YES
The TRS could be proven terminating. The proof took 1520 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (301ms).
| – Problem 2 was processed with processor SubtermCriterion (23ms).
| – 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 PolynomialLinearRange4iUR (767ms).
| | – Problem 9 was processed with processor PolynomialLinearRange4iUR (349ms).
| – Problem 7 was processed with processor SubtermCriterion (1ms).
| | – Problem 8 was processed with processor SubtermCriterion (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| top#(ok(X)) | → | top#(active(X)) | | 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#(and(X1, X2)) | → | and#(active(X1), X2) |
| proper#(and(X1, X2)) | → | proper#(X2) | | plus#(X1, mark(X2)) | → | plus#(X1, X2) |
| proper#(plus(X1, X2)) | → | proper#(X1) | | proper#(plus(X1, X2)) | → | proper#(X2) |
| top#(mark(X)) | → | proper#(X) | | proper#(plus(X1, X2)) | → | plus#(proper(X1), proper(X2)) |
| active#(plus(N, s(M))) | → | plus#(N, M) | | plus#(ok(X1), ok(X2)) | → | plus#(X1, X2) |
| top#(mark(X)) | → | top#(proper(X)) | | and#(mark(X1), X2) | → | and#(X1, X2) |
| active#(s(X)) | → | s#(active(X)) | | s#(ok(X)) | → | s#(X) |
| active#(plus(N, s(M))) | → | s#(plus(N, M)) | | s#(mark(X)) | → | s#(X) |
| active#(plus(X1, X2)) | → | plus#(X1, active(X2)) | | proper#(s(X)) | → | proper#(X) |
| active#(plus(X1, X2)) | → | active#(X1) | | active#(s(X)) | → | active#(X) |
| active#(plus(X1, X2)) | → | plus#(active(X1), X2) | | proper#(s(X)) | → | s#(proper(X)) |
| active#(plus(X1, X2)) | → | active#(X2) | | active#(and(X1, X2)) | → | active#(X1) |
| plus#(mark(X1), X2) | → | plus#(X1, X2) |
Rewrite Rules
| active(and(tt, X)) | → | mark(X) | | active(plus(N, 0)) | → | mark(N) |
| active(plus(N, s(M))) | → | mark(s(plus(N, M))) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(plus(X1, X2)) | → | plus(active(X1), X2) | | active(plus(X1, X2)) | → | plus(X1, active(X2)) |
| active(s(X)) | → | s(active(X)) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| plus(mark(X1), X2) | → | mark(plus(X1, X2)) | | plus(X1, mark(X2)) | → | mark(plus(X1, X2)) |
| s(mark(X)) | → | mark(s(X)) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | proper(plus(X1, X2)) | → | plus(proper(X1), proper(X2)) |
| proper(0) | → | ok(0) | | proper(s(X)) | → | s(proper(X)) |
| and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) | | plus(ok(X1), ok(X2)) | → | ok(plus(X1, X2)) |
| 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: plus, 0, s, tt, active, mark, ok, proper, and, top
Strategy
The following SCCs where found
| plus#(ok(X1), ok(X2)) → plus#(X1, X2) | plus#(X1, mark(X2)) → plus#(X1, X2) |
| plus#(mark(X1), X2) → plus#(X1, X2) |
| active#(plus(X1, X2)) → active#(X1) | active#(s(X)) → active#(X) |
| active#(plus(X1, X2)) → active#(X2) | active#(and(X1, X2)) → active#(X1) |
| proper#(s(X)) → proper#(X) | proper#(and(X1, X2)) → proper#(X2) |
| proper#(plus(X1, X2)) → proper#(X1) | proper#(plus(X1, X2)) → proper#(X2) |
| proper#(and(X1, X2)) → proper#(X1) |
| and#(ok(X1), ok(X2)) → and#(X1, X2) | and#(mark(X1), X2) → and#(X1, X2) |
| s#(mark(X)) → s#(X) | s#(ok(X)) → s#(X) |
| top#(mark(X)) → top#(proper(X)) | top#(ok(X)) → top#(active(X)) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| active#(plus(X1, X2)) | → | active#(X1) | | active#(s(X)) | → | active#(X) |
| active#(plus(X1, X2)) | → | active#(X2) | | active#(and(X1, X2)) | → | active#(X1) |
Rewrite Rules
| active(and(tt, X)) | → | mark(X) | | active(plus(N, 0)) | → | mark(N) |
