true, y, xs, ys, x)sumList#(xs, y)
sumList#(ys, x)if#(isEmpty(ys), isZero(head(ys)), x, tail(ys), cons(p(head(ys)), tail(ys)), inc(x))sumList#(_xs, _y)if#(isEmpty(_xs), isZero(head(_xs)), _y, tail(_xs), cons(p(head(_xs)), tail(_xs)), inc(_y))

Rewrite Rules

isEmpty(cons(x, xs))falseisEmpty(nil)true
isZero(0)trueisZero(s(x))false
head(cons(x, xs))xtail(cons(x, xs))xs
tail(nil)nilp(s(s(x)))s(p(s(x)))
p(s(0))0p(0)0
inc(s(x))s(inc(x))inc(0)s(0)
sumList(xs, y)if(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))if(true, b, y, xs, ys, x)y
if(false, true, y, xs, ys, x)sumList(xs, y)if(false, false, y, xs, ys, x)sumList(ys, x)
sum(xs)sumList(xs, 0)

Original Signature

Termination of terms over the following signature is verified: sumList, sum, true, tail, isEmpty, 0, s, inc, if, p, false, head, cons, nil, isZero

Strategy


Instantiation

For all potential successors l → r of the rule if#(false, true, y, xs, ys, x) → sumList#(xs, y) on dependency pair chains it holds that: Thus, if#(false, true, y, xs, ys, x) → sumList#(xs, y) is replaced by instances determined through the above matching. These instances are:
if#(false, true, y, xs, xs, y) → sumList#(xs, y)if#(false, true, y, xs, ys, x) → sumList#(xs, y)

Instantiation

For all potential successors l → r of the rule if#(false, false, y, xs, ys, x) → sumList#(ys, x) on dependency pair chains it holds that: Thus, if#(false, false, y, xs, ys, x) → sumList#(ys, x) is replaced by instances determined through the above matching. These instances are:
if#(false, false, y, xs, ys, x) → sumList#(ys, x)

Problem 6: BackwardInstantiation



Dependency Pair Problem

Dependency Pairs

if#(false, true, y, xs, xs, y)sumList#(xs, y)if#(false, false, y, xs, ys, x)sumList#(ys, x)
if#(false, true, y, xs, ys, x)sumList#(xs, y)sumList#(ys, x)if#(isEmpty(ys), isZero(head(ys)), x, tail(ys), cons(p(head(ys)), tail(ys)), inc(x))
sumList#(_xs, _y)if#(isEmpty(_xs), isZero(head(_xs)), _y, tail(_xs), cons(p(head(_xs)), tail(_xs)), inc(_y))

Rewrite Rules

isEmpty(cons(x, xs))falseisEmpty(nil)true
isZero(0)trueisZero(s(x))false
head(cons(x, xs))xtail(cons(x, xs))xs
tail(nil)nilp(s(s(x)))s(p(s(x)))
p(s(0))0p(0)0
inc(s(x))s(inc(x))inc(0)s(0)
sumList(xs, y)if(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))if(true, b, y, xs, ys, x)y
if(false, true, y, xs, ys, x)sumList(xs, y)if(false, false, y, xs, ys, x)sumList(ys, x)
sum(xs)sumList(xs, 0)

Original Signature

Termination of terms over the following signature is verified: sumList, sum, true, tail, isEmpty, 0, s, inc, if, p, false, head, isZero, nil, cons

Strategy


Instantiation

For all potential predecessors l → r of the rule sumList#(ys, x) → if#(isEmpty(ys), isZero(head(ys)), x, tail(ys), cons(p(head(ys)), tail(ys)), inc(x)) on dependency pair chains it holds that: Thus, sumList#(ys, x) → if#(isEmpty(ys), isZero(head(ys)), x, tail(ys), cons(p(head(ys)), tail(ys)), inc(x)) is replaced by instances determined through the above matching. These instances are:
sumList#(_ys, _x) → if#(isEmpty(_ys), isZero(head(_ys)), _x, tail(_ys), cons(p(head(_ys)), tail(_ys)), inc(_x))sumList#(xs, y) → if#(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))

Instantiation

For all potential predecessors l → r of the rule sumList#(_xs, _y) → if#(isEmpty(_xs), isZero(head(_xs)), _y, tail(_xs), cons(p(head(_xs)), tail(_xs)), inc(_y)) on dependency pair chains it holds that: Thus, sumList#(_xs, _y) → if#(isEmpty(_xs), isZero(head(_xs)), _y, tail(_xs), cons(p(head(_xs)), tail(_xs)), inc(_y)) is replaced by instances determined through the above matching. These instances are:
sumList#(xs, y) → if#(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))sumList#(ys, x) → if#(isEmpty(ys), isZero(head(ys)), x, tail(ys), cons(p(head(ys)), tail(ys)), inc(x))

