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 (558ms).
| Problem 2 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (629ms), DependencyGraph (2ms), PolynomialLinearRange8NegiUR (3759ms), DependencyGraph (1ms), ReductionPairSAT (1951ms), DependencyGraph (1ms), SizeChangePrinciple (timeout)].
| Problem 3 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (2ms), PolynomialLinearRange4iUR (1293ms), DependencyGraph (1ms), PolynomialLinearRange8NegiUR (7500ms), DependencyGraph (1ms), ReductionPairSAT (2394ms), DependencyGraph (0ms)].
| Problem 4 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (219ms), PolynomialLinearRange4iUR (2530ms), DependencyGraph (159ms), PolynomialLinearRange8NegiUR (43ms), DependencyGraph (168ms), ReductionPairSAT (2543ms), DependencyGraph (186ms)].
| Problem 5 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (2ms), PolynomialLinearRange4iUR (4ms), DependencyGraph (1ms), PolynomialLinearRange8NegiUR (0ms), DependencyGraph (1ms), ReductionPairSAT (1253ms), DependencyGraph (1ms)].
The following open problems remain:
Open Dependency Pair Problem 2
Dependency Pairs
dbl1#(s(X)) | → | dbl1#(activate(X)) |
Rewrite Rules
dbl(0) | → | 0 | | dbl(s(X)) | → | s(n__s(n__dbl(activate(X)))) |
dbls(nil) | → | nil | | dbls(cons(X, Y)) | → | cons(n__dbl(activate(X)), n__dbls(activate(Y))) |
sel(0, cons(X, Y)) | → | activate(X) | | sel(s(X), cons(Y, Z)) | → | sel(activate(X), activate(Z)) |
indx(nil, X) | → | nil | | indx(cons(X, Y), Z) | → | cons(n__sel(activate(X), activate(Z)), n__indx(activate(Y), activate(Z))) |
from(X) | → | cons(activate(X), n__from(n__s(activate(X)))) | | dbl1(0) | → | 01 |
dbl1(s(X)) | → | s1(s1(dbl1(activate(X)))) | | sel1(0, cons(X, Y)) | → | activate(X) |
sel1(s(X), cons(Y, Z)) | → | sel1(activate(X), activate(Z)) | | quote(0) | → | 01 |
quote(s(X)) | → | s1(quote(activate(X))) | | quote(dbl(X)) | → | dbl1(X) |
quote(sel(X, Y)) | → | sel1(X, Y) | | s(X) | → | n__s(X) |
dbl(X) | → | n__dbl(X) | | dbls(X) | → | n__dbls(X) |
sel(X1, X2) | → | n__sel(X1, X2) | | indx(X1, X2) | → | n__indx(X1, X2) |
from(X) | → | n__from(X) | | activate(n__s(X)) | → | s(X) |
activate(n__dbl(X)) | → | dbl(activate(X)) | | activate(n__dbls(X)) | → | dbls(activate(X)) |
activate(n__sel(X1, X2)) | → | sel(activate(X1), activate(X2)) | | activate(n__indx(X1, X2)) | → | indx(activate(X1), X2) |
activate(n__from(X)) | → | from(X) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__indx, s1, n__from, dbl1, dbl, from, 01, n__sel, dbls, activate, n__s, n__dbls, 0, sel1, s, indx, quote, n__dbl, sel, nil, cons
Open Dependency Pair Problem 3
Dependency Pairs
sel1#(s(X), cons(Y, Z)) | → | sel1#(activate(X), activate(Z)) |
Rewrite Rules
dbl(0) | → | 0 | | dbl(s(X)) | → | s(n__s(n__dbl(activate(X)))) |
dbls(nil) | → | nil | | dbls(cons(X, Y)) | → | cons(n__dbl(activate(X)), n__dbls(activate(Y))) |
sel(0, cons(X, Y)) | → | activate(X) | | sel(s(X), cons(Y, Z)) | → | sel(activate(X), activate(Z)) |