| active(plus(N, s(M))) | → | mark(s(plus(N, M))) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(plus(X1, X2)) | → | plus(active(X1), X2) | | active(plus(X1, X2)) | → | plus(X1, active(X2)) |
| active(s(X)) | → | s(active(X)) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| plus(mark(X1), X2) | → | mark(plus(X1, X2)) | | plus(X1, mark(X2)) | → | mark(plus(X1, X2)) |
| s(mark(X)) | → | mark(s(X)) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | proper(plus(X1, X2)) | → | plus(proper(X1), proper(X2)) |
| proper(0) | → | ok(0) | | proper(s(X)) | → | s(proper(X)) |
| and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) | | plus(ok(X1), ok(X2)) | → | ok(plus(X1, X2)) |
| 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: plus, 0, s, tt, active, mark, ok, proper, and, top
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| active#(plus(X1, X2)) | → | active#(X1) | | active#(s(X)) | → | active#(X) |
| active#(plus(X1, X2)) | → | active#(X2) | | active#(and(X1, X2)) | → | active#(X1) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| s#(mark(X)) | → | s#(X) | | s#(ok(X)) | → | s#(X) |
Rewrite Rules
| active(and(tt, X)) | → | mark(X) | | active(plus(N, 0)) | → | mark(N) |
| active(plus(N, s(M))) | → | mark(s(plus(N, M))) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(plus(X1, X2)) | → | plus(active(X1), X2) | | active(plus(X1, X2)) | → | plus(X1, active(X2)) |
| active(s(X)) | → | s(active(X)) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| plus(mark(X1), X2) | → | mark(plus(X1, X2)) | | plus(X1, mark(X2)) | → | mark(plus(X1, X2)) |
| s(mark(X)) | → | mark(s(X)) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | proper(plus(X1, X2)) | → | plus(proper(X1), proper(X2)) |
| proper(0) | → | ok(0) | | proper(s(X)) | → | s(proper(X)) |
| and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) | | plus(ok(X1), ok(X2)) | → | ok(plus(X1, X2)) |
| 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: plus, 0, s, tt, active, mark, ok, proper, and, 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 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| and#(ok(X1), ok(X2)) | → | and#(X1, X2) | | and#(mark(X1), X2) | → | and#(X1, X2) |
Rewrite Rules
| active(and(tt, X)) | → | mark(X) | | active(plus(N, 0)) | → | mark(N) |
| active(plus(N, s(M))) | → | mark(s(plus(N, M))) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(plus(X1, X2)) | → | plus(active(X1), X2) | | active(plus(X1, X2)) | → | plus(X1, active(X2)) |
| active(s(X)) | → | s(active(X)) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| plus(mark(X1), X2) | → | mark(plus(X1, X2)) | | plus(X1, mark(X2)) | → | mark(plus(X1, X2)) |
| s(mark(X)) | → | mark(s(X)) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | proper(plus(X1, X2)) | → | plus(proper(X1), proper(X2)) |
| proper(0) | → | ok(0) | | proper(s(X)) | → | s(proper(X)) |
| and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) | | plus(ok(X1), ok(X2)) | → | ok(plus(X1, X2)) |
| 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: plus, 0, s, tt, active, mark, ok, proper, and, top
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 5: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| proper#(s(X)) | → | proper#(X) | | proper#(and(X1, X2)) | → | proper#(X2) |
| proper#(plus(X1, X2)) | → | proper#(X1) | | proper#(plus(X1, X2)) | → | proper#(X2) |
| proper#(and(X1, X2)) | → | proper#(X1) |
Rewrite Rules
| active(and(tt, X)) | → | mark(X) | | active(plus(N, 0)) | → | mark(N) |
| active(plus(N, s(M))) | → | mark(s(plus(N, M))) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(plus(X1, X2)) | → | plus(active(X1), X2) | | active(plus(X1, X2)) | → | plus(X1, active(X2)) |