Problem 7: ForwardInstantiation



Dependency Pair Problem

Dependency Pairs

sumList#(_ys, _x)if#(isEmpty(_ys), isZero(head(_ys)), _x, tail(_ys), cons(p(head(_ys)), tail(_ys)), inc(_x))sumList#(xs, y)if#(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))
if#(false, true, y, xs, xs, y)sumList#(xs, y)if#(false, true, y, xs, ys, x)sumList#(xs, y)
if#(false, false, y, xs, ys, x)sumList#(ys, x)sumList#(ys, x)if#(isEmpty(ys), isZero(head(ys)), x, tail(ys), cons(p(head(ys)), tail(ys)), inc(x))

Rewrite Rules

isEmpty(cons(x, xs))falseisEmpty(nil)true
isZero(0)trueisZero(s(x))false
head(cons(x, xs))xtail(cons(x, xs))xs
tail(nil)nilp(s(s(x)))s(p(s(x)))
p(s(0))0p(0)0
inc(s(x))s(inc(x))inc(0)s(0)
sumList(xs, y)if(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))if(true, b, y, xs, ys, x)y
if(false, true, y, xs, ys, x)sumList(xs, y)if(false, false, y, xs, ys, x)sumList(ys, x)
sum(xs)sumList(xs, 0)

Original Signature

Termination of terms over the following signature is verified: sumList, sum, true, tail, isEmpty, 0, s, inc, if, p, false, head, cons, nil, isZero

Strategy


Instantiation

For all potential successors l → r of the rule if#(false, true, y, xs, xs, y) → sumList#(xs, y) on dependency pair chains it holds that: Thus, if#(false, true, y, xs, xs, y) → sumList#(xs, y) is replaced by instances determined through the above matching. These instances are:
if#(false, true, y, xs, xs, y) → sumList#(xs, y)

Instantiation

For all potential successors l → r of the rule if#(false, false, y, xs, ys, x) → sumList#(ys, x) on dependency pair chains it holds that: Thus, if#(false, false, y, xs, ys, x) → sumList#(ys, x) is replaced by instances determined through the above matching. These instances are:
if#(false, false, x, ys, ys, x) → sumList#(ys, x)if#(false, false, y, xs, ys, x) → sumList#(ys, x)

Instantiation

For all potential successors l → r of the rule if#(false, true, y, xs, ys, x) → sumList#(xs, y) on dependency pair chains it holds that: Thus, if#(false, true, y, xs, ys, x) → sumList#(xs, y) is replaced by instances determined through the above matching. These instances are:
if#(false, true, y, xs, xs, y) → sumList#(xs, y)if#(false, true, y, xs, ys, x) → sumList#(xs, y)

Problem 8: BackwardInstantiation



Dependency Pair Problem

Dependency Pairs

sumList#(xs, y)if#(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))sumList#(_ys, _x)if#(isEmpty(_ys), isZero(head(_ys)), _x, tail(_ys), cons(p(head(_ys)), tail(_ys)), inc(_x))
if#(false, true, y, xs, xs, y)sumList#(xs, y)if#(false, true, y, xs, ys, x)sumList#(xs, y)
if#(false, false, x, ys, ys, x)sumList#(ys, x)if#(false, false, y, xs, ys, x)sumList#(ys, x)
sumList#(ys, x)if#(isEmpty(ys), isZero(head(ys)), x, tail(ys), cons(p(head(ys)), tail(ys)), inc(x))

Rewrite Rules

isEmpty(cons(x, xs))falseisEmpty(nil)true
isZero(0)trueisZero(s(x))false
head(cons(x, xs))xtail(cons(x, xs))xs
tail(nil)nilp(s(s(x)))s(p(s(x)))
p(s(0))0p(0)0
inc(s(x))s(inc(x))inc(0)s(0)
sumList(xs, y)if(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))if(true, b, y, xs, ys, x)y
if(false, true, y, xs, ys, x)sumList(xs, y)if(false, false, y, xs, ys, x)sumList(ys, x)
sum(xs)sumList(xs, 0)

Original Signature

Termination of terms over the following signature is verified: sumList, sum, true, tail, isEmpty, 0, s, inc, if, p, false, head, isZero, nil, cons

Strategy


Instantiation

For all potential predecessors l → r of the rule sumList#(_ys, _x) → if#(isEmpty(_ys), isZero(head(_ys)), _x, tail(_ys), cons(p(head(_ys)), tail(_ys)), inc(_x)) on dependency pair chains it holds that: Thus, sumList#(_ys, _x) → if#(isEmpty(_ys), isZero(head(_ys)), _x, tail(_ys), cons(p(head(_ys)), tail(_ys)), inc(_x)) is replaced by instances determined through the above matching. These instances are:
sumList#(xs, y) → if#(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))sumList#(ys, x) → if#(isEmpty(ys), isZero(head(ys)), x, tail(ys), cons(p(head(ys)), tail(ys)), inc(x))

Instantiation

For all potential predecessors l → r of the rule sumList#(xs, y) → if#(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y)) on dependency pair chains it holds that: Thus, sumList#(xs, y) → if#(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y)) is replaced by instances determined through the above matching. These instances are:
sumList#(ys, x) → if#(isEmpty(ys), isZero(head(ys)), x, tail(ys), cons(p(head(ys)), tail(ys)), inc(x))sumList#(_xs, _y) → if#(isEmpty(_xs), isZero(head(_xs)), _y, tail(_xs), cons(p(head(_xs)), tail(_xs)), inc(_y))

Instantiation

For all potential predecessors l → r of the rule sumList#(ys, x) → if#(isEmpty(ys), isZero(head(ys)), x, tail(ys), cons(p(head(ys)), tail(ys)), inc(x)) on dependency pair chains it holds that: Thus, sumList#(ys, x) → if#(isEmpty(ys), isZero(head(ys)), x, tail(ys), cons(p(head(ys)), tail(ys)), inc(x)) is replaced by instances determined through the above matching. These instances are:
sumList#(_ys, _x) → if#(isEmpty(_ys), isZero(head(_ys)), _x, tail(_ys), cons(p(head(_ys)), tail(_ys)), inc(_x))sumList#(xs, y) → if#(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))

Problem 9: Propagation



Dependency Pair Problem

Dependency Pairs

sumList#(_ys, _x)if#(isEmpty(_ys), isZero(head(_ys)), _x, tail(_ys), cons(p(head(_ys)), tail(_ys)), inc(_x))sumList#(xs, y)if#(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))
if#(false, true, y, xs, xs, y)sumList#(xs, y)if#(false, false, y, xs, ys, x)sumList#(ys, x)
if#(false, false, x, ys, ys, x)sumList#(ys, x)if#(false, true, y, xs, ys, x)sumList#(xs, y)
sumList#(ys, x)if#(isEmpty(ys), isZero(head(ys)), x, tail(ys), cons(p(head(ys)), tail(ys)), inc(x))sumList#(_xs, _y)if#(isEmpty(_xs), isZero(head(_xs)), _y, tail(_xs), cons(p(head(_xs)), tail(_xs)), inc(_y))

Rewrite Rules

isEmpty(cons(x, xs))falseisEmpty(nil)true
isZero(0)trueisZero(s(x))false
head(cons(x, xs))xtail(cons(x, xs))xs
tail(nil)nilp(s(s(x)))s(p(s(x)))
p(s(0))0p(0)0
inc(s(x))s(inc(x))inc(0)s(0)
sumList(xs, y)if(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))if(true, b, y, xs, ys, x)y
if(false, true, y, xs, ys, x)sumList(xs, y)if(false, false, y, xs, ys, x)sumList(ys, x)
sum(xs)sumList(xs, 0)

Original Signature

Termination of terms over the following signature is verified: sumList, sum, true, tail, isEmpty, 0, s, inc, if, p, false, head, cons, nil, isZero

Strategy


The dependency pairs if#(false, true, y, xs, xs, y) → sumList#(xs, y) and sumList#(xs, y) → if#(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y)) are consolidated into the rule if#(false, true, y, xs, xs, y) → if#(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y)) .

This is possible as

The dependency pairs if#(false, true, y, xs, ys, x) → sumList#(xs, y) and sumList#(xs, y) → if#(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y)) are consolidated into the rule if#(false, true, y, xs, ys, x) → if#(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y)) .

This is possible as


Summary

Removed Dependency PairsAdded Dependency Pairs
sumList#(xs, y) → if#(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))if#(false, true, y, xs, ys, x) → if#(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))
if#(false, true, y, xs, xs, y) → sumList#(xs, y)if#(false, true, y, xs, xs, y) → if#(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))
if#(false, true, y, xs, ys, x) → sumList#(xs, y) 

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

inc#(s(x))inc#(x)

Rewrite Rules

isEmpty(cons(x, xs))falseisEmpty(nil)true
isZero(0)trueisZero(s(x))false
head(cons(x, xs))xtail(cons(x, xs))xs
tail(nil)nilp(s(s(x)))s(p(s(x)))
p(s(0))0p(0)0
inc(s(x))s(inc(x))inc(0)s(0)
sumList(xs, y)if(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))if(true, b, y, xs, ys, x)y
if(false, true, y, xs, ys, x)sumList(xs, y)if(false, false, y, xs, ys, x)sumList(ys, x)
sum(xs)sumList(xs, 0)

Original Signature

Termination of terms over the following signature is verified: sumList, sum, true, tail, isEmpty, 0, s, inc, if, p, false, head, isZero, nil, cons

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

inc#(s(x))inc#(x)