YES
The TRS could be proven terminating. The proof took 84 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (6ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor PolynomialLinearRange4iUR (58ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| times#(X, s(Y)) | → | times#(Y, X) | | plus#(plus(X, Y), Z) | → | plus#(Y, Z) |
| times#(X, s(Y)) | → | plus#(X, times(Y, X)) | | plus#(plus(X, Y), Z) | → | plus#(X, plus(Y, Z)) |
Rewrite Rules
| plus(plus(X, Y), Z) | → | plus(X, plus(Y, Z)) | | times(X, s(Y)) | → | plus(X, times(Y, X)) |
Original Signature
Termination of terms over the following signature is verified: plus, s, times
Strategy
The following SCCs where found
| times#(X, s(Y)) → times#(Y, X) |
| plus#(plus(X, Y), Z) → plus#(Y, Z) | plus#(plus(X, Y), Z) → plus#(X, plus(Y, Z)) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| plus#(plus(X, Y), Z) | → | plus#(Y, Z) | | plus#(plus(X, Y), Z) | → | plus#(X, plus(Y, Z)) |
Rewrite Rules
| plus(plus(X, Y), Z) | → | plus(X, plus(Y, Z)) | | times(X, s(Y)) | → | plus(X, times(Y, X)) |
Original Signature
Termination of terms over the following signature is verified: plus, s, times
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| plus#(plus(X, Y), Z) | → | plus#(Y, Z) | | plus#(plus(X, Y), Z) | → | plus#(X, plus(Y, Z)) |
Problem 3: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| times#(X, s(Y)) | → | times#(Y, X) |
Rewrite Rules
| plus(plus(X, Y), Z) | → | plus(X, plus(Y, Z)) | | times(X, s(Y)) | → | plus(X, times(Y, X)) |
Original Signature
Termination of terms over the following signature is verified: plus, s, times
Strategy
Polynomial Interpretation
- plus(x,y): 0
- s(x): x + 1
- times(x,y): 0
- times#(x,y): y + 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:
| times#(X, s(Y)) | → | times#(Y, X) |