indx(nil, X) | → | nil | | indx(cons(X, Y), Z) | → | cons(n__sel(activate(X), activate(Z)), n__indx(activate(Y), activate(Z))) |
from(X) | → | cons(activate(X), n__from(n__s(activate(X)))) | | dbl1(0) | → | 01 |
dbl1(s(X)) | → | s1(s1(dbl1(activate(X)))) | | sel1(0, cons(X, Y)) | → | activate(X) |
sel1(s(X), cons(Y, Z)) | → | sel1(activate(X), activate(Z)) | | quote(0) | → | 01 |
quote(s(X)) | → | s1(quote(activate(X))) | | quote(dbl(X)) | → | dbl1(X) |
quote(sel(X, Y)) | → | sel1(X, Y) | | s(X) | → | n__s(X) |
dbl(X) | → | n__dbl(X) | | dbls(X) | → | n__dbls(X) |
sel(X1, X2) | → | n__sel(X1, X2) | | indx(X1, X2) | → | n__indx(X1, X2) |
from(X) | → | n__from(X) | | activate(n__s(X)) | → | s(X) |
activate(n__dbl(X)) | → | dbl(activate(X)) | | activate(n__dbls(X)) | → | dbls(activate(X)) |
activate(n__sel(X1, X2)) | → | sel(activate(X1), activate(X2)) | | activate(n__indx(X1, X2)) | → | indx(activate(X1), X2) |
activate(n__from(X)) | → | from(X) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__indx, s1, n__from, dbl1, dbl, from, 01, n__sel, dbls, activate, n__s, n__dbls, 0, sel1, s, indx, quote, n__dbl, sel, nil, cons
Open Dependency Pair Problem 4
Dependency Pairs
activate#(n__sel(X1, X2)) | → | sel#(activate(X1), activate(X2)) | | activate#(n__sel(X1, X2)) | → | activate#(X2) |
dbl#(s(X)) | → | activate#(X) | | sel#(s(X), cons(Y, Z)) | → | sel#(activate(X), activate(Z)) |
activate#(n__sel(X1, X2)) | → | activate#(X1) | | activate#(n__dbl(X)) | → | dbl#(activate(X)) |
activate#(n__from(X)) | → | from#(X) | | activate#(n__indx(X1, X2)) | → | activate#(X1) |
activate#(n__dbls(X)) | → | dbls#(activate(X)) | | activate#(n__indx(X1, X2)) | → | indx#(activate(X1), X2) |
dbls#(cons(X, Y)) | → | activate#(X) | | from#(X) | → | activate#(X) |
activate#(n__dbls(X)) | → | activate#(X) | | sel#(s(X), cons(Y, Z)) | → | activate#(Z) |
indx#(cons(X, Y), Z) | → | activate#(Y) | | dbls#(cons(X, Y)) | → | activate#(Y) |
sel#(0, cons(X, Y)) | → | activate#(X) | | indx#(cons(X, Y), Z) | → | activate#(X) |
sel#(s(X), cons(Y, Z)) | → | activate#(X) | | indx#(cons(X, Y), Z) | → | activate#(Z) |
activate#(n__dbl(X)) | → | activate#(X) |
Rewrite Rules
dbl(0) | → | 0 | | dbl(s(X)) | → | s(n__s(n__dbl(activate(X)))) |
dbls(nil) | → | nil | | dbls(cons(X, Y)) | → | cons(n__dbl(activate(X)), n__dbls(activate(Y))) |
sel(0, cons(X, Y)) | → | activate(X) | | sel(s(X), cons(Y, Z)) | → | sel(activate(X), activate(Z)) |
indx(nil, X) | → | nil | | indx(cons(X, Y), Z) | → | cons(n__sel(activate(X), activate(Z)), n__indx(activate(Y), activate(Z))) |
from(X) | → | cons(activate(X), n__from(n__s(activate(X)))) | | dbl1(0) | → | 01 |
dbl1(s(X)) | → | s1(s1(dbl1(activate(X)))) | | sel1(0, cons(X, Y)) | → | activate(X) |
sel1(s(X), cons(Y, Z)) | → | sel1(activate(X), activate(Z)) | | quote(0) | → | 01 |
quote(s(X)) | → | s1(quote(activate(X))) | | quote(dbl(X)) | → | dbl1(X) |
quote(sel(X, Y)) | → | sel1(X, Y) | | s(X) | → | n__s(X) |
dbl(X) | → | n__dbl(X) | | dbls(X) | → | n__dbls(X) |
sel(X1, X2) | → | n__sel(X1, X2) | | indx(X1, X2) | → | n__indx(X1, X2) |
