YES
The TRS could be proven terminating. The proof took 467 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (34ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (252ms).
| – Problem 3 was processed with processor SubtermCriterion (3ms).
| | – Problem 4 was processed with processor SubtermCriterion (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(g(f(x))) | → | f#(h(s(0), x)) | | +#(x, s(y)) | → | +#(x, y) |
| f#(h(x, h(y, z))) | → | +#(x, y) | | +#(x, +(y, z)) | → | +#(+(x, y), z) |
| f#(g(h(x, y))) | → | f#(h(s(x), y)) | | +#(s(x), y) | → | +#(x, y) |
| +#(x, +(y, z)) | → | +#(x, y) | | f#(h(x, h(y, z))) | → | f#(h(+(x, y), z)) |
Rewrite Rules
| +(x, 0) | → | x | | +(x, s(y)) | → | s(+(x, y)) |
| +(0, y) | → | y | | +(s(x), y) | → | s(+(x, y)) |
| +(x, +(y, z)) | → | +(+(x, y), z) | | f(g(f(x))) | → | f(h(s(0), x)) |
| f(g(h(x, y))) | → | f(h(s(x), y)) | | f(h(x, h(y, z))) | → | f(h(+(x, y), z)) |
Original Signature
Termination of terms over the following signature is verified: f, g, 0, s, +, h
Strategy
The following SCCs where found
| +#(x, s(y)) → +#(x, y) | +#(x, +(y, z)) → +#(+(x, y), z) |
| +#(s(x), y) → +#(x, y) | +#(x, +(y, z)) → +#(x, y) |
| f#(h(x, h(y, z))) → f#(h(+(x, y), z)) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| f#(h(x, h(y, z))) | → | f#(h(+(x, y), z)) |
Rewrite Rules
| +(x, 0) | → | x | | +(x, s(y)) | → | s(+(x, y)) |
| +(0, y) | → | y | | +(s(x), y) | → | s(+(x, y)) |
| +(x, +(y, z)) | → | +(+(x, y), z) | | f(g(f(x))) | → | f(h(s(0), x)) |
| f(g(h(x, y))) | → | f(h(s(x), y)) | | f(h(x, h(y, z))) | → | f(h(+(x, y), z)) |
Original Signature
Termination of terms over the following signature is verified: f, g, 0, s, +, h
Strategy
Polynomial Interpretation
- +(x,y): 1
- 0: 2
- f(x): 0
- f#(x): x + 1
- g(x): 0
- h(x,y): y + 1
- s(x): 1
Improved Usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| f#(h(x, h(y, z))) | → | f#(h(+(x, y), z)) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| +#(x, s(y)) | → | +#(x, y) | | +#(x, +(y, z)) | → | +#(+(x, y), z) |
| +#(s(x), y) | → | +#(x, y) | | +#(x, +(y, z)) | → | +#(x, y) |
Rewrite Rules
| +(x, 0) | → | x | | +(x, s(y)) | → | s(+(x, y)) |
| +(0, y) | → | y | | +(s(x), y) | → | s(+(x, y)) |
| +(x, +(y, z)) | → | +(+(x, y), z) | | f(g(f(x))) | → | f(h(s(0), x)) |
| f(g(h(x, y))) | → | f(h(s(x), y)) | | f(h(x, h(y, z))) | → | f(h(+(x, y), z)) |
Original Signature
Termination of terms over the following signature is verified: f, g, 0, s, +, h
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| +#(x, s(y)) | → | +#(x, y) | | +#(x, +(y, z)) | → | +#(+(x, y), z) |
| +#(x, +(y, z)) | → | +#(x, y) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| +(x, 0) | → | x | | +(x, s(y)) | → | s(+(x, y)) |
| +(0, y) | → | y | | +(s(x), y) | → | s(+(x, y)) |
| +(x, +(y, z)) | → | +(+(x, y), z) | | f(g(f(x))) | → | f(h(s(0), x)) |
| f(g(h(x, y))) | → | f(h(s(x), y)) | | f(h(x, h(y, z))) | → | f(h(+(x, y), z)) |
Original Signature
Termination of terms over the following signature is verified: f, g, 0, s, +, h
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed: