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 (1254ms).
| – Problem 2 was processed with processor SubtermCriterion (2ms).
| – Problem 3 was processed with processor SubtermCriterion (3ms).
| | – Problem 10 was processed with processor PolynomialLinearRange4iUR (120ms).
| – Problem 4 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (6ms), PolynomialLinearRange4iUR (3539ms), DependencyGraph (4ms), PolynomialLinearRange4iUR (2812ms), DependencyGraph (3ms), PolynomialLinearRange8NegiUR (timeout), PolynomialLinearRange8NegiUR (-6ms), ReductionPairSAT (3915ms), DependencyGraph (2ms), SizeChangePrinciple (timeout)].
| – Problem 5 was processed with processor SubtermCriterion (1ms).
| – Problem 6 was processed with processor SubtermCriterion (0ms).
| – Problem 7 was processed with processor SubtermCriterion (1ms).
| – Problem 8 was processed with processor SubtermCriterion (1ms).
| – Problem 9 was processed with processor SubtermCriterion (1ms).
The following open problems remain:
Open Dependency Pair Problem 4
Dependency Pairs
| top#(mark(X)) | → | top#(proper(X)) | | top#(ok(X)) | → | top#(active(X)) |
Rewrite Rules
| active(zeros) | → | mark(cons(0, zeros)) | | active(and(tt, X)) | → | mark(X) |
| active(length(nil)) | → | mark(0) | | active(length(cons(N, L))) | → | mark(s(length(L))) |
| active(take(0, IL)) | → | mark(nil) | | active(take(s(M), cons(N, IL))) | → | mark(cons(N, take(M, IL))) |
| active(cons(X1, X2)) | → | cons(active(X1), X2) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(length(X)) | → | length(active(X)) | | active(s(X)) | → | s(active(X)) |
| active(take(X1, X2)) | → | take(active(X1), X2) | | active(take(X1, X2)) | → | take(X1, active(X2)) |
| cons(mark(X1), X2) | → | mark(cons(X1, X2)) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| length(mark(X)) | → | mark(length(X)) | | s(mark(X)) | → | mark(s(X)) |
| take(mark(X1), X2) | → | mark(take(X1, X2)) | | take(X1, mark(X2)) | → | mark(take(X1, X2)) |
| proper(zeros) | → | ok(zeros) | | proper(cons(X1, X2)) | → | cons(proper(X1), proper(X2)) |
| proper(0) | → | ok(0) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | proper(length(X)) | → | length(proper(X)) |
| proper(nil) | → | ok(nil) | | proper(s(X)) | → | s(proper(X)) |
| proper(take(X1, X2)) | → | take(proper(X1), proper(X2)) | | cons(ok(X1), ok(X2)) | → | ok(cons(X1, X2)) |
| and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) | | length(ok(X)) | → | ok(length(X)) |
| s(ok(X)) | → | ok(s(X)) | | take(ok(X1), ok(X2)) | → | ok(take(X1, X2)) |
| top(mark(X)) | → | top(proper(X)) | | top(ok(X)) | → | top(active(X)) |
Original Signature
Termination of terms over the following signature is verified: mark, and, 0, s, zeros, tt, take, length, active, ok, proper, top, cons, nil
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| proper#(cons(X1, X2)) | → | proper#(X1) | | take#(mark(X1), X2) | → | take#(X1, X2) |
| proper#(length(X)) | → | length#(proper(X)) | | top#(ok(X)) | → | top#(active(X)) |
| cons#(mark(X1), X2) | → | cons#(X1, X2) | | cons#(ok(X1), ok(X2)) | → | cons#(X1, X2) |
| and#(ok(X1), ok(X2)) | → | and#(X1, X2) | | proper#(and(X1, X2)) | → | proper#(X1) |
| proper#(and(X1, X2)) | → | and#(proper(X1), proper(X2)) | | top#(ok(X)) | → | active#(X) |
| active#(cons(X1, X2)) | → | cons#(active(X1), X2) | | active#(take(X1, X2)) | → | take#(active(X1), X2) |
