YES
The TRS could be proven terminating. The proof took 838 ms.
The following DP Processors were used
Problem 1 was processed with processor PolynomialLinearRange4iUR (442ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (168ms).
Problem 1: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| +#(f(x), f(y)) | → | +#(x, y) | | +#(+(x, y), z) | → | +#(y, z) |
| +#(f(x), +(f(y), z)) | → | +#(x, y) | | +#(f(x), +(f(y), z)) | → | +#(f(+(x, y)), z) |
| +#(+(x, y), z) | → | +#(x, +(y, z)) |
Rewrite Rules
| +(+(x, y), z) | → | +(x, +(y, z)) | | +(f(x), f(y)) | → | f(+(x, y)) |
| +(f(x), +(f(y), z)) | → | +(f(+(x, y)), z) |
Original Signature
Termination of terms over the following signature is verified: f, +
Strategy
Polynomial Interpretation
- +(x,y): y + x
- +#(x,y): y + x
- f(x): 2x + 1
Improved Usable rules
| +(+(x, y), z) | → | +(x, +(y, z)) | | +(f(x), f(y)) | → | f(+(x, y)) |
| +(f(x), +(f(y), z)) | → | +(f(+(x, y)), z) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| +#(f(x), f(y)) | → | +#(x, y) | | +#(f(x), +(f(y), z)) | → | +#(x, y) |
| +#(f(x), +(f(y), z)) | → | +#(f(+(x, y)), z) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| +#(+(x, y), z) | → | +#(y, z) | | +#(+(x, y), z) | → | +#(x, +(y, z)) |
Rewrite Rules
| +(+(x, y), z) | → | +(x, +(y, z)) | | +(f(x), f(y)) | → | f(+(x, y)) |
| +(f(x), +(f(y), z)) | → | +(f(+(x, y)), z) |
Original Signature
Termination of terms over the following signature is verified: f, +
Strategy
Polynomial Interpretation
- +(x,y): 2y + 2x + 1
- +#(x,y): x
- f(x): x
Improved Usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| +#(+(x, y), z) | → | +#(y, z) | | +#(+(x, y), z) | → | +#(x, +(y, z)) |