YES
The TRS could be proven terminating. The proof took 205 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (19ms).
| – Problem 2 was processed with processor SubtermCriterion (2ms).
| – Problem 3 was processed with processor PolynomialLinearRange4iUR (156ms).
| – Problem 4 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| div#(s(x), s(y)) | → | lt#(x, y) | | div#(s(x), s(y)) | → | if#(lt(x, y), 0, s(div(-(x, y), s(y)))) |
| div#(s(x), s(y)) | → | -#(x, y) | | lt#(s(x), s(y)) | → | lt#(x, y) |
| -#(s(x), s(y)) | → | -#(x, y) | | div#(s(x), s(y)) | → | div#(-(x, y), s(y)) |
Rewrite Rules
| -(x, 0) | → | x | | -(0, s(y)) | → | 0 |
| -(s(x), s(y)) | → | -(x, y) | | lt(x, 0) | → | false |
| lt(0, s(y)) | → | true | | lt(s(x), s(y)) | → | lt(x, y) |
| if(true, x, y) | → | x | | if(false, x, y) | → | y |
| div(x, 0) | → | 0 | | div(0, y) | → | 0 |
| div(s(x), s(y)) | → | if(lt(x, y), 0, s(div(-(x, y), s(y)))) |
Original Signature
Termination of terms over the following signature is verified: 0, s, if, div, false, true, lt, -
Strategy
The following SCCs where found
| lt#(s(x), s(y)) → lt#(x, y) |
| div#(s(x), s(y)) → div#(-(x, y), s(y)) |
| -#(s(x), s(y)) → -#(x, y) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| lt#(s(x), s(y)) | → | lt#(x, y) |
Rewrite Rules
| -(x, 0) | → | x | | -(0, s(y)) | → | 0 |
| -(s(x), s(y)) | → | -(x, y) | | lt(x, 0) | → | false |
| lt(0, s(y)) | → | true | | lt(s(x), s(y)) | → | lt(x, y) |
| if(true, x, y) | → | x | | if(false, x, y) | → | y |
| div(x, 0) | → | 0 | | div(0, y) | → | 0 |
| div(s(x), s(y)) | → | if(lt(x, y), 0, s(div(-(x, y), s(y)))) |
Original Signature
Termination of terms over the following signature is verified: 0, s, if, div, false, true, lt, -
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| lt#(s(x), s(y)) | → | lt#(x, y) |
Problem 3: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| div#(s(x), s(y)) | → | div#(-(x, y), s(y)) |
Rewrite Rules
| -(x, 0) | → | x | | -(0, s(y)) | → | 0 |
| -(s(x), s(y)) | → | -(x, y) | | lt(x, 0) | → | false |
| lt(0, s(y)) | → | true | | lt(s(x), s(y)) | → | lt(x, y) |
| if(true, x, y) | → | x | | if(false, x, y) | → | y |
| div(x, 0) | → | 0 | | div(0, y) | → | 0 |
| div(s(x), s(y)) | → | if(lt(x, y), 0, s(div(-(x, y), s(y)))) |
Original Signature
Termination of terms over the following signature is verified: 0, s, if, div, false, true, lt, -
Strategy
Polynomial Interpretation
- -(x,y): x + 1
- 0: 0
- div(x,y): 0
- div#(x,y): x
- false: 0
- if(x,y,z): 0
- lt(x,y): 0
- s(x): x + 2
- true: 0
Improved Usable rules
| -(s(x), s(y)) | → | -(x, y) | | -(x, 0) | → | x |
| -(0, s(y)) | → | 0 |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| div#(s(x), s(y)) | → | div#(-(x, y), s(y)) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| -#(s(x), s(y)) | → | -#(x, y) |
Rewrite Rules
| -(x, 0) | → | x | | -(0, s(y)) | → | 0 |
| -(s(x), s(y)) | → | -(x, y) | | lt(x, 0) | → | false |
| lt(0, s(y)) | → | true | | lt(s(x), s(y)) | → | lt(x, y) |
| if(true, x, y) | → | x | | if(false, x, y) | → | y |
| div(x, 0) | → | 0 | | div(0, y) | → | 0 |
| div(s(x), s(y)) | → | if(lt(x, y), 0, s(div(-(x, y), s(y)))) |
Original Signature
Termination of terms over the following signature is verified: 0, s, if, div, false, true, lt, -
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| -#(s(x), s(y)) | → | -#(x, y) |