YES
The TRS could be proven terminating. The proof took 38 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (18ms).
| – Problem 2 was processed with processor SubtermCriterion (0ms).
| – Problem 3 was processed with processor SubtermCriterion (2ms).
| | – Problem 4 was processed with processor SubtermCriterion (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| rev#(++(x, y)) | → | ++#(rev(y), rev(x)) | | rev#(++(x, y)) | → | rev#(x) |
| rev#(++(x, y)) | → | rev#(y) | | ++#(x, ++(y, z)) | → | ++#(++(x, y), z) |
| ++#(.(x, y), z) | → | ++#(y, z) | | ++#(x, ++(y, z)) | → | ++#(x, y) |
Rewrite Rules
| rev(nil) | → | nil | | rev(rev(x)) | → | x |
| rev(++(x, y)) | → | ++(rev(y), rev(x)) | | ++(nil, y) | → | y |
| ++(x, nil) | → | x | | ++(.(x, y), z) | → | .(x, ++(y, z)) |
| ++(x, ++(y, z)) | → | ++(++(x, y), z) | | make(x) | → | .(x, nil) |
Original Signature
Termination of terms over the following signature is verified: rev, ++, ., make, nil
Strategy
The following SCCs where found
| ++#(x, ++(y, z)) → ++#(++(x, y), z) | ++#(.(x, y), z) → ++#(y, z) |
| ++#(x, ++(y, z)) → ++#(x, y) |
| rev#(++(x, y)) → rev#(x) | rev#(++(x, y)) → rev#(y) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| rev#(++(x, y)) | → | rev#(x) | | rev#(++(x, y)) | → | rev#(y) |
Rewrite Rules
| rev(nil) | → | nil | | rev(rev(x)) | → | x |
| rev(++(x, y)) | → | ++(rev(y), rev(x)) | | ++(nil, y) | → | y |
| ++(x, nil) | → | x | | ++(.(x, y), z) | → | .(x, ++(y, z)) |
| ++(x, ++(y, z)) | → | ++(++(x, y), z) | | make(x) | → | .(x, nil) |
Original Signature
Termination of terms over the following signature is verified: rev, ++, ., make, 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) |
| ++#(x, ++(y, z)) | → | ++#(x, y) |
Rewrite Rules
| rev(nil) | → | nil | | rev(rev(x)) | → | x |
| rev(++(x, y)) | → | ++(rev(y), rev(x)) | | ++(nil, y) | → | y |
| ++(x, nil) | → | x | | ++(.(x, y), z) | → | .(x, ++(y, z)) |
| ++(x, ++(y, z)) | → | ++(++(x, y), z) | | make(x) | → | .(x, nil) |
Original Signature
Termination of terms over the following signature is verified: rev, ++, ., make, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| ++#(x, ++(y, z)) | → | ++#(++(x, y), z) | | ++#(x, ++(y, z)) | → | ++#(x, y) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| ++#(.(x, y), z) | → | ++#(y, z) |
Rewrite Rules
| rev(nil) | → | nil | | rev(rev(x)) | → | x |
| rev(++(x, y)) | → | ++(rev(y), rev(x)) | | ++(nil, y) | → | y |
| ++(x, nil) | → | x | | ++(.(x, y), z) | → | .(x, ++(y, z)) |
| ++(x, ++(y, z)) | → | ++(++(x, y), z) | | make(x) | → | .(x, nil) |
Original Signature
Termination of terms over the following signature is verified: rev, ., ++, make, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| ++#(.(x, y), z) | → | ++#(y, z) |