YES
The TRS could be proven terminating. The proof took 1514 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (288ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
| – Problem 4 was processed with processor PolynomialLinearRange4iUR (640ms).
| | – Problem 9 was processed with processor PolynomialLinearRange4iUR (425ms).
| – Problem 5 was processed with processor SubtermCriterion (1ms).
| – Problem 6 was processed with processor SubtermCriterion (1ms).
| | – Problem 8 was processed with processor SubtermCriterion (1ms).
| – Problem 7 was processed with processor SubtermCriterion (2ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| top#(ok(X)) | → | top#(active(X)) | | active#(__(__(X, Y), Z)) | → | __#(X, __(Y, Z)) |
| 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) |
| proper#(__(X1, X2)) | → | __#(proper(X1), proper(X2)) | | active#(and(X1, X2)) | → | and#(active(X1), X2) |
| proper#(and(X1, X2)) | → | proper#(X2) | | active#(__(X1, X2)) | → | active#(X2) |
| active#(__(X1, X2)) | → | __#(X1, active(X2)) | | active#(isNePal(X)) | → | isNePal#(active(X)) |
| top#(mark(X)) | → | proper#(X) | | active#(__(X1, X2)) | → | __#(active(X1), X2) |
| proper#(__(X1, X2)) | → | proper#(X1) | | __#(mark(X1), X2) | → | __#(X1, X2) |
| active#(isNePal(X)) | → | active#(X) | | top#(mark(X)) | → | top#(proper(X)) |
| active#(__(X1, X2)) | → | active#(X1) | | __#(ok(X1), ok(X2)) | → | __#(X1, X2) |
| proper#(isNePal(X)) | → | proper#(X) | | __#(X1, mark(X2)) | → | __#(X1, X2) |
| active#(__(__(X, Y), Z)) | → | __#(Y, Z) | | isNePal#(mark(X)) | → | isNePal#(X) |
| proper#(__(X1, X2)) | → | proper#(X2) | | and#(mark(X1), X2) | → | and#(X1, X2) |
| isNePal#(ok(X)) | → | isNePal#(X) | | proper#(isNePal(X)) | → | isNePal#(proper(X)) |
| active#(and(X1, X2)) | → | active#(X1) |
Rewrite Rules
| active(__(__(X, Y), Z)) | → | mark(__(X, __(Y, Z))) | | active(__(X, nil)) | → | mark(X) |
| active(__(nil, X)) | → | mark(X) | | active(and(tt, X)) | → | mark(X) |
| active(isNePal(__(I, __(P, I)))) | → | mark(tt) | | active(__(X1, X2)) | → | __(active(X1), X2) |
| active(__(X1, X2)) | → | __(X1, active(X2)) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(isNePal(X)) | → | isNePal(active(X)) | | __(mark(X1), X2) | → | mark(__(X1, X2)) |
| __(X1, mark(X2)) | → | mark(__(X1, X2)) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| isNePal(mark(X)) | → | mark(isNePal(X)) | | proper(__(X1, X2)) | → | __(proper(X1), proper(X2)) |
| proper(nil) | → | ok(nil) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | proper(isNePal(X)) | → | isNePal(proper(X)) |
| __(ok(X1), ok(X2)) | → | ok(__(X1, X2)) | | and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) |
| isNePal(ok(X)) | → | ok(isNePal(X)) | | top(mark(X)) | → | top(proper(X)) |
| top(ok(X)) | → | top(active(X)) |
Original Signature
Termination of terms over the following signature is verified: tt, isNePal, active, __, mark, ok, proper, nil, and, top
Strategy
The following SCCs where found
| __#(mark(X1), X2) → __#(X1, X2) | __#(ok(X1), ok(X2)) → __#(X1, X2) |
| __#(X1, mark(X2)) → __#(X1, X2) |
| active#(isNePal(X)) → active#(X) | active#(__(X1, X2)) → active#(X1) |
| active#(__(X1, X2)) → active#(X2) | active#(and(X1, X2)) → active#(X1) |
| proper#(__(X1, X2)) → proper#(X1) | proper#(and(X1, X2)) → proper#(X2) |
| proper#(isNePal(X)) → proper#(X) | proper#(__(X1, X2)) → proper#(X2) |
| proper#(and(X1, X2)) → proper#(X1) |
| isNePal#(mark(X)) → isNePal#(X) | isNePal#(ok(X)) → isNePal#(X) |
| 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)) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| and#(ok(X1), ok(X2)) | → | and#(X1, X2) | | and#(mark(X1), X2) | → | and#(X1, X2) |
Rewrite Rules
| active(__(__(X, Y), Z)) | → | mark(__(X, __(Y, Z))) | | active(__(X, nil)) | → | mark(X) |
| active(__(nil, X)) | → | mark(X) | | active(and(tt, X)) | → | mark(X) |
| active(isNePal(__(I, __(P, I)))) | → | mark(tt) | | active(__(X1, X2)) | → | __(active(X1), X2) |
| active(__(X1, X2)) | → | __(X1, active(X2)) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(isNePal(X)) | → | isNePal(active(X)) | | __(mark(X1), X2) | → | mark(__(X1, X2)) |
| __(X1, mark(X2)) | → | mark(__(X1, X2)) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| isNePal(mark(X)) | → | mark(isNePal(X)) | | proper(__(X1, X2)) | → | __(proper(X1), proper(X2)) |
| proper(nil) | → | ok(nil) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | proper(isNePal(X)) | → | isNePal(proper(X)) |
| __(ok(X1), ok(X2)) | → | ok(__(X1, X2)) | | and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) |
| isNePal(ok(X)) | → | ok(isNePal(X)) | | top(mark(X)) | → | top(proper(X)) |
| top(ok(X)) | → | top(active(X)) |
Original Signature
Termination of terms over the following signature is verified: tt, isNePal, active, __, mark, ok, proper, nil, 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 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| isNePal#(mark(X)) | → | isNePal#(X) | | isNePal#(ok(X)) | → | isNePal#(X) |
Rewrite Rules
| active(__(__(X, Y), Z)) | → | mark(__(X, __(Y, Z))) | | active(__(X, nil)) | → | mark(X) |
| active(__(nil, X)) | → | mark(X) | | active(and(tt, X)) | → | mark(X) |
| active(isNePal(__(I, __(P, I)))) | → | mark(tt) | | active(__(X1, X2)) | → | __(active(X1), X2) |
| active(__(X1, X2)) | → | __(X1, active(X2)) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(isNePal(X)) | → | isNePal(active(X)) | | __(mark(X1), X2) | → | mark(__(X1, X2)) |
| __(X1, mark(X2)) | → | mark(__(X1, X2)) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| isNePal(mark(X)) | → | mark(isNePal(X)) | | proper(__(X1, X2)) | → | __(proper(X1), proper(X2)) |
| proper(nil) | → | ok(nil) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | proper(isNePal(X)) | → | isNePal(proper(X)) |
| __(ok(X1), ok(X2)) | → | ok(__(X1, X2)) | | and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) |
| isNePal(ok(X)) | → | ok(isNePal(X)) | | top(mark(X)) | → | top(proper(X)) |
| top(ok(X)) | → | top(active(X)) |
Original Signature
Termination of terms over the following signature is verified: tt, isNePal, active, __, mark, ok, proper, nil, and, top
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| isNePal#(mark(X)) | → | isNePal#(X) | | isNePal#(ok(X)) | → | isNePal#(X) |
Problem 4: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| top#(mark(X)) | → | top#(proper(X)) | | top#(ok(X)) | → | top#(active(X)) |
Rewrite Rules
| active(__(__(X, Y), Z)) | → | mark(__(X, __(Y, Z))) | | active(__(X, nil)) | → | mark(X) |
| active(__(nil, X)) | → | mark(X) | | active(and(tt, X)) | → | mark(X) |
| active(isNePal(__(I, __(P, I)))) | → | mark(tt) | | active(__(X1, X2)) | → | __(active(X1), X2) |
| active(__(X1, X2)) | → | __(X1, active(X2)) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(isNePal(X)) | → | isNePal(active(X)) | | __(mark(X1), X2) | → | mark(__(X1, X2)) |
| __(X1, mark(X2)) | → | mark(__(X1, X2)) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| isNePal(mark(X)) | → | mark(isNePal(X)) | | proper(__(X1, X2)) | → | __(proper(X1), proper(X2)) |
| proper(nil) | → | ok(nil) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | proper(isNePal(X)) | → | isNePal(proper(X)) |
| __(ok(X1), ok(X2)) | → | ok(__(X1, X2)) | | and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) |
| isNePal(ok(X)) | → | ok(isNePal(X)) | | top(mark(X)) | → | top(proper(X)) |
| top(ok(X)) | → | top(active(X)) |
Original Signature
Termination of terms over the following signature is verified: tt, isNePal, active, __, mark, ok, proper, nil, and, top
Strategy
Polynomial Interpretation
- __(x,y): y + 2x + 1
- active(x): x
- and(x,y): y + x + 1
- isNePal(x): x + 1
- mark(x): x + 1
- nil: 0
- ok(x): x
- proper(x): x
- top(x): 0
- top#(x): 2x
- tt: 2
Improved Usable rules
| active(__(X1, X2)) | → | __(active(X1), X2) | | __(ok(X1), ok(X2)) | → | ok(__(X1, X2)) |
| active(__(nil, X)) | → | mark(X) | | and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) |
| active(__(X, nil)) | → | mark(X) | | proper(__(X1, X2)) | → | __(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | active(isNePal(X)) | → | isNePal(active(X)) |
| active(and(tt, X)) | → | mark(X) | | proper(isNePal(X)) | → | isNePal(proper(X)) |
| __(mark(X1), X2) | → | mark(__(X1, X2)) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| active(__(__(X, Y), Z)) | → | mark(__(X, __(Y, Z))) | | proper(nil) | → | ok(nil) |
| and(mark(X1), X2) | → | mark(and(X1, X2)) | | isNePal(ok(X)) | → | ok(isNePal(X)) |
| active(__(X1, X2)) | → | __(X1, active(X2)) | | isNePal(mark(X)) | → | mark(isNePal(X)) |
| __(X1, mark(X2)) | → | mark(__(X1, X2)) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(isNePal(__(I, __(P, I)))) | → | mark(tt) |
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(__(__(X, Y), Z)) | → | mark(__(X, __(Y, Z))) | | active(__(X, nil)) | → | mark(X) |
| active(__(nil, X)) | → | mark(X) | | active(and(tt, X)) | → | mark(X) |
| active(isNePal(__(I, __(P, I)))) | → | mark(tt) | | active(__(X1, X2)) | → | __(active(X1), X2) |
| active(__(X1, X2)) | → | __(X1, active(X2)) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(isNePal(X)) | → | isNePal(active(X)) | | __(mark(X1), X2) | → | mark(__(X1, X2)) |
| __(X1, mark(X2)) | → | mark(__(X1, X2)) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| isNePal(mark(X)) | → | mark(isNePal(X)) | | proper(__(X1, X2)) | → | __(proper(X1), proper(X2)) |
| proper(nil) | → | ok(nil) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | proper(isNePal(X)) | → | isNePal(proper(X)) |
| __(ok(X1), ok(X2)) | → | ok(__(X1, X2)) | | and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) |
| isNePal(ok(X)) | → | ok(isNePal(X)) | | top(mark(X)) | → | top(proper(X)) |
| top(ok(X)) | → | top(active(X)) |
Original Signature
Termination of terms over the following signature is verified: tt, isNePal, __, active, ok, mark, proper, top, and, nil
Strategy
Polynomial Interpretation
- __(x,y): x + 1
- active(x): x
- and(x,y): x
- isNePal(x): 2x
- mark(x): 1
- nil: 0
- ok(x): x + 1
- proper(x): 0
- top(x): 0
- top#(x): x + 1
- tt: 2
Improved Usable rules
| active(__(X1, X2)) | → | __(active(X1), X2) | | __(ok(X1), ok(X2)) | → | ok(__(X1, X2)) |
| active(__(nil, X)) | → | mark(X) | | and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) |
| active(__(X, nil)) | → | mark(X) | | active(isNePal(X)) | → | isNePal(active(X)) |
| active(and(tt, X)) | → | mark(X) | | __(mark(X1), X2) | → | mark(__(X1, X2)) |
| active(__(__(X, Y), Z)) | → | mark(__(X, __(Y, Z))) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| isNePal(ok(X)) | → | ok(isNePal(X)) | | active(__(X1, X2)) | → | __(X1, active(X2)) |
| isNePal(mark(X)) | → | mark(isNePal(X)) | | __(X1, mark(X2)) | → | mark(__(X1, X2)) |
| active(isNePal(__(I, __(P, I)))) | → | mark(tt) | | 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 5: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| proper#(__(X1, X2)) | → | proper#(X1) | | proper#(and(X1, X2)) | → | proper#(X2) |
| proper#(isNePal(X)) | → | proper#(X) | | proper#(__(X1, X2)) | → | proper#(X2) |
| proper#(and(X1, X2)) | → | proper#(X1) |
Rewrite Rules
| active(__(__(X, Y), Z)) | → | mark(__(X, __(Y, Z))) | | active(__(X, nil)) | → | mark(X) |
| active(__(nil, X)) | → | mark(X) | | active(and(tt, X)) | → | mark(X) |
| active(isNePal(__(I, __(P, I)))) | → | mark(tt) | | active(__(X1, X2)) | → | __(active(X1), X2) |
| active(__(X1, X2)) | → | __(X1, active(X2)) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(isNePal(X)) | → | isNePal(active(X)) | | __(mark(X1), X2) | → | mark(__(X1, X2)) |
| __(X1, mark(X2)) | → | mark(__(X1, X2)) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| isNePal(mark(X)) | → | mark(isNePal(X)) | | proper(__(X1, X2)) | → | __(proper(X1), proper(X2)) |
| proper(nil) | → | ok(nil) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | proper(isNePal(X)) | → | isNePal(proper(X)) |
| __(ok(X1), ok(X2)) | → | ok(__(X1, X2)) | | and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) |
| isNePal(ok(X)) | → | ok(isNePal(X)) | | top(mark(X)) | → | top(proper(X)) |
| top(ok(X)) | → | top(active(X)) |
Original Signature
Termination of terms over the following signature is verified: tt, isNePal, active, __, mark, ok, proper, nil, and, top
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| proper#(__(X1, X2)) | → | proper#(X1) | | proper#(and(X1, X2)) | → | proper#(X2) |
| proper#(isNePal(X)) | → | proper#(X) | | proper#(__(X1, X2)) | → | proper#(X2) |
| proper#(and(X1, X2)) | → | proper#(X1) |
Problem 6: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| __#(mark(X1), X2) | → | __#(X1, X2) | | __#(ok(X1), ok(X2)) | → | __#(X1, X2) |
| __#(X1, mark(X2)) | → | __#(X1, X2) |
Rewrite Rules
| active(__(__(X, Y), Z)) | → | mark(__(X, __(Y, Z))) | | active(__(X, nil)) | → | mark(X) |
| active(__(nil, X)) | → | mark(X) | | active(and(tt, X)) | → | mark(X) |
| active(isNePal(__(I, __(P, I)))) | → | mark(tt) | | active(__(X1, X2)) | → | __(active(X1), X2) |
| active(__(X1, X2)) | → | __(X1, active(X2)) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(isNePal(X)) | → | isNePal(active(X)) | | __(mark(X1), X2) | → | mark(__(X1, X2)) |
| __(X1, mark(X2)) | → | mark(__(X1, X2)) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| isNePal(mark(X)) | → | mark(isNePal(X)) | | proper(__(X1, X2)) | → | __(proper(X1), proper(X2)) |
| proper(nil) | → | ok(nil) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | proper(isNePal(X)) | → | isNePal(proper(X)) |
| __(ok(X1), ok(X2)) | → | ok(__(X1, X2)) | | and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) |
| isNePal(ok(X)) | → | ok(isNePal(X)) | | top(mark(X)) | → | top(proper(X)) |
| top(ok(X)) | → | top(active(X)) |
Original Signature
Termination of terms over the following signature is verified: tt, isNePal, active, __, mark, ok, proper, nil, and, top
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| __#(mark(X1), X2) | → | __#(X1, X2) | | __#(ok(X1), ok(X2)) | → | __#(X1, X2) |
Problem 8: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| __#(X1, mark(X2)) | → | __#(X1, X2) |
Rewrite Rules
| active(__(__(X, Y), Z)) | → | mark(__(X, __(Y, Z))) | | active(__(X, nil)) | → | mark(X) |
| active(__(nil, X)) | → | mark(X) | | active(and(tt, X)) | → | mark(X) |
| active(isNePal(__(I, __(P, I)))) | → | mark(tt) | | active(__(X1, X2)) | → | __(active(X1), X2) |
| active(__(X1, X2)) | → | __(X1, active(X2)) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(isNePal(X)) | → | isNePal(active(X)) | | __(mark(X1), X2) | → | mark(__(X1, X2)) |
| __(X1, mark(X2)) | → | mark(__(X1, X2)) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| isNePal(mark(X)) | → | mark(isNePal(X)) | | proper(__(X1, X2)) | → | __(proper(X1), proper(X2)) |
| proper(nil) | → | ok(nil) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | proper(isNePal(X)) | → | isNePal(proper(X)) |
| __(ok(X1), ok(X2)) | → | ok(__(X1, X2)) | | and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) |
| isNePal(ok(X)) | → | ok(isNePal(X)) | | top(mark(X)) | → | top(proper(X)) |
| top(ok(X)) | → | top(active(X)) |
Original Signature
Termination of terms over the following signature is verified: tt, isNePal, __, active, ok, mark, proper, top, and, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| __#(X1, mark(X2)) | → | __#(X1, X2) |
Problem 7: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| active#(isNePal(X)) | → | active#(X) | | active#(__(X1, X2)) | → | active#(X1) |
| active#(__(X1, X2)) | → | active#(X2) | | active#(and(X1, X2)) | → | active#(X1) |
Rewrite Rules
| active(__(__(X, Y), Z)) | → | mark(__(X, __(Y, Z))) | | active(__(X, nil)) | → | mark(X) |
| active(__(nil, X)) | → | mark(X) | | active(and(tt, X)) | → | mark(X) |
| active(isNePal(__(I, __(P, I)))) | → | mark(tt) | | active(__(X1, X2)) | → | __(active(X1), X2) |
| active(__(X1, X2)) | → | __(X1, active(X2)) | | active(and(X1, X2)) | → | and(active(X1), X2) |
| active(isNePal(X)) | → | isNePal(active(X)) | | __(mark(X1), X2) | → | mark(__(X1, X2)) |
| __(X1, mark(X2)) | → | mark(__(X1, X2)) | | and(mark(X1), X2) | → | mark(and(X1, X2)) |
| isNePal(mark(X)) | → | mark(isNePal(X)) | | proper(__(X1, X2)) | → | __(proper(X1), proper(X2)) |
| proper(nil) | → | ok(nil) | | proper(and(X1, X2)) | → | and(proper(X1), proper(X2)) |
| proper(tt) | → | ok(tt) | | proper(isNePal(X)) | → | isNePal(proper(X)) |
| __(ok(X1), ok(X2)) | → | ok(__(X1, X2)) | | and(ok(X1), ok(X2)) | → | ok(and(X1, X2)) |
| isNePal(ok(X)) | → | ok(isNePal(X)) | | top(mark(X)) | → | top(proper(X)) |
| top(ok(X)) | → | top(active(X)) |
Original Signature
Termination of terms over the following signature is verified: tt, isNePal, active, __, mark, ok, proper, nil, and, top
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| active#(isNePal(X)) | → | active#(X) | | active#(__(X1, X2)) | → | active#(X1) |
| active#(__(X1, X2)) | → | active#(X2) | | active#(and(X1, X2)) | → | active#(X1) |