YES
The TRS could be proven terminating. The proof took 143 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (74ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
| – Problem 4 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| rev#(++(x, y)) | → | ++#(rev(y), rev(x)) | | ++#(++(x, y), z) | → | ++#(x, ++(y, z)) |
| ++#(++(x, y), z) | → | ++#(y, z) | | flatten#(++(x, y)) | → | flatten#(y) |
| flatten#(++(x, y)) | → | flatten#(x) | | flatten#(++(unit(x), y)) | → | ++#(flatten(x), flatten(y)) |
| flatten#(++(unit(x), y)) | → | flatten#(y) | | rev#(++(x, y)) | → | rev#(x) |
| rev#(++(x, y)) | → | rev#(y) | | flatten#(flatten(x)) | → | flatten#(x) |
| flatten#(++(x, y)) | → | ++#(flatten(x), flatten(y)) | | flatten#(unit(x)) | → | flatten#(x) |
| flatten#(++(unit(x), y)) | → | flatten#(x) |
Rewrite Rules
| flatten(nil) | → | nil | | flatten(unit(x)) | → | flatten(x) |
| flatten(++(x, y)) | → | ++(flatten(x), flatten(y)) | | flatten(++(unit(x), y)) | → | ++(flatten(x), flatten(y)) |
| flatten(flatten(x)) | → | flatten(x) | | rev(nil) | → | nil |
| rev(unit(x)) | → | unit(x) | | rev(++(x, y)) | → | ++(rev(y), rev(x)) |
| rev(rev(x)) | → | x | | ++(x, nil) | → | x |
| ++(nil, y) | → | y | | ++(++(x, y), z) | → | ++(x, ++(y, z)) |
Original Signature
Termination of terms over the following signature is verified: unit, flatten, rev, ++, nil
Strategy
The following SCCs where found
| rev#(++(x, y)) → rev#(x) | rev#(++(x, y)) → rev#(y) |
| ++#(++(x, y), z) → ++#(x, ++(y, z)) | ++#(++(x, y), z) → ++#(y, z) |
| flatten#(++(unit(x), y)) → flatten#(y) | flatten#(flatten(x)) → flatten#(x) |
| flatten#(++(x, y)) → flatten#(y) | flatten#(unit(x)) → flatten#(x) |
| flatten#(++(x, y)) → flatten#(x) | flatten#(++(unit(x), y)) → flatten#(x) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| rev#(++(x, y)) | → | rev#(x) | | rev#(++(x, y)) | → | rev#(y) |
Rewrite Rules
| flatten(nil) | → | nil | | flatten(unit(x)) | → | flatten(x) |
| flatten(++(x, y)) | → | ++(flatten(x), flatten(y)) | | flatten(++(unit(x), y)) | → | ++(flatten(x), flatten(y)) |
| flatten(flatten(x)) | → | flatten(x) | | rev(nil) | → | nil |
| rev(unit(x)) | → | unit(x) | | rev(++(x, y)) | → | ++(rev(y), rev(x)) |
| rev(rev(x)) | → | x | | ++(x, nil) | → | x |
| ++(nil, y) | → | y | | ++(++(x, y), z) | → | ++(x, ++(y, z)) |
Original Signature
Termination of terms over the following signature is verified: unit, flatten, rev, ++, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| rev#(++(x, y)) | → | rev#(x) | | rev#(++(x, y)) | → | rev#(y) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| ++#(++(x, y), z) | → | ++#(x, ++(y, z)) | | ++#(++(x, y), z) | → | ++#(y, z) |
Rewrite Rules
| flatten(nil) | → | nil | | flatten(unit(x)) | → | flatten(x) |
| flatten(++(x, y)) | → | ++(flatten(x), flatten(y)) | | flatten(++(unit(x), y)) | → | ++(flatten(x), flatten(y)) |
| flatten(flatten(x)) | → | flatten(x) | | rev(nil) | → | nil |
| rev(unit(x)) | → | unit(x) | | rev(++(x, y)) | → | ++(rev(y), rev(x)) |
| rev(rev(x)) | → | x | | ++(x, nil) | → | x |
| ++(nil, y) | → | y | | ++(++(x, y), z) | → | ++(x, ++(y, z)) |
Original Signature
Termination of terms over the following signature is verified: unit, flatten, rev, ++, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| ++#(++(x, y), z) | → | ++#(x, ++(y, z)) | | ++#(++(x, y), z) | → | ++#(y, z) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| flatten#(++(unit(x), y)) | → | flatten#(y) | | flatten#(flatten(x)) | → | flatten#(x) |
| flatten#(++(x, y)) | → | flatten#(y) | | flatten#(unit(x)) | → | flatten#(x) |
| flatten#(++(x, y)) | → | flatten#(x) | | flatten#(++(unit(x), y)) | → | flatten#(x) |
Rewrite Rules
| flatten(nil) | → | nil | | flatten(unit(x)) | → | flatten(x) |
| flatten(++(x, y)) | → | ++(flatten(x), flatten(y)) | | flatten(++(unit(x), y)) | → | ++(flatten(x), flatten(y)) |
| flatten(flatten(x)) | → | flatten(x) | | rev(nil) | → | nil |
| rev(unit(x)) | → | unit(x) | | rev(++(x, y)) | → | ++(rev(y), rev(x)) |
| rev(rev(x)) | → | x | | ++(x, nil) | → | x |
| ++(nil, y) | → | y | | ++(++(x, y), z) | → | ++(x, ++(y, z)) |
Original Signature
Termination of terms over the following signature is verified: unit, flatten, rev, ++, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| flatten#(++(unit(x), y)) | → | flatten#(y) | | flatten#(flatten(x)) | → | flatten#(x) |
| flatten#(++(x, y)) | → | flatten#(y) | | flatten#(unit(x)) | → | flatten#(x) |
| flatten#(++(x, y)) | → | flatten#(x) | | flatten#(++(unit(x), y)) | → | flatten#(x) |