YES
The TRS could be proven terminating. The proof took 28 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (10ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| | – Problem 4 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| merge#(.(x, y), .(u, v)) | → | if#(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v))) | | merge#(.(x, y), .(u, v)) | → | merge#(.(x, y), v) |
| ++#(.(x, y), z) | → | ++#(y, z) | | merge#(.(x, y), .(u, v)) | → | merge#(y, .(u, v)) |
Rewrite Rules
| merge(nil, y) | → | y | | merge(x, nil) | → | x |
| merge(.(x, y), .(u, v)) | → | if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v))) | | ++(nil, y) | → | y |
| ++(.(x, y), z) | → | .(x, ++(y, z)) | | if(true, x, y) | → | x |
| if(false, x, y) | → | x |
Original Signature
Termination of terms over the following signature is verified: if, merge, true, false, ., ++, nil, <
Strategy
The following SCCs where found
| ++#(.(x, y), z) → ++#(y, z) |
| merge#(.(x, y), .(u, v)) → merge#(.(x, y), v) | merge#(.(x, y), .(u, v)) → merge#(y, .(u, v)) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| merge#(.(x, y), .(u, v)) | → | merge#(.(x, y), v) | | merge#(.(x, y), .(u, v)) | → | merge#(y, .(u, v)) |
Rewrite Rules
| merge(nil, y) | → | y | | merge(x, nil) | → | x |
| merge(.(x, y), .(u, v)) | → | if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v))) | | ++(nil, y) | → | y |
| ++(.(x, y), z) | → | .(x, ++(y, z)) | | if(true, x, y) | → | x |
| if(false, x, y) | → | x |
Original Signature
Termination of terms over the following signature is verified: if, merge, true, false, ., ++, nil, <
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| merge#(.(x, y), .(u, v)) | → | merge#(y, .(u, v)) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| merge#(.(x, y), .(u, v)) | → | merge#(.(x, y), v) |
Rewrite Rules
| merge(nil, y) | → | y | | merge(x, nil) | → | x |
| merge(.(x, y), .(u, v)) | → | if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v))) | | ++(nil, y) | → | y |
| ++(.(x, y), z) | → | .(x, ++(y, z)) | | if(true, x, y) | → | x |
| if(false, x, y) | → | x |
Original Signature
Termination of terms over the following signature is verified: if, merge, false, true, ++, ., <, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| merge#(.(x, y), .(u, v)) | → | merge#(.(x, y), v) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| ++#(.(x, y), z) | → | ++#(y, z) |
Rewrite Rules
| merge(nil, y) | → | y | | merge(x, nil) | → | x |
| merge(.(x, y), .(u, v)) | → | if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v))) | | ++(nil, y) | → | y |
| ++(.(x, y), z) | → | .(x, ++(y, z)) | | if(true, x, y) | → | x |
| if(false, x, y) | → | x |
Original Signature
Termination of terms over the following signature is verified: if, merge, true, false, ., ++, nil, <
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| ++#(.(x, y), z) | → | ++#(y, z) |