| active(s(X)) | → | s(active(X)) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| plus(mark(X1), X2) | → | mark(plus(X1, X2)) | | plus(X1, mark(X2)) | → | mark(plus(X1, X2)) |
| s(mark(X)) | → | mark(s(X)) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | proper(plus(X1, X2)) | → | plus(proper(X1), proper(X2)) |
| proper(0) | → | ok(0) | | proper(s(X)) | → | s(proper(X)) |
| and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) | | plus(ok(X1), ok(X2)) | → | ok(plus(X1, X2)) |
| 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: plus, 0, s, tt, active, mark, ok, proper, and, top
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| proper#(s(X)) | → | proper#(X) | | proper#(and(X1, X2)) | → | proper#(X2) |
| proper#(plus(X1, X2)) | → | proper#(X1) | | proper#(plus(X1, X2)) | → | proper#(X2) |
| proper#(and(X1, X2)) | → | proper#(X1) |
Problem 6: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| top#(mark(X)) | → | top#(proper(X)) | | top#(ok(X)) | → | top#(active(X)) |
Rewrite Rules
| active(and(tt, X)) | → | mark(X) | | active(plus(N, 0)) | → | mark(N) |
| active(plus(N, s(M))) | → | mark(s(plus(N, M))) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(plus(X1, X2)) | → | plus(active(X1), X2) | | active(plus(X1, X2)) | → | plus(X1, active(X2)) |
| active(s(X)) | → | s(active(X)) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| plus(mark(X1), X2) | → | mark(plus(X1, X2)) | | plus(X1, mark(X2)) | → | mark(plus(X1, X2)) |
| s(mark(X)) | → | mark(s(X)) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | proper(plus(X1, X2)) | → | plus(proper(X1), proper(X2)) |
| proper(0) | → | ok(0) | | proper(s(X)) | → | s(proper(X)) |
| and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) | | plus(ok(X1), ok(X2)) | → | ok(plus(X1, X2)) |
| 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: plus, 0, s, tt, active, mark, ok, proper, and, top
Strategy
Polynomial Interpretation
- 0: 1
- active(x): x
- and(x,y): y + x + 2
- mark(x): x + 1
- ok(x): x
- plus(x,y): 2y + x
- proper(x): x
- s(x): x + 1
- top(x): 0
- top#(x): 2x
- tt: 0
Improved Usable rules
| plus(X1, mark(X2)) | → | mark(plus(X1, X2)) | | active(plus(X1, X2)) | → | plus(X1, active(X2)) |
| active(plus(X1, X2)) | → | plus(active(X1), X2) | | active(s(X)) | → | s(active(X)) |
| and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) | | proper(plus(X1, X2)) | → | plus(proper(X1), proper(X2)) |
| s(ok(X)) | → | ok(s(X)) | | plus(mark(X1), X2) | → | mark(plus(X1, X2)) |
| proper(tt) | → | ok(tt) | | active(plus(N, s(M))) | → | mark(s(plus(N, M))) |
| active(and(tt, X)) | → | mark(X) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| plus(ok(X1), ok(X2)) | → | ok(plus(X1, X2)) | | s(mark(X)) | → | mark(s(X)) |
| proper(s(X)) | → | s(proper(X)) | | active(plus(N, 0)) | → | mark(N) |
| and(mark(X1), X2) | → | mark(and(X1, X2)) | | proper(0) | → | ok(0) |
| active(and(X1, X2)) | → | and(active(X1), X2) |
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(and(tt, X)) | → | mark(X) | | active(plus(N, 0)) | → | mark(N) |
| active(plus(N, s(M))) | → | mark(s(plus(N, M))) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(plus(X1, X2)) | → | plus(active(X1), X2) | | active(plus(X1, X2)) | → | plus(X1, active(X2)) |
| active(s(X)) | → | s(active(X)) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| plus(mark(X1), X2) | → | mark(plus(X1, X2)) | | plus(X1, mark(X2)) | → | mark(plus(X1, X2)) |