| active#(and(X1, X2)) | → | and#(active(X1), X2) | | proper#(and(X1, X2)) | → | proper#(X2) |
| proper#(take(X1, X2)) | → | proper#(X1) | | length#(ok(X)) | → | length#(X) |
| active#(take(s(M), cons(N, IL))) | → | take#(M, IL) | | active#(take(s(M), cons(N, IL))) | → | cons#(N, take(M, IL)) |
| take#(ok(X1), ok(X2)) | → | take#(X1, X2) | | top#(mark(X)) | → | proper#(X) |
| length#(mark(X)) | → | length#(X) | | top#(mark(X)) | → | top#(proper(X)) |
| active#(length(X)) | → | length#(active(X)) | | proper#(cons(X1, X2)) | → | proper#(X2) |
| active#(take(X1, X2)) | → | active#(X2) | | active#(take(X1, X2)) | → | active#(X1) |
| active#(length(X)) | → | active#(X) | | proper#(take(X1, X2)) | → | proper#(X2) |
| take#(X1, mark(X2)) | → | take#(X1, X2) | | and#(mark(X1), X2) | → | and#(X1, X2) |
| active#(s(X)) | → | s#(active(X)) | | active#(length(cons(N, L))) | → | s#(length(L)) |
| active#(length(cons(N, L))) | → | length#(L) | | s#(ok(X)) | → | s#(X) |
| s#(mark(X)) | → | s#(X) | | proper#(length(X)) | → | proper#(X) |
| proper#(s(X)) | → | proper#(X) | | proper#(cons(X1, X2)) | → | cons#(proper(X1), proper(X2)) |
| active#(take(X1, X2)) | → | take#(X1, active(X2)) | | active#(s(X)) | → | active#(X) |
| proper#(s(X)) | → | s#(proper(X)) | | active#(zeros) | → | cons#(0, zeros) |
| proper#(take(X1, X2)) | → | take#(proper(X1), proper(X2)) | | active#(and(X1, X2)) | → | active#(X1) |
| active#(cons(X1, X2)) | → | active#(X1) |
Rewrite Rules
| active(zeros) | → | mark(cons(0, zeros)) | | active(and(tt, X)) | → | mark(X) |
| active(length(nil)) | → | mark(0) | | active(length(cons(N, L))) | → | mark(s(length(L))) |
| active(take(0, IL)) | → | mark(nil) | | active(take(s(M), cons(N, IL))) | → | mark(cons(N, take(M, IL))) |
| active(cons(X1, X2)) | → | cons(active(X1), X2) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(length(X)) | → | length(active(X)) | | active(s(X)) | → | s(active(X)) |
| active(take(X1, X2)) | → | take(active(X1), X2) | | active(take(X1, X2)) | → | take(X1, active(X2)) |
| cons(mark(X1), X2) | → | mark(cons(X1, X2)) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| length(mark(X)) | → | mark(length(X)) | | s(mark(X)) | → | mark(s(X)) |
| take(mark(X1), X2) | → | mark(take(X1, X2)) | | take(X1, mark(X2)) | → | mark(take(X1, X2)) |
| proper(zeros) | → | ok(zeros) | | proper(cons(X1, X2)) | → | cons(proper(X1), proper(X2)) |
| proper(0) | → | ok(0) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | proper(length(X)) | → | length(proper(X)) |
| proper(nil) | → | ok(nil) | | proper(s(X)) | → | s(proper(X)) |
| proper(take(X1, X2)) | → | take(proper(X1), proper(X2)) | | cons(ok(X1), ok(X2)) | → | ok(cons(X1, X2)) |
| and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) | | length(ok(X)) | → | ok(length(X)) |
| s(ok(X)) | → | ok(s(X)) | | take(ok(X1), ok(X2)) | → | ok(take(X1, X2)) |
| top(mark(X)) | → | top(proper(X)) | | top(ok(X)) | → | top(active(X)) |
Original Signature
Termination of terms over the following signature is verified: mark, and, 0, s, zeros, tt, take, length, active, ok, proper, nil, top, cons
Strategy
The following SCCs where found
| active#(s(X)) → active#(X) | active#(take(X1, X2)) → active#(X2) |
| active#(take(X1, X2)) → active#(X1) | active#(length(X)) → active#(X) |
| active#(and(X1, X2)) → active#(X1) | active#(cons(X1, X2)) → active#(X1) |
| cons#(mark(X1), X2) → cons#(X1, X2) | cons#(ok(X1), ok(X2)) → cons#(X1, X2) |
