YES
The TRS could be proven terminating. The proof took 46 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (27ms).
| – Problem 2 was processed with processor SubtermCriterion (0ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
| – Problem 4 was processed with processor SubtermCriterion (2ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(+(x, y), z) | → | +#(f(x, z), f(y, z)) | | +#(a, +(b, z)) | → | +#(a, z) |
| +#(a, b) | → | +#(b, a) | | f#(+(x, y), z) | → | f#(y, z) |
| +#(+(x, y), z) | → | +#(y, z) | | f#(+(x, y), z) | → | f#(x, z) |
| +#(a, +(b, z)) | → | +#(b, +(a, z)) | | +#(+(x, y), z) | → | +#(x, +(y, z)) |
Rewrite Rules
| +(a, b) | → | +(b, a) | | +(a, +(b, z)) | → | +(b, +(a, z)) |
| +(+(x, y), z) | → | +(x, +(y, z)) | | f(a, y) | → | a |
| f(b, y) | → | b | | f(+(x, y), z) | → | +(f(x, z), f(y, z)) |
Original Signature
Termination of terms over the following signature is verified: f, b, a, +
Strategy
The following SCCs where found
| +#(a, +(b, z)) → +#(a, z) |
| f#(+(x, y), z) → f#(y, z) | f#(+(x, y), z) → f#(x, z) |
| +#(+(x, y), z) → +#(y, z) | +#(+(x, y), z) → +#(x, +(y, z)) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| +#(a, +(b, z)) | → | +#(a, z) |
Rewrite Rules
| +(a, b) | → | +(b, a) | | +(a, +(b, z)) | → | +(b, +(a, z)) |
| +(+(x, y), z) | → | +(x, +(y, z)) | | f(a, y) | → | a |
| f(b, y) | → | b | | f(+(x, y), z) | → | +(f(x, z), f(y, z)) |
Original Signature
Termination of terms over the following signature is verified: f, b, a, +
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| +#(a, +(b, z)) | → | +#(a, z) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| f#(+(x, y), z) | → | f#(y, z) | | f#(+(x, y), z) | → | f#(x, z) |
Rewrite Rules
| +(a, b) | → | +(b, a) | | +(a, +(b, z)) | → | +(b, +(a, z)) |
| +(+(x, y), z) | → | +(x, +(y, z)) | | f(a, y) | → | a |
| f(b, y) | → | b | | f(+(x, y), z) | → | +(f(x, z), f(y, z)) |
Original Signature
Termination of terms over the following signature is verified: f, b, a, +
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| f#(+(x, y), z) | → | f#(y, z) | | f#(+(x, y), z) | → | f#(x, z) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| +#(+(x, y), z) | → | +#(y, z) | | +#(+(x, y), z) | → | +#(x, +(y, z)) |
Rewrite Rules
| +(a, b) | → | +(b, a) | | +(a, +(b, z)) | → | +(b, +(a, z)) |
| +(+(x, y), z) | → | +(x, +(y, z)) | | f(a, y) | → | a |
| f(b, y) | → | b | | f(+(x, y), z) | → | +(f(x, z), f(y, z)) |
Original Signature
Termination of terms over the following signature is verified: f, b, a, +
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)) |