The TRS could be proven terminating. The proof took 113 ms.
Problem 1 was processed with processor DependencyGraph (6ms). | – Problem 2 was processed with processor PolynomialLinearRange4 (47ms). | – Problem 3 was processed with processor PolynomialLinearRange4 (25ms).
| and#(tt, X) | → | T(X) | x#(N, s(M)) | → | plus#(x(N, M), N) | |
| x#(N, s(M)) | → | x#(N, M) | plus#(N, s(M)) | → | plus#(N, M) |
| and(tt, X) | → | X | plus(N, 0) | → | N | |
| plus(N, s(M)) | → | s(plus(N, M)) | x(N, 0) | → | 0 | |
| x(N, s(M)) | → | plus(x(N, M), N) |
Termination of terms over the following signature is verified: plus, 0, s, tt, and, x
Context-sensitive strategy:
μ(T) = μ(0) = μ(tt) = ∅
μ(s) = μ(and#) = μ(and) = {1}
μ(plus) = μ(plus#) = μ(x#) = μ(x) = {1, 2}
| x#(N, s(M)) → x#(N, M) |
| plus#(N, s(M)) → plus#(N, M) |
| x#(N, s(M)) | → | x#(N, M) |
| and(tt, X) | → | X | plus(N, 0) | → | N | |
| plus(N, s(M)) | → | s(plus(N, M)) | x(N, 0) | → | 0 | |
| x(N, s(M)) | → | plus(x(N, M), N) |
Termination of terms over the following signature is verified: plus, 0, s, tt, and, x
Context-sensitive strategy:
μ(T) = μ(0) = μ(tt) = ∅
μ(s) = μ(and#) = μ(and) = {1}
μ(plus) = μ(plus#) = μ(x#) = μ(x) = {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:
| x#(N, s(M)) | → | x#(N, M) |
| plus#(N, s(M)) | → | plus#(N, M) |
| and(tt, X) | → | X | plus(N, 0) | → | N | |
| plus(N, s(M)) | → | s(plus(N, M)) | x(N, 0) | → | 0 | |
| x(N, s(M)) | → | plus(x(N, M), N) |
Termination of terms over the following signature is verified: plus, 0, s, tt, and, x
Context-sensitive strategy:
μ(T) = μ(0) = μ(tt) = ∅
μ(s) = μ(and#) = μ(and) = {1}
μ(plus) = μ(plus#) = μ(x#) = μ(x) = {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#(N, s(M)) | → | plus#(N, M) |