The TRS could be proven terminating. The proof took 99 ms.
Problem 1 was processed with processor DependencyGraph (12ms). | – Problem 2 was processed with processor PolynomialLinearRange4 (55ms).
| f#(x, y) | → | U21#(plus(x, y), y, x) | U21#(plus(z, z), y, x) | → | T(y) | |
| f#(x, y) | → | plus#(x, y) | U21#(plus(z, z), y, x) | → | plus#(y, z) | |
| plus#(s(x), y) | → | plus#(x, s(y)) |
| plus(0, y) | → | y | plus(s(x), y) | → | plus(x, s(y)) | |
| f(x, y) | → | U21(plus(x, y), y, x) | U21(plus(z, z), y, x) | → | plus(y, z) |
Termination of terms over the following signature is verified: f, plus, 0, s, z
Context-sensitive strategy:
μ(T) = μ(0) = μ(z) = ∅
μ(s) = μ(U21#) = μ(U21) = {1}
μ(f) = μ(plus) = μ(f#) = μ(plus#) = {1, 2}
| plus#(s(x), y) → plus#(x, s(y)) |
| plus#(s(x), y) | → | plus#(x, s(y)) |
| plus(0, y) | → | y | plus(s(x), y) | → | plus(x, s(y)) | |
| f(x, y) | → | U21(plus(x, y), y, x) | U21(plus(z, z), y, x) | → | plus(y, z) |
Termination of terms over the following signature is verified: f, plus, 0, s, z
Context-sensitive strategy:
μ(T) = μ(0) = μ(z) = ∅
μ(s) = μ(U21#) = μ(U21) = {1}
μ(f) = μ(plus) = μ(f#) = μ(plus#) = {1, 2}
There are no usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| plus#(s(x), y) | → | plus#(x, s(y)) |