| length#(mark(X)) → length#(X) | length#(ok(X)) → length#(X) |
| proper#(s(X)) → proper#(X) | proper#(length(X)) → proper#(X) |
| proper#(cons(X1, X2)) → proper#(X1) | proper#(cons(X1, X2)) → proper#(X2) |
| proper#(and(X1, X2)) → proper#(X2) | proper#(take(X1, X2)) → proper#(X1) |
| proper#(take(X1, X2)) → proper#(X2) | proper#(and(X1, X2)) → proper#(X1) |
| take#(mark(X1), X2) → take#(X1, X2) | take#(X1, mark(X2)) → take#(X1, X2) |
| take#(ok(X1), ok(X2)) → take#(X1, X2) |
| and#(ok(X1), ok(X2)) → and#(X1, X2) | and#(mark(X1), X2) → and#(X1, X2) |
| top#(mark(X)) → top#(proper(X)) | top#(ok(X)) → top#(active(X)) |
| s#(mark(X)) → s#(X) | s#(ok(X)) → s#(X) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| cons#(mark(X1), X2) | → | cons#(X1, X2) | | cons#(ok(X1), ok(X2)) | → | cons#(X1, X2) |
Rewrite Rules
| active(zeros) | → | mark(cons(0, zeros)) | | active(and(tt, X)) | → | mark(X) |
| active(length(nil)) | → | mark(0) | | active(length(cons(N, L))) | → | mark(s(length(L))) |
| active(take(0, IL)) | → | mark(nil) | | active(take(s(M), cons(N, IL))) | → | mark(cons(N, take(M, IL))) |
| active(cons(X1, X2)) | → | cons(active(X1), X2) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(length(X)) | → | length(active(X)) | | active(s(X)) | → | s(active(X)) |
| active(take(X1, X2)) | → | take(active(X1), X2) | | active(take(X1, X2)) | → | take(X1, active(X2)) |
| cons(mark(X1), X2) | → | mark(cons(X1, X2)) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| length(mark(X)) | → | mark(length(X)) | | s(mark(X)) | → | mark(s(X)) |
| take(mark(X1), X2) | → | mark(take(X1, X2)) | | take(X1, mark(X2)) | → | mark(take(X1, X2)) |
| proper(zeros) | → | ok(zeros) | | proper(cons(X1, X2)) | → | cons(proper(X1), proper(X2)) |
| proper(0) | → | ok(0) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | proper(length(X)) | → | length(proper(X)) |
| proper(nil) | → | ok(nil) | | proper(s(X)) | → | s(proper(X)) |
| proper(take(X1, X2)) | → | take(proper(X1), proper(X2)) | | cons(ok(X1), ok(X2)) | → | ok(cons(X1, X2)) |
| and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) | | length(ok(X)) | → | ok(length(X)) |
| s(ok(X)) | → | ok(s(X)) | | take(ok(X1), ok(X2)) | → | ok(take(X1, X2)) |
| top(mark(X)) | → | top(proper(X)) | | top(ok(X)) | → | top(active(X)) |
Original Signature
Termination of terms over the following signature is verified: mark, and, 0, s, zeros, tt, take, length, active, ok, proper, nil, top, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| cons#(mark(X1), X2) | → | cons#(X1, X2) | | cons#(ok(X1), ok(X2)) | → | cons#(X1, X2) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| take#(mark(X1), X2) | → | take#(X1, X2) | | take#(X1, mark(X2)) | → | take#(X1, X2) |
| take#(ok(X1), ok(X2)) | → | take#(X1, X2) |
Rewrite Rules
| active(zeros) | → | mark(cons(0, zeros)) | | active(and(tt, X)) | → | mark(X) |
| active(length(nil)) | → | mark(0) | | active(length(cons(N, L))) | → | mark(s(length(L))) |
| active(take(0, IL)) | → | mark(nil) | | active(take(s(M), cons(N, IL))) | → | mark(cons(N, take(M, IL))) |
| active(cons(X1, X2)) | → | cons(active(X1), X2) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(length(X)) | → | length(active(X)) | | active(s(X)) | → | s(active(X)) |
| active(take(X1, X2)) | → | take(active(X1), X2) | | active(take(X1, X2)) | → | take(X1, active(X2)) |
