The TRS could be proven terminating. The proof took 84 ms.
Problem 1 was processed with processor DependencyGraph (4ms). | – Problem 2 was processed with processor PolynomialLinearRange4 (58ms).
| plus#(s(x), y) | → | plus#(x, y) | plus#(s(x), y) | → | U21#(plus(x, y), y, x) |
| plus(x, 0) | → | x | plus(0, x) | → | x | |
| plus(s(x), y) | → | U21(plus(x, y), y, x) | U21(z, y, x) | → | s(z) |
Termination of terms over the following signature is verified: plus, 0, s
Context-sensitive strategy:
μ(T) = μ(0) = ∅
μ(s) = μ(U21#) = μ(U21) = {1}
μ(plus) = μ(plus#) = {1, 2}
| plus#(s(x), y) → plus#(x, y) |
| plus#(s(x), y) | → | plus#(x, y) |
| plus(x, 0) | → | x | plus(0, x) | → | x | |
| plus(s(x), y) | → | U21(plus(x, y), y, x) | U21(z, y, x) | → | s(z) |
Termination of terms over the following signature is verified: plus, 0, s
Context-sensitive strategy:
μ(T) = μ(0) = ∅
μ(s) = μ(U21#) = μ(U21) = {1}
μ(plus) = 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, y) |