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 (126ms).
 | – Problem 2 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (2ms), PolynomialLinearRange4iUR (1864ms), DependencyGraph (1ms), PolynomialLinearRange8NegiUR (9223ms), DependencyGraph (1ms), ReductionPairSAT (timeout)].
 | – Problem 3 was processed with processor SubtermCriterion (1ms).
 | – Problem 4 was processed with processor SubtermCriterion (1ms).
 | – Problem 5 was processed with processor SubtermCriterion (1ms).
 | – Problem 6 was processed with processor SubtermCriterion (2ms).
 | – Problem 7 was processed with processor SubtermCriterion (1ms).

The following open problems remain:



Open Dependency Pair Problem 2

Dependency Pairs

if#(true, c, xs, ys, z) → if#(le(s(c), min(len(xs), len(ys))), s(c), xs, ys, cons(sum(take(c, xs), take(c, ys)), z))

Rewrite Rules

min(0, y) → 0min(s(x), 0) → 0
min(s(x), s(y)) → min(x, y)len(nil) → 0
len(cons(x, xs)) → s(len(xs))sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))le(0, x) → true
le(s(x), 0) → falsele(s(x), s(y)) → le(x, y)
take(0, cons(y, ys)) → ytake(s(x), cons(y, ys)) → take(x, ys)
addList(x, y) → if(le(0, min(len(x), len(y))), 0, x, y, nil)if(false, c, x, y, z) → z
if(true, c, xs, ys, z) → if(le(s(c), min(len(xs), len(ys))), s(c), xs, ys, cons(sum(take(c, xs), take(c, ys)), z))

Original Signature

Termination of terms over the following signature is verified: min, sum, true, len, 0, s, le, take, if, false, nil, cons, addList


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

if#(true, c, xs, ys, z) → min#(len(xs), len(ys))sum#(x, s(y)) → sum#(x, y)
addList#(x, y) → len#(y)if#(true, c, xs, ys, z) → le#(s(c), min(len(xs), len(ys)))
if#(true, c, xs, ys, z) → take#(c, ys)addList#(x, y) → len#(x)
if#(true, c, xs, ys, z) → len#(xs)if#(true, c, xs, ys, z) → len#(ys)
min#(s(x), s(y)) → min#(x, y)len#(cons(x, xs)) → len#(xs)
le#(s(x), s(y)) → le#(x, y)addList#(x, y) → le#(0, min(len(x), len(y)))
addList#(x, y) → min#(len(x), len(y))if#(true, c, xs, ys, z) → take#(c, xs)
take#(s(x), cons(y, ys)) → take#(x, ys)if#(true, c, xs, ys, z) → sum#(take(c, xs), take(c, ys))
if#(true, c, xs, ys, z) → if#(le(s(c), min(len(xs), len(ys))), s(c), xs, ys, cons(sum(take(c, xs), take(c, ys)), z))addList#(x, y) → if#(le(0, min(len(x), len(y))), 0, x, y, nil)

Rewrite Rules

min(0, y) → 0min(s(x), 0) → 0
min(s(x), s(y)) → min(x, y)len(nil) → 0
len(cons(x, xs)) → s(len(xs))sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))le(0, x) → true
le(s(x), 0) → falsele(s(x), s(y)) → le(x, y)
take(0, cons(y, ys)) → ytake(s(x), cons(y, ys)) → take(x, ys)
addList(x, y) → if(le(0, min(len(x), len(y))), 0, x, y, nil)if(false, c, x, y, z) → z
if(true, c, xs, ys, z) → if(le(s(c), min(len(xs), len(ys))), s(c), xs, ys, cons(sum(take(c, xs), take(c, ys)), z))

Original Signature

Termination of terms over the following signature is verified: min, sum, true, len, 0, s, le, take, if, false, nil, cons, addList

Strategy


The following SCCs where found

le#(s(x), s(y)) → le#(x, y)

len#(cons(x, xs)) → len#(xs)

sum#(x, s(y)) → sum#(x, y)

min#(s(x), s(y)) → min#(x, y)

take#(s(x), cons(y, ys)) → take#(x, ys)

if#(true, c, xs, ys, z) → if#(le(s(c), min(len(xs), len(ys))), s(c), xs, ys, cons(sum(take(c, xs), take(c, ys)), z))

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

le#(s(x), s(y)) → le#(x, y)

Rewrite Rules

min(0, y) → 0min(s(x), 0) → 0
min(s(x), s(y)) → min(x, y)len(nil) → 0
len(cons(x, xs)) → s(len(xs))sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))le(0, x) → true
le(s(x), 0) → falsele(s(x), s(y)) → le(x, y)
take(0, cons(y, ys)) → ytake(s(x), cons(y, ys)) → take(x, ys)
addList(x, y) → if(le(0, min(len(x), len(y))), 0, x, y, nil)if(false, c, x, y, z) → z
if(true, c, xs, ys, z) → if(le(s(c), min(len(xs), len(ys))), s(c), xs, ys, cons(sum(take(c, xs), take(c, ys)), z))

Original Signature

Termination of terms over the following signature is verified: min, sum, true, len, 0, s, le, take, if, false, nil, cons, addList

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

le#(s(x), s(y)) → le#(x, y)

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

min#(s(x), s(y)) → min#(x, y)

Rewrite Rules

min(0, y) → 0min(s(x), 0) → 0
min(s(x), s(y)) → min(x, y)len(nil) → 0
len(cons(x, xs)) → s(len(xs))sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))le(0, x) → true
le(s(x), 0) → falsele(s(x), s(y)) → le(x, y)
take(0, cons(y, ys)) → ytake(s(x), cons(y, ys)) → take(x, ys)
addList(x, y) → if(le(0, min(len(x), len(y))), 0, x, y, nil)if(false, c, x, y, z) → z
if(true, c, xs, ys, z) → if(le(s(c), min(len(xs), len(ys))), s(c), xs, ys, cons(sum(take(c, xs), take(c, ys)), z))

Original Signature

Termination of terms over the following signature is verified: min, sum, true, len, 0, s, le, take, if, false, nil, cons, addList

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

min#(s(x), s(y)) → min#(x, y)

Problem 5: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

sum#(x, s(y)) → sum#(x, y)

Rewrite Rules

min(0, y) → 0min(s(x), 0) → 0
min(s(x), s(y)) → min(x, y)len(nil) → 0
len(cons(x, xs)) → s(len(xs))sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))le(0, x) → true
le(s(x), 0) → falsele(s(x), s(y)) → le(x, y)
take(0, cons(y, ys)) → ytake(s(x), cons(y, ys)) → take(x, ys)
addList(x, y) → if(le(0, min(len(x), len(y))), 0, x, y, nil)if(false, c, x, y, z) → z
if(true, c, xs, ys, z) → if(le(s(c), min(len(xs), len(ys))), s(c), xs, ys, cons(sum(take(c, xs), take(c, ys)), z))

Original Signature

Termination of terms over the following signature is verified: min, sum, true, len, 0, s, le, take, if, false, nil, cons, addList

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

sum#(x, s(y)) → sum#(x, y)

Problem 6: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

take#(s(x), cons(y, ys)) → take#(x, ys)

Rewrite Rules

min(0, y) → 0min(s(x), 0) → 0
min(s(x), s(y)) → min(x, y)len(nil) → 0
len(cons(x, xs)) → s(len(xs))sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))le(0, x) → true
le(s(x), 0) → falsele(s(x), s(y)) → le(x, y)
take(0, cons(y, ys)) → ytake(s(x), cons(y, ys)) → take(x, ys)
addList(x, y) → if(le(0, min(len(x), len(y))), 0, x, y, nil)if(false, c, x, y, z) → z
if(true, c, xs, ys, z) → if(le(s(c), min(len(xs), len(ys))), s(c), xs, ys, cons(sum(take(c, xs), take(c, ys)), z))

Original Signature

Termination of terms over the following signature is verified: min, sum, true, len, 0, s, le, take, if, false, nil, cons, addList

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

take#(s(x), cons(y, ys)) → take#(x, ys)

Problem 7: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

len#(cons(x, xs)) → len#(xs)

Rewrite Rules

min(0, y) → 0min(s(x), 0) → 0
min(s(x), s(y)) → min(x, y)len(nil) → 0
len(cons(x, xs)) → s(len(xs))sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))le(0, x) → true
le(s(x), 0) → falsele(s(x), s(y)) → le(x, y)
take(0, cons(y, ys)) → ytake(s(x), cons(y, ys)) → take(x, ys)
addList(x, y) → if(le(0, min(len(x), len(y))), 0, x, y, nil)if(false, c, x, y, z) → z
if(true, c, xs, ys, z) → if(le(s(c), min(len(xs), len(ys))), s(c), xs, ys, cons(sum(take(c, xs), take(c, ys)), z))

Original Signature

Termination of terms over the following signature is verified: min, sum, true, len, 0, s, le, take, if, false, nil, cons, addList

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

len#(cons(x, xs)) → len#(xs)