| cons(mark(X1), X2) | → | mark(cons(X1, X2)) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| length(mark(X)) | → | mark(length(X)) | | s(mark(X)) | → | mark(s(X)) |
| take(mark(X1), X2) | → | mark(take(X1, X2)) | | take(X1, mark(X2)) | → | mark(take(X1, X2)) |
| proper(zeros) | → | ok(zeros) | | proper(cons(X1, X2)) | → | cons(proper(X1), proper(X2)) |
| proper(0) | → | ok(0) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | proper(length(X)) | → | length(proper(X)) |
| proper(nil) | → | ok(nil) | | proper(s(X)) | → | s(proper(X)) |
| proper(take(X1, X2)) | → | take(proper(X1), proper(X2)) | | cons(ok(X1), ok(X2)) | → | ok(cons(X1, X2)) |
| and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) | | length(ok(X)) | → | ok(length(X)) |
| s(ok(X)) | → | ok(s(X)) | | take(ok(X1), ok(X2)) | → | ok(take(X1, X2)) |
| top(mark(X)) | → | top(proper(X)) | | top(ok(X)) | → | top(active(X)) |
Original Signature
Termination of terms over the following signature is verified: mark, and, 0, s, zeros, tt, take, length, active, ok, proper, nil, top, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| take#(mark(X1), X2) | → | take#(X1, X2) | | take#(ok(X1), ok(X2)) | → | take#(X1, X2) |
Problem 10: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| take#(X1, mark(X2)) | → | take#(X1, X2) |
Rewrite Rules
| active(zeros) | → | mark(cons(0, zeros)) | | active(and(tt, X)) | → | mark(X) |
| active(length(nil)) | → | mark(0) | | active(length(cons(N, L))) | → | mark(s(length(L))) |
| active(take(0, IL)) | → | mark(nil) | | active(take(s(M), cons(N, IL))) | → | mark(cons(N, take(M, IL))) |
| active(cons(X1, X2)) | → | cons(active(X1), X2) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(length(X)) | → | length(active(X)) | | active(s(X)) | → | s(active(X)) |
| active(take(X1, X2)) | → | take(active(X1), X2) | | active(take(X1, X2)) | → | take(X1, active(X2)) |
| cons(mark(X1), X2) | → | mark(cons(X1, X2)) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| length(mark(X)) | → | mark(length(X)) | | s(mark(X)) | → | mark(s(X)) |
| take(mark(X1), X2) | → | mark(take(X1, X2)) | | take(X1, mark(X2)) | → | mark(take(X1, X2)) |
| proper(zeros) | → | ok(zeros) | | proper(cons(X1, X2)) | → | cons(proper(X1), proper(X2)) |
| proper(0) | → | ok(0) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | proper(length(X)) | → | length(proper(X)) |
| proper(nil) | → | ok(nil) | | proper(s(X)) | → | s(proper(X)) |
| proper(take(X1, X2)) | → | take(proper(X1), proper(X2)) | | cons(ok(X1), ok(X2)) | → | ok(cons(X1, X2)) |
| and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) | | length(ok(X)) | → | ok(length(X)) |
| s(ok(X)) | → | ok(s(X)) | | take(ok(X1), ok(X2)) | → | ok(take(X1, X2)) |
| top(mark(X)) | → | top(proper(X)) | | top(ok(X)) | → | top(active(X)) |
Original Signature
Termination of terms over the following signature is verified: mark, and, 0, s, zeros, tt, take, length, active, ok, proper, cons, nil, top
Strategy
Polynomial Interpretation
- 0: 0
- active(x): 0
- and(x,y): 0
- cons(x,y): 0
- length(x): 0
- mark(x): x + 2
- nil: 0
- ok(x): 0
- proper(x): 0
- s(x): 0
- take(x,y): 0
- take#(x,y): y + x + 1
- top(x): 0
- tt: 0
- zeros: 0
There are no usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| take#(X1, mark(X2)) | → | take#(X1, X2) |
Problem 5: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| proper#(s(X)) | → | proper#(X) | | proper#(length(X)) | → | proper#(X) |
