YES
The TRS could be proven terminating. The proof took 248 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (7ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (193ms).
| – Problem 3 was processed with processor SubtermCriterion (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| max#(N(L(x), N(y, z))) | → | max#(N(L(x), L(max(N(y, z))))) | | max#(N(L(x), N(y, z))) | → | max#(N(y, z)) |
| max#(N(L(s(x)), L(s(y)))) | → | max#(N(L(x), L(y))) |
Rewrite Rules
| max(L(x)) | → | x | | max(N(L(0), L(y))) | → | y |
| max(N(L(s(x)), L(s(y)))) | → | s(max(N(L(x), L(y)))) | | max(N(L(x), N(y, z))) | → | max(N(L(x), L(max(N(y, z))))) |
Original Signature
Termination of terms over the following signature is verified: max, 0, s, L, N
Strategy
The following SCCs where found
| max#(N(L(x), N(y, z))) → max#(N(y, z)) |
| max#(N(L(s(x)), L(s(y)))) → max#(N(L(x), L(y))) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| max#(N(L(s(x)), L(s(y)))) | → | max#(N(L(x), L(y))) |
Rewrite Rules
| max(L(x)) | → | x | | max(N(L(0), L(y))) | → | y |
| max(N(L(s(x)), L(s(y)))) | → | s(max(N(L(x), L(y)))) | | max(N(L(x), N(y, z))) | → | max(N(L(x), L(max(N(y, z))))) |
Original Signature
Termination of terms over the following signature is verified: max, 0, s, L, N
Strategy
Polynomial Interpretation
- 0: 0
- L(x): x
- N(x,y): x
- max(x): 0
- max#(x): 2x
- s(x): x + 1
There are no usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| max#(N(L(s(x)), L(s(y)))) | → | max#(N(L(x), L(y))) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| max#(N(L(x), N(y, z))) | → | max#(N(y, z)) |
Rewrite Rules
| max(L(x)) | → | x | | max(N(L(0), L(y))) | → | y |
| max(N(L(s(x)), L(s(y)))) | → | s(max(N(L(x), L(y)))) | | max(N(L(x), N(y, z))) | → | max(N(L(x), L(max(N(y, z))))) |
Original Signature
Termination of terms over the following signature is verified: max, 0, s, L, N
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| max#(N(L(x), N(y, z))) | → | max#(N(y, z)) |