TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60559 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (226ms).
 | – Problem 2 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (53ms), DependencyGraph (1ms), PolynomialOrderingProcessor (0ms), DependencyGraph (0ms), PolynomialLinearRange4 (201ms), DependencyGraph (1ms), PolynomialLinearRange4 (101ms), DependencyGraph (0ms), PolynomialLinearRange4 (72ms), DependencyGraph (0ms), PolynomialLinearRange4 (68ms), DependencyGraph (0ms), ReductionPairSAT (265ms), DependencyGraph (0ms), SizeChangePrinciple (timeout)].
 | – Problem 3 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (11ms), PolynomialLinearRange4iUR (0ms), DependencyGraph (0ms), PolynomialOrderingProcessor (0ms), DependencyGraph (1ms), PolynomialLinearRange4 (153ms), DependencyGraph (0ms), PolynomialLinearRange4 (95ms), DependencyGraph (13ms), PolynomialLinearRange4 (91ms), DependencyGraph (1ms), PolynomialLinearRange4 (69ms), DependencyGraph (1ms), ReductionPairSAT (395ms), DependencyGraph (0ms)].
 | – Problem 4 was processed with processor PolynomialLinearRange4 (13ms).
 | – Problem 5 was processed with processor PolynomialLinearRange4 (10ms).
 | – Problem 6 was processed with processor PolynomialLinearRange4 (127ms).
 |    | – Problem 9 was processed with processor PolynomialLinearRange4 (37ms).
 |    |    | – Problem 10 was processed with processor PolynomialLinearRange4 (14ms).
 | – Problem 7 was processed with processor PolynomialLinearRange4 (94ms).
 | – Problem 8 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (18ms), PolynomialLinearRange4iUR (17ms), DependencyGraph (2ms), PolynomialOrderingProcessor (0ms), DependencyGraph (1ms), PolynomialLinearRange4 (126ms), DependencyGraph (2ms), PolynomialLinearRange4 (99ms), DependencyGraph (1ms), PolynomialLinearRange4 (89ms), DependencyGraph (1ms), PolynomialLinearRange4 (85ms), DependencyGraph (2ms), ReductionPairSAT (261ms), DependencyGraph (4ms)].

The following open problems remain:



Open Dependency Pair Problem 2

Dependency Pairs

sel#(s(X), cons(Y, Z))sel#(X, Z)

Rewrite Rules

sel(s(X), cons(Y, Z))sel(X, Z)sel(0, cons(X, Z))X
first(0, Z)nilfirst(s(X), cons(Y, Z))cons(Y, first(X, Z))
from(X)cons(X, from(s(X)))sel1(s(X), cons(Y, Z))sel1(X, Z)
sel1(0, cons(X, Z))quote(X)first1(0, Z)nil1
first1(s(X), cons(Y, Z))cons1(quote(Y), first1(X, Z))quote(0)01
quote1(cons(X, Z))cons1(quote(X), quote1(Z))quote1(nil)nil1
quote(s(X))s1(quote(X))quote(sel(X, Z))sel1(X, Z)
quote1(first(X, Z))first1(X, Z)unquote(01)0
unquote(s1(X))s(unquote(X))unquote1(nil1)nil
unquote1(cons1(X, Z))fcons(unquote(X), unquote1(Z))fcons(X, Z)cons(X, Z)

Original Signature

Termination of terms over the following signature is verified: s1, cons1, fcons, from, 01, first1, quote1, unquote, 0, s, sel1, unquote1, quote, nil1, sel, first, cons, nil




Open Dependency Pair Problem 3

Dependency Pairs

first1#(s(X), cons(Y, Z))first1#(X, Z)

Rewrite Rules

sel(s(X), cons(Y, Z))sel(X, Z)sel(0, cons(X, Z))X
first(0, Z)nilfirst(s(X), cons(Y, Z))cons(Y, first(X, Z))
from(X)cons(X, from(s(X)))sel1(s(X), cons(Y, Z))sel1(X, Z)
sel1(0, cons(X, Z))quote(X)first1(0, Z)nil1
first1(s(X), cons(Y, Z))cons1(quote(Y), first1(X, Z))quote(0)01
quote1(cons(X, Z))cons1(quote(X), quote1(Z))quote1(nil)nil1
quote(s(X))s1(quote(X))quote(sel(X, Z))sel1(X, Z)
quote1(first(X, Z))first1(X, Z)unquote(01)0
unquote(s1(X))s(unquote(X))unquote1(nil1)nil
unquote1(cons1(X, Z))fcons(unquote(X), unquote1(Z))fcons(X, Z)cons(X, Z)

