YES
The TRS could be proven terminating. The proof took 30 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (13ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
| – Problem 4 was processed with processor SubtermCriterion (3ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| sqr#(s(x)) | → | double#(x) | | sqr#(s(x)) | → | +#(sqr(x), s(double(x))) |
| +#(x, s(y)) | → | +#(x, y) | | sqr#(s(x)) | → | sqr#(x) |
| double#(s(x)) | → | double#(x) | | sqr#(s(x)) | → | +#(sqr(x), double(x)) |
Rewrite Rules
| sqr(0) | → | 0 | | sqr(s(x)) | → | +(sqr(x), s(double(x))) |
| double(0) | → | 0 | | double(s(x)) | → | s(s(double(x))) |
| +(x, 0) | → | x | | +(x, s(y)) | → | s(+(x, y)) |
| sqr(s(x)) | → | s(+(sqr(x), double(x))) |
Original Signature
Termination of terms over the following signature is verified: 0, s, sqr, +, double
Strategy
The following SCCs where found
| double#(s(x)) → double#(x) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| double#(s(x)) | → | double#(x) |
Rewrite Rules
| sqr(0) | → | 0 | | sqr(s(x)) | → | +(sqr(x), s(double(x))) |
| double(0) | → | 0 | | double(s(x)) | → | s(s(double(x))) |
| +(x, 0) | → | x | | +(x, s(y)) | → | s(+(x, y)) |
| sqr(s(x)) | → | s(+(sqr(x), double(x))) |
Original Signature
Termination of terms over the following signature is verified: 0, s, sqr, +, double
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| double#(s(x)) | → | double#(x) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| sqr(0) | → | 0 | | sqr(s(x)) | → | +(sqr(x), s(double(x))) |
| double(0) | → | 0 | | double(s(x)) | → | s(s(double(x))) |
| +(x, 0) | → | x | | +(x, s(y)) | → | s(+(x, y)) |
| sqr(s(x)) | → | s(+(sqr(x), double(x))) |
Original Signature
Termination of terms over the following signature is verified: 0, s, sqr, +, double
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| sqr(0) | → | 0 | | sqr(s(x)) | → | +(sqr(x), s(double(x))) |
| double(0) | → | 0 | | double(s(x)) | → | s(s(double(x))) |
| +(x, 0) | → | x | | +(x, s(y)) | → | s(+(x, y)) |
| sqr(s(x)) | → | s(+(sqr(x), double(x))) |
Original Signature
Termination of terms over the following signature is verified: 0, s, sqr, +, double
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed: