YES
The TRS could be proven terminating. The proof took 1536 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (313ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor SubtermCriterion (0ms).
| – Problem 4 was processed with processor PolynomialLinearRange4iUR (785ms).
| | – Problem 9 was processed with processor PolynomialLinearRange4iUR (336ms).
| – Problem 5 was processed with processor SubtermCriterion (1ms).
| – Problem 6 was processed with processor SubtermCriterion (3ms).
| – Problem 7 was processed with processor SubtermCriterion (1ms).
| – Problem 8 was processed with processor SubtermCriterion (2ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| proper#(cons(X1, X2)) | → | proper#(X1) | | top#(ok(X)) | → | top#(active(X)) |
| active#(2nd(X)) | → | 2nd#(active(X)) | | cons#(mark(X1), X2) | → | cons#(X1, X2) |
| 2nd#(mark(X)) | → | 2nd#(X) | | active#(from(X)) | → | from#(active(X)) |
| cons#(ok(X1), ok(X2)) | → | cons#(X1, X2) | | from#(mark(X)) | → | from#(X) |
| from#(ok(X)) | → | from#(X) | | top#(ok(X)) | → | active#(X) |
| active#(cons(X1, X2)) | → | cons#(active(X1), X2) | | 2nd#(ok(X)) | → | 2nd#(X) |
| proper#(2nd(X)) | → | proper#(X) | | active#(from(X)) | → | cons#(X, from(s(X))) |
| proper#(from(X)) | → | from#(proper(X)) | | top#(mark(X)) | → | proper#(X) |
| proper#(from(X)) | → | proper#(X) | | active#(2nd(X)) | → | active#(X) |
| top#(mark(X)) | → | top#(proper(X)) | | proper#(cons(X1, X2)) | → | proper#(X2) |
| active#(from(X)) | → | active#(X) | | proper#(2nd(X)) | → | 2nd#(proper(X)) |
| active#(s(X)) | → | s#(active(X)) | | s#(ok(X)) | → | s#(X) |
| s#(mark(X)) | → | s#(X) | | active#(from(X)) | → | s#(X) |
| proper#(s(X)) | → | proper#(X) | | proper#(cons(X1, X2)) | → | cons#(proper(X1), proper(X2)) |
| active#(s(X)) | → | active#(X) | | proper#(s(X)) | → | s#(proper(X)) |
| active#(from(X)) | → | from#(s(X)) | | active#(cons(X1, X2)) | → | active#(X1) |
Rewrite Rules
| active(2nd(cons(X, cons(Y, Z)))) | → | mark(Y) | | active(from(X)) | → | mark(cons(X, from(s(X)))) |
| active(2nd(X)) | → | 2nd(active(X)) | | active(cons(X1, X2)) | → | cons(active(X1), X2) |
| active(from(X)) | → | from(active(X)) | | active(s(X)) | → | s(active(X)) |
| 2nd(mark(X)) | → | mark(2nd(X)) | | cons(mark(X1), X2) | → | mark(cons(X1, X2)) |
| from(mark(X)) | → | mark(from(X)) | | s(mark(X)) | → | mark(s(X)) |
| proper(2nd(X)) | → | 2nd(proper(X)) | | proper(cons(X1, X2)) | → | cons(proper(X1), proper(X2)) |
| proper(from(X)) | → | from(proper(X)) | | proper(s(X)) | → | s(proper(X)) |
| 2nd(ok(X)) | → | ok(2nd(X)) | | cons(ok(X1), ok(X2)) | → | ok(cons(X1, X2)) |
| from(ok(X)) | → | ok(from(X)) | | s(ok(X)) | → | ok(s(X)) |
| top(mark(X)) | → | top(proper(X)) | | top(ok(X)) | → | top(active(X)) |
Original Signature
Termination of terms over the following signature is verified: 2nd, s, active, mark, ok, from, proper, cons, top
Strategy
The following SCCs where found
| cons#(mark(X1), X2) → cons#(X1, X2) | cons#(ok(X1), ok(X2)) → cons#(X1, X2) |
