YES
The TRS could be proven terminating. The proof took 174 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (71ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| | – Problem 6 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
| | – Problem 7 was processed with processor SubtermCriterion (0ms).
| – Problem 4 was processed with processor SubtermCriterion (1ms).
| – Problem 5 was processed with processor SubtermCriterion (2ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(s(x), s(y), s(z)) | → | f#(x, y, f(s(x), s(y), z)) | | f#(0, s(s(y)), s(s(z))) | → | f#(0, s(s(y)), s(z)) |
| f#(s(x), s(y), s(z)) | → | f#(s(x), s(y), z) | | f#(0, s(s(y)), s(s(z))) | → | f#(0, y, f(0, s(s(y)), s(z))) |
| f#(s(x), 0, s(z)) | → | f#(x, s(0), z) | | f#(s(0), y, z) | → | f#(0, s(y), s(z)) |
| f#(s(x), s(y), 0) | → | f#(x, y, s(0)) | | f#(0, s(0), s(s(z))) | → | f#(0, s(0), z) |
| f#(0, s(s(y)), s(0)) | → | f#(0, y, s(0)) |
Rewrite Rules
| f(x, 0, 0) | → | s(x) | | f(0, y, 0) | → | s(y) |
| f(0, 0, z) | → | s(z) | | f(s(0), y, z) | → | f(0, s(y), s(z)) |
| f(s(x), s(y), 0) | → | f(x, y, s(0)) | | f(s(x), 0, s(z)) | → | f(x, s(0), z) |
| f(0, s(0), s(0)) | → | s(s(0)) | | f(s(x), s(y), s(z)) | → | f(x, y, f(s(x), s(y), z)) |
| f(0, s(s(y)), s(0)) | → | f(0, y, s(0)) | | f(0, s(0), s(s(z))) | → | f(0, s(0), z) |
| f(0, s(s(y)), s(s(z))) | → | f(0, y, f(0, s(s(y)), s(z))) |
Original Signature
Termination of terms over the following signature is verified: f, 0, s
Strategy
The following SCCs where found
| f#(0, s(s(y)), s(s(z))) → f#(0, s(s(y)), s(z)) | f#(0, s(s(y)), s(s(z))) → f#(0, y, f(0, s(s(y)), s(z))) |
| f#(s(x), s(y), s(z)) → f#(x, y, f(s(x), s(y), z)) | f#(s(x), s(y), s(z)) → f#(s(x), s(y), z) |
| f#(s(x), 0, s(z)) → f#(x, s(0), z) | f#(s(x), s(y), 0) → f#(x, y, s(0)) |
| f#(0, s(0), s(s(z))) → f#(0, s(0), z) |
| f#(0, s(s(y)), s(0)) → f#(0, y, s(0)) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| f#(s(x), s(y), s(z)) | → | f#(x, y, f(s(x), s(y), z)) | | f#(s(x), s(y), s(z)) | → | f#(s(x), s(y), z) |
| f#(s(x), 0, s(z)) | → | f#(x, s(0), z) | | f#(s(x), s(y), 0) | → | f#(x, y, s(0)) |
Rewrite Rules
| f(x, 0, 0) | → | s(x) | | f(0, y, 0) | → | s(y) |
| f(0, 0, z) | → | s(z) | | f(s(0), y, z) | → | f(0, s(y), s(z)) |
| f(s(x), s(y), 0) | → | f(x, y, s(0)) | | f(s(x), 0, s(z)) | → | f(x, s(0), z) |
| f(0, s(0), s(0)) | → | s(s(0)) | | f(s(x), s(y), s(z)) | → | f(x, y, f(s(x), s(y), z)) |
| f(0, s(s(y)), s(0)) | → | f(0, y, s(0)) | | f(0, s(0), s(s(z))) | → | f(0, s(0), z) |
| f(0, s(s(y)), s(s(z))) | → | f(0, y, f(0, s(s(y)), s(z))) |
Original Signature
Termination of terms over the following signature is verified: f, 0, s
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| f#(s(x), s(y), s(z)) | → | f#(x, y, f(s(x), s(y), z)) | | f#(s(x), 0, s(z)) | → | f#(x, s(0), z) |
| f#(s(x), s(y), 0) | → | f#(x, y, s(0)) |
Problem 6: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| f#(s(x), s(y), s(z)) | → | f#(s(x), s(y), z) |
Rewrite Rules
| f(x, 0, 0) | → | s(x) | | f(0, y, 0) | → | s(y) |
| f(0, 0, z) | → | s(z) | | f(s(0), y, z) | → | f(0, s(y), s(z)) |
| f(s(x), s(y), 0) | → | f(x, y, s(0)) | | f(s(x), 0, s(z)) | → | f(x, s(0), z) |
| f(0, s(0), s(0)) | → | s(s(0)) | | f(s(x), s(y), s(z)) | → | f(x, y, f(s(x), s(y), z)) |
| f(0, s(s(y)), s(0)) | → | f(0, y, s(0)) | | f(0, s(0), s(s(z))) | → | f(0, s(0), z) |
| f(0, s(s(y)), s(s(z))) | → | f(0, y, f(0, s(s(y)), s(z))) |
Original Signature
Termination of terms over the following signature is verified: f, 0, s
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| f#(s(x), s(y), s(z)) | → | f#(s(x), s(y), z) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| f#(0, s(s(y)), s(s(z))) | → | f#(0, s(s(y)), s(z)) | | f#(0, s(s(y)), s(s(z))) | → | f#(0, y, f(0, s(s(y)), s(z))) |
Rewrite Rules
| f(x, 0, 0) | → | s(x) | | f(0, y, 0) | → | s(y) |
| f(0, 0, z) | → | s(z) | | f(s(0), y, z) | → | f(0, s(y), s(z)) |
| f(s(x), s(y), 0) | → | f(x, y, s(0)) | | f(s(x), 0, s(z)) | → | f(x, s(0), z) |
| f(0, s(0), s(0)) | → | s(s(0)) | | f(s(x), s(y), s(z)) | → | f(x, y, f(s(x), s(y), z)) |
| f(0, s(s(y)), s(0)) | → | f(0, y, s(0)) | | f(0, s(0), s(s(z))) | → | f(0, s(0), z) |
| f(0, s(s(y)), s(s(z))) | → | f(0, y, f(0, s(s(y)), s(z))) |
Original Signature
Termination of terms over the following signature is verified: f, 0, s
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| f#(0, s(s(y)), s(s(z))) | → | f#(0, y, f(0, s(s(y)), s(z))) |
Problem 7: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| f#(0, s(s(y)), s(s(z))) | → | f#(0, s(s(y)), s(z)) |
Rewrite Rules
| f(x, 0, 0) | → | s(x) | | f(0, y, 0) | → | s(y) |
| f(0, 0, z) | → | s(z) | | f(s(0), y, z) | → | f(0, s(y), s(z)) |
| f(s(x), s(y), 0) | → | f(x, y, s(0)) | | f(s(x), 0, s(z)) | → | f(x, s(0), z) |
| f(0, s(0), s(0)) | → | s(s(0)) | | f(s(x), s(y), s(z)) | → | f(x, y, f(s(x), s(y), z)) |
| f(0, s(s(y)), s(0)) | → | f(0, y, s(0)) | | f(0, s(0), s(s(z))) | → | f(0, s(0), z) |
| f(0, s(s(y)), s(s(z))) | → | f(0, y, f(0, s(s(y)), s(z))) |
Original Signature
Termination of terms over the following signature is verified: f, 0, s
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| f#(0, s(s(y)), s(s(z))) | → | f#(0, s(s(y)), s(z)) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| f#(0, s(0), s(s(z))) | → | f#(0, s(0), z) |
Rewrite Rules
| f(x, 0, 0) | → | s(x) | | f(0, y, 0) | → | s(y) |
| f(0, 0, z) | → | s(z) | | f(s(0), y, z) | → | f(0, s(y), s(z)) |
| f(s(x), s(y), 0) | → | f(x, y, s(0)) | | f(s(x), 0, s(z)) | → | f(x, s(0), z) |
| f(0, s(0), s(0)) | → | s(s(0)) | | f(s(x), s(y), s(z)) | → | f(x, y, f(s(x), s(y), z)) |
| f(0, s(s(y)), s(0)) | → | f(0, y, s(0)) | | f(0, s(0), s(s(z))) | → | f(0, s(0), z) |
| f(0, s(s(y)), s(s(z))) | → | f(0, y, f(0, s(s(y)), s(z))) |
Original Signature
Termination of terms over the following signature is verified: f, 0, s
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| f#(0, s(0), s(s(z))) | → | f#(0, s(0), z) |
Problem 5: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| f#(0, s(s(y)), s(0)) | → | f#(0, y, s(0)) |
Rewrite Rules
| f(x, 0, 0) | → | s(x) | | f(0, y, 0) | → | s(y) |
| f(0, 0, z) | → | s(z) | | f(s(0), y, z) | → | f(0, s(y), s(z)) |
| f(s(x), s(y), 0) | → | f(x, y, s(0)) | | f(s(x), 0, s(z)) | → | f(x, s(0), z) |
| f(0, s(0), s(0)) | → | s(s(0)) | | f(s(x), s(y), s(z)) | → | f(x, y, f(s(x), s(y), z)) |
| f(0, s(s(y)), s(0)) | → | f(0, y, s(0)) | | f(0, s(0), s(s(z))) | → | f(0, s(0), z) |
| f(0, s(s(y)), s(s(z))) | → | f(0, y, f(0, s(s(y)), s(z))) |
Original Signature
Termination of terms over the following signature is verified: f, 0, s
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| f#(0, s(s(y)), s(0)) | → | f#(0, y, s(0)) |