from(X) | → | n__from(X) | | activate(n__s(X)) | → | s(X) |
activate(n__dbl(X)) | → | dbl(activate(X)) | | activate(n__dbls(X)) | → | dbls(activate(X)) |
activate(n__sel(X1, X2)) | → | sel(activate(X1), activate(X2)) | | activate(n__indx(X1, X2)) | → | indx(activate(X1), X2) |
activate(n__from(X)) | → | from(X) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__indx, s1, n__from, dbl1, dbl, from, 01, n__sel, dbls, activate, n__s, n__dbls, 0, sel1, s, indx, quote, n__dbl, sel, nil, cons
Open Dependency Pair Problem 5
Dependency Pairs
quote#(s(X)) | → | quote#(activate(X)) |
Rewrite Rules
dbl(0) | → | 0 | | dbl(s(X)) | → | s(n__s(n__dbl(activate(X)))) |
dbls(nil) | → | nil | | dbls(cons(X, Y)) | → | cons(n__dbl(activate(X)), n__dbls(activate(Y))) |
sel(0, cons(X, Y)) | → | activate(X) | | sel(s(X), cons(Y, Z)) | → | sel(activate(X), activate(Z)) |
indx(nil, X) | → | nil | | indx(cons(X, Y), Z) | → | cons(n__sel(activate(X), activate(Z)), n__indx(activate(Y), activate(Z))) |
from(X) | → | cons(activate(X), n__from(n__s(activate(X)))) | | dbl1(0) | → | 01 |
dbl1(s(X)) | → | s1(s1(dbl1(activate(X)))) | | sel1(0, cons(X, Y)) | → | activate(X) |
sel1(s(X), cons(Y, Z)) | → | sel1(activate(X), activate(Z)) | | quote(0) | → | 01 |
quote(s(X)) | → | s1(quote(activate(X))) | | quote(dbl(X)) | → | dbl1(X) |
quote(sel(X, Y)) | → | sel1(X, Y) | | s(X) | → | n__s(X) |
dbl(X) | → | n__dbl(X) | | dbls(X) | → | n__dbls(X) |
sel(X1, X2) | → | n__sel(X1, X2) | | indx(X1, X2) | → | n__indx(X1, X2) |
from(X) | → | n__from(X) | | activate(n__s(X)) | → | s(X) |
activate(n__dbl(X)) | → | dbl(activate(X)) | | activate(n__dbls(X)) | → | dbls(activate(X)) |
activate(n__sel(X1, X2)) | → | sel(activate(X1), activate(X2)) | | activate(n__indx(X1, X2)) | → | indx(activate(X1), X2) |
activate(n__from(X)) | → | from(X) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__indx, s1, n__from, dbl1, dbl, from, 01, n__sel, dbls, activate, n__s, n__dbls, 0, sel1, s, indx, quote, n__dbl, sel, nil, cons
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
dbl#(s(X)) | → | s#(n__s(n__dbl(activate(X)))) | | activate#(n__sel(X1, X2)) | → | activate#(X2) |
dbl#(s(X)) | → | activate#(X) | | activate#(n__sel(X1, X2)) | → | activate#(X1) |
activate#(n__dbl(X)) | → | dbl#(activate(X)) | | quote#(s(X)) | → | activate#(X) |
activate#(n__indx(X1, X2)) | → | activate#(X1) | | quote#(sel(X, Y)) | → | sel1#(X, Y) |
quote#(dbl(X)) | → | dbl1#(X) | | sel#(s(X), cons(Y, Z)) | → | activate#(Z) |
indx#(cons(X, Y), Z) | → | activate#(Y) | | dbls#(cons(X, Y)) | → | activate#(Y) |
sel#(0, cons(X, Y)) | → | activate#(X) | | indx#(cons(X, Y), Z) | → | activate#(X) |
sel#(s(X), cons(Y, Z)) | → | activate#(X) | | indx#(cons(X, Y), Z) | → | activate#(Z) |
quote#(s(X)) | → | quote#(activate(X)) | | activate#(n__dbl(X)) | → | activate#(X) |
activate#(n__sel(X1, X2)) | → | sel#(activate(X1), activate(X2)) | | sel#(s(X), cons(Y, Z)) | → | sel#(activate(X), activate(Z)) |