| proper#(s(X)) → proper#(X) | proper#(2nd(X)) → proper#(X) |
| proper#(cons(X1, X2)) → proper#(X1) | proper#(cons(X1, X2)) → proper#(X2) |
| proper#(from(X)) → proper#(X) |
| active#(from(X)) → active#(X) | active#(s(X)) → active#(X) |
| active#(cons(X1, X2)) → active#(X1) | active#(2nd(X)) → active#(X) |
| from#(mark(X)) → from#(X) | from#(ok(X)) → from#(X) |
| top#(mark(X)) → top#(proper(X)) | top#(ok(X)) → top#(active(X)) |
| s#(mark(X)) → s#(X) | s#(ok(X)) → s#(X) |
| 2nd#(ok(X)) → 2nd#(X) | 2nd#(mark(X)) → 2nd#(X) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| proper#(s(X)) | → | proper#(X) | | proper#(2nd(X)) | → | proper#(X) |
| proper#(cons(X1, X2)) | → | proper#(X1) | | proper#(cons(X1, X2)) | → | proper#(X2) |
| proper#(from(X)) | → | proper#(X) |
Rewrite Rules
| active(2nd(cons(X, cons(Y, Z)))) | → | mark(Y) | | active(from(X)) | → | mark(cons(X, from(s(X)))) |
| active(2nd(X)) | → | 2nd(active(X)) | | active(cons(X1, X2)) | → | cons(active(X1), X2) |
| active(from(X)) | → | from(active(X)) | | active(s(X)) | → | s(active(X)) |
| 2nd(mark(X)) | → | mark(2nd(X)) | | cons(mark(X1), X2) | → | mark(cons(X1, X2)) |
| from(mark(X)) | → | mark(from(X)) | | s(mark(X)) | → | mark(s(X)) |
| proper(2nd(X)) | → | 2nd(proper(X)) | | proper(cons(X1, X2)) | → | cons(proper(X1), proper(X2)) |
| proper(from(X)) | → | from(proper(X)) | | proper(s(X)) | → | s(proper(X)) |
| 2nd(ok(X)) | → | ok(2nd(X)) | | cons(ok(X1), ok(X2)) | → | ok(cons(X1, X2)) |
| from(ok(X)) | → | ok(from(X)) | | s(ok(X)) | → | ok(s(X)) |
| top(mark(X)) | → | top(proper(X)) | | top(ok(X)) | → | top(active(X)) |
Original Signature
Termination of terms over the following signature is verified: 2nd, s, active, mark, ok, from, proper, cons, top
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| proper#(s(X)) | → | proper#(X) | | proper#(2nd(X)) | → | proper#(X) |
| proper#(cons(X1, X2)) | → | proper#(X1) | | proper#(cons(X1, X2)) | → | proper#(X2) |
| proper#(from(X)) | → | proper#(X) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| active#(from(X)) | → | active#(X) | | active#(s(X)) | → | active#(X) |
| active#(cons(X1, X2)) | → | active#(X1) | | active#(2nd(X)) | → | active#(X) |
Rewrite Rules
| active(2nd(cons(X, cons(Y, Z)))) | → | mark(Y) | | active(from(X)) | → | mark(cons(X, from(s(X)))) |
| active(2nd(X)) | → | 2nd(active(X)) | | active(cons(X1, X2)) | → | cons(active(X1), X2) |
| active(from(X)) | → | from(active(X)) | | active(s(X)) | → | s(active(X)) |
| 2nd(mark(X)) | → | mark(2nd(X)) | | cons(mark(X1), X2) | → | mark(cons(X1, X2)) |
| from(mark(X)) | → | mark(from(X)) | | s(mark(X)) | → | mark(s(X)) |
| proper(2nd(X)) | → | 2nd(proper(X)) | | proper(cons(X1, X2)) | → | cons(proper(X1), proper(X2)) |
| proper(from(X)) | → | from(proper(X)) | | proper(s(X)) | → | s(proper(X)) |
| 2nd(ok(X)) | → | ok(2nd(X)) | | cons(ok(X1), ok(X2)) | → | ok(cons(X1, X2)) |
| from(ok(X)) | → | ok(from(X)) | | s(ok(X)) | → | ok(s(X)) |
| top(mark(X)) | → | top(proper(X)) | | top(ok(X)) | → | top(active(X)) |