Original Signature

Termination of terms over the following signature is verified: s1, cons1, fcons, from, 01, first1, quote1, unquote, 0, s, sel1, unquote1, quote, nil1, sel, first, cons, nil




Open Dependency Pair Problem 8

Dependency Pairs

quote#(sel(X, Z))sel1#(X, Z)sel1#(0, cons(X, Z))quote#(X)
quote#(s(X))quote#(X)sel1#(s(X), cons(Y, Z))sel1#(X, Z)

Rewrite Rules

sel(s(X), cons(Y, Z))sel(X, Z)sel(0, cons(X, Z))X
first(0, Z)nilfirst(s(X), cons(Y, Z))cons(Y, first(X, Z))
from(X)cons(X, from(s(X)))sel1(s(X), cons(Y, Z))sel1(X, Z)
sel1(0, cons(X, Z))quote(X)first1(0, Z)nil1
first1(s(X), cons(Y, Z))cons1(quote(Y), first1(X, Z))quote(0)01
quote1(cons(X, Z))cons1(quote(X), quote1(Z))quote1(nil)nil1
quote(s(X))s1(quote(X))quote(sel(X, Z))sel1(X, Z)
quote1(first(X, Z))first1(X, Z)unquote(01)0
unquote(s1(X))s(unquote(X))unquote1(nil1)nil
unquote1(cons1(X, Z))fcons(unquote(X), unquote1(Z))fcons(X, Z)cons(X, Z)

Original Signature

Termination of terms over the following signature is verified: s1, cons1, fcons, from, 01, first1, quote1, unquote, 0, s, sel1, unquote1, quote, nil1, sel, first, cons, nil


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

first1#(s(X), cons(Y, Z))quote#(Y)T(from(x_1))T(x_1)
quote#(s(X))quote#(X)first1#(s(X), cons(Y, Z))first1#(X, Z)
quote#(sel(X, Z))T(X)quote1#(first(X, Z))first1#(X, Z)
sel1#(s(X), cons(Y, Z))T(Z)unquote1#(cons1(X, Z))fcons#(unquote(X), unquote1(Z))
unquote1#(cons1(X, Z))unquote1#(Z)T(s(x_1))T(x_1)
quote1#(first(X, Z))T(X)T(first(x_1, x_2))T(x_2)
quote1#(cons(X, Z))quote#(X)quote#(sel(X, Z))sel1#(X, Z)
T(first(X, Z))first#(X, Z)sel#(s(X), cons(Y, Z))sel#(X, Z)
unquote#(s1(X))unquote#(X)first1#(s(X), cons(Y, Z))T(Z)
quote#(sel(X, Z))T(Z)unquote1#(cons1(X, Z))unquote#(X)
T(first(x_1, x_2))T(x_1)sel1#(s(X), cons(Y, Z))sel1#(X, Z)
quote1#(cons(X, Z))quote1#(Z)sel1#(0, cons(X, Z))quote#(X)
quote1#(first(X, Z))T(Z)sel#(s(X), cons(Y, Z))T(Z)
T(from(s(X)))from#(s(X))

Rewrite Rules

sel(s(X), cons(Y, Z))sel(X, Z)sel(0, cons(X, Z))X
first(0, Z)nilfirst(s(X), cons(Y, Z))cons(Y, first(X, Z))
from(X)cons(X, from(s(X)))sel1(s(X), cons(Y, Z))sel1(X, Z)
sel1(0, cons(X, Z))quote(X)first1(0, Z)nil1
first1(s(X), cons(Y, Z))cons1(quote(Y), first1(X, Z))quote(0)01
quote1(cons(X, Z))cons1(quote(X), quote1(Z))quote1(nil)nil1
quote(s(X))s1(quote(X))quote(sel(X, Z))sel1(X, Z)
quote1(first(X, Z))first1(X, Z)unquote(01)0
unquote(s1(X))s(unquote(X))unquote1(nil1)nil
unquote1(cons1(X, Z))fcons(unquote(X), unquote1(Z))fcons(X, Z)cons(X, Z)

Original Signature

Termination of terms over the following signature is verified: s1, cons1, fcons, from, 01, first1, quote1, unquote, 0, s, sel1, unquote1, quote, nil1, first, sel, nil, cons