| proper#(cons(X1, X2)) | → | proper#(X1) | | proper#(cons(X1, X2)) | → | proper#(X2) |
| proper#(and(X1, X2)) | → | proper#(X2) | | proper#(take(X1, X2)) | → | proper#(X1) |
| proper#(take(X1, X2)) | → | proper#(X2) | | proper#(and(X1, X2)) | → | proper#(X1) |
Rewrite Rules
| active(zeros) | → | mark(cons(0, zeros)) | | active(and(tt, X)) | → | mark(X) |
| active(length(nil)) | → | mark(0) | | active(length(cons(N, L))) | → | mark(s(length(L))) |
| active(take(0, IL)) | → | mark(nil) | | active(take(s(M), cons(N, IL))) | → | mark(cons(N, take(M, IL))) |
| active(cons(X1, X2)) | → | cons(active(X1), X2) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(length(X)) | → | length(active(X)) | | active(s(X)) | → | s(active(X)) |
| active(take(X1, X2)) | → | take(active(X1), X2) | | active(take(X1, X2)) | → | take(X1, active(X2)) |
| cons(mark(X1), X2) | → | mark(cons(X1, X2)) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| length(mark(X)) | → | mark(length(X)) | | s(mark(X)) | → | mark(s(X)) |
| take(mark(X1), X2) | → | mark(take(X1, X2)) | | take(X1, mark(X2)) | → | mark(take(X1, X2)) |
| proper(zeros) | → | ok(zeros) | | proper(cons(X1, X2)) | → | cons(proper(X1), proper(X2)) |
| proper(0) | → | ok(0) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | proper(length(X)) | → | length(proper(X)) |
| proper(nil) | → | ok(nil) | | proper(s(X)) | → | s(proper(X)) |
| proper(take(X1, X2)) | → | take(proper(X1), proper(X2)) | | cons(ok(X1), ok(X2)) | → | ok(cons(X1, X2)) |
| and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) | | length(ok(X)) | → | ok(length(X)) |
| s(ok(X)) | → | ok(s(X)) | | take(ok(X1), ok(X2)) | → | ok(take(X1, X2)) |
| top(mark(X)) | → | top(proper(X)) | | top(ok(X)) | → | top(active(X)) |
Original Signature
Termination of terms over the following signature is verified: mark, and, 0, s, zeros, tt, take, length, active, ok, proper, nil, top, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| proper#(length(X)) | → | proper#(X) | | proper#(s(X)) | → | proper#(X) |
| proper#(cons(X1, X2)) | → | proper#(X1) | | proper#(cons(X1, X2)) | → | proper#(X2) |
| proper#(and(X1, X2)) | → | proper#(X2) | | proper#(take(X1, X2)) | → | proper#(X1) |
| proper#(take(X1, X2)) | → | proper#(X2) | | proper#(and(X1, X2)) | → | proper#(X1) |
Problem 6: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| length#(mark(X)) | → | length#(X) | | length#(ok(X)) | → | length#(X) |
Rewrite Rules
| active(zeros) | → | mark(cons(0, zeros)) | | active(and(tt, X)) | → | mark(X) |
| active(length(nil)) | → | mark(0) | | active(length(cons(N, L))) | → | mark(s(length(L))) |
| active(take(0, IL)) | → | mark(nil) | | active(take(s(M), cons(N, IL))) | → | mark(cons(N, take(M, IL))) |
| active(cons(X1, X2)) | → | cons(active(X1), X2) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(length(X)) | → | length(active(X)) | | active(s(X)) | → | s(active(X)) |
| active(take(X1, X2)) | → | take(active(X1), X2) | | active(take(X1, X2)) | → | take(X1, active(X2)) |
| cons(mark(X1), X2) | → | mark(cons(X1, X2)) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| length(mark(X)) | → | mark(length(X)) | | s(mark(X)) | → | mark(s(X)) |
| take(mark(X1), X2) | → | mark(take(X1, X2)) | | take(X1, mark(X2)) | → | mark(take(X1, X2)) |
| proper(zeros) | → | ok(zeros) | | proper(cons(X1, X2)) | → | cons(proper(X1), proper(X2)) |
