YES
The TRS could be proven terminating. The proof took 100 ms.
Problem 1 was processed with processor DependencyGraph (5ms). | Problem 2 was processed with processor PolynomialLinearRange4 (75ms).
f#(b) | → | a# | f#(b) | → | f#(a) | |
a# | → | b# | a# | → | U21#(b) |
b | → | c | f(b) | → | f(a) | |
a | → | U21(b) | U21(c) | → | c |
Termination of terms over the following signature is verified: f, b, c, a
Context-sensitive strategy:
μ(T) = μ(b) = μ(c) = μ(a) = μ(b#) = μ(a#) = ∅
μ(f) = μ(f#) = μ(U21#) = μ(U21) = {1}
f#(b) → f#(a) |
f#(b) | → | f#(a) |
b | → | c | f(b) | → | f(a) | |
a | → | U21(b) | U21(c) | → | c |
Termination of terms over the following signature is verified: f, b, c, a
Context-sensitive strategy:
μ(T) = μ(b) = μ(c) = μ(a) = μ(b#) = μ(a#) = ∅
μ(f) = μ(f#) = μ(U21#) = μ(U21) = {1}
U21(c) | → | c | b | → | c | |
a | → | U21(b) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
f#(b) | → | f#(a) |