Strategy

Context-sensitive strategy:
μ(T) = μ(quote1#) = μ(quote) = μ(quote#) = μ(01) = μ(quote1) = μ(0) = μ(nil1) = μ(nil) = ∅
μ(from#) = μ(from) = μ(unquote) = μ(unquote1) = μ(unquote#) = μ(cons) = μ(unquote1#) = μ(s1) = μ(s) = {1}
μ(fcons#) = μ(cons1) = μ(sel#) = μ(sel1#) = μ(first#) = μ(first1#) = μ(sel1) = μ(fcons) = μ(first1) = μ(sel) = μ(first) = {1, 2}


The following SCCs where found

quote1#(cons(X, Z)) → quote1#(Z)

quote#(sel(X, Z)) → sel1#(X, Z)sel1#(0, cons(X, Z)) → quote#(X)
quote#(s(X)) → quote#(X)sel1#(s(X), cons(Y, Z)) → sel1#(X, Z)

unquote1#(cons1(X, Z)) → unquote1#(Z)

sel#(s(X), cons(Y, Z)) → sel#(X, Z)

unquote#(s1(X)) → unquote#(X)

first1#(s(X), cons(Y, Z)) → first1#(X, Z)

T(first(x_1, x_2)) → T(x_2)T(s(x_1)) → T(x_1)
T(from(x_1)) → T(x_1)T(first(x_1, x_2)) → T(x_1)

Problem 4: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

quote1#(cons(X, Z))quote1#(Z)

Rewrite Rules

sel(s(X), cons(Y, Z))sel(X, Z)sel(0, cons(X, Z))X
first(0, Z)nilfirst(s(X), cons(Y, Z))cons(Y, first(X, Z))
from(X)cons(X, from(s(X)))sel1(s(X), cons(Y, Z))sel1(X, Z)
sel1(0, cons(X, Z))quote(X)first1(0, Z)nil1
first1(s(X), cons(Y, Z))cons1(quote(Y), first1(X, Z))quote(0)01
quote1(cons(X, Z))cons1(quote(X), quote1(Z))quote1(nil)nil1
quote(s(X))s1(quote(X))quote(sel(X, Z))sel1(X, Z)
quote1(first(X, Z))first1(X, Z)unquote(01)0
unquote(s1(X))s(unquote(X))unquote1(nil1)nil
unquote1(cons1(X, Z))fcons(unquote(X), unquote1(Z))fcons(X, Z)cons(X, Z)

Original Signature

Termination of terms over the following signature is verified: s1, cons1, fcons, from, 01, first1, quote1, unquote, 0, s, sel1, unquote1, quote, nil1, first, sel, nil, cons

Strategy

Context-sensitive strategy:
μ(T) = μ(quote1#) = μ(quote) = μ(quote#) = μ(01) = μ(quote1) = μ(0) = μ(nil1) = μ(nil) = ∅
μ(from#) = μ(from) = μ(unquote) = μ(unquote1) = μ(unquote#) = μ(cons) = μ(unquote1#) = μ(s1) = μ(s) = {1}
μ(cons1) = μ(fcons#) = μ(sel#) = μ(sel1#) = μ(first#) = μ(first1#) = μ(sel1) = μ(fcons) = μ(first1) = μ(first) = μ(sel) = {1, 2}


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:

quote1#(cons(X, Z))quote1#(Z)

Problem 5: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

unquote1#(cons1(X, Z))unquote1#(Z)

Rewrite Rules

sel(s(X), cons(Y, Z))sel(X, Z)sel(0, cons(X, Z))X
first(0, Z)nilfirst(s(X), cons(Y, Z))cons(Y, first(X, Z))
from(X)cons(X, from(s(X)))sel1(s(X), cons(Y, Z))sel1(X, Z)
sel1(0, cons(X, Z))quote(X)first1(0, Z)nil1
first1(s(X), cons(Y, Z))cons1(quote(Y), first1(X, Z))quote(0)01
quote1(cons(X, Z))cons1(quote(X), quote1(Z))quote1(nil)nil1
quote(s(X))s1(quote(X))quote(sel(X, Z))sel1(X, Z)
quote1(first(X, Z))first1(X, Z)unquote(01)0
unquote(s1(X))s(unquote(X))unquote1(nil1)nil
unquote1(cons1(X, Z))fcons(unquote(X), unquote1(Z))fcons(X, Z)cons(X, Z)

