YES
The TRS could be proven terminating. The proof took 19 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (5ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| rev#(.(x, y)) | → | rev#(y) | | ++#(.(x, y), z) | → | ++#(y, z) |
| rev#(.(x, y)) | → | ++#(rev(y), .(x, nil)) |
Rewrite Rules
| rev(nil) | → | nil | | rev(.(x, y)) | → | ++(rev(y), .(x, nil)) |
| car(.(x, y)) | → | x | | cdr(.(x, y)) | → | y |
| null(nil) | → | true | | null(.(x, y)) | → | false |
| ++(nil, y) | → | y | | ++(.(x, y), z) | → | .(x, ++(y, z)) |
Original Signature
Termination of terms over the following signature is verified: car, rev, true, false, cdr, ++, ., null, nil
Strategy
The following SCCs where found
| ++#(.(x, y), z) → ++#(y, z) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| rev(nil) | → | nil | | rev(.(x, y)) | → | ++(rev(y), .(x, nil)) |
| car(.(x, y)) | → | x | | cdr(.(x, y)) | → | y |
| null(nil) | → | true | | null(.(x, y)) | → | false |
| ++(nil, y) | → | y | | ++(.(x, y), z) | → | .(x, ++(y, z)) |
Original Signature
Termination of terms over the following signature is verified: car, rev, true, false, cdr, ++, ., null, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| ++#(.(x, y), z) | → | ++#(y, z) |
Rewrite Rules
| rev(nil) | → | nil | | rev(.(x, y)) | → | ++(rev(y), .(x, nil)) |
| car(.(x, y)) | → | x | | cdr(.(x, y)) | → | y |
| null(nil) | → | true | | null(.(x, y)) | → | false |
| ++(nil, y) | → | y | | ++(.(x, y), z) | → | .(x, ++(y, z)) |
Original Signature
Termination of terms over the following signature is verified: car, rev, true, false, cdr, ++, ., null, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| ++#(.(x, y), z) | → | ++#(y, z) |