Original Signature
Termination of terms over the following signature is verified: 2nd, s, active, mark, ok, from, proper, cons, top
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| active#(s(X)) | → | active#(X) | | active#(from(X)) | → | active#(X) |
| active#(2nd(X)) | → | active#(X) | | active#(cons(X1, X2)) | → | active#(X1) |
Problem 4: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| top#(mark(X)) | → | top#(proper(X)) | | top#(ok(X)) | → | top#(active(X)) |
Rewrite Rules
| active(2nd(cons(X, cons(Y, Z)))) | → | mark(Y) | | active(from(X)) | → | mark(cons(X, from(s(X)))) |
| active(2nd(X)) | → | 2nd(active(X)) | | active(cons(X1, X2)) | → | cons(active(X1), X2) |
| active(from(X)) | → | from(active(X)) | | active(s(X)) | → | s(active(X)) |
| 2nd(mark(X)) | → | mark(2nd(X)) | | cons(mark(X1), X2) | → | mark(cons(X1, X2)) |
| from(mark(X)) | → | mark(from(X)) | | s(mark(X)) | → | mark(s(X)) |
| proper(2nd(X)) | → | 2nd(proper(X)) | | proper(cons(X1, X2)) | → | cons(proper(X1), proper(X2)) |
| proper(from(X)) | → | from(proper(X)) | | proper(s(X)) | → | s(proper(X)) |
| 2nd(ok(X)) | → | ok(2nd(X)) | | cons(ok(X1), ok(X2)) | → | ok(cons(X1, X2)) |
| from(ok(X)) | → | ok(from(X)) | | s(ok(X)) | → | ok(s(X)) |
| top(mark(X)) | → | top(proper(X)) | | top(ok(X)) | → | top(active(X)) |
Original Signature
Termination of terms over the following signature is verified: 2nd, s, active, mark, ok, from, proper, cons, top
Strategy
Polynomial Interpretation
- 2nd(x): x
- active(x): 2
- cons(x,y): x
- from(x): x
- mark(x): 2
- ok(x): 2x + 2
- proper(x): 0
- s(x): x
- top(x): 0
- top#(x): x
Improved Usable rules
| active(s(X)) | → | s(active(X)) | | 2nd(mark(X)) | → | mark(2nd(X)) |
| active(2nd(cons(X, cons(Y, Z)))) | → | mark(Y) | | s(ok(X)) | → | ok(s(X)) |
| cons(mark(X1), X2) | → | mark(cons(X1, X2)) | | proper(2nd(X)) | → | 2nd(proper(X)) |
| from(ok(X)) | → | ok(from(X)) | | active(from(X)) | → | from(active(X)) |
| cons(ok(X1), ok(X2)) | → | ok(cons(X1, X2)) | | from(mark(X)) | → | mark(from(X)) |
| active(cons(X1, X2)) | → | cons(active(X1), X2) | | s(mark(X)) | → | mark(s(X)) |
| proper(cons(X1, X2)) | → | cons(proper(X1), proper(X2)) | | 2nd(ok(X)) | → | ok(2nd(X)) |
| proper(s(X)) | → | s(proper(X)) | | active(2nd(X)) | → | 2nd(active(X)) |
| active(from(X)) | → | mark(cons(X, from(s(X)))) | | proper(from(X)) | → | from(proper(X)) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| top#(mark(X)) | → | top#(proper(X)) |
Problem 9: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| top#(ok(X)) | → | top#(active(X)) |
Rewrite Rules
| active(2nd(cons(X, cons(Y, Z)))) | → | mark(Y) | | active(from(X)) | → | mark(cons(X, from(s(X)))) |
| active(2nd(X)) | → | 2nd(active(X)) | | active(cons(X1, X2)) | → | cons(active(X1), X2) |
| active(from(X)) | → | from(active(X)) | | active(s(X)) | → | s(active(X)) |
| 2nd(mark(X)) | → | mark(2nd(X)) | | cons(mark(X1), X2) | → | mark(cons(X1, X2)) |
| from(mark(X)) | → | mark(from(X)) | | s(mark(X)) | → | mark(s(X)) |
