YES
The TRS could be proven terminating. The proof took 284 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (9ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
| | – Problem 4 was processed with processor PolynomialLinearRange4iUR (164ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| plus#(s(x), y) | → | plus#(x, y) | | quot#(x, 0, s(z)) | → | quot#(x, plus(z, s(0)), s(z)) |
| quot#(x, 0, s(z)) | → | plus#(z, s(0)) | | quot#(s(x), s(y), z) | → | quot#(x, y, z) |
Rewrite Rules
| quot(0, s(y), s(z)) | → | 0 | | quot(s(x), s(y), z) | → | quot(x, y, z) |
| plus(0, y) | → | y | | plus(s(x), y) | → | s(plus(x, y)) |
| quot(x, 0, s(z)) | → | s(quot(x, plus(z, s(0)), s(z))) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, quot
Strategy
The following SCCs where found
| quot#(x, 0, s(z)) → quot#(x, plus(z, s(0)), s(z)) | quot#(s(x), s(y), z) → quot#(x, y, z) |
| plus#(s(x), y) → plus#(x, y) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| plus#(s(x), y) | → | plus#(x, y) |
Rewrite Rules
| quot(0, s(y), s(z)) | → | 0 | | quot(s(x), s(y), z) | → | quot(x, y, z) |
| plus(0, y) | → | y | | plus(s(x), y) | → | s(plus(x, y)) |
| quot(x, 0, s(z)) | → | s(quot(x, plus(z, s(0)), s(z))) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, quot
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| plus#(s(x), y) | → | plus#(x, y) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| quot#(x, 0, s(z)) | → | quot#(x, plus(z, s(0)), s(z)) | | quot#(s(x), s(y), z) | → | quot#(x, y, z) |
Rewrite Rules
| quot(0, s(y), s(z)) | → | 0 | | quot(s(x), s(y), z) | → | quot(x, y, z) |
| plus(0, y) | → | y | | plus(s(x), y) | → | s(plus(x, y)) |
| quot(x, 0, s(z)) | → | s(quot(x, plus(z, s(0)), s(z))) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, quot
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| quot#(s(x), s(y), z) | → | quot#(x, y, z) |
Problem 4: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| quot#(x, 0, s(z)) | → | quot#(x, plus(z, s(0)), s(z)) |
Rewrite Rules
| quot(0, s(y), s(z)) | → | 0 | | quot(s(x), s(y), z) | → | quot(x, y, z) |
| plus(0, y) | → | y | | plus(s(x), y) | → | s(plus(x, y)) |
| quot(x, 0, s(z)) | → | s(quot(x, plus(z, s(0)), s(z))) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, quot
Strategy
Polynomial Interpretation
- 0: 1
- plus(x,y): y
- quot(x,y,z): 0
- quot#(x,y,z): y + x + 1
- s(x): 0
Improved Usable rules
| plus(s(x), y) | → | s(plus(x, y)) | | plus(0, y) | → | y |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| quot#(x, 0, s(z)) | → | quot#(x, plus(z, s(0)), s(z)) |