| s(mark(X)) | → | mark(s(X)) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | proper(plus(X1, X2)) | → | plus(proper(X1), proper(X2)) |
| proper(0) | → | ok(0) | | proper(s(X)) | → | s(proper(X)) |
| and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) | | plus(ok(X1), ok(X2)) | → | ok(plus(X1, X2)) |
| 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: plus, 0, s, tt, active, ok, mark, proper, top, and
Strategy
Polynomial Interpretation
- 0: 2
- active(x): x
- and(x,y): x
- mark(x): 0
- ok(x): x + 1
- plus(x,y): x
- proper(x): 0
- s(x): 2x + 1
- top(x): 0
- top#(x): 2x + 1
- tt: 0
Improved Usable rules
| active(plus(X1, X2)) | → | plus(X1, active(X2)) | | plus(X1, mark(X2)) | → | mark(plus(X1, X2)) |
| active(plus(X1, X2)) | → | plus(active(X1), X2) | | active(s(X)) | → | s(active(X)) |
| and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) | | s(ok(X)) | → | ok(s(X)) |
| plus(mark(X1), X2) | → | mark(plus(X1, X2)) | | active(plus(N, s(M))) | → | mark(s(plus(N, M))) |
| active(and(tt, X)) | → | mark(X) | | s(mark(X)) | → | mark(s(X)) |
| plus(ok(X1), ok(X2)) | → | ok(plus(X1, X2)) | | active(plus(N, 0)) | → | mark(N) |
| and(mark(X1), X2) | → | mark(and(X1, X2)) | | active(and(X1, X2)) | → | and(active(X1), X2) |
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 7: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| plus#(ok(X1), ok(X2)) | → | plus#(X1, X2) | | plus#(X1, mark(X2)) | → | plus#(X1, X2) |
| plus#(mark(X1), X2) | → | plus#(X1, X2) |
Rewrite Rules
| active(and(tt, X)) | → | mark(X) | | active(plus(N, 0)) | → | mark(N) |
| active(plus(N, s(M))) | → | mark(s(plus(N, M))) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(plus(X1, X2)) | → | plus(active(X1), X2) | | active(plus(X1, X2)) | → | plus(X1, active(X2)) |
| active(s(X)) | → | s(active(X)) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| plus(mark(X1), X2) | → | mark(plus(X1, X2)) | | plus(X1, mark(X2)) | → | mark(plus(X1, X2)) |
| s(mark(X)) | → | mark(s(X)) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | proper(plus(X1, X2)) | → | plus(proper(X1), proper(X2)) |
| proper(0) | → | ok(0) | | proper(s(X)) | → | s(proper(X)) |
| and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) | | plus(ok(X1), ok(X2)) | → | ok(plus(X1, X2)) |
| 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: plus, 0, s, tt, active, mark, ok, proper, and, top
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| plus#(ok(X1), ok(X2)) | → | plus#(X1, X2) | | plus#(mark(X1), X2) | → | plus#(X1, X2) |
Problem 8: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| plus#(X1, mark(X2)) | → | plus#(X1, X2) |
Rewrite Rules
| active(and(tt, X)) | → | mark(X) | | active(plus(N, 0)) | → | mark(N) |
| active(plus(N, s(M))) | → | mark(s(plus(N, M))) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(plus(X1, X2)) | → | plus(active(X1), X2) | | active(plus(X1, X2)) | → | plus(X1, active(X2)) |
| active(s(X)) | → | s(active(X)) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| plus(mark(X1), X2) | → | mark(plus(X1, X2)) | | plus(X1, mark(X2)) | → | mark(plus(X1, X2)) |
| s(mark(X)) | → | mark(s(X)) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | proper(plus(X1, X2)) | → | plus(proper(X1), proper(X2)) |
| proper(0) | → | ok(0) | | proper(s(X)) | → | s(proper(X)) |
| and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) | | plus(ok(X1), ok(X2)) | → | ok(plus(X1, X2)) |
| 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: plus, 0, s, tt, active, ok, mark, proper, top, and
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| plus#(X1, mark(X2)) | → | plus#(X1, X2) |