| proper(0) | → | ok(0) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | proper(length(X)) | → | length(proper(X)) |
| proper(nil) | → | ok(nil) | | proper(s(X)) | → | s(proper(X)) |
| proper(take(X1, X2)) | → | take(proper(X1), proper(X2)) | | cons(ok(X1), ok(X2)) | → | ok(cons(X1, X2)) |
| and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) | | length(ok(X)) | → | ok(length(X)) |
| s(ok(X)) | → | ok(s(X)) | | take(ok(X1), ok(X2)) | → | ok(take(X1, X2)) |
| top(mark(X)) | → | top(proper(X)) | | top(ok(X)) | → | top(active(X)) |
Original Signature
Termination of terms over the following signature is verified: mark, and, 0, s, zeros, tt, take, length, active, ok, proper, nil, top, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| length#(mark(X)) | → | length#(X) | | length#(ok(X)) | → | length#(X) |
Problem 7: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| s#(mark(X)) | → | s#(X) | | s#(ok(X)) | → | s#(X) |
Rewrite Rules
| active(zeros) | → | mark(cons(0, zeros)) | | active(and(tt, X)) | → | mark(X) |
| active(length(nil)) | → | mark(0) | | active(length(cons(N, L))) | → | mark(s(length(L))) |
| active(take(0, IL)) | → | mark(nil) | | active(take(s(M), cons(N, IL))) | → | mark(cons(N, take(M, IL))) |
| active(cons(X1, X2)) | → | cons(active(X1), X2) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(length(X)) | → | length(active(X)) | | active(s(X)) | → | s(active(X)) |
| active(take(X1, X2)) | → | take(active(X1), X2) | | active(take(X1, X2)) | → | take(X1, active(X2)) |
| cons(mark(X1), X2) | → | mark(cons(X1, X2)) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| length(mark(X)) | → | mark(length(X)) | | s(mark(X)) | → | mark(s(X)) |
| take(mark(X1), X2) | → | mark(take(X1, X2)) | | take(X1, mark(X2)) | → | mark(take(X1, X2)) |
| proper(zeros) | → | ok(zeros) | | proper(cons(X1, X2)) | → | cons(proper(X1), proper(X2)) |
| proper(0) | → | ok(0) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | proper(length(X)) | → | length(proper(X)) |
| proper(nil) | → | ok(nil) | | proper(s(X)) | → | s(proper(X)) |
| proper(take(X1, X2)) | → | take(proper(X1), proper(X2)) | | cons(ok(X1), ok(X2)) | → | ok(cons(X1, X2)) |
| and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) | | length(ok(X)) | → | ok(length(X)) |
| s(ok(X)) | → | ok(s(X)) | | take(ok(X1), ok(X2)) | → | ok(take(X1, X2)) |
| top(mark(X)) | → | top(proper(X)) | | top(ok(X)) | → | top(active(X)) |
Original Signature
Termination of terms over the following signature is verified: mark, and, 0, s, zeros, tt, take, length, active, ok, proper, nil, top, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| s#(mark(X)) | → | s#(X) | | s#(ok(X)) | → | s#(X) |
Problem 8: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| and#(ok(X1), ok(X2)) | → | and#(X1, X2) | | and#(mark(X1), X2) | → | and#(X1, X2) |
Rewrite Rules
| active(zeros) | → | mark(cons(0, zeros)) | | active(and(tt, X)) | → | mark(X) |
| active(length(nil)) | → | mark(0) | | active(length(cons(N, L))) | → | mark(s(length(L))) |
| active(take(0, IL)) | → | mark(nil) | | active(take(s(M), cons(N, IL))) | → | mark(cons(N, take(M, IL))) |
| active(cons(X1, X2)) | → | cons(active(X1), X2) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(length(X)) | → | length(active(X)) | | active(s(X)) | → | s(active(X)) |
| active(take(X1, X2)) | → | take(active(X1), X2) | | active(take(X1, X2)) | → | take(X1, active(X2)) |
