YES
The TRS could be proven terminating. The proof took 307 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (52ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor SubtermCriterion (0ms).
| – Problem 4 was processed with processor SubtermCriterion (2ms).
| | – Problem 5 was processed with processor PolynomialLinearRange4iUR (151ms).
| | | – Problem 6 was processed with processor DependencyGraph (13ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| times#(s(x), y) | → | plus#(y, times(x, y)) | | times#(s(x), y) | → | times#(x, y) |
| plus#(s(x), y) | → | plus#(x, y) | | quot#(x, 0, s(z)) | → | div#(x, s(z)) |
| div#(div(x, y), z) | → | times#(y, z) | | div#(x, y) | → | quot#(x, y, y) |
| div#(div(x, y), z) | → | div#(x, times(y, z)) | | quot#(s(x), s(y), z) | → | quot#(x, y, z) |
Rewrite Rules
| plus(x, 0) | → | x | | plus(0, y) | → | y |
| plus(s(x), y) | → | s(plus(x, y)) | | times(0, y) | → | 0 |
| times(s(0), y) | → | y | | times(s(x), y) | → | plus(y, times(x, y)) |
| div(0, y) | → | 0 | | div(x, y) | → | quot(x, y, y) |
| quot(0, s(y), z) | → | 0 | | quot(s(x), s(y), z) | → | quot(x, y, z) |
| quot(x, 0, s(z)) | → | s(div(x, s(z))) | | div(div(x, y), z) | → | div(x, times(y, z)) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, times, div, quot
Strategy
The following SCCs where found
| times#(s(x), y) → times#(x, y) |
| plus#(s(x), y) → plus#(x, y) |
| quot#(x, 0, s(z)) → div#(x, s(z)) | div#(x, y) → quot#(x, y, y) |
| div#(div(x, y), z) → div#(x, times(y, z)) | quot#(s(x), s(y), z) → quot#(x, y, z) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| times#(s(x), y) | → | times#(x, y) |
Rewrite Rules
| plus(x, 0) | → | x | | plus(0, y) | → | y |
| plus(s(x), y) | → | s(plus(x, y)) | | times(0, y) | → | 0 |
| times(s(0), y) | → | y | | times(s(x), y) | → | plus(y, times(x, y)) |
| div(0, y) | → | 0 | | div(x, y) | → | quot(x, y, y) |
| quot(0, s(y), z) | → | 0 | | quot(s(x), s(y), z) | → | quot(x, y, z) |
| quot(x, 0, s(z)) | → | s(div(x, s(z))) | | div(div(x, y), z) | → | div(x, times(y, z)) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, times, div, quot
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| times#(s(x), y) | → | times#(x, y) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| plus#(s(x), y) | → | plus#(x, y) |
Rewrite Rules
| plus(x, 0) | → | x | | plus(0, y) | → | y |
| plus(s(x), y) | → | s(plus(x, y)) | | times(0, y) | → | 0 |
| times(s(0), y) | → | y | | times(s(x), y) | → | plus(y, times(x, y)) |
| div(0, y) | → | 0 | | div(x, y) | → | quot(x, y, y) |
| quot(0, s(y), z) | → | 0 | | quot(s(x), s(y), z) | → | quot(x, y, z) |
| quot(x, 0, s(z)) | → | s(div(x, s(z))) | | div(div(x, y), z) | → | div(x, times(y, z)) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, times, div, quot
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| plus#(s(x), y) | → | plus#(x, y) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| quot#(x, 0, s(z)) | → | div#(x, s(z)) | | div#(x, y) | → | quot#(x, y, y) |
| div#(div(x, y), z) | → | div#(x, times(y, z)) | | quot#(s(x), s(y), z) | → | quot#(x, y, z) |
Rewrite Rules
| plus(x, 0) | → | x | | plus(0, y) | → | y |
| plus(s(x), y) | → | s(plus(x, y)) | | times(0, y) | → | 0 |
| times(s(0), y) | → | y | | times(s(x), y) | → | plus(y, times(x, y)) |
| div(0, y) | → | 0 | | div(x, y) | → | quot(x, y, y) |
| quot(0, s(y), z) | → | 0 | | quot(s(x), s(y), z) | → | quot(x, y, z) |
| quot(x, 0, s(z)) | → | s(div(x, s(z))) | | div(div(x, y), z) | → | div(x, times(y, z)) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, times, div, quot
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| div#(div(x, y), z) | → | div#(x, times(y, z)) | | quot#(s(x), s(y), z) | → | quot#(x, y, z) |
Problem 5: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| quot#(x, 0, s(z)) | → | div#(x, s(z)) | | div#(x, y) | → | quot#(x, y, y) |
Rewrite Rules
| plus(x, 0) | → | x | | plus(0, y) | → | y |
| plus(s(x), y) | → | s(plus(x, y)) | | times(0, y) | → | 0 |
| times(s(0), y) | → | y | | times(s(x), y) | → | plus(y, times(x, y)) |
| div(0, y) | → | 0 | | div(x, y) | → | quot(x, y, y) |
| quot(0, s(y), z) | → | 0 | | quot(s(x), s(y), z) | → | quot(x, y, z) |
| quot(x, 0, s(z)) | → | s(div(x, s(z))) | | div(div(x, y), z) | → | div(x, times(y, z)) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, times, div, quot
Strategy
Polynomial Interpretation
- 0: 2
- div(x,y): 0
- div#(x,y): 2y
- plus(x,y): 0
- quot(x,y,z): 0
- quot#(x,y,z): y
- s(x): 0
- times(x,y): 0
There are no usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| quot#(x, 0, s(z)) | → | div#(x, s(z)) |
Problem 6: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| div#(x, y) | → | quot#(x, y, y) |
Rewrite Rules
| plus(x, 0) | → | x | | plus(0, y) | → | y |
| plus(s(x), y) | → | s(plus(x, y)) | | times(0, y) | → | 0 |
| times(s(0), y) | → | y | | times(s(x), y) | → | plus(y, times(x, y)) |
| div(0, y) | → | 0 | | div(x, y) | → | quot(x, y, y) |
| quot(0, s(y), z) | → | 0 | | quot(s(x), s(y), z) | → | quot(x, y, z) |
| quot(x, 0, s(z)) | → | s(div(x, s(z))) | | div(div(x, y), z) | → | div(x, times(y, z)) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, times, div, quot
Strategy
There are no SCCs!