YES
The TRS could be proven terminating. The proof took 54 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (30ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| | – Problem 4 was processed with processor SubtermCriterion (2ms).
| – Problem 3 was processed with processor SubtermCriterion (2ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| *#(x, +(y, z)) | → | *#(x, z) | | *#(+(x, y), z) | → | *#(x, z) |
| *#(+(x, y), z) | → | +#(*(x, z), *(y, z)) | | *#(x, +(y, z)) | → | +#(*(x, y), *(x, z)) |
| +#(+(x, y), z) | → | +#(y, z) | | *#(x, +(y, z)) | → | *#(x, y) |
| *#(+(x, y), z) | → | *#(y, z) | | +#(+(x, y), z) | → | +#(x, +(y, z)) |
Rewrite Rules
| +(x, 0) | → | x | | +(x, i(x)) | → | 0 |
| +(+(x, y), z) | → | +(x, +(y, z)) | | *(x, +(y, z)) | → | +(*(x, y), *(x, z)) |
| *(+(x, y), z) | → | +(*(x, z), *(y, z)) |
Original Signature
Termination of terms over the following signature is verified: 0, *, +, i
Strategy
The following SCCs where found
| *#(x, +(y, z)) → *#(x, z) | *#(+(x, y), z) → *#(x, z) |
| *#(x, +(y, z)) → *#(x, y) | *#(+(x, y), z) → *#(y, z) |
| +#(+(x, y), z) → +#(y, z) | +#(+(x, y), z) → +#(x, +(y, z)) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| *#(x, +(y, z)) | → | *#(x, z) | | *#(+(x, y), z) | → | *#(x, z) |
| *#(x, +(y, z)) | → | *#(x, y) | | *#(+(x, y), z) | → | *#(y, z) |
Rewrite Rules
| +(x, 0) | → | x | | +(x, i(x)) | → | 0 |
| +(+(x, y), z) | → | +(x, +(y, z)) | | *(x, +(y, z)) | → | +(*(x, y), *(x, z)) |
| *(+(x, y), z) | → | +(*(x, z), *(y, z)) |
Original Signature
Termination of terms over the following signature is verified: 0, *, +, i
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| *#(+(x, y), z) | → | *#(x, z) | | *#(+(x, y), z) | → | *#(y, z) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| *#(x, +(y, z)) | → | *#(x, z) | | *#(x, +(y, z)) | → | *#(x, y) |
Rewrite Rules
| +(x, 0) | → | x | | +(x, i(x)) | → | 0 |
| +(+(x, y), z) | → | +(x, +(y, z)) | | *(x, +(y, z)) | → | +(*(x, y), *(x, z)) |
| *(+(x, y), z) | → | +(*(x, z), *(y, z)) |
Original Signature
Termination of terms over the following signature is verified: 0, *, +, i
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| *#(x, +(y, z)) | → | *#(x, z) | | *#(x, +(y, z)) | → | *#(x, y) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| +#(+(x, y), z) | → | +#(y, z) | | +#(+(x, y), z) | → | +#(x, +(y, z)) |
Rewrite Rules
| +(x, 0) | → | x | | +(x, i(x)) | → | 0 |
| +(+(x, y), z) | → | +(x, +(y, z)) | | *(x, +(y, z)) | → | +(*(x, y), *(x, z)) |
| *(+(x, y), z) | → | +(*(x, z), *(y, z)) |
Original Signature
Termination of terms over the following signature is verified: 0, *, +, i
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| +#(+(x, y), z) | → | +#(y, z) | | +#(+(x, y), z) | → | +#(x, +(y, z)) |