YES
The TRS could be proven terminating. The proof took 178 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (22ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (123ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
| | – Problem 4 was processed with processor SubtermCriterion (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| plus#(x, s(s(y))) | → | plus#(s(x), y) | | ack#(s(x), s(y)) | → | plus#(y, ack(s(x), y)) |
| ack#(s(x), s(y)) | → | ack#(x, plus(y, ack(s(x), y))) | | ack#(s(x), s(y)) | → | ack#(s(x), y) |
| plus#(s(s(x)), y) | → | plus#(x, s(y)) | | ack#(s(x), 0) | → | ack#(x, s(0)) |
Rewrite Rules
| plus(s(s(x)), y) | → | s(plus(x, s(y))) | | plus(x, s(s(y))) | → | s(plus(s(x), y)) |
| plus(s(0), y) | → | s(y) | | plus(0, y) | → | y |
| ack(0, y) | → | s(y) | | ack(s(x), 0) | → | ack(x, s(0)) |
| ack(s(x), s(y)) | → | ack(x, plus(y, ack(s(x), y))) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, ack
Strategy
The following SCCs where found
| plus#(x, s(s(y))) → plus#(s(x), y) | plus#(s(s(x)), y) → plus#(x, s(y)) |
| ack#(s(x), s(y)) → ack#(x, plus(y, ack(s(x), y))) | ack#(s(x), s(y)) → ack#(s(x), y) |
| ack#(s(x), 0) → ack#(x, s(0)) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| plus#(x, s(s(y))) | → | plus#(s(x), y) | | plus#(s(s(x)), y) | → | plus#(x, s(y)) |
Rewrite Rules
| plus(s(s(x)), y) | → | s(plus(x, s(y))) | | plus(x, s(s(y))) | → | s(plus(s(x), y)) |
| plus(s(0), y) | → | s(y) | | plus(0, y) | → | y |
| ack(0, y) | → | s(y) | | ack(s(x), 0) | → | ack(x, s(0)) |
| ack(s(x), s(y)) | → | ack(x, plus(y, ack(s(x), y))) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, ack
Strategy
Polynomial Interpretation
- 0: 0
- ack(x,y): 0
- plus(x,y): 0
- plus#(x,y): y + x + 1
- 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:
| plus#(x, s(s(y))) | → | plus#(s(x), y) | | plus#(s(s(x)), y) | → | plus#(x, s(y)) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| ack#(s(x), s(y)) | → | ack#(x, plus(y, ack(s(x), y))) | | ack#(s(x), s(y)) | → | ack#(s(x), y) |
| ack#(s(x), 0) | → | ack#(x, s(0)) |
Rewrite Rules
| plus(s(s(x)), y) | → | s(plus(x, s(y))) | | plus(x, s(s(y))) | → | s(plus(s(x), y)) |
| plus(s(0), y) | → | s(y) | | plus(0, y) | → | y |
| ack(0, y) | → | s(y) | | ack(s(x), 0) | → | ack(x, s(0)) |
| ack(s(x), s(y)) | → | ack(x, plus(y, ack(s(x), y))) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, ack
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| ack#(s(x), s(y)) | → | ack#(x, plus(y, ack(s(x), y))) | | ack#(s(x), 0) | → | ack#(x, s(0)) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| ack#(s(x), s(y)) | → | ack#(s(x), y) |
Rewrite Rules
| plus(s(s(x)), y) | → | s(plus(x, s(y))) | | plus(x, s(s(y))) | → | s(plus(s(x), y)) |
| plus(s(0), y) | → | s(y) | | plus(0, y) | → | y |
| ack(0, y) | → | s(y) | | ack(s(x), 0) | → | ack(x, s(0)) |
| ack(s(x), s(y)) | → | ack(x, plus(y, ack(s(x), y))) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, ack
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| ack#(s(x), s(y)) | → | ack#(s(x), y) |