Original Signature

Termination of terms over the following signature is verified: s1, cons1, fcons, from, 01, first1, quote1, unquote, 0, s, sel1, unquote1, quote, nil1, first, sel, nil, cons

Strategy

Context-sensitive strategy:
μ(T) = μ(quote1#) = μ(quote) = μ(quote#) = μ(01) = μ(quote1) = μ(0) = μ(nil1) = μ(nil) = ∅
μ(from#) = μ(from) = μ(unquote) = μ(unquote1) = μ(unquote#) = μ(cons) = μ(unquote1#) = μ(s1) = μ(s) = {1}
μ(cons1) = μ(fcons#) = μ(sel#) = μ(sel1#) = μ(first#) = μ(first1#) = μ(sel1) = μ(fcons) = μ(first1) = μ(first) = μ(sel) = {1, 2}


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:

unquote1#(cons1(X, Z))unquote1#(Z)

Problem 6: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

T(first(x_1, x_2))T(x_2)T(s(x_1))T(x_1)
T(from(x_1))T(x_1)T(first(x_1, x_2))T(x_1)

Rewrite Rules

sel(s(X), cons(Y, Z))sel(X, Z)sel(0, cons(X, Z))X
first(0, Z)nilfirst(s(X), cons(Y, Z))cons(Y, first(X, Z))
from(X)cons(X, from(s(X)))sel1(s(X), cons(Y, Z))sel1(X, Z)
sel1(0, cons(X, Z))quote(X)first1(0, Z)nil1
first1(s(X), cons(Y, Z))cons1(quote(Y), first1(X, Z))quote(0)01
quote1(cons(X, Z))cons1(quote(X), quote1(Z))quote1(nil)nil1
quote(s(X))s1(quote(X))quote(sel(X, Z))sel1(X, Z)
quote1(first(X, Z))first1(X, Z)unquote(01)0
unquote(s1(X))s(unquote(X))unquote1(nil1)nil
unquote1(cons1(X, Z))fcons(unquote(X), unquote1(Z))fcons(X, Z)cons(X, Z)

Original Signature

Termination of terms over the following signature is verified: s1, cons1, fcons, from, 01, first1, quote1, unquote, 0, s, sel1, unquote1, quote, nil1, first, sel, nil, cons

Strategy

Context-sensitive strategy:
μ(T) = μ(quote1#) = μ(quote) = μ(quote#) = μ(01) = μ(quote1) = μ(0) = μ(nil1) = μ(nil) = ∅
μ(from#) = μ(from) = μ(unquote) = μ(unquote1) = μ(unquote#) = μ(cons) = μ(unquote1#) = μ(s1) = μ(s) = {1}
μ(cons1) = μ(fcons#) = μ(sel#) = μ(sel1#) = μ(first#) = μ(first1#) = μ(sel1) = μ(fcons) = μ(first1) = μ(first) = μ(sel) = {1, 2}


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:

T(from(x_1))T(x_1)

Problem 9: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

T(first(x_1, x_2))T(x_2)T(s(x_1))T(x_1)
T(first(x_1, x_2))T(x_1)

Rewrite Rules

sel(s(X), cons(Y, Z))sel(X, Z)sel(0, cons(X, Z))X
first(0, Z)nilfirst(s(X), cons(Y, Z))cons(Y, first(X, Z))
from(X)cons(X, from(s(X)))sel1(s(X), cons(Y, Z))sel1(X, Z)
sel1(0, cons(X, Z))quote(X)first1(0, Z)nil1
first1(s(X), cons(Y, Z))cons1(quote(Y), first1(X, Z))quote(0)01
quote1(cons(X, Z))cons1(quote(X), quote1(Z))quote1(nil)nil1
quote(s(X))s1(quote(X))quote(sel(X, Z))sel1(X, Z)
quote1(first(X, Z))first1(X, Z)unquote(01)0
unquote(s1(X))s(unquote(X))unquote1(nil1)nil
unquote1(cons1(X, Z))fcons(unquote(X), unquote1(Z))fcons(X, Z)cons(X, Z)

Original Signature

Termination of terms over the following signature is verified: s1, cons1, fcons, from, 01, first1, quote1, unquote, 0, s, sel1, unquote1, quote, nil1, sel, first, cons, nil