sel1#(s(X), cons(Y, Z)) | → | activate#(Z) | | dbl1#(s(X)) | → | dbl1#(activate(X)) |
sel1#(s(X), cons(Y, Z)) | → | activate#(X) | | activate#(n__from(X)) | → | from#(X) |
sel1#(0, cons(X, Y)) | → | activate#(X) | | activate#(n__s(X)) | → | s#(X) |
activate#(n__dbls(X)) | → | dbls#(activate(X)) | | activate#(n__indx(X1, X2)) | → | indx#(activate(X1), X2) |
dbls#(cons(X, Y)) | → | activate#(X) | | from#(X) | → | activate#(X) |
activate#(n__dbls(X)) | → | activate#(X) | | sel1#(s(X), cons(Y, Z)) | → | sel1#(activate(X), activate(Z)) |
dbl1#(s(X)) | → | activate#(X) |
Rewrite Rules
dbl(0) | → | 0 | | dbl(s(X)) | → | s(n__s(n__dbl(activate(X)))) |
dbls(nil) | → | nil | | dbls(cons(X, Y)) | → | cons(n__dbl(activate(X)), n__dbls(activate(Y))) |
sel(0, cons(X, Y)) | → | activate(X) | | sel(s(X), cons(Y, Z)) | → | sel(activate(X), activate(Z)) |
indx(nil, X) | → | nil | | indx(cons(X, Y), Z) | → | cons(n__sel(activate(X), activate(Z)), n__indx(activate(Y), activate(Z))) |
from(X) | → | cons(activate(X), n__from(n__s(activate(X)))) | | dbl1(0) | → | 01 |
dbl1(s(X)) | → | s1(s1(dbl1(activate(X)))) | | sel1(0, cons(X, Y)) | → | activate(X) |
sel1(s(X), cons(Y, Z)) | → | sel1(activate(X), activate(Z)) | | quote(0) | → | 01 |
quote(s(X)) | → | s1(quote(activate(X))) | | quote(dbl(X)) | → | dbl1(X) |
quote(sel(X, Y)) | → | sel1(X, Y) | | s(X) | → | n__s(X) |
dbl(X) | → | n__dbl(X) | | dbls(X) | → | n__dbls(X) |
sel(X1, X2) | → | n__sel(X1, X2) | | indx(X1, X2) | → | n__indx(X1, X2) |
from(X) | → | n__from(X) | | activate(n__s(X)) | → | s(X) |
activate(n__dbl(X)) | → | dbl(activate(X)) | | activate(n__dbls(X)) | → | dbls(activate(X)) |
activate(n__sel(X1, X2)) | → | sel(activate(X1), activate(X2)) | | activate(n__indx(X1, X2)) | → | indx(activate(X1), X2) |
activate(n__from(X)) | → | from(X) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__indx, s1, n__from, dbl1, dbl, from, 01, n__sel, dbls, activate, n__s, n__dbls, 0, indx, s, sel1, quote, sel, n__dbl, cons, nil
Strategy
The following SCCs where found
sel1#(s(X), cons(Y, Z)) → sel1#(activate(X), activate(Z)) |
dbl1#(s(X)) → dbl1#(activate(X)) |
activate#(n__sel(X1, X2)) → sel#(activate(X1), activate(X2)) | activate#(n__sel(X1, X2)) → activate#(X2) |
sel#(s(X), cons(Y, Z)) → sel#(activate(X), activate(Z)) | dbl#(s(X)) → activate#(X) |
activate#(n__sel(X1, X2)) → activate#(X1) | activate#(n__dbl(X)) → dbl#(activate(X)) |
activate#(n__from(X)) → from#(X) | activate#(n__indx(X1, X2)) → activate#(X1) |
activate#(n__dbls(X)) → dbls#(activate(X)) | activate#(n__indx(X1, X2)) → indx#(activate(X1), X2) |
dbls#(cons(X, Y)) → activate#(X) | from#(X) → activate#(X) |
activate#(n__dbls(X)) → activate#(X) | sel#(s(X), cons(Y, Z)) → activate#(Z) |
indx#(cons(X, Y), Z) → activate#(Y) | dbls#(cons(X, Y)) → activate#(Y) |
sel#(0, cons(X, Y)) → activate#(X) | indx#(cons(X, Y), Z) → activate#(X) |
sel#(s(X), cons(Y, Z)) → activate#(X) | indx#(cons(X, Y), Z) → activate#(Z) |
activate#(n__dbl(X)) → activate#(X) |
quote#(s(X)) → quote#(activate(X)) |