| proper(2nd(X)) | → | 2nd(proper(X)) | | proper(cons(X1, X2)) | → | cons(proper(X1), proper(X2)) |
| proper(from(X)) | → | from(proper(X)) | | proper(s(X)) | → | s(proper(X)) |
| 2nd(ok(X)) | → | ok(2nd(X)) | | cons(ok(X1), ok(X2)) | → | ok(cons(X1, X2)) |
| from(ok(X)) | → | ok(from(X)) | | s(ok(X)) | → | ok(s(X)) |
| top(mark(X)) | → | top(proper(X)) | | top(ok(X)) | → | top(active(X)) |
Original Signature
Termination of terms over the following signature is verified: 2nd, s, active, ok, mark, proper, from, top, cons
Strategy
Polynomial Interpretation
- 2nd(x): x
- active(x): 0
- cons(x,y): x
- from(x): x
- mark(x): 0
- ok(x): 1
- proper(x): 0
- s(x): x
- top(x): 0
- top#(x): 2x
Improved Usable rules
| active(s(X)) | → | s(active(X)) | | 2nd(mark(X)) | → | mark(2nd(X)) |
| active(2nd(cons(X, cons(Y, Z)))) | → | mark(Y) | | s(ok(X)) | → | ok(s(X)) |
| cons(mark(X1), X2) | → | mark(cons(X1, X2)) | | from(ok(X)) | → | ok(from(X)) |
| active(from(X)) | → | from(active(X)) | | cons(ok(X1), ok(X2)) | → | ok(cons(X1, X2)) |
| from(mark(X)) | → | mark(from(X)) | | active(cons(X1, X2)) | → | cons(active(X1), X2) |
| s(mark(X)) | → | mark(s(X)) | | 2nd(ok(X)) | → | ok(2nd(X)) |
| active(2nd(X)) | → | 2nd(active(X)) | | active(from(X)) | → | mark(cons(X, from(s(X)))) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| top#(ok(X)) | → | top#(active(X)) |
Problem 5: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| from#(mark(X)) | → | from#(X) | | from#(ok(X)) | → | from#(X) |
Rewrite Rules
| active(2nd(cons(X, cons(Y, Z)))) | → | mark(Y) | | active(from(X)) | → | mark(cons(X, from(s(X)))) |
| active(2nd(X)) | → | 2nd(active(X)) | | active(cons(X1, X2)) | → | cons(active(X1), X2) |
| active(from(X)) | → | from(active(X)) | | active(s(X)) | → | s(active(X)) |
| 2nd(mark(X)) | → | mark(2nd(X)) | | cons(mark(X1), X2) | → | mark(cons(X1, X2)) |
| from(mark(X)) | → | mark(from(X)) | | s(mark(X)) | → | mark(s(X)) |
| proper(2nd(X)) | → | 2nd(proper(X)) | | proper(cons(X1, X2)) | → | cons(proper(X1), proper(X2)) |
| proper(from(X)) | → | from(proper(X)) | | proper(s(X)) | → | s(proper(X)) |
| 2nd(ok(X)) | → | ok(2nd(X)) | | cons(ok(X1), ok(X2)) | → | ok(cons(X1, X2)) |
| from(ok(X)) | → | ok(from(X)) | | s(ok(X)) | → | ok(s(X)) |
| top(mark(X)) | → | top(proper(X)) | | top(ok(X)) | → | top(active(X)) |
Original Signature
Termination of terms over the following signature is verified: 2nd, s, active, mark, ok, from, proper, cons, top
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| from#(mark(X)) | → | from#(X) | | from#(ok(X)) | → | from#(X) |
Problem 6: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| cons#(mark(X1), X2) | → | cons#(X1, X2) | | cons#(ok(X1), ok(X2)) | → | cons#(X1, X2) |
Rewrite Rules
| active(2nd(cons(X, cons(Y, Z)))) | → | mark(Y) | | active(from(X)) | → | mark(cons(X, from(s(X)))) |
| active(2nd(X)) | → | 2nd(active(X)) | | active(cons(X1, X2)) | → | cons(active(X1), X2) |
| active(from(X)) | → | from(active(X)) | | active(s(X)) | → | s(active(X)) |