Strategy

Context-sensitive strategy:
μ(T) = μ(quote1#) = μ(quote) = μ(quote#) = μ(01) = μ(quote1) = μ(0) = μ(nil1) = μ(nil) = ∅
μ(from#) = μ(from) = μ(unquote) = μ(unquote1) = μ(unquote#) = μ(cons) = μ(unquote1#) = μ(s1) = μ(s) = {1}
μ(fcons#) = μ(cons1) = μ(sel#) = μ(sel1#) = μ(first#) = μ(first1#) = μ(sel1) = μ(fcons) = μ(first1) = μ(sel) = μ(first) = {1, 2}


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:

T(first(x_1, x_2))T(x_2)T(first(x_1, x_2))T(x_1)

Problem 10: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

T(s(x_1))T(x_1)

Rewrite Rules

sel(s(X), cons(Y, Z))sel(X, Z)sel(0, cons(X, Z))X
first(0, Z)nilfirst(s(X), cons(Y, Z))cons(Y, first(X, Z))
from(X)cons(X, from(s(X)))sel1(s(X), cons(Y, Z))sel1(X, Z)
sel1(0, cons(X, Z))quote(X)first1(0, Z)nil1
first1(s(X), cons(Y, Z))cons1(quote(Y), first1(X, Z))quote(0)01
quote1(cons(X, Z))cons1(quote(X), quote1(Z))quote1(nil)nil1
quote(s(X))s1(quote(X))quote(sel(X, Z))sel1(X, Z)
quote1(first(X, Z))first1(X, Z)unquote(01)0
unquote(s1(X))s(unquote(X))unquote1(nil1)nil
unquote1(cons1(X, Z))fcons(unquote(X), unquote1(Z))fcons(X, Z)cons(X, Z)

Original Signature

Termination of terms over the following signature is verified: s1, cons1, fcons, from, 01, first1, quote1, unquote, 0, s, sel1, unquote1, quote, nil1, first, sel, nil, cons

Strategy

Context-sensitive strategy:
μ(T) = μ(quote1#) = μ(quote) = μ(quote#) = μ(01) = μ(quote1) = μ(0) = μ(nil1) = μ(nil) = ∅
μ(from#) = μ(from) = μ(unquote) = μ(unquote1) = μ(unquote#) = μ(cons) = μ(unquote1#) = μ(s1) = μ(s) = {1}
μ(cons1) = μ(fcons#) = μ(sel#) = μ(sel1#) = μ(first#) = μ(first1#) = μ(sel1) = μ(fcons) = μ(first1) = μ(first) = μ(sel) = {1, 2}


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:

T(s(x_1))T(x_1)

Problem 7: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

unquote#(s1(X))unquote#(X)

Rewrite Rules

sel(s(X), cons(Y, Z))sel(X, Z)sel(0, cons(X, Z))X
first(0, Z)nilfirst(s(X), cons(Y, Z))cons(Y, first(X, Z))
from(X)cons(X, from(s(X)))sel1(s(X), cons(Y, Z))sel1(X, Z)
sel1(0, cons(X, Z))quote(X)first1(0, Z)nil1
first1(s(X), cons(Y, Z))cons1(quote(Y), first1(X, Z))quote(0)01
quote1(cons(X, Z))cons1(quote(X), quote1(Z))quote1(nil)nil1
quote(s(X))s1(quote(X))quote(sel(X, Z))sel1(X, Z)
quote1(first(X, Z))first1(X, Z)unquote(01)0
unquote(s1(X))s(unquote(X))unquote1(nil1)nil
unquote1(cons1(X, Z))fcons(unquote(X), unquote1(Z))fcons(X, Z)cons(X, Z)

Original Signature

Termination of terms over the following signature is verified: s1, cons1, fcons, from, 01, first1, quote1, unquote, 0, s, sel1, unquote1, quote, nil1, first, sel, nil, cons

Strategy

Context-sensitive strategy:
μ(T) = μ(quote1#) = μ(quote) = μ(quote#) = μ(01) = μ(quote1) = μ(0) = μ(nil1) = μ(nil) = ∅
μ(from#) = μ(from) = μ(unquote) = μ(unquote1) = μ(unquote#) = μ(cons) = μ(unquote1#) = μ(s1) = μ(s) = {1}
μ(cons1) = μ(fcons#) = μ(sel#) = μ(sel1#) = μ(first#) = μ(first1#) = μ(sel1) = μ(fcons) = μ(first1) = μ(first) = μ(sel) = {1, 2}


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:

unquote#(s1(X))unquote#(X)