| cons(mark(X1), X2) | → | mark(cons(X1, X2)) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| length(mark(X)) | → | mark(length(X)) | | s(mark(X)) | → | mark(s(X)) |
| take(mark(X1), X2) | → | mark(take(X1, X2)) | | take(X1, mark(X2)) | → | mark(take(X1, X2)) |
| proper(zeros) | → | ok(zeros) | | proper(cons(X1, X2)) | → | cons(proper(X1), proper(X2)) |
| proper(0) | → | ok(0) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | proper(length(X)) | → | length(proper(X)) |
| proper(nil) | → | ok(nil) | | proper(s(X)) | → | s(proper(X)) |
| proper(take(X1, X2)) | → | take(proper(X1), proper(X2)) | | cons(ok(X1), ok(X2)) | → | ok(cons(X1, X2)) |
| and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) | | length(ok(X)) | → | ok(length(X)) |
| s(ok(X)) | → | ok(s(X)) | | take(ok(X1), ok(X2)) | → | ok(take(X1, X2)) |
| top(mark(X)) | → | top(proper(X)) | | top(ok(X)) | → | top(active(X)) |
Original Signature
Termination of terms over the following signature is verified: mark, and, 0, s, zeros, tt, take, length, active, ok, proper, nil, top, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| and#(ok(X1), ok(X2)) | → | and#(X1, X2) | | and#(mark(X1), X2) | → | and#(X1, X2) |
Problem 9: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| active#(s(X)) | → | active#(X) | | active#(take(X1, X2)) | → | active#(X2) |
| active#(take(X1, X2)) | → | active#(X1) | | active#(length(X)) | → | active#(X) |
| active#(and(X1, X2)) | → | active#(X1) | | active#(cons(X1, X2)) | → | active#(X1) |
Rewrite Rules
| active(zeros) | → | mark(cons(0, zeros)) | | active(and(tt, X)) | → | mark(X) |
| active(length(nil)) | → | mark(0) | | active(length(cons(N, L))) | → | mark(s(length(L))) |
| active(take(0, IL)) | → | mark(nil) | | active(take(s(M), cons(N, IL))) | → | mark(cons(N, take(M, IL))) |
| active(cons(X1, X2)) | → | cons(active(X1), X2) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(length(X)) | → | length(active(X)) | | active(s(X)) | → | s(active(X)) |
| active(take(X1, X2)) | → | take(active(X1), X2) | | active(take(X1, X2)) | → | take(X1, active(X2)) |
| cons(mark(X1), X2) | → | mark(cons(X1, X2)) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| length(mark(X)) | → | mark(length(X)) | | s(mark(X)) | → | mark(s(X)) |
| take(mark(X1), X2) | → | mark(take(X1, X2)) | | take(X1, mark(X2)) | → | mark(take(X1, X2)) |
| proper(zeros) | → | ok(zeros) | | proper(cons(X1, X2)) | → | cons(proper(X1), proper(X2)) |
| proper(0) | → | ok(0) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | proper(length(X)) | → | length(proper(X)) |
| proper(nil) | → | ok(nil) | | proper(s(X)) | → | s(proper(X)) |
| proper(take(X1, X2)) | → | take(proper(X1), proper(X2)) | | cons(ok(X1), ok(X2)) | → | ok(cons(X1, X2)) |
| and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) | | length(ok(X)) | → | ok(length(X)) |
| s(ok(X)) | → | ok(s(X)) | | take(ok(X1), ok(X2)) | → | ok(take(X1, X2)) |
| top(mark(X)) | → | top(proper(X)) | | top(ok(X)) | → | top(active(X)) |
Original Signature
Termination of terms over the following signature is verified: mark, and, 0, s, zeros, tt, take, length, active, ok, proper, nil, top, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| active#(s(X)) | → | active#(X) | | active#(take(X1, X2)) | → | active#(X2) |
| active#(length(X)) | → | active#(X) | | active#(take(X1, X2)) | → | active#(X1) |
| active#(and(X1, X2)) | → | active#(X1) | | active#(cons(X1, X2)) | → | active#(X1) |