| 2nd(mark(X)) | → | mark(2nd(X)) | | cons(mark(X1), X2) | → | mark(cons(X1, X2)) |
| from(mark(X)) | → | mark(from(X)) | | s(mark(X)) | → | mark(s(X)) |
| proper(2nd(X)) | → | 2nd(proper(X)) | | proper(cons(X1, X2)) | → | cons(proper(X1), proper(X2)) |
| proper(from(X)) | → | from(proper(X)) | | proper(s(X)) | → | s(proper(X)) |
| 2nd(ok(X)) | → | ok(2nd(X)) | | cons(ok(X1), ok(X2)) | → | ok(cons(X1, X2)) |
| from(ok(X)) | → | ok(from(X)) | | s(ok(X)) | → | ok(s(X)) |
| top(mark(X)) | → | top(proper(X)) | | top(ok(X)) | → | top(active(X)) |
Original Signature
Termination of terms over the following signature is verified: 2nd, s, active, mark, ok, from, proper, cons, top
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 7: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| s#(mark(X)) | → | s#(X) | | s#(ok(X)) | → | s#(X) |
Rewrite Rules
| active(2nd(cons(X, cons(Y, Z)))) | → | mark(Y) | | active(from(X)) | → | mark(cons(X, from(s(X)))) |
| active(2nd(X)) | → | 2nd(active(X)) | | active(cons(X1, X2)) | → | cons(active(X1), X2) |
| active(from(X)) | → | from(active(X)) | | active(s(X)) | → | s(active(X)) |
| 2nd(mark(X)) | → | mark(2nd(X)) | | cons(mark(X1), X2) | → | mark(cons(X1, X2)) |
| from(mark(X)) | → | mark(from(X)) | | s(mark(X)) | → | mark(s(X)) |
| proper(2nd(X)) | → | 2nd(proper(X)) | | proper(cons(X1, X2)) | → | cons(proper(X1), proper(X2)) |
| proper(from(X)) | → | from(proper(X)) | | proper(s(X)) | → | s(proper(X)) |
| 2nd(ok(X)) | → | ok(2nd(X)) | | cons(ok(X1), ok(X2)) | → | ok(cons(X1, X2)) |
| from(ok(X)) | → | ok(from(X)) | | s(ok(X)) | → | ok(s(X)) |
| top(mark(X)) | → | top(proper(X)) | | top(ok(X)) | → | top(active(X)) |
Original Signature
Termination of terms over the following signature is verified: 2nd, s, active, mark, ok, from, proper, cons, top
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
| 2nd#(ok(X)) | → | 2nd#(X) | | 2nd#(mark(X)) | → | 2nd#(X) |
Rewrite Rules
| active(2nd(cons(X, cons(Y, Z)))) | → | mark(Y) | | active(from(X)) | → | mark(cons(X, from(s(X)))) |
| active(2nd(X)) | → | 2nd(active(X)) | | active(cons(X1, X2)) | → | cons(active(X1), X2) |
| active(from(X)) | → | from(active(X)) | | active(s(X)) | → | s(active(X)) |
| 2nd(mark(X)) | → | mark(2nd(X)) | | cons(mark(X1), X2) | → | mark(cons(X1, X2)) |
| from(mark(X)) | → | mark(from(X)) | | s(mark(X)) | → | mark(s(X)) |
| proper(2nd(X)) | → | 2nd(proper(X)) | | proper(cons(X1, X2)) | → | cons(proper(X1), proper(X2)) |
| proper(from(X)) | → | from(proper(X)) | | proper(s(X)) | → | s(proper(X)) |
| 2nd(ok(X)) | → | ok(2nd(X)) | | cons(ok(X1), ok(X2)) | → | ok(cons(X1, X2)) |
| from(ok(X)) | → | ok(from(X)) | | s(ok(X)) | → | ok(s(X)) |
| top(mark(X)) | → | top(proper(X)) | | top(ok(X)) | → | top(active(X)) |
Original Signature
Termination of terms over the following signature is verified: 2nd, s, active, mark, ok, from, proper, cons, top
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| 2nd#(ok(X)) | → | 2nd#(X) | | 2nd#(mark(X)